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Integer solutions to $x^3+y^3+z^3 = x+y+z = 8$ Find all integers $x,y,z$ that satisfy $$x^3+y^3+z^3 = x+y+z = 8$$ Let $a = y+z, b = x+z, c = x+y$. Then $8 = x^3+y^3+z^3 = (x+y+z)^3-3abc$ and therefore $abc = 168$ and $a+b+c = 16$. Then do I just use the prime factorization of $168$?

Hint: Taking from where you left off: $ab \mid 168 \implies ab = \pm 1, \pm 2, \pm 4, \pm 6, \pm 7, \pm 8, \pm 12, \pm 14, \pm 21, \pm 24, \pm 28, \pm 42, \pm 56, \pm 84, \pm 168$. Even though it looks cumbersome, it is easy to solve. For example, $ab = 6 \implies c = \dfrac{168}{6} = 28 \implies a+b = 16-c = 16 - 28 = -12$. This case yields no solution. But other cases yield some solutions as you go through them you might find the work is not that tedious and may be fun indeed.

Algebraic solution for the value of $x$. I solved this problem the fifteen years ago without numerically solving equations of degree 4, I was happy in a substitution that I avoid directly attacking equations of degree 4. Today my nephew, who is an enthusiastic student of mathematics, proposes me the same problem. It occurs to me that I am a very rusted to algebraic substitutions exhaustively. I tried for about 3 hours. How disappointed do not want my nephew to his attempts (and not me) ask for help to the ME community. And of course I will give all the credits to ME. My attempt. Note that $ \cos \alpha = \frac{1}{x}$, $\cos \alpha = \frac{y}{1}$, $\cos \alpha = \frac{y+1}{x+1}$. Then we have $$ xy=1. $$ By Pythagoras Theorem we have $(y+1)^2+1^2=(x+1)^2 \Longleftrightarrow y^2+2y+1+1=x^2+2x+1$, i.e. $$ x^2-y^2+2(x-y)-1=0. $$ Update [ July 26 2016 ] I remember the time I solved this question, I tried something like \begin{align} x^2-y^2+2(x-y)-1=0 & \Longleftrightarrow (x-y)[x+2+y]=1 \\ & \Longleftrightarrow (x-y)[x+2\cdot 1+y]=1 \\ & \Longleftrightarrow (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})[x+2\sqrt{x}\sqrt{y}+y]=1 \\ & \Longleftrightarrow (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})^3=1 \\ & \Longleftrightarrow (\sqrt{x}+\sqrt{y})^3=\frac{1}{(\sqrt{x}-\sqrt{y})} \\ & \Longleftrightarrow (\sqrt{x}+\sqrt{y})^3=\frac{(\sqrt{x}+\sqrt{y})}{(x-y)} \\ & \Longleftrightarrow (\sqrt{x}+\sqrt{y})^2=\frac{1}{(x-y)} \\ & \Longleftrightarrow (x+y+2)=\frac{1}{(x-y)} \end{align} We then have two ways to tackle the problem: $$ \left\{\begin{array}{rl}\sqrt{x}\sqrt{y}=&1 \\ (\sqrt{x}+\sqrt{y})^3=&\frac{1}{(\sqrt{x}-\sqrt{y})} \end{array}\right. \quad\mbox{ or } \quad \left\{\begin{array}{rl}xy=&1 \\ (x+y+2)=&\frac{1}{x-y} \end{array}\right. $$

From $y = 1/x,$ then multiplying by $x^2,$ i got $$ x^4 + 2 x^3 - x^2 - 2 x - 1. $$ This looks bad. However, set $$ x = t - \frac{1}{2} $$ and you get rid of the cubic term, always worth a try. I was pleased to discover that the linear term also vanished, giving $$ t^4 - \frac{5}{2} t^2 - \frac{7}{16}, $$ and you can solve for $t^2$ with the Quadratic Formula. I get $$ t^2 = \frac{5 \pm \sqrt {32}}{4} $$ with two pure imaginary roots, a real negative, and a real positive for $t$ itself. Then $x$ is that minus 1/2. I get $1.132241883$ as $x.$ One good habit is to simply draw a graph of the function. I do them by hand with a calculator to find points. I have appended a good online graph. Notice that the graph appears to be symmetric across the vertical line $x = -\frac{1}{2}.$ We could confirm this by taking $ f(x) = x^4 + 2 x^3 - x^2 - 2 x - 1 $ and then checking whether $f(-1-x) = f(x)??$ In turn, this confirmation would tell us that the translation I tried would, in fact, give a graph symmetric across the y axis, meaning all even exponents. Calculus ideas that are, at least, consistent with the symmetry notion include $$ f'(x) = 2 (2x+1) \left(x^2 + x -1 \right) $$ and $$ f''(x) = 2 \left(6 x^2 + 6 x - 1 \right), $$ so that $x=-1/2$ gives a local maximum, while the inflection points are symmetric around $x = -1/2$ by the quadratic formula. Huh. Turns out the local minima really are along $y = -2,$ since $$ f(x) + 2 = \left( x^2 + x - 1 \right)^2. $$ Go Figure. ========================= parisize = 4000000, primelimit = 500509 ? x = -1 - w %1 = -w - 1 ? x^4 + 2 * x^3 - x^2 - 2 * x - 1 %2 = w^4 + 2*w^3 - w^2 - 2*w - 1 ? ================================== parisize = 4000000, primelimit = 500509 ? factor( x^4 + 2 * x^3 - x^2 - 2 * x - 1 ) %1 = [x^4 + 2*x^3 - x^2 - 2*x - 1 1] ? x = t - (1/2) %2 = t - 1/2 ? p = x^4 + 2 * x^3 - x^2 - 2 * x - 1 %3 = t^4 - 5/2*t^2 - 7/16 ? ==============================================

Joint probability density function $(X^2,Y^2)$ Let $X$ and $Y$ be random variables having the following joint probability density function $f(x,y)=\begin{cases} \frac{3}{8}xy & x\geq0,\,y\geq0,\:x+y\leq2,\\ 0 & \mbox{otherwise}. \end{cases}$ Find the joint probability density function of $X^2$ and $Y^2$. This is my solution: $z=x^2, \Longrightarrow x=\sqrt z$, because $x\geq0$ $w=y^2, \Longrightarrow y=\sqrt w$, because $y\geq0$. The Jacobian $J$ of the inverse transformation would then equal: $J={ \left|\begin{array}{cc} \frac{1}{2\sqrt{z}} & 0\\ 0 & \frac{1}{2\sqrt{w}} \end{array}\right|}=\frac{1}{4\sqrt{wz}}$ so, the joint probability density function \begin{align} g(z,w)&=\begin{cases} \frac{3}{8}\sqrt z \sqrt w \frac{1}{4\sqrt wz}& w\geq0,\,z\geq0,\:\sqrt w+\sqrt z\leq2\\ 0 & \mbox{otherwise} \end{cases} \\ &=\begin{cases} \frac{3}{32}\,\,\,\,& w\geq0,\,z\geq0,\:\sqrt w+\sqrt z\leq2\\ 0 & \mbox{otherwise} \end{cases} \end{align} but I have a problem, $\int_{0}^{4}\int_{0}^{w-4\sqrt{w}+4}\frac{3}{32}\,dz\,dw=\frac{1}{4}\neq1$. Please help me at my mistake

I think the issue is the original pdf: $$ \int_{\mathbb{R}^2}f(x,y)\;dydx=\frac{3}{8}\int_0^2\int_{0}^{2-x}xy\;dydx=\frac{3}{8}\int_0^2\frac{x(2-x)^2}{2}\;dx$$ $$ =\frac{3}{8}\int_0^2\frac{(2-x)x^2}{2}\;dx=\frac{3}{8}\cdot\frac{2}{3}=\frac{1}{4}$$

Integral over $0\leq x,y,z\leq 1$ and $x+y+z\leq 2$ What is $$\int_{S}(x+y+z)dS,$$ where $S$ is the region $0\leq x,y,z\leq 1$ and $x+y+z\leq 2$? We can change the region to $0\leq x,y,z\leq 1$ and $x+y+z\geq 2$, because the total of the two integrals is just $$\int_0^1\int_0^1\int_0^1(x+y+z)dxdydz=3\int_0^1xdxdydz=\frac{3}{2}.$$ Now, can we write the new integral as $$\int_0^1\int_{\min(2-x,1)}^1\int_{\min(2-x-y,1)}^1(x+y+z)dzdydx?$$ This gets more involved since we have to divide into cases whether $2-x-y\leq 1$ or $\geq 1$. Is there a simpler way?

You could write \begin{equation} \int_S x \, dx dy dz = \int_0^1 \left( \int_{y+z \leq 2-x; \, 0\leq y,z \leq 1} dy dz \right) x dx \end{equation} Now, you can interpret $y+z \leq 2-x$ with $y,z \geq 0$ as a triangle in the plane, whose area is $\frac{(2-x)^2}{2}$. From this triangle, you subtract two smaller triangles to account for the fact that $0 \leq y,z \leq 1$. You can write the area of these smaller triangles as $2\times \frac{(2-x-1)^2}{2} = (1-x)^2$. In total, we have \begin{equation} \int_{y+z \leq 2-x; \, 0\leq y,z \leq 1} dy dz = \frac{(2-x)^2}{2} - (1-x)^2 = 1 - \frac{x^2}{2} \end{equation} So in total, \begin{equation} \int_0^1 x \left(1 - \frac{x^2}{2} \right) dy dz = \frac{3}{8} \end{equation} You have three of those integrals, so $\int_S x+y+z \, dx dx dz = \frac{9}{8}$.

Spivak Calculus - Chapter 1 Question 4.6 In Spivak's Calculus, Chapter 1 Question 4.6: Find all the numbers $x$ for which $x^2+x+1>2$ The chapter focuses on using the following properties of numbers to prove solutions are correct: Based on those properties, I am able to perform the following algebra: $ \begin{align} x^2 + x + 1 &> 2 & \text{Given}\\ x (x + 1) + 1 &> 2 & \text{P9}\\ x (x+1) &> 1 & \text{P3 P2 and Addition} \end{align} $ And from there, I can note that: $ \begin{align} x &\neq (x+1)^{-1}\\ x^{-1} &\neq (x+1)\\ \end{align} $ By P6, because $x (x+1) > 1$ and $x (x+1) \neq 1$. However, in his book Spivak is able to find the following: $ \begin{align} x &> \frac{-1+\sqrt{5}}{2} \text{ or}\\ x &< \frac{-1-\sqrt{5}}{2} \end{align} $ How does he come to that conclusion using only the properties listed above?

Complete the Square $ \begin{align} x^2+x+1&>2 & \text{Given}\\ x^2+x+1+0&>2+0 & \text{By Addition}\\ x^2+x+1+0&>2 & \text{By P2}\\ x^2+x+0+1&>2 & \text{By P4}\\ x^2+x+\left( \frac{1}{2} \right)^2+(-1)\left( \frac{1}{2} \right)^2+1 &>2 & \text{By P3}\\ \left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+(-1)\left( \frac{1}{2} \right)^2+1 &>2 & \text{By P9}\\ \left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+ (-1)\left( \frac{1}{4} \right) + 1 &> 2 & \text{By Multiplication}\\ \left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right) &> \left( \frac{5}{4} \right) & \text{By Addition, P3, and P2}\\ \end{align} $ Spivak doesn't formally define exponents in Chapter 1, so it's a little difficult to finish the proof using only the properties listed in the chapter. But it is at least clear how to get to Spivak's result from there.

Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$ $$3x^2dx = \sec^2(y)dy$$ $$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$ How should I proceed after this? EDITED: Corrected the final integral and the limit from $45$ to $\pi/4$

Continuing where MK12 left off, we proceed as follows: $$\frac{2-x^2}{x^4-x^2+1} = \frac{1}{2} \times \frac{4-2x^2}{x^4-x^2+1} = \frac{1}{2} \times \frac{1+x^2 + 3(1-x^2)}{x^4-x^2+1}$$ Split the sub-integral into two parts. For one, make the substitution $u=x+\frac{1}{x}$ and the other $v = x-\frac{1}{x}$ Then the rest should be fairly trivial, provided you are meticulously careful with the limits of integration.

Solution of $ydx-xdy+3x^2y^2e^{x^2}dx=0$ Find the solution of given differential equation: $$ydx-xdy+3x^2y^2e^{x^2}dx=0$$ I am not able to solve this because of $e^{x^2}$. Could someone help me with this one?

By dividing both side to $y^2$ we get $$ydx-xdy+3x^{ 2 }y^{ 2 }e^{ x^{ 2 } }dx=0\\ \frac { ydx-xdy }{ { y }^{ 2 } } +3x^{ 2 }e^{ x^{ 2 } }=0\\ d\left( \frac { x }{ y } \right) =3x^{ 2 }e^{ x^{ 2 } }\\ \int { d\left( \frac { x }{ y } \right) =\int { 3x^{ 2 }e^{ x^{ 2 } }dx } } =\frac { 3 }{ 2 } \int { x } d{ e }^{ { x }^{ 2 } }=\frac { 3 }{ 2 } \left( x{ e }^{ { x }^{ 2 } }-\int { { e }^{ { x }^{ 2 } }dx } \right) \\ \frac { x }{ y } =\frac { 3 }{ 2 } \left( x{ e }^{ { x }^{ 2 } }-\int { { e }^{ { x }^{ 2 } }dx } \right) +C\\ y=\frac { 2x }{ 3\left( \left( x{ e }^{ { x }^{ 2 } }-\int { { e }^{ { x }^{ 2 } }dx } \right) +C \right) } $$

Evaluate the reciprocal of the following infinite product I hae to evaluate the reciprocal of the following product to infinity $$\frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 17 \cdot 19}{2 \cdot 2 \cdot 6 \cdot 6 \cdot 10 \cdot 10 \cdot 14 \cdot 14 \cdot 18 \cdot 18}\cdot\ldots $$ I am guessing you express each number as a sum. Example, the first $\frac{1}{2}$ can be expressed as $\frac{2-1}{2}$ and the next $\frac{3}{2}$ as $\frac{2+1}{2}$, hence their product equals $1-\frac{1}{2^2}$.

Note that $$1-\frac{1}{\left(4n-2\right)^{2}}=\frac{\left(4n-3\right)\left(4n-1\right)}{\left(4n-2\right)\left(4n-2\right)} $$ and $$P=\prod_{n\geq1}\left(1-\frac{1}{\left(4n-2\right)^{2}}\right)^{-1}=\prod_{n\geq1}\frac{\left(4n-2\right)\left(4n-2\right)}{\left(4n-3\right)\left(4n-1\right)} $$ $$=\prod_{n\geq0}\frac{\left(4n+2\right)\left(4n+2\right)}{\left(4n+1\right)\left(4n+3\right)}=\prod_{n\geq0}\frac{\left(n+1/2\right)\left(n+1/2\right)}{\left(n+1/4\right)\left(n+3/4\right)}$$ and now we can use the well known identity $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\, a+b=c+d $$ which follows from the representation of the Gamma function in the form $$\Gamma\left(z\right)=\lim_{n\rightarrow\infty}\frac{n^{z-1}n!}{z\left(z+1\right)\cdots\left(z+n-1\right)}. $$ Hence $$P=\frac{\Gamma\left(1/4\right)\Gamma\left(3/4\right)}{\Gamma\left(1/2\right)^{2}}=\color{red}{\sqrt{2}}.$$

Solve $2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$ Solve differential equation $$2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$$ I know that we have to use some smart substitution here, so that the equation becomes linear. The only thing I came up with is a smart guessed particular solution: $y = x^2$. If we plug this function in, we get: $$2\cdot2\cdot x^2 - 3(2x)^2 +8x^2 = 4x^2 - 12x^2 + 8x^2= 0$$ I made a mistake. The coefficients where different in the exam: $$ \begin{cases} 3\ddot{y}y + 3(\dot{y})^2 - 2x^2 = 0, \\ y(0) = 1, \\ \dot{y}(0) = 0. \end{cases} $$ Does it make the solution easier?

Hint: Let $y=\dfrac{1}{u^2}$ , Then $y'=-\dfrac{2u'}{u^3}$ $y''=\dfrac{6(u')^2}{u^4}-\dfrac{2u''}{u^3}$ $\therefore\dfrac{2}{u^2}\left(\dfrac{6(u')^2}{u^4}-\dfrac{2u''}{u^3}\right)-3\left(-\dfrac{2u'}{u^3}\right)^2+8x^2=0$ $\dfrac{12(u')^2}{u^6}-\dfrac{4u''}{u^5}-\dfrac{12(u')^2}{u^6}=-8x^2$ $\dfrac{4u''}{u^5}=8x^2$ $u''=2x^2u^5$ This reduces to a special case of Emden-Fowler equation.

Stuck in integration: $\int {\frac{dx}{( 1+\sqrt {x})\sqrt{(x-{x}^2)}}}$ $\displaystyle\int {\frac{dx}{( 1+\sqrt {x})\sqrt{(x-{x}^2)}}}$ $=\displaystyle\int\frac{(1-\sqrt x)}{(1+x)\sqrt{x-x^2}}\,dx$ $=\displaystyle\int\frac{(1-\sqrt x+x-x)}{(1+x)\sqrt{x-x^2}}\,dx$ $=\displaystyle\int\frac{\,dx}{\sqrt{x-x^2}}-\displaystyle\int\frac{\sqrt x(1+\sqrt x)}{\sqrt x(1+x)\sqrt{1-x}}\,dx$ I have tried the above integration a lot . But not able to solve further . Please give me some hint how to proceed or some other method

By subtsitution twice we get $t=\sqrt { x } \Rightarrow dt=\frac { dx }{ 2\sqrt { x } } $ $$\int { \frac { dx }{ \left( 1+\sqrt { x } \right) \sqrt { x-{ x }^{ 2 } } } } =2\int { \frac { d\sqrt { x } }{ \left( 1+\sqrt { x } \right) \sqrt { 1-x } } = } 2\int { \frac { dt }{ \left( 1+t \right) \sqrt { 1-{ t }^{ 2 } } } } \\ \\ $$ $t=\sin { z } \Rightarrow dt=\cos { z } dz$ $$\int { \frac { \cos { z } dz }{ \left( 1+\sin { z } \right) \cos { z } } } =2\int { \frac { dz }{ 1+\sin { z } } } \overset { multiplying\quad both\quad (1-\sin { z) } \quad }{ = } 2\int { \frac { 1-\sin { z } }{ \cos ^{ 2 }{ z } } } dz=2\left( \int { \frac { dz }{ \cos ^{ 2 }{ z } } -\int { \frac { \sin { z } dz }{ \cos ^{ 2 }{ z } } } } \right) =2\left( \tan { z } -\frac { 1 }{ \cos { z } } \right) +C=\\ =2\left( \tan { \left( \arcsin { \left( \sqrt { x } \right) } \right) -\frac { 1 }{ \cos { \left( \arcsin { \left( \sqrt { x } \right) } \right) } } } \right) +C=2\left( \frac { \sqrt { x } }{ \sqrt { 1-x } } -\frac { 1 }{ \sqrt { 1-x } } \right) +C=2\frac { \sqrt { x } -1 }{ \sqrt { 1-x } } +C$$

Help solving $1 < \frac{x + 3}{x - 2} < 2$ I worked a lot of inequalities here in MSE and that greatly helped me. I've seen a similar inequality [here] [1], but the one I have today is significantly different, in that I'll end up with a division by 0, which is not possible. [1] [Simple inequality $$1 < \frac{x + 3}{x - 2} < 2$$ $$\implies 1 < \frac{x - 2 + 2 + 3}{x - 2}$$ $$\implies 1 < 1 + \frac{5}{x - 2} < 2$$ $$\implies 0 < \frac{5}{x - 2} < 1$$ $$\implies 0 < \frac{1}{x - 2} < \frac{1}{5}$$ I can't continue without falling into the division by 0. So my many thanks for your help.

Break it into two separate inequalities when you get to here: $$ 0 < \frac{5}{x-2} < 1$$ First let's consider $0 < \dfrac{5}{x-2}$. Since $5 > 0$, this inequality is satisfied when $x-2 > 0$, i.e., when $x > 2$. Now let's consider $\dfrac{5}{x-2} < 1$. Subtract $1$ from both sides and get a common denominator to get $$\dfrac{5}{x-2} - \dfrac{x-2}{x-2} < 0.$$ Combine and we have $$ \dfrac{5 - (x-2)}{x-2} < 0.$$ Simplify to get $$\frac{7-x}{x-2} < 0.$$ This inequality is satisfied when either of the following two conditions hold: * *$7-x > 0$ and $x-2 < 0$ *$7-x < 0$ and $x-2 > 0$ Can you take it from here?

Find the fourier series representation of a function Consider the function $f(x) = \begin{cases} \frac{\pi}{2}+x & & x \in (-\pi, 0] \\ \frac{\pi}{2}-x & & x \in (0, \pi]\\ \end{cases}$ extended 2$\pi$ periodically to $\mathbb{R}$. Calculate $a_0, a_n, b_n$ I understand how to work out a fourier series but I am unsure what to set for $f(x)$ due to the way its set out. Would I have $a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\pi}{2}+x dx$ due to splitting it into odd and even parts?

Divide it two parts and calculate $$a_{ 0 }=\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi } f\left( x \right) dx=\frac { 1 }{ \pi } \int _{ -\pi }^{ 0 }{ \left( \frac { \pi }{ 2 } +x \right) dx } +\frac { 1 }{ \pi } \int _{ 0 }^{ \pi }{ \left( \frac { \pi }{ 2 } -x \right) dx } =\\ ={ \left( \frac { \pi }{ 2 } x+\frac { { x }^{ 2 } }{ 2 } \right) }_{ -\pi }^{ 0 }{ +\left( \frac { \pi }{ 2 } x-\frac { { x }^{ 2 } }{ 2 } \right) }_{ -\pi }^{ 0 }=0\\ { a }_{ n }=\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi } f\left( x \right) \cos { \left( nx \right) } dx=\frac { 1 }{ \pi } \left( \int _{ -\pi }^{ 0 }{ \left( \frac { \pi }{ 2 } +x \right) \cos { \left( nx \right) } dx } +\int _{ 0 }^{ \pi }{ \left( \frac { \pi }{ 2 } -x \right) \cos { \left( nx \right) } dx } \right) \\ { b }_{ n }=\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi } f\left( x \right) \sin { \left( nx \right) } dx=\frac { 1 }{ \pi } \left( \int _{ -\pi }^{ 0 }{ \left( \frac { \pi }{ 2 } +x \right) \sin { \left( nx \right) } dx } +\int _{ 0 }^{ \pi }{ \left( \frac { \pi }{ 2 } -x \right) \sin { \left( nx \right) } dx } \right) $$ Can you take from here?

Inequality on non-negative reals For non-negative $x,y,z$ satisfy $\frac{1}{2x^2+1}+\frac{1}{2y^2+1}+\frac{1}{2z^2+1}=1$ then show that $x^2+y^2+z^2+6\geq 3(x+y+z)$ Idea how to handle the constraint? I'm unaware .

Yes, it's true for all reals. $\sum\limits_{cyc}(x^2-3x+2)=\sum\limits_{cyc}\left(x^2-3x+2-\frac{9}{4}\left(\frac{1}{2x^2+1}-\frac{1}{3}\right)\right)=\sum\limits_{cyc}\frac{(x-1)^2(2x-1)^2}{2(2x^2+1)}\geq0$. Done!

Two lines through a point, tangent to a curve We are looking for two lines through $(2,8)$ tangent to $y=x^3$. Let's denote the intersection point as $(a, a^3)$ and use the slope equation together with the derivative to get $\frac{a^3-8}{a-2}=3a^2$. This yields a cubic equation. Of course, one of the lines is tangent to $y=x^3$ at $(2,8)$, so we can get $a=2$ and the first line almost immediately. Knowing this, we write the cubic equation as $(a-2)(a^2+pa+q)$, find $p$ and $q$ from the original cubic equation and get $(a-2)(a-2)(a+1)=0$, which gives the solution and the lines. Is there a quicker way?

The tangents must be of the form $$y-8=m(x-2),$$ and they intersect the cubic $\color{blue}{y=x^3}$ when $$x^3-8=m(x-2).$$ This equation must have a double root, so that differentiating on $x$, we also have $$3x^2=m.$$ With the obvious solution $x=2$, we deduce $m=12$ and $$\color{green}{y-8=12(x-2)}.$$ Otherwise, we may simplify to get $$x^2+2x+4=m=3x^2.$$ This gives another solution $x=-1$, then $m=1$ and $$\color{magenta}{y-8=3(x-2)}.$$

How would I write a function for the following pattern? \begin{align} Y(0) ={}& 1\\ Y(1) ={}& 2.5\\ Y(2) ={}& 2.5\cdot2.3\\ Y(3) ={}& 2.5\cdot 2.3\cdot 2.1\\ Y(4) ={}& 2.5\cdot 2.3\cdot 2.1\cdot 1.9\\ \vdots\,\,\, \end{align} How would I solve for something like $Y(1.3)$ or $Y(2.7)$? How would a function for $Y(x)$ be defined?

Notice that \begin{align*} Y(x) &= \frac{25}{10} \cdot \frac{23}{10} \cdot \frac{21}{10} \cdots \frac{27 - 2x}{10} \\[5pt] &= \frac{1}{10^x} \cdot \frac{(26)(25)(24)(23)(22)(21) \cdots (28-2x)(27-2x)}{(26)(24)(22)\cdots (28-2x)} \\ &= \frac{1}{10^x} \cdot \frac{(26)(25)(24)(23)(22)(21) \cdots (2)(1)}{(26)(24)(22)\cdots (4)(2)} \cdot \frac{(26-2x)(24-2x)(22-2x)\cdots (4)(2)} {(26-2x)(25-2x)(24-2x)(23-2x)\cdots (2)(1)} \\ &= \frac{1}{10^x} \cdot \frac{26!}{2^{13} 13!} \cdot \frac{2^{13-x} (13-x)!} {(26-2x)!} \\ &= \boxed{\frac{1}{20^x} \cdot \frac{26!}{13!} \cdot \frac{(13-x)!}{(26-2x)!}}. \end{align*} In particular, this gives you a closed form for integer $x$ when $x = 0, 1, 2, 3, \ldots, 13$. But we'd like this to work for $x > 13$ as well (when the above formula is undefined), and for real $x$ instead of just integer $x$. In fact, the factorial function extends to a function $\Pi(z)$ for all complex $z$ except negative integers ($z! = \Pi(z) = \Gamma(z+1)$ where $\Gamma$ is the Gamma function). Then we get $$ Y(x) = \frac{1}{20^x} \cdot \frac{26!}{13!} \cdot \frac{\Pi(13-x)}{\Pi(26-2x)} $$ We still have a problem when $x > 13$ is an integer: $\Pi(13-x)$ and $\Pi(26-2x)$ are undefined but we expect $Y(x)$ to still have a formula. We can employ the duplication formula which says that $$ \Pi(2z) = \frac{1}{\sqrt{\pi}} 4^z \Pi(z) \Pi(z - \tfrac12) $$ with $z = 13-x$ to obtain \begin{align*} Y(x) &= \frac{1}{20^x} \cdot \frac{26!}{13!} \cdot \frac{\Pi(13-x)}{(\tfrac{1}{\sqrt{\pi}})4^{13-x} \Pi(13-x)\Pi(\tfrac{25}{2} - x)} \\ &= \boxed{\left( \frac{26! \sqrt{\pi}}{13! \;4^{13}} \right) \left(\frac{1}{5^x}\right) \left(\frac{1}{\Pi(\tfrac{25}{2} - x)}\right)}. \end{align*} The constant term in front is of course $\Pi(\tfrac{25}{2}) = \Gamma(\tfrac{27}{2})$, so this agrees with Raymond Manzoni's answer.

If $(a + b\sqrt c)^n = d + e\sqrt c$, then $(a - b\sqrt c)^n = d - e\sqrt c$ I think that: If $a,b,(c\ge0$ not a prefect square$),d,e,f\in\mathbb Z$ such that for some $n \ge 1$, $(a + b\sqrt c)^n = d + e\sqrt c$, then $(a - b\sqrt c)^n = d - e\sqrt c$ Is this true? Can someone provide a proof or give a hint for how to prove this? (preferably without induction if it doesn't provide insight)

Consider the set \begin{align*} A = \{ x + y\sqrt{c}: x, y \in \mathbb{Z} \} \end{align*} Consider the map $f: A \rightarrow A$ defined by $f(x+y\sqrt{c}) = x - y \sqrt{c}$. It is easy to see that $f$ is well defined (since $c$ is not a perfect square) and for any two $a_1, a_2 \in A$, \begin{align*} f(a_1+a_2) &= f(a_1) + f(a_2)\\ f(a_1 \cdot a_2) &= f(a_1)\cdot f(a_2) \end{align*} Thus if $(a + b\sqrt c)^n = d + e\sqrt c$, then \begin{align*} f((a + b\sqrt c)^n) &= f(d + e\sqrt c) \\ (f(a+b\sqrt{c}))^n &= d-e\sqrt{c} \\ (a-b\sqrt{c})^n &= d-e\sqrt{c} \end{align*}

Suppose we have two dice one fair and one tha brings $6$ with quintuple probability.Find the probability to throw randomly one die and show $6$ Suppose we have two dice one fair and one that brings $6$ with quintuple probability than the other numbers.We get a die randomly and we throw it.What is the probability to have $6$?In the same problem if we know that we have $6$ what is the probability that we have throwed the second die? Any ideas for these parts especially the second one?

The probability of rolling 6 on the fair die is obviously $\frac{1}{6}$. Let $x$ denote the probability of rolling 6 on the non-fair die: * *Then $\frac{1}{5}x$ is the probability of rolling each one of the other $5$ values *Therefore $x+5\cdot\frac{1}{5}x=1$, therefore $2x=1$, therefore $x=\frac{1}{2}$ So the probability of rolling 6 on the non-fair die is $\frac{1}{2}$. What is the probability of rolling 6? Split it into disjoint events, and then add up their probabilities: * *The probability of choosing the fair die and then rolling 6 on that die is $\frac{1}{2}\cdot\frac{1}{6}=\frac{1}{12}$ *The probability of choosing the non-fair die and then rolling 6 on that die is $\frac12\cdot\frac{1}{2}=\frac{1}{4}$ So the probability of rolling 6 is $\frac{1}{12}+\frac{1}{4}=\frac{1}{3}$. If we know that we rolled 6, what is the probability that it was on the non-fair die? Use Bayes formula for conditional probability: * *Let $A$ denote the event of rolling the non-fair die *Let $B$ denote the event of rolling 6 So $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{12}+\frac{1}{4}}=\frac{3}{4}$.

Squaring Infinite Series Expansion Of e^x $Fact$:$$\lim\limits_{n \to \infty}\frac{x^0}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots+\frac{x^n}{n!}=e^x$$ so $$\lim\limits_{n \to \infty}\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}=e$$ also $$\lim\limits_{n \to \infty}e^2=\frac{2^0}{0!}+\frac{2}{1!}+\frac{4}{2!}+\frac{8}{3!}+\dots+\frac{2^n}{n!}$$ can I safely say... $$L.H.S=(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots)^2=\frac{2^0}{0!}+\frac{2}{2!}+\frac{4}{2!}+\frac{8}{3!}+\dots=R.H.S$$ if yes, how to prove L.H.S=R.H.S (without using the above fact)? It is not a problem from textbook I was just thinking about it.

Since $e^ne^m = e^{n+m}$ you can simply let $x = 2t$, giving you $$(e^{t})^2 = e^{2t} = 1 + 2t + \frac{4t^2}{2!}+\frac{8t^3}{3!} + \cdots$$ Now, let $t = 1$ and you have $$e^2 = 1 + 2 + \frac{4}{2!} + \frac{8}{3!} + \cdots = (1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots )^2 = (e^1)^2.$$

Find the second smallest integer such that its square's last two digits are $ 44 $ Given that the last two digits of $ 12^2 = 144 $ are $ 44, $ find the next integer that have this property. My approach is two solve the equation $ n^2 \equiv 44 \pmod{100}, $ but I do not know how to proceed to solve that equation. I try a different path by letting $ n = 10x + y $ for some integers $ x, y, $ where $ 0 \le y \le9. $ Then $ n^2 \equiv 44 \; \pmod{100} $ can be reduced to $ 20xy + y^2 \equiv 44 \pmod{100}. $ At this point I let $ x $ run from $ 0, 1, 2, \dots $ and find the integer $ y \in \mathbb{Z}_{100} $ such that $ y^2 + 20xy - 44 = 0. $ My question is is there an alternative way to tackle this problem without having to try each $ x $ and $ y? $ Maybe try to solve the initial congruence equation $ n^2 \equiv 44 \pmod{100}. $

If $x^2$ ends with $44$ then $x$ is even. Let $y=2x$. We are trying to solve $$(2y)^2\equiv 44\pmod{100}$$ and this equation is equivalent to $$y^2\equiv 11\pmod{25}$$ Since $6^2\equiv 11\pmod{25}$ this equation can be written as $$(y-6)(y+6)\equiv 0\pmod{25}$$ It is not possible that both $y-6$ and $y+6$ are multiples of $5$, so the solutions are $y\equiv\pm6\pmod{25}$. The first positive values for $y$ are $$6,19,31,44,56,69,81,94,106,\ldots$$ Multiply these numbers by $2$ to get the solutions for $x$.

Finding $a$ in quadratic equation $2x^2 - (a+1)x + (a-1)=0$ so that difference of two roots is equal to its product Given equation: $$2x^2 - (a+1)x + (a-1)=0$$ I have to find when the difference of two roots is equal to its product, i.e.: $$x_1x_2 = x_1 - x_2.$$ From Vieta's formulas we know that: $$x_1 + x_2 = \frac{a + 1}{2},$$ $$x_1x_2 = \frac{a-1}{2} = x_1 - x_2.$$ Then, solving system of equations: $$x_1 + x_2 = \frac{a + 1}{2}$$ $$x_1 - x_2 = \frac{a-1}{2}$$ we get that $x_1 = \frac{a}{2}$ and $x_2 = \frac{1}{2}$. Then, plugging it into equation $x_1x_2 = x_1 - x_2$ we get: $$\frac{a}{4} = \frac{a - 1}{2}$$ $$4a - 4 = 2a$$ $$2a = 4$$ $$a = 2.$$ Plugging $2$ into previous equation we get that $\frac{1}{2} = \frac{1}{2}$, so solution have to be true. Is my approach correct? If so, are there another ways to solve those kind of problems?

Using by the formula $${ \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+4{ x }_{ 1 }{ x }_{ 2 }={ \left( { x }_{ 1 }+{ x }_{ 2 } \right) }^{ 2 }$$ make be easy

Is the difference of two irrationals which are each contained under a single square root irrational? Is $ x^\frac{1}{3} - y^\frac{1}{3}$ irrational, given that both $x$ and $y$ are not perfect cubes, are distinct and are integers (i.e. the two cube roots are yield irrational answers)? I understand that the sum/difference of two irrationals can be rational (see this thread: Is the sum and difference of two irrationals always irrational?). However, if my irrationals are contained under one root (so for example $3^\frac{1}{3}$ and not $2^\frac{1}{2} + 1$), can one generalise to show that $ x^\frac{1}{p} - y^\frac{1}{q} $ is irrational, where of course $x$ and $y$ are not powers of $p$ and $q$ respectively?

Denote the cube roots by $X,Y$, so that $X^3=x$ and $Y^3=y$ with $x,y\in \mathbb Z$. Suppose, with slightly greater generality, that we have $X-Y-R=0$ where $X^3,Y^3,R\in \mathbb Q$. We first want to argue that $XY\in \mathbb Q$. To do so, observe the identity: $$X^3-Y^3-R^3-3XYR=(X-Y-R)(X^2+Y^2+R^2+XY+XR-YR)$$ This immediately implies that $$3XYR=X^3-Y^3-R^3$$ which, as desired, implies that $XY\in \mathbb Q$. But then we have two real numbers, $X,Y$ such that both $X-Y$ and $XY$ are rational. It follows that $X,-Y$ both satisfy a quadratic equation with rational coefficients, namely \begin{align} X^2 - (X-Y)X -XY &= 0,& Y^2 +(X-Y)Y -XY&=0. \end{align} But this is not possible, as $Z^3-x=0$ is the minimal polynomial for the cube root of $x$ over $\mathbb Q$.

Evaluate the integral $\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$ This integral looks a lot like an elliptic integral, but with cubes instead of squares: $$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$$ Let's consider $a,b>0$ for now. $$I(a,a)=\int_0^\infty \frac{dx}{x^3+a^3}=\frac{2 \pi}{3 \sqrt{3} a^2}$$ I obtained the general series solution the following way. Choose $a,b$ such that $a \geq b$, then: $$I(a,b)=\frac{1}{a^2} \int_0^\infty \frac{dt}{\sqrt{(t^3+1)(t^3+b^3/a^3)}}=\frac{1}{a^2} I \left(1, \frac{b}{a} \right)$$ $$\frac{b^3}{a^3}=p, \qquad I \left(1, \frac{b}{a} \right)=I_1(p)$$ $$I_1(p)=\int_0^\infty\frac{dt}{\sqrt{(t^3+1)(t^3+p)}}=2 \frac{d}{dp} J(p)$$ $$J(p)=\int_0^\infty\sqrt{\frac{t^3+p}{t^3+1}}dt=\int_0^\infty\sqrt{1+\frac{p-1}{t^3+1}}dt=$$ $$|p-1| \leq 1$$ $$=\sum_{k=0}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) (p-1)^k \int_0^\infty \frac{dt}{(t^3+1)^k}$$ Now this is the most problematic part. The first integral of this series diverges. However, it's a constant in $p$, so if we differentiate, it formally disappears: $$I_1(p)=2 \sum_{k=1}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) k (p-1)^{k-1} \int_0^\infty \frac{dt}{(t^3+1)^k}$$ Now, every integral in this series converges. The integtals can be computed using the Beta function, if we substitute: $$t^3=\frac{1}{u}-1$$ Finally, we rewrite: $$I_1(p)=\frac{\Gamma (1/3)}{3 \sqrt{\pi}} \sum_{k=1}^\infty \frac{k^2}{k!^2} \Gamma \left(k- \frac{1}{2}\right) \Gamma \left(k- \frac{1}{3}\right) (1-p)^{k-1}$$ Or, using the Pochhammer symbol: $$I_1(p)=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{(k+1)^2}{(k+1)!^2} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k (1-p)^k$$ My questions are: Is the method I used valid (see the 'problematic part')? How to get this series into a Hypergeometric function form? Is there any 'arithmetic-geometric mean'-like transformation (Landen's transformation) for this integral? How to go about finding it? If the method I used is correct, it can be used for any integral of the form ($m \geq 2$): $$I_m(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^m+a^m)(x^m+b^m)}}$$

More generally, with $|p-1|<1$, some experimentation shows that, $$\int_0^\infty \frac{dt}{\sqrt{(t^m+1)(t^m+p)}} = \pi\,\frac{\,_2F_1\big(\tfrac12,\tfrac{m-1}{m};1;1-p\big)}{m\sin\big(\tfrac{\pi}{m}\big)}$$ where the question was just the case $m=3$.

Multi-index sum property Exercise 1.2.3.29 in Donald Knuth's The Art of Computer Programming (3e) states the following property of a multi-indexed sum: $$ \sum_{i=0}^n \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k = \frac{1}{3}S_3 + \frac{1}{2}S_1S_2 + \frac{1}{6}S_1^3, $$ where $S_r = \sum_{i=0}^n a_i^r$. I tried to prove it and failed. What I found are the following identities: $$ \sum_{i=0}^n \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k = \frac{1}{2} \sum_{i=0}^n a_i \left( \left( \sum_{j=0}^i a_j \right)^2 + \sum_{j=0}^i a_j^2 \right), $$ $$ S_1S_2 = \sum_{i=0}^n \sum_{j=0}^n a_i a_j^2, $$ $$ S_1^3 = \sum_{i=0}^n \sum_{j=0}^n \sum_{k=0}^n a_i a_j a_k. $$ Can anybody help in completing the proof? Based on grand_chat's answer, the identity can be proven inductively. $$ \begin{align} \sum_{i=0}^{n+1} \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k & = \sum_{i=0}^n \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k + a_{n+1} \sum_{i=0}^{n+1} \sum_{j=0}^i a_ia_j \\ & \stackrel{hyp}{=} \underbrace{\frac{1}{3}S_{n,3} + \frac{1}{2}S_{n,1}S_{n,2} + \frac{1}{6}S_{n,1}^3}_{=:S_n} + \frac{1}{2} a_{n+1} \left( \left( \sum_{i=0}^{n+1} a_i \right)^2 + \sum_{i=0}^{n+1} a_i^2 \right) \\ & = S_n + \frac{1}{2} a_{n+1} \left( \left( \sum_{i=0}^n a_i + a_{n+1} \right)^2 + \sum_{i=0}^n a_i^2 + a_{n+1}^2 \right) \\ & = S_n + \frac{1}{2} a_{n+1} \left( \left( \sum_{i=0}^n a_i \right)^2 + 2 \sum_{i=0}^n a_i a_{n+1} + a_{n+1}^2 + \sum_{i=0}^n a_i^2 + a_{n+1}^2 \right) \\ & = S_n + a_{n+1}^3 + \sum_{i=0}^n a_i a_{n+1}^2 + \frac{1}{2} \left( \sum_{i=0}^n a_i \right)^2 a_{n+1} + \frac{1}{2} \sum_{i=0}^n a_i^2 a_{n+1} \\ & = S_n + a_{n+1}^3 \left( \frac{1}{3} + \frac{1}{2} + \frac{1}{6} \right) + \sum_{i=0}^n a_i a_{n+1}^2 \left( \frac{1}{2} + \frac{3}{6} \right) + \frac{3}{6} \left( \sum_{i=0}^n a_i \right)^2 a_{n+1} + \frac{1}{2} \sum_{i=0}^n a_i^2 a_{n+1} \\ & = \frac{1}{3} \left( S_{n,3} + a_{n+1}^3 \right) + \frac{1}{2} \left( S_{n,1}S_{n,2} + \sum_{i=0}^n a_i a_{n+1}^2 + \sum_{i=0}^n a_i^2 a_{n+1} + a_{n+1}^3 \right) + \frac{1}{6} \left( S_{n,1}^3 + 3 \left( \sum_{i=0}^n a_i \right)^2 a_{n+1} + 3 \sum_{i=0}^n a_i a_{n+1}^2 + a_{n+1}^3 \right) \\ & = \frac{1}{3}S_{n+1,3} + \frac{1}{2}S_{n+1,1}S_{n+1,2} + \frac{1}{6}S_{n+1,1}^3. \end{align} $$ The problem here is that the hypothesis is taken from thin air.

For future reference this is the sum over all multisets of size three chosen from the variables $A_0$ to $A_n$ and evaluated at $a_q.$ Therefore by the Polya Enumeration Theorem it is given by $$\left.Z(S_3)\left(\sum_{q=0}^n A_q\right)\right|_{A_q=a_q}$$ where $Z(S_3)$ is the cycle index of the symmetric group $Z(S_3).$ Now the permutations are $(1)(2)(3)$, $(12)(3)$, $(13)(2)$, $(23)(1)$ and $(123)$ and $(132).$ Therefore the cycle index is given by $$Z(S_3)= \frac{1}{6} (s_1^3 + 3 s_1 s_2 + 2 s_3) = \frac{1}{6} s_1^3 + \frac{1}{2} s_1 s_2 + \frac{1}{3} s_3.$$ This gives for the sum $$ \left.\frac{1}{6} s_1^3 + \frac{1}{2} s_1 s_2 + \frac{1}{3} s_3 \right|_{s_r=S_r} = \frac{1}{6} S_1^3 + \frac{1}{2} S_1 S_2 + \frac{1}{3} S_3$$ where we have used the standard cycle index substitution $$s_r = \left.\sum_{q=0}^n A_q^r\right|_{A_q=a_q} = S_r.$$

Find the value of $\tan A + \tan B$, given values of $\frac{\sin (A)}{\sin (B)}$ and $\frac{\cos (A)}{\cos (B)}$ Given $$\frac{\sin (A)}{\sin (B)} = \frac{\sqrt{3}}{2}$$ $$\frac{\cos (A)}{\cos (B)} = \frac{\sqrt{5}}{3}$$ Find $\tan A + \tan B$. Approach Dividing the equations, we get the relation between $\tan A$ and $\tan B$ but that doesn't help in getting the value of $\tan A + \tan B$. The value comes in terms of $\tan A$ or $\tan B$ but the expected answer is independent of any variable . Also $$\frac{\sin(A)\cdot\cos(B) + \sin(B)\cdot\cos(A)}{\cos(A)\cdot\cos(B)} = \tan(A) + \tan(B)$$ We could get a value only if instead of $\cos A$ there was $\sin B$ in the relation(which we get on adding the ratios)

Although there is something wrong with this question,there is a way which I think maybe a little bit easier to solve this kind of problem. $$\tan A + \tan B = \frac{\sin(A)\cdot\cos(B) + \sin(B)\cdot\cos(A)}{\cos(A)\cdot\cos(B)} = \frac{{\sin A \over \sin B}+ {\cos A \over \cos B}}{\cos A \over \sin B} $$ then to get the value of $\cos A \over \sin B $ : $$ {\sin A \over \sin B }= {\sqrt{3} \over 2} \Rightarrow {(1-\cos^2 A) \over \sin^2 B} = {3 \over 4} \Rightarrow {1 \over \cos^2A} = 1+{{3\cdot \sin^2 B} \over 4\cdot \cos^2A }$$ $${\cos A \over \cos B }= {\sqrt{5} \over 3} \Rightarrow {\cos^2A \over (1-\sin^2 B)} = {5 \over 9} \Rightarrow {1 \over \cos^2A} = {9 \over 5}+{{\sin^2 B} \over \cos^2A }$$ then $$ {\sin^2B \over \cos^2A }={-16 \over 5} $$ this is impossible. But if the question is right and then you can get the value of $\cos A \over \sin B $ ,and just get the answer with the first equation.

Product of roots of $ax^2 + (a+3)x + a-3 = 0$ when these are positive integers There is only one real value of $'a'$ for which the quadratic equation $$ax^2 + (a+3)x + a-3 = 0$$ has two positive integral solutions.The product of these two solutions is : Since the solutions are positive, therefore the product of roots and sum of root will be positive. This will give us two inequalities in $a$. Substitute the values of $a$ in the quadratic but I'm not getting my answer correct. The two inequalities will be $$\frac{a-3}{a} > 0$$ and the other one will be $$\frac {a+3}{a} < 0$$ From the first one we get $a>3$ and from the second one we get $a< -3$. I don't know how to proceed after this. Kindly help.

We need $$\frac{a+3}{a}\in\Bbb{Z}\ , \frac{a-3}{a}\in\Bbb{Z}$$ or $$1+\frac{3}{a}\in\Bbb{Z}\ , \ 1-\frac{3}{a}\in\Bbb{Z}$$ thus $\displaystyle \frac{3}{a}\in\Bbb{Z}$, means that $\displaystyle a=\frac{3}{m}$ where $m\in\Bbb{Z}$. Now we can write the equation as $$\frac{3}{m}x^2+\left(\frac{3}{m}+3\right)x+\frac{3}{m}-3=0\implies x^2+(m+1)x+1-m=0$$ Using quadratic formula we have $$x_{1,2}=\frac{-m-1\pm\sqrt{m^2+6m-3}}{2}=\frac{-m-1\pm\sqrt{(m+3)^2-12}}{2}$$We need the expression under the square root to be an integer. There are only two squares with difference $12$ (those are $4,16$), hence we want $(m+3)^2=16\implies m=1\text{ or }m=-7$, thus $$a=3 \text{ or } a=-\frac{3}{7}$$ and as $a<0$ we have $$a=-\frac{3}{7}$$

If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$. Let $a,b,c$ be the lengths of the sides of a triangle. Prove that $$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$ Attempt. By clearing the denominators, the required inequality is equivalent to $$a^2(b+c)+b^2(c+a)+c^2(a+b)>a^3+b^3+c^3\,.$$ Since $b+c>a$, $c+a>b$, and $a+b>c$, the inequality above is true. Is there a better, non-bruteforce way?

\begin{align*} \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} & = \frac{2a}{2(b+c)}+\frac{2b}{2(c+a)}+\frac{2c}{2(a+b)} \\ &< \frac{2a}{a+b+c} + \frac{2b}{c+a+b} + \frac{2c}{a+b+c} \\ &= 2 \end{align*}

Is A276175 integer-only? The terms of the sequence A276123, defined by $a_0=a_1=a_2=1$ and $$a_n=\dfrac{(a_{n-1}+1)(a_{n-2}+1)}{a_{n-3}}\;,$$ are all integers (it's easy to prove that for all $n\geq2$, $a_n=\frac{9-3(-1)^n}{2}a_{n-1}-a_{n-2}-1$). But is it also true for the sequence A276175 defined by $a_0=a_1=a_2=a_3=1$ and $$a_n=\dfrac{(a_{n-1}+1)(a_{n-2}+1)(a_{n-3}+1)}{a_{n-4}} \;\;?$$ Update : I crossposted to MO.

Yes, $(a_n)$ is a sequence of integers. To prove this we first need to study some auxiliary sequences that satisfy a polynomial recurrence relation (unlike $(a_n)$ which has a rational fraction as its recurrence). Consider the sequences $(b_n)$ of positive reals satisfying the recurrence relation $b_nb_{n+4} = b_{n+1}b_{n+2}b_{n+3} + 1$. It turns out we can express $b_{n+8}$ as a polynomial in $b_n, \ldots, b_{n+7}$ : Since $b_{n+1}b_{n+5} \equiv b_{n+2}b_{n+6} \equiv b_{n+3}b_{n+7} \equiv 1 \pmod {b_{n+4}}$ and $b_{n+1}b_{n+2}b_{n+3} \equiv -1 \pmod {b_{n+4}}$, we have $b_{n+5}b_{n+6}b_{n+7} \equiv -1 \pmod {b_{n+4}}$, which suggests the existence of a formula for $b_{n+8}$. With this roadmap, we can write $(b_{n+1}b_{n+2}b_{n+3})(b_{n+5}b_{n+6}b_{n+7}+1) \\ = (b_{n+1}b_{n+5})(b_{n+2}b_{n+6})(b_{n+3}b_{n+7}) + (b_{n+1}b_{n+2}b_{n+3}) \\ = (b_{n+2}b_{n+3}b_{n+4}+1)(b_{n+3}b_{n+4}b_{n+5}+1)(b_{n+4}b_{n+5}b_{n+6}+1)+(b_nb_{n+4}-1) \\ = b_{n+4}.F(b_{n+i})$ where $F$ is some big polynomial. And finally, $(b_{n+5}b_{n+6}b_{n+7}+1) = (b_{n+5}b_{n+6}b_{n+7}+1)(b_nb_{n+4} - b_{n+1}b_{n+2}b_{n+3}) \\ = b_{n+4}(b_nb_{n+5}b_{n+6}b_{n+7}+b_n - F(b_{n+i})) = b_{n+4} G(b_{n+i})$. And so, $b_{n+8} = G(b_{n+i})$. This means that if $b_0, \ldots, b_7 \in R$ for some subring $R$ of $\Bbb R$, then the whole sequence is in $R$. Now to link back to the original sequence. Given such a sequence $(b_n)$, we define a sequence $(a_n)$ by $a_n = b_nb_{n+1}b_{n+2}$. This sequence satisfies $a_na_{n+4} = (b_n b_{n+1}b_{n+2})(b_{n+4}b_{n+5}b_{n+6}) \\ = (b_n b_{n+4})(b_{n+1} b_{n+5})(b_{n+2} b_{n+6}) = (b_{n+1}b_{n+2}b_{n+3}+1)(b_{n+2}b_{n+3}b_{n+4}+1)(b_{n+3}b_{n+4}b_{n+5}+1) \\ = (a_{n+1}+1)(a_{n+2}+1)(a_{n+3}+1)$. Finally, taking $b_0 \ldots b_7 = \frac 12, 4, \frac 12, \frac 12, 4, \frac 12, 4, 18$, we obtain a sequence $(b_n)$ with terms in $\Bbb Z[\frac 12]$, with the corresponding $(a_n)$ sequence $1,1,1,1,8,36, \ldots$ Since the recurrence relation is symmetric, it can go backwards as well as forward, hence the ring $R_n = \Bbb Z[b_n, \ldots, b_{n+7}]$ is independant of $n$. There is no hope of finding $8$ consecutive integer values in our sequence $b_n$. If we look at the sequence $(b_n)$ modulo $8$, from our first octuplet and by applying the polynomial transformation, we can get to $17225$ different octuplets mod $8$, and none of those correspond to any noninteger $a_n$. This computation proves that $a_n$ is an integer forall $n$ (be careful, one step can go from one octuplet to several octuplets, because precision can be lost sometimes). Note that using this definition, $a_na_{n+2}/a_{n+1}(a_{n+1}+1) = b_nb_{n+2}b_{n+4}/(b_{n+1}b_{n+2}b_{n+3}+1) = b_{n+2}$, and so to go in the other direction you have to define $(b_n)$ from $(a_n)$ with $b_n = a_{n-2}a_n/a_{n-1}(a_{n-1}+1)$. Then, once again the recurrence relation of $(b_n)$ follows from that of $(a_n)$. This shows that for any such rational sequence $(a_n)$, there is a corresponding rational sequence $(b_n)$, and so $(a_n)$ is in a finitely generated subring of $\Bbb Q$.

Trouble understanding proof of the inequality - $(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1) \ge 64 $, for $a,b,c > 0$ and $a+b+c = 1$ I was looking into this problem in a book discussing inequalities, However I found the proof quite hard to understand.The problem is as follows: Let $a,b,c$ be positive numbers with $a+b+c=1$, prove that $$\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right) \ge 64$$ and the proof provided was the following: Note that $$ abc \le (\frac{a+b+c}{3})^3 = \frac{1}{27} \tag{1}$$ by AM-GM inequality. Then $$(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc} \tag{2}$$ $$\ge 1+\frac{3}{\sqrt[3]{abc}}+\frac{3}{\sqrt[3]{(abc)^2}} +\frac {1}{abc} \tag{3}$$ $$=(1+\frac{1}{\sqrt[3]{abc}})^3 \ge 4^3 \tag{4}$$ Steps 1 and 2 are easy for me to understand, but if someone could help me with steps 3 and 4 ,I would be very thankful.

using for $$\frac{1}{a},\frac{1}{b},\frac{1}{c}$$ the AM-GM inequality we obtain $$\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq \sqrt[3]{\frac{1}{abc}}$$ and for $$\frac{1}{ab},\frac{1}{bc},\frac{1}{ca}$$ the same we get $(3)$

Prove $\int_0^\infty \frac{dx}{\sqrt{(x^4+a^4)(x^4+b^4)}}=\frac{\pi}{2 \sqrt2 a b} ( \text{agm} (\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} ))^{-1}$ The following definite integral turns out to be expressible as the Arithmetic-Geometric Mean: $$I_4(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^4+a^4)(x^4+b^4)}}=\frac{\pi}{2 \sqrt2 a b} \left( \text{agm} \left(\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} \right)\right)^{-1}$$ $$I_4(1,1)=\frac{\pi}{2 \sqrt2}$$ I would like to remind you that: $$I_2(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}}=\frac{\pi}{2} \left( \text{agm} \left(a,b \right)\right)^{-1}$$ $$I_2(1,1)=\frac{\pi}{2}$$ Which is why I have two questions: How do we prove the identity for $I_4(a,b)$? Is it possible to also express other integrals of this type using agm? Such as $I_8(a,b)$? Because of the relation of the agm to elliptic integrals, we can also write: $$I_4(a,b)=\frac{1}{a b \sqrt{a^2+b^2}} K \left( \frac{a-b}{\sqrt{2(a^2+b^2)}} \right)$$ Here the parameter convention is $$K(k)=\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2 t^2)}}$$ This seems to be the best way to prove the identity, but I don't know which substitution to use. Another way to express this integral would be through the hypergeometric function: $$I_4(a,b)=\frac{\pi}{2 \sqrt2 a^3} {_2F_1} \left(\frac{1}{2},\frac{3}{4};1;1-\frac{b^4}{a^4} \right)$$ And for every integral of this type we have: $$I_m(a,b)=\frac{I_m(1,1)}{a^{m-1}} {_2F_1} \left(\frac{1}{2},\frac{m-1}{m};1;1-\frac{b^m}{a^m} \right)$$ The outline for the proof can be found in this question for $m=3$ and is easily adapted to the general case. Here we assume $a \geq b$. As for the arithmetic geometric mean, I tried to get it into a simpler form, but the only transformation I was able to achieve is this: $$\text{agm} \left(\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} \right)=\text{agm} \left(\frac{a+b}{2}+i \frac{a-b}{2},\frac{a+b}{2}-i \frac{a-b}{2} \right)$$ Since we have complex conjugates, it's quite obvious, that they will give real numbers at first iteration, an it would give the left hand side.

Substitition is sufficient. Let $$\displaystyle z=x-\frac1x,w=x+\frac1x$$ then $$\displaystyle \frac{\mathrm dx}{\sqrt{x^8+p x^4+1}}=\frac12\left( \frac{\mathrm dz}{\sqrt{z^4+4z^2+2+p}}+\frac{\mathrm dw}{\sqrt{w^4-4w^2+2+p}}\right) $$ So $$f(p)=\displaystyle \int_0^\infty \frac{\mathrm dx}{\sqrt{x^8+p x^4+1}} =\frac12\int_{-\infty}^\infty \frac{\mathrm dx}{\sqrt{x^4+4x^2+2+p}} =\int_0^\infty \frac{\mathrm dx}{\sqrt{x^4+4x^2+2+p}} $$ let $$x^4+4x^2+2+p= (x^2+u^2)(x^2+v^2) $$ We know that $$\displaystyle \operatorname{agm}(u,v)= \frac{\pi}2\left( \int_0^\infty \frac{\mathrm dx}{\sqrt{(x^2+u^2)(x^2+v^2)}}\right)^{-1} =\frac{\pi}2 f(p)^{-1} $$ With the identity of agm, $$ \operatorname{agm}(u,v)=\operatorname{agm}\left(\frac{u+v}2,\sqrt{uv}\right) =\operatorname{agm}\left(\sqrt{\frac{\sqrt{p+2}}2+1},\sqrt[4]{p+2}\right) $$ So $$ \int_0^\infty \frac{\mathrm dx}{\sqrt{(x^4+a^4)(x^4+b^4)}} =\frac1{ab\sqrt{ab}}f\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right) =\frac\pi{2ab\sqrt{ab}}\left(\operatorname{agm}\left( \frac{a+b}{\sqrt{2ab}},\sqrt{\frac{a^2+b^2}{ab}} \right)\right)^{-1} =\frac\pi{2\sqrt2 ab}\left(\operatorname{agm}\left( \frac{a+b}2,\sqrt{\frac{a^2+b^2}2} \right)\right)^{-1} $$

If $a+b+c=0$ then the roots of $ax^2+bx+c=0$ are rational? If $a+b+c=0$ then the roots of $ax^2+bx+c=0$ are rational ? Is it a "If and only if " statement or "only if " statement ? For $a,b,c \in \mathbb Q$ , I think it is a "if and only if" statement . Am I correct ? I can prove that if $a+b+c=0$ and $a,b,c \in \mathbb Q$ , then roots are rational. But I can not prove that if roots are rational and $a,b,c \in \mathbb Q$ , then $a+b+c=0$. Any help ? or any conditions that $ax^2 +bx+c =0 $ should satisfy in order to have rational roots ? My clarifications : 1) What is a rational number ? It is a number which can be written in the form $\frac{p}{q} $ , where $q \neq 0$ and $p , q \in \mathbb Z$ 2) In this case : Let's take the quadratic equation $ax^2+bx+c=0$ where $a \neq 0$. We all know that the roots are given by , $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ Case 1 : Suppose that $a,b,c \in \mathbb Q$. Then $x$ is rational if and only if $b^2-4ac$ is a perfect square or zero $(0,1,4,9,16,...)$. Now we need to make $b^2-4ac$ a perfect square ! Now observe that if $a+b+c=0$ , then $b^2-4ac = (-a-c)^2-4ac=(a-c)^2$ That is if $a,b,c \in \mathbb Q$ and $a+b+c=0$ then the solutions are rational. Case 2 : Suppose that $a,b,c \in \mathbb Q$. If $c=0$ , then all the roots are rational. (This is easy if all $a,b,c$ are rationals) Case 3 : Suppose that $b,c \in \mathbb R- \mathbb Q$.(If $a$ is irrational we can divide by $a$ ) $a+b+c=0$ condition does not satisfy. Ex : $(1-\sqrt{2})x^2-2x+(1+\sqrt{2})=0$

If $c = -a-b$, then the discriminant is $$b^2-4ac = b^2 +4a(a+b) = b^2 +4ab +4a^2 = (b+2a)^2.$$ Since the discriminant is a perfect square, then the roots are always rational. Their values are $$\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} =\dfrac{-b \pm (b+2a)}{2a} \in \left\{ 1, -\dfrac{a+b}{a} \right\}$$ Of course, now that you know this, you can say that $$(x-1)(ax+a+b) = ax^2 + bx -(a+b) = ax^2+ bx + c$$

Evaluate the integral $\int_0^\pi \sin{(x \cos{t}})\cos{t}\; dt$ How to evaluate: $\int \sin{(x \cos{t}})\cos{t}\; dt$ or: $\int_0^\pi \sin{(x \cos{t}})\cos{t}\; dt$

$\int\sin(x\cos t)\cos t~dt=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}\cos^{2n+2}t}{(2n+1)!}~dt$ For $n$ is any non-negative integer, $\int\cos^{2n+2}t~dt=\dfrac{(2n+2)!t}{4^{n+1}((n+1)!)^2}+\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin t\cos^{2k+1}t}{4^{n-k+1}((n+1)!)^2(2k+1)!}+C$ This result can be done by successive integration by parts. $\therefore\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}\cos^{2n+2}t}{(2n+1)!}~dt$ $=\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}t}{2^{2n+1}n!(n+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nx^{2n+1}(k!)^2\sin t\cos^{2k+1}t}{2^{2n-2k+1}n!(n+1)!(2k+1)!}+C$ $\therefore\int_0^\pi\sin(x\cos t)\cos t~dt$ $=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}t}{2^{2n+1}n!(n+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nx^{2n+1}(k!)^2\sin t\cos^{2k+1}t}{2^{2n-2k+1}n!(n+1)!(2k+1)!}\right]_0^\pi$ $=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\pi x^{2n+1}}{2^{2n+1}n!(n+1)!}$ $=\pi J_1(x)$ Specifically for $\int_0^\pi\sin(x\cos t)\cos t~dt$ , $\int_0^\pi\sin(x\cos t)\cos t~dt$ $=\int_0^\frac{\pi}{2}\sin(x\cos t)\cos t~dt+\int_\frac{\pi}{2}^\pi\sin(x\cos t)\cos t~dt$ $=\int_0^\frac{\pi}{2}\sin(x\cos t)\cos t~dt+\int_\frac{\pi}{2}^0\sin(x\cos(\pi-t))\cos(\pi-t)~d(\pi-t)$ $=\int_0^\frac{\pi}{2}\sin(x\cos t)\cos t~dt+\int_0^\frac{\pi}{2}\sin(x\cos t)\cos t~dt$ $=2\int_0^\frac{\pi}{2}\sin(x\cos t)\cos t~dt$ $=2\int_\frac{\pi}{2}^0\sin\left(x\cos\left(\dfrac{\pi}{2}-t\right)\right)\cos\left(\dfrac{\pi}{2}-t\right)~d\left(\dfrac{\pi}{2}-t\right)$ $=2\int_0^\frac{\pi}{2}\sin(x\sin t)\sin t~dt$ $=\int_0^\frac{\pi}{2}\cos(x\sin t-t)~dt-\int_0^\frac{\pi}{2}\cos(x\sin t+t)~dt$ $=\int_0^\frac{\pi}{2}\cos(x\sin t-t)~dt-\int_\pi^\frac{\pi}{2}\cos(x\sin(\pi-t)+\pi-t)~d(\pi-t)$ $=\int_0^\frac{\pi}{2}\cos(x\sin t-t)~dt+\int_\frac{\pi}{2}^\pi\cos(x\sin t-t)~dt$ $=\int_0^\pi\cos(x\sin t-t)~dt$ $=\pi J_1(x)$

Show that any integer $n>7$ can be written as the sum of $3$s and $5$s exclusively Show that any integer $n>7$ can be written as the sum of $3$s and $5$s exclusively, i.e., $$ 8= 5+3 \\ 9=3+3+3 \\ 10 = 5+5 \\ 11 = 5+3+3 \\ 12 = 3+3+3+3 $$ So I've started in a couple directions without progress. I think it makes sense to write $$n=3x+5y$$ for $x,y \geq 0$, observing that either $3\mid n$ or $5\mid n$. It also seems that if you are at $n$, you can increment to $n+1$ by replacing a $5$ with two $3$s. Or you can take $n-1$ and replace a $3$ with a $5$. In this way, it seems like you should be able to keep incrementing after $n=8$, but these ideas aren't formalizing into anything...

$2*3 - 5 = 1$ $2n*3 - 5n = n$. $3(2n - 5k) + 5(3k - n)= n$. To assure that $2n - 5k \ge 0$ and $3k - n > 0$... If $n = 3m - r; r = 0, 1,2$ then $k$ can be anything equal or greater than $m$ so long as $2n - 5k \ge 0$ i.e. $6m - 2r - 5k \ge 0\implies k \le 6m/5 - 2r/5= m + \frac{m-2r}5$. So long as $m \ge 4$ we will always be able to find such $k$. i.e. so long as $n \ge 3*4 -2 = 10$. If $m = 3$ we'll be able find such $k$ if $r \le 1$ i.e. if $n = 8$ or $9$. We will not be able to find any such $k$ for $n =7$ (where $m = 3; r= 2$). ======= Or another way: If we can find $n = 3a + 5b$ we can find $n+1 = 3(a-3) + 5(b+2) = 3(a+2) + 5(b-1)$ so long as either $a \ge 3$ or $b \ge 1$. If we have $a + b \ge 2$ and $n=3a+5b$ we can find $n+1$ and in doing so either $a$ will increase by $2$ and $b$ (which was at least $1$) decrease by $1$ and $(a+2) + (b-1)\ge 2$ or $a$ (which was at least $3$) will decrease by $3$ but $b$ will in by $2$ so $(a - 3) + (b+2)\ge 2$. Thus by induction if $n = 3a + 5b; a+b \ge 2$ then $n+1 = 3a' + 5b'; a'+b' \ge 2$. Base case: $n = 8 = 3*1 + 5*1$. So is possible for all $n \ge 8$.

Counting using permutation and combination How many solutions are there to the equation $x_1+x_2+x_3+x_4+x_5=21$, where $x_i,i=1,2,3,4,5$, is a nonnegative integer such that $ 0 ≤ x_1 ≤ 3$, $1 ≤ x_2 < 4$, and $x_3 ≥ 15$? I tried it .My Approach-: $ x_3=x_3'+15 \implies x_1+x_2+x_3'+15+x_4+x_5=21 \implies x_1+x_2+x_3'+x_4+x_5=6 \implies C(5+6-1,6)$ but stuck at finding $0 ≤ x_1 ≤3$ $1 ≤ x_2 ≤4$ please help!!!

Let $y_3 = x_3 -15$. We need the number of solutions to $x_1+x_2+y_3+x_4+x_5 = 21 - 15$ with $y_3 \geq 0$, $0 \leq x_1 \leq 3$ and $1 \leq x_2 <4$. The number of solutions is the coefficient of $x^6$ in \begin{align*} (1+x+x^2+x^3)&(x+x^2+x^3)(1+x+x^2+\cdots)(1+x+x^2+\cdots)(1+x+x^2+\cdots)\\ &= x(1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)^3 \\ &= x\left(\frac{1-x^4}{1-x}\right)(1+x+x^2)\left(\frac{1}{1-x}\right)^3\\ &= x(1-x^4)(1+x+x^2)(1-x)^{-4} \\ &=x(1-x^4)(1+x+x^2)\left(1+4x+\binom{5}{2}x^2+\binom{6}{3}x^3+ \cdots\right)\\ &= x(1+x+x^2-x^4-x^5-x^6)\left(1+4x+\binom{5}{2}x^2+\binom{6}{3}x^3+ \cdots\right)\\ \end{align*} Hence the coefficient is \begin{align*} -1-4+\binom{6}{3}+\binom{7}{4}+\binom{8}{5} = 106 \end{align*}

Show $A^T$ has an eigenvector with all components rational Matrix $A$ is a $5 \times 5$ matrix with rational entries such that $(1, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5})^T$ is an eigenvector of A. Show that $A^T$ has eigenvector with all components rational. My idea is: let the eigenvalue associated with the above eigenvector be $λ$. Since all matrix entries are rational numbers so an irrational number will be linearly independent. Use this $2(a_{11} + 2a_{14}) = a_{41} + 2a_{44}$ $a_{21}+2a_{24} = 0$ $a_{31}+2a_{34} = 0$ $a_{51}+2a_{54} = 0$ but I can't find transposed matrix's eigenvector.

Let $v = (1, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5})^T$ and assume $A v = \lambda v$. From the first row we get $$a_{1,1} + 2 a_{1,4} + a_{1,2} \sqrt{2} + a_{1,3} \sqrt{3} + a_{1,5} \sqrt{5} = \lambda$$ From the second row we get: $$a_{2,1} + 2 a_{2,4} + a_{2,2} \sqrt{2} + a_{2,3} \sqrt{3} + a_{2,5} \sqrt{5} = \lambda \sqrt{2}$$ Now substitute $\lambda$: $$a_{2,1} + 2 a_{2,4} + a_{2,2} \sqrt{2} + a_{2,3} \sqrt{3} + a_{2,5} \sqrt{5} = (a_{1,1} + 2 a_{1,4} + a_{1,2} \sqrt{2} + a_{1,3} \sqrt{3} + a_{1,5} \sqrt{5}) \sqrt{2}$$ Multiply and rearrange the terms: $$a_{2,1} + 2 a_{2,4} - 2a_{1,2} + (a_{2,2}-a_{1,1}-2a_{1,4}) \sqrt{2} + a_{2,3} \sqrt{3} + a_{2,5} \sqrt{5} - a_{1,3} \sqrt{6} - a_{1,5} \sqrt{10} = 0$$ Since the roots of the squarefree positive integers are linearly independent over $\mathbb{Q}$, we obtain in particular $a_{1,3} = 0$ and $a_{1,5} = 0$, so $$\lambda = a_{1,1} + 2a_{1,4} + a_{1,2} \sqrt{2}.$$ From the third row we get: $$a_{3,1} + 2 a_{3,4} + a_{3,2} \sqrt{2} + a_{3,3} \sqrt{3} + a_{3,5} \sqrt{5} = \lambda \sqrt{3}$$ Again, substitute $\lambda$: $$a_{3,1} + 2 a_{3,4} + a_{3,2} \sqrt{2} + a_{3,3} \sqrt{3} + a_{3,5} \sqrt{5} = (a_{1,1} + 2a_{1,4} + a_{1,2} \sqrt{2}) \sqrt{3}$$ Multiply and rearrange the terms: $$a_{3,1} + 2 a_{3,4} + a_{3,2} \sqrt{2} + (a_{3,3}-a_{1,1}-2a_{1,4}) \sqrt{3} + a_{3,5} \sqrt{5} - a_{1,2} \sqrt{6} = 0$$ So we obtain $a_{1,2} = 0$ and $$\lambda = a_{1,1} + 2 a_{1,4}$$ Thus $\lambda \in \mathbb{Q}$. Since $A^T$ and $A$ have the same set of eigenvalues, $A^T$ is a matrix with rational components and a rational eigenvalue. Therefore $A^T$ has an eigenvector with rational components.

How would you solve this polynomial? Is there a way to find the roots of equations such as $x^3-9\sqrt[3]{2}+9=0$? I've just been using Wolfram Alpha to factor it into $(x-\sqrt[3]{4}+\sqrt[3]{2}-1)(x^2+(1-\sqrt[3]{2}+\sqrt[3]{4})x+3\sqrt[3]{4}-3)$. But for harder equations such as $$x^3-63\sqrt[3]{20}+9=0$$, Wolfram Alpha just factors it into $$(x-\text{root of }x^9+27x^6+243x^3-5000211)(x-\text{same thing})(x-\text{same thing})$$ Which is kind of problematic, because I would like the exact values of the factors. So is there a way to factor such polynomials into multiple factors? Or some way to find its roots! Anything helps. Note: I would not like nested radicals such as $x=\sqrt[3]{3-\sqrt[4]{4}}$ because I then have to go through the process of finding there simplified surds.

We can find $\sqrt[3]{9(\sqrt[3]2-1)}$ by the following way without WA. Indeed, let $\sqrt[3]2=x$. Hence, $$x^3=2$$ or $$9(x^3-1)=9$$ or $$9(x-1)=\frac{9}{1+x+x^2}$$ or $$9(x-1)=\frac{27}{x^3+3x^2+3x+1}$$ or $$\sqrt[3]{9(\sqrt[3]2-1)}=\frac{3}{\sqrt[3]2+1}$$ or $$\sqrt[3]{9(\sqrt[3]2-1)}=\sqrt[3]4-\sqrt[3]2+1$$ and we are done!

The integral $\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy$ What is $$\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy?$$ I split it as $\frac{y^{2}}{(y^2+1)^{0.5}} + \sqrt{y^2+1}.$ Now I substituted $y^{2}=u $ thus $2y\,dy=du$ so we get $0.5 \sqrt{\frac{u}{u + 1}} + 0.5 \sqrt{\frac{1 + u}{u}}$ but now what to do? Another idea was doing $+1-1$ in original question but that too doesn't lead anywhere. Now $y=\tan{x} $ as suggested below is an easy way but I am seeking for a purely algebraic way. Thanks.

$\displaystyle\int\frac{4y^2+2}{\sqrt{y^2+1}}dy=\int\frac{2y^2+2}{\sqrt{y^2+1}}dy+\int\frac{2y^2}{\sqrt{y^2+1}}dy=2\int\sqrt{y^2+1}dy+\int\frac{2y^2}{\sqrt{y^2+1}}dy.$ Now use $\displaystyle u=2y,\;dv=\frac{y}{\sqrt{y^2+1}}dy\;$ so $\;du=2dy,\;v=\sqrt{y^2+1}$ in the 2nd integral to obtain $\displaystyle2\int\sqrt{y^2+1}dy+2y\sqrt{y^2+1}-2\int\sqrt{y^2+1}dy=\color{red}{2y\sqrt{y^2+1}+C}$

Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that: $$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$ My Approach: $$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$ $$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$ $$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$ Now, please help me to continue from here.

use that $$\cos(120^{\circ}+x)=-1/2\,\cos \left( x \right) -1/2\,\sqrt {3}\sin \left( x \right) $$ and $$\cos(240^{\circ}+x)=-1/2\,\cos \left( x \right) +1/2\,\sqrt {3}\sin \left( x \right) $$

Proof of quadratic inequality using AM-GM Proof of quadratic inequality using AM-GM

we have to show that $$x^2y^2+x^2+y^2+4-6xy\geq 0$$ dividing by $x^2+1$ gives $$y^2-\frac{6xy}{x^2+1}+\frac{4+x^2}{x^2+1}\geq 0$$ this is equivalent to $$\left(y-\frac{3x}{x^2+1}\right)^2+\frac{(4+x^2)(x^2+1)-9x^2}{(x^2+1)^2}\geq 0$$ and this is equivalent to $$\left(y-\frac{3x}{x^2+1}\right)^{ 2 }+\frac{(x^2-2)^2}{(x^2+1)^2}\geq 0$$ which is true.

Let A be an $n*n$ matrix. Prove that if $rank(A) = 1$, then $det(A + E) = 1 + trace(A)$ I feel like I've got the answer, but I've never been good at putting what I think into words. $\begin{vmatrix} n_{11} & n_{12} \\ n_{21} & n_{22} \end{vmatrix} = 0 = n_{11}n_{22} - n_{12}n_{21}$ $\begin{vmatrix} n_{11} + 1 & n_{12} \\ n_{21} & n_{22} + 1 \end{vmatrix} = n_{11}n_{22} + n_{11} + n_{22} + 1 - n_{12}n_{21} = \begin{vmatrix} n_{11} & n_{12} \\ n_{21} & n_{22} \end{vmatrix} + n_{11} + n_{22} + 1 = n_{11} + n_{22} + 1$ Which shows that it's true for a 2*2 matrix. Looking at other 2*2 matrices $\begin{vmatrix} n_{11} + 1 & n_{12} \\ n_{21} & n_{22} \end{vmatrix} = n_{11}n_{22} + n_{22} - n_{12}n_{21} = \begin{vmatrix} n_{11} & n_{12} \\ n_{21} & n_{22} \end{vmatrix} + n_{22} = n_{22}$ Similarly: $\begin{vmatrix} n_{11} & n_{12} \\ n_{21} & n_{22} + 1 \end{vmatrix} = n_{11}$ $\begin{vmatrix} n_{11} & n_{12} + 1 \\ n_{21} & n_{22} \end{vmatrix} = -n_{21}$ $\begin{vmatrix} n_{11} & n_{12} \\ n_{21} + 1 & n_{22} \end{vmatrix} = -n_{12}$ Positive if on the main diagonal, negative if on the side diagonal. Using that, we can show that $\begin{vmatrix} n_{11} + 1 & n_{12} & n_{13} \\ n_{21} & n_{22} + 1 & n_{23} \\ n_{31} & n_{32} & n_{33} + 1 \end{vmatrix} = (n_{11} + 1)(n_{22} + n_{33} + 1) - n_{12}n_{21} - n_{13}n_{31} = n_{11}n_{22} - n_{12}n_{21} + n_{11}n_{33} - n_{13}n_{31} + n_{11} + n_{22} + n_{33} + 1 = n_{11} + n_{22} + n_{33} + 1$ And in a similar fashion we can apply this to higher order matrices. But I haven't the slightest clue how to word or show this "similar fashion".

The value of a determinant as well of a trace is independent of the choice of basis. So suppose that the image of $A$ is generated by a vector $v_1$. Complement this vector with $v_2,...,v_n$ to form a base. In this base the matrix of $A$ takes the form: $$ \underline{A} = \left( \begin{matrix} a_{11} & a_{12} & ... & a_{1n} \\ 0 & 0 & ... & 0 \\ \\ . & . & ... & . \\ \\ 0 & 0 & ... & 0 \end{matrix} \right) $$ and that of $E+A$: $$ \underline{1+A} = \left( \begin{matrix} 1+a_{11} & a_{12} & ... & a_{1n} \\ 0 & 1 & ... & 0 \\ \\ . & . & ... & . \\ \\ 0 & 0 & ... & 1 \end{matrix} \right) $$ Then clearly ${\rm tr} \ A=a_{11}\ $ and $\ \det (1+A)=1+a_{11}$. In less abstract terms, as $A$ has rank 1, we may write $A= u v^T$ where $u$ and $v$ are column vectors. Suppose that $e_1^T u\neq 0$. Then $u,e_2,...,e_n$ forms a basis. Carrying out the products one verifies: $$ (1+ A) \left[ \begin{matrix} u & e_2 & ... & e_n \end{matrix} \right] = \left[ \begin{matrix} u & e_2 & ... & e_n \end{matrix} \right] \left[ \begin{matrix} 1+v^T u & v_2 & ... & v_n\\ 0 & 1 & ... & 0 \\ . & . & ... & . \\ 0 & 0 & ... & 1 \end{matrix} \right]$$ so $1+A$ is conjugated to the matrix on the right which verifies the claimed identity. And determinant and trace is invariant under conjugation.

How to factorize $5$ in $\mathbb{Z}[\root 3 \of 2]$? Since $5$ has a norm of $125$ in this domain, and $N(1 + (\root 3 \of 2)^2) = 5$, it seems like a sensible proposition that $5 = (1 + (\root 3 \of 2)^2) \pi_2 \pi_3$, where $\pi_2, \pi_3$ are two other numbers in this domain having norms of $5$ or $-5$. This is supposed to be a unique factorization domain, right? I am encouraged by the fact that $$N\left(\frac{5}{1 + (\root 3 \of 2)^2}\right) = N(1 + 2 \root 3 \of 2 - (\root 3 \of 2)^2) = 25.$$ But I am discouraged by the fact that $$\frac{5}{(-1 - (\root 3 \of 2)^2)(1 + (\root 3 \of 2)^2)} = \frac{7 - 6 \root 3 \of 2 - 2 (\root 3 \of 2)^2}{5}$$ is an algebraic number but not an algebraic integer. I have found a couple of other numbers with norms of $5$ or $-5$ but can't get them to multiply to $5$ in any of the combinations of three of them that I have tried. I feel like I'm going around in circles.

We have $$ \mathbf Z[\sqrt[3]{2}]/(5) \cong \mathbf Z[x]/(x^3 - 2, 5) \cong \mathbf Z_5[x]/(x^3 - 2) \cong \mathbf Z_5[x]/(x+2) \times \mathbf Z_5[x]/(x^2 + 3x + 4) $$ so that the ideal $ (5) $ factors as $ (5) = \mathfrak p_1 \mathfrak p_2 $. To find the ideals $ \mathfrak p_1 $ and $ \mathfrak p_2 $, note that they correspond to the maximal ideals in the ring $ \mathbf Z_5[x]/(x^3 - 2) $. We therefore have the following factorization: $$ (5) = (\sqrt[3]{2} + 2, 5)(\sqrt[3]{4} + 3 \sqrt[3]{2} + 4, 5) $$ Since the ring $ \mathbf Z[\sqrt[3]{2}] $ is a principal ideal domain, these ideals are principally generated by prime elements of norm $ 5 $ and $ 25 $, respectively. Some easy trial and error by hand with the norm form yields that the first ideal is generated by $ 2\sqrt[3]{4} - 3 $, and dividing $ 5 $ by this number finally gives the factorization $$ 5 = (2\sqrt[3]{4} - 3)(6 \sqrt[3]{4} + 8 \sqrt[3]{2} + 9) $$ Here's some more detail on the trial and error part: we know that $ \mathfrak p_1 $ is a prime ideal of norm $ 5 $, and we know that $ (5) = \mathfrak p_1 \mathfrak p_2 $. Since any number with norm $ 5 $ generates a prime ideal, we can conclude by unique factorization of ideals that any such element would have to generate $ \mathfrak p_1 $. After that, we just look for solutions to the equation $ x^3 + 2y^3 + 4z^3 - 6xyz = 5 $ (the norm form), and it is easily seen that $ x = -3 $ and $ z = 2 $ do the trick. Note that the factorization you are asking for is impossible: the prime $ 5 $ does not factor as the product of three prime ideals, but as two.

Cayley graph interpretation D3 I am trying to understand the Cayley graph for the group $D_3$, which from Mathematica, I got: I tried to get the multiplication table in Mathematica: $\left( \begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 1 & 4 & 3 & 6 & 5 \\ 3 & 5 & 1 & 6 & 2 & 4 \\ 4 & 6 & 2 & 5 & 1 & 3 \\ 5 & 3 & 6 & 1 & 4 & 2 \\ 6 & 4 & 5 & 2 & 3 & 1 \\ \end{array} \right)$ where the first row and column are the table headings. How does this multiplication table relate to the graph? Thanks!

The table and graph are related as follows. Let $X=\{1,2,3,4,5,6\}$ and let $S=\{2,4\}$. The table is missing its "headings" so that the actual table looks like this $$ \begin{array}{c|cccccc} & 1 & 2& 3& 4& 5&6\\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 2 & 1 & 4 & 3 & 6 & 5 \\ 3 & 3 & 5 & 1 & 6 & 2 & 4 \\ 4 & 4 & 6 & 2 & 5 & 1 & 3 \\ 5 & 5 & 3 & 6 & 1 & 4 & 2 \\ 6 & 6 & 4 & 5 & 2 & 3 & 1 \\ \end{array} $$ where $i\cdot j$ is equal to the number in the $i$th row and $j$th column (so for example $4\cdot 2 =6$ and $2\cdot 4=3$). Next notice that $S$ generates $X$ under the above multiplication. To see this note that $2^2=1,2,4,4^2=5,2\cdot 4=3,2\cdot 4^2=6$ are all the elements of $X$. Now let us turn to the graph, the vertices are the elements of $X$ and there is a red edge from $x$ to $y$ if $x\cdot 2 =y$. Similarly there is purple edge from $x$ to $y$ if $x\cdot 4=y$.

Find $\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^4}dx$ I was asked to prove that $$\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^{4}}dx=\pi\frac{14325195794+(2815367209\sqrt{26})}{14623232(9+2\sqrt{26})^\frac{7}{2}}$$ I checked the result numerically and the first digits correct using W|F $$\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^4}dx\approx 0.19874620328$$ I tried to start with trig substitution but the high power in the integral make it more complicated. Is there any way to evaluate this integral?

Hint. A route. One may recall the following result, which goes back at least to G. Boole (1857). Proposition. Let $f \in L^1(\mathbb{R})$ and let $f$ be an even function. Then $$ \int_{-\infty}^{+\infty}x^{2n}f\left(x-\frac1x\right) dx=\sum_{k=0}^n \frac{(n+k)!}{(2k)!(n-k)!}\int_{-\infty}^{+\infty} x^{2k}f(x)\: dx. \tag1 $$ Then one may write $$ \begin{align} &\int_0^{+\infty}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^{4}}\:dx \\\\&=\int_0^{+\infty}\frac{x^{-16}}{\left(\left(x^4+\dfrac1{x^4}\right)+5\left(x^2+\dfrac1{x^2}\right)+14\right)^{4}}\:dx \\\\&=\int_0^{+\infty}\frac{x^{-16}}{\left(\left[\left(x-\dfrac1x\right)^2+2\right]^2+5\left(x-\dfrac1x\right)^2+22\right)^{4}}\:dx \\\\&=\int_0^{+\infty}\frac{x^{14}}{\left(\left[\left(x-\dfrac1x\right)^2+2\right]^2+5\left(x-\dfrac1x\right)^2+22\right)^{4}}\:dx \qquad \left(x \to \dfrac1x \right) \\\\&=\frac12\int\limits^{\infty}_{-\infty}\frac{x^{14}}{\left(\left[\left(x-\dfrac1x\right)^2+2\right]^2+5\left(x-\dfrac1x\right)^2+22\right)^{4}}\:dx \\\\&=\frac12\sum_{k=0}^7 \frac{(7+k)!}{(2k)!(7-k)!}\int_{-\infty}^{+\infty} \frac{x^{2k}}{\left(\left(x^2+2\right)^2+5x^2+22\right)^{4}}\:dx \qquad (\text{using}\,\,(1)) \\\\&=\sum_{k=0}^7 \frac{(7+k)!}{(2k)!(7-k)!}\int_0^{+\infty} \frac{x^{2k}}{\left(x^4+6x^2+26\right)^{4}}\:dx \\\\&=\pi\:\frac{14325195794+(2815367209\sqrt{26})}{14623232(9+2\sqrt{26})^\frac{7}{2}} \end{align} $$ where we have concluded by using Theorem $3.1$ (p.$6$) here, in G. Boros and V. Moll's paper.

Proving $\angle QAP=45^\circ$ if $ABCD$ is a square with points $P$ in $BC$, $Q$ in $CD$ satisfying $\overline{BP}+\overline{DQ}=\overline{PQ}$ Here is the problem: Let $ABCD$ be a square with points $P$ in $BC$, $Q$ in $CD$ satisfying $\overline{BP}+\overline{DQ}=\overline{PQ}$. Prove that $\angle QAP=45^\circ$. So far I have been trying to show that $\overline{BP}=\overline{DQ}$ so that the sum of angles on both sides of $\angle QAP$ is $45^\circ$ Any hint or guidance will be great, Thanks in advance.

Using the cosine rule:$$|AP|^2+|AQ|^2-2|AP||AQ|cos\alpha=|PQ|^2$$Replacing $|AP|,|AQ|$ and $|PQ|$:$$|AP|=\sqrt{a^2+r^2}$$$$|PQ|=\sqrt{b^2+r^2}$$$$|PQ|=a+b$$ Here is $a=|BP|, b=|DQ|$:$$a^2+r^2+b^2+r^2-2\sqrt{a^2+r^2}\sqrt{b^2+r^2}cos\alpha=(a+b)^2$$$$\implies2r^2-2\sqrt{a^2+r^2}\sqrt{b^2+r^2}cos\alpha=2ab$$Note that $b =\frac{r^2-ar}{r+a}$. Substituting $b$ and simplifying further:$$cos\alpha=\frac{r^2(r+a)-a(r^2-ar)}{\sqrt{a^2+r^2}\sqrt{2r^2(a^2+r^2)}}\implies cos\alpha=\frac{1}{\sqrt{2}}$$$$\implies \alpha=45°$$

Can completing the square apply to more higher degree? I have some trouble about completing the square lessons. As we know , minimum value of $\ \ 3x^{2} + 4x + 1 $ is $-\frac{1}{3}$ when $x=-\frac{2}{3}$ by completing the square. Then , how about minimum value of $\ \ x^5 +4x^4 + 3x^3 +2x^2 + x + 1$ when $(-2<x<2)$ ? According to Wolfram Alpha , minimum value is $12 - 5\sqrt{5}$ when $x=-\frac{3}{2} + \frac{\sqrt{5}}{2}$ I try to put 5th degree to form $(x-a)^5$ or any nth degree but it doesn't work. I'm just curious how completing the square can apply to higher deree equations to find their minimum or maximum value. Thank you for every help comments.

Yes we have something same. But if we look to the topic like following. $3x^2+4x+1$ has a complete square $3x^2+4x+\frac43$ corresponding to it in which their difference, $-\frac13$, is of degree maximum two degrees less than the original. For any polynomial in one variable we have a unique multiple of a power of a $(x-a)$ in which the difference of the two is of degree maximum two degrees less than the degrees of them. Before showing what I said, I try to give an algorithm to find that unique multiply of a complete power. It is easy: If the initial polynomial is of the form $ax^n+bx^{n-1}+\dots+c$, then the unique multiply of a complete power is $$a(x+\frac{b}{na})^n,$$ which it would be expanded as $$a\left(x^n+n({ \frac{b}{na}})x^{n-1}+\dots+\frac{b^n}{n^na^n}\right)=a(x^n+\frac b a x^{n-1}+\dots+\frac {b^n}{n^na^n})=$$ $$ax^n+bx^{n-1}+\dots+\frac{b^n}{n^na^{n-1}}$$. So by this: The multiple of complete power for $3x^2+4x+1$ is $3(x+\frac{4}{2\cdot 3})^2$, i.e. $3(x+\frac{2}{3})^2=3(x^2+\frac{2\cdot2}{3}+\frac{4}{9})=3x^2+4x+\frac43$. And so for the quintic, $5x^5+4x^4+3x^3+2x^2+x+1$, the unique multiple of the complete power of a would be $5(x+\frac{4}{5\cdot5})^5$, i.e. $5(x+\frac4{25})^5$, which has an expanded form like: $5x^5+4x^4+\frac{32}{25}x^3+\frac{128}{625}x^2+\frac{256}{15625}x+\frac{1024}{1953125}$ which their difference is (of course of two degrees lesser): $-\frac{43}{25}x^3 -\frac{1122}{625}x^2 -\frac{15369}{15625}x -\frac{1952101}{1953125}$. I wish you got something of what you where looking for!

Convergence/divergence of $\sum_{k=1}^\infty\frac{2\times 4\times 6\times\cdots\times(2k)}{1\times 3\times 5\times\cdots\times(2k-1)}$ A problem asks me to determine if the series $$\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$$ converges or diverges. (from the textbook Calculus by Laura Taalman and Peter Kohn (2014 edition); section 7.6, p. 639, problem 33) I am allowed to use the ratio test first and then any other convergence/divergence test if the former test does not work. In my original work, I attempted the ratio test and it was rendered inconclusive. $$ a_k = \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)} $$ $$ a_{k + 1} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2(k+1))}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2(k+1)-1)} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)} $$ $$ \frac{a_{k + 1}}{a_k} = \frac{\frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)}}{\frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)} \times \frac{1 \times 3 \times 5 \times \cdots \times (2k-1)}{2 \times 4 \times 6 \times \cdots \times (2k)} = \frac{2k+2}{2k+1}$$ Evaluating $\rho = \lim_{x \to \infty} \frac{a_{k + 1}}{a_k}$ will determine if $\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$ converges or diverges. The conclusions for the ratio test are as follows: $\circ$ If $\rho < 1$, then $\sum_{k=1}^\infty a_k$ converges. $\circ$ If $\rho > 1$, then $\sum_{k=1}^\infty a_k$ diverges. $\circ$ If $\rho = 1$, then the test is inconclusive. $$ \rho = \lim_{x \to \infty} \frac{a_{k + 1}}{a_k} = \lim_{x \to \infty} \frac{2k+2}{2k+1} = 1$$ Since $\rho = 1$, the ratio test is rendered inconclusive, as I stated earlier. I will have to use other convergence/divergence tests to solve the problem. My issue is that I'm not sure which other convergence/divergence test to use. Any suggestions? Many thanks for the help.

$$\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}=$$ $$\sum_{k=1}^\infty \frac{\prod_{j=1}^k(2j)}{\prod_{j=1}^k(2j-1)}=$$ $$\sum_{k=1}^\infty \prod_{j=1}^k\frac{(2j)}{(2j-1)}$$ The product $a_k = \prod_{j=1}^k\frac{(2j)}{(2j-1)}$ is $$\left(1+\dfrac11\right)\left(1+\dfrac13\right)\left(1+\dfrac15\right)\cdots \left(1+\dfrac1{2n-1}\right)$$ An infinite product $\lim_{k \to \infty} \left(1+a_k\right)$ converges to a non-zero number if one of the $\sum_{k \to \infty} \vert a_k \vert$ converges. Conclude what you want from this.

Characteristic equation of a recurrence relation? I am trying to find the general term of the following recurrence relation: $$a_{n + 1} = \frac{1}{2}(a_{n} + \frac{1}{a_{n}})$$ where $a_1 = 3$. I'm failing to write the characteristic equation.

Here is what I came upon: If we have: $a_{n} = \frac{1}{2}(a_{n} + \frac{\lambda^2}{a_{n}})$, then it holds that: $\frac{a_{n} - \lambda}{a_{n} + \lambda} = (\frac{a_{1} - \lambda}{a_{1} + \lambda})^{2^{n-1}}$, solving for $a_{n}$ from the above example we have: $$\frac{a_{n} - 2}{a_{n} + 2} = (\frac{3 - 2}{3 + 2})^{2^{n-1}}$$ $$a_n - 2 = \frac{a_n + 2}{5^{2^{n-1}}}$$ $$a_n5^{2^{n-1}} - a_n - 2.5^{2^{n-1}} - 2 = 0$$ $$a_n(5^{2^{n-1}} - 1) - 2(5^{2^{n-1}} + 1) = 0$$ $$a_n = 2\frac{5^{2^{n-1}} + 1}{5^{2^{n-1}} - 1}$$

Prove the inequality $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$ Let $a,b,c -$ triangle side and $a+b+c=1$. Prove the inequality $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$$ My work so far: 1) $a^2+b^2=c^2-2ab\cos \gamma \ge c^2-2ab$ 2) $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}\le3\sqrt{\frac{2(a^2+b^2+c^2)}3}$$

Try to use Lagrange multipliers to solve this problem. max $[\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}+\lambda(a+b+c-1)]$

Find the equation of two ... Find the single equation of two straight lines that pass through the point $(2,3)$ and parallel to the line $x^2 - 6xy + 8y^2 = 0$. My Attempt: Let, $a_1x+b_1y=0$ and $a_2x+b_2y=0$ be the two lines represented by $x^2-6xy+8y^2=0$. then, $$(a_1x+b_1y)(a_2x+b_2y)=0$$ $$(a_1a_2)x^2+(a_1b_2+b_1a_2)xy+(b_1b_2)y^2=0$$ Comparing with $x^2-6xy+8y^2=0$, $a_1a_2=1, -(a_1b_2+b_1a_2)=6, b_1b_2=8$. I got stuck at here. Please help me to continue this.

The lines $$(x-2)^2 - 6(x-2)(y-3) + 8(y-3)^2 = 0$$ on transfer of origin to $(2,3)$, using the transformations $X = x-2, Y= y-2$ becomes $$X^2 - 6XY + 8Y^2 = 0$$ Hence the required equation is $$(x-2)^2 - 6(x-2)(y-3) + 8(y-3)^2 = 0$$

Solve $\iint_D\sqrt{9-x^2-y^2}$ Where $D$ is the positive side of a circle of radius 3 Solve $\displaystyle\iint_D\sqrt{9-x^2-y^2}$ Where $D$ is the positive side of a circle of radius 3 ($x^2+y^2=9,x\ge0,y\ge0$) I tried to subsitute variables to $r$ & $\theta$: $$x = r\cos\theta$$ $$y = r\sin\theta$$ $$E = \{0\le r\le3,0\le\theta\le\pi\}$$ $$\displaystyle\iint_D\sqrt{9-x^2-y^2} = \displaystyle\iint_EJ\sqrt{9-(r\cos\theta)^2-(r\sin\theta)^2}=\displaystyle\iint_Er\sqrt{9-(r\cos\theta)^2-(r\sin\theta)^2}$$ But have no clue on how to solve this new integral.

It just is the half of the capacity of the ball: ($x^2+y^2+z^2=9)$.

Finding the gradient of a given function Given the function $\phi=B_0/r=B_0/\vert \mathbf{x}\vert$ in a spherical axisymmetric geometry, where $B_0$ is a constant. Find $\nabla\phi$. The given answer is $$\phi=\frac{B_0}{r}=\frac{B_0}{\vert \mathbf{x}\vert}\implies \nabla\phi = -\frac{B_0\mathbf{x}}{\vert \mathbf{x}\vert^3}=-\frac{B_0\mathbf{e}_r}{r^2}$$ However I'm confused on how this implication has been achieved. I have attempted: $$\nabla\phi = \frac{\partial \phi}{\partial x_i}\mathbf{x}_i = B_0\frac{\partial}{\partial x_i}\left(\frac{1}{\vert\mathbf{x}\vert}\right)\mathbf{x}_i=B_0\frac{\partial}{\partial x_i}\left((\sum_{i=1}^3 x_i^2)^{-1/2}\right)\mathbf{x}_i$$ But I have no idea how to find this derivative. Can anyone help?

1º Note that r=$\sqrt{x^2+y^2+z^2}$, then ∇ϕ=($\frac{∂}{∂x}\frac{B_0}{\sqrt{x^2+y^2+z^2}}$,$\frac{∂}{∂x}\frac{B_0}{\sqrt{x^2+y^2+z^2}}$,$\frac{∂}{∂x}\frac{B_0}{\sqrt{x^2+y^2+z^2}}$). 2º$\frac{∂}{∂x}\frac{B_0}{\sqrt{x^2+y^2+z^2}}$=$-\frac{B_0}{2(\sqrt{x^2+y^2+z^2})^3} 2x$. 3º ∇ϕ=($-\frac{B_0}{\sqrt{x^2+y^2+z^2})^3} x,-\frac{B_0}{\sqrt{x^2+y^2+z^2})^3} y,-\frac{B_0}{\sqrt{x^2+y^2+z^2})^3} z$)=$-\frac{B_0}{(\sqrt{x^2+y^2+z^2})^3}e_r$, onde $e_r$=(x,y,z)

Prove $\lim_{(x,y)\to(1,1)} x^2+xy+y=3$ Prove that $$\lim_{(x,y)\to(1,1)} x^2 + xy + y = 3$$ using the epsilon-delta definition. What I have tried: Let $\epsilon > 0$ be arbitrary. We must show that for every $\epsilon$ we can find $\delta>0$ such that $$0 < \|(x,y) - (1,1)\| < \delta \implies \|f(x,y) - 3\| < \epsilon$$. Or equivalently, $$0 < \sqrt{(x-1)^2 + (y-1)^2} < \delta \implies |x^2+xy+y-3| < \epsilon$$ The problem is finding the $\delta$. I have been trying to manipulate $|x^2+xy+y-3|$ with no success. $|x^2+xy+y-3|$ $=|(x^2-1)+y(x+1)-2|$ $=|(x-1)(x+1)+y(x+1)-2|$ $=|(x+1)[(x-1)+y]-2|$ $=|(x+1)||[(x-1)+(y-1)-1|$ Any help is appreciated!

\begin{align} x^2+xy+y-3 &= (x-1)^2+2x-1+(x-1)(y-1)+x+y-1+(y-1)+1-3 \\ &=(x-1)^2+(x-1)(y-1)+(y-1)+3x+y-4 \\ &=(x-1)^2+(x-1)(y-1)+(y-1)+3(x-1)+(y-1)\\ &=(x-1)^2+(x-1)(y-1)+2(y-1)+3(x-1)\\ \end{align} Let $\delta= \min(1, \frac{\epsilon}7),$ Then \begin{align} |x^2+xy+y-3| &\leq |x-1|^2+|x-1||y-1|+2|y-1|+3|x-1| \\ &\leq 2\delta^2+5\delta \\ & \leq 7 \delta \\ & \leq \epsilon \end{align} Alternatively, let me work from where you left off, there is a mistake at the last line of your equation, it should be \begin{align} (x+1)[(x-1)+(y)]-2 &=(x+1)[(x-1)+(y-1+1)]-2 \\ &=(x+1)[(x-1)+(y-1)]+x+1-2 \\ &=(x+1)[(x-1)+(y-1)]+(x-1) \end{align} Choose $\delta = \min(1, \frac{\epsilon}7)$ Then $|x-1|\leq \delta$ implies $1-\delta \leq x \leq 1+\delta$ and hence $|x+1|\leq 3,$ Hence, \begin{align} |(x+1)[(x-1)+(y-1)]+(x-1)| &\leq 3 [|x-1|+|y-1|]+|x-1|\\ & \leq 3(2\delta)+\delta \\ & = 7\delta \\ & \leq \epsilon \end{align}

Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by $$ \large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right) $$ I had to find $f'$ and $f''$. Below, I have calculated them. But, isn't there a better and more convenient way to do this? My method: $$ {f'(x)}=\left [\ln \left (\frac{x}{(x^2+1)^\frac{1}{2}} \right) \right ]'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right)\left (\frac{x}{(x^2+1)^\frac{1}{2}} \right)'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[(x^2+1)^\frac{1}{2}]'}{[(x^2+1)^\frac{1}{2}]^2} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(x^2+1)']}{\left | x^2+1 \right |} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)]}{x^2+1} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x^2(x^+1)^{-\frac{1}{2}}}{x^2+1} \right)=\frac{(x^2+1)^{(\frac{1}{2}+\frac{1}{2})}-x^2(x^2+1)^{\frac{1}{2}+-\frac{1}{2}{}}}{x(x^2+1)}=-\frac{x^2}{x}=-x $$ and $$ f''(x)=(-x)'=-1\ $$ This took me much more than 1.5 hours just to type into LaTex :'(

By implicit differentiation: Let $$ y(x) = \log\left[\frac{x}{\sqrt{x^2 + 1}}\right]. $$ Then $$ (x^2 + 1)e^{2 y(x)} = x^2. $$ Differentiating both sides, $$ (x^2 + 1)e^{2y(x)}y'(x) + x e^{2y(x)} = x. $$ Solving for $y'(x)$, $$ y'(x) = \frac{x(e^{-2y(x)}-1)}{(x^2 + 1)} = \frac{1}{x(x^2+1)}. $$

Prove that $a/(p-a) + b/(p-b) + c/(p-c) \ge 6$ Prove that $a/(p-a) + b/(p-b) + c/(p-c) \ge 6$ , where $a,b,c$ are the sides of a triangle and $p$ is the semi-perimeter .

By C-S $\sum\limits_{cyc}\frac{a}{p-a}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(pa-a^2)}=\frac{2(a+b+c)^2}{(a+b+c)^2-2(a^2+b^2+c^2)}\geq\frac{2(a+b+c)^2}{(a+b+c)^2-\frac{2}{3}(a+b+c)^2}=6$

Solving system of $9$ linear equations in $9$ variables I have a system of $9$ linear equations in $9$ variables: \begin{array}{rl} -c_{1}x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\ x_{1} - c_{2}x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\ x_{1} + x_{2} - c_{3}x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\ x_{1} + x_{2} + x_{3} - c_{4}x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\ x_{1} + x_{2} + x_{3} + x_{4} - c_{5}x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} - c_{6}x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} - c_{7}x_{7} + x_{8} + x_{9} &= 0 \\ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} - c_{8}x_{8} + x_{9} &= 0 \\ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} - c_{9}x_{9} &= 0 \end{array} I want to find a general non-trivial solution for it. What would be the easiest and least time consuming way to find it by hand? I don't have a lot of background in maths, so I would very much appreciate if you actually found the solution and explained briefly. Thanks in advance! EDIT: Very important to mention is that always any $c_{i} > 1$ and any $x_{i} \geq 20$. Also it would be nice if someone posted how would a general non-trivial solution look in the form of $$S = \left \{( x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9}\right )\}$$

Let $$\mathrm A := 1_n 1_n^T - \mbox{diag} (1 + c_1, \dots, 1 + c_n)$$ where $c_i \neq -1$ for all $i \in \{1,2,\dots,n\}$. Using the matrix determinant lemma, $$\det (\mathrm A) = \left( 1 - \sum_{i=1}^n \frac{1}{1 + c_i} \right) (-1)^n \left( \prod_{i=1}^n (1+c_i)\right)$$ We want the homogeneous linear system $\mathrm A \mathrm x = \mathrm 0_n$ to have non-trivial solutions. Thus, we impose the equality constraint $\det (\mathrm A) = 0$, or, equivalently, $$\sum_{i=1}^n \frac{1}{1 + c_i} = 1$$ If this constraint is satisfied, using visual inspection, we conclude that all points on the line $$\left\{ \gamma \begin{bmatrix} \frac{1}{1 + c_1}\\ \vdots\\ \frac{1}{1 + c_n}\end{bmatrix} : \gamma \in \mathbb R \right\}$$ are solutions to the aforementioned homogeneous linear system.

Determinant involving function of $x$ If $f(x)$ is a polynomial satisfying $$f(x)=\frac{1}{2} \begin{vmatrix} f(x) & f(\frac{1}{x})-f(x) \\ 1 & f(\frac{1}{x}) \end{vmatrix} $$ and $f(3)=244$ then $f(2)$ is what? My attempt— Replacing $x$ by $\frac{1}{x}$ we get $$f\left(\frac{1}{x}\right)=\frac12\begin{vmatrix} f\left(\frac{1}{x}\right) & -(f(\frac{1}{x})-f(x)) \\ 1 & f(x) \end{vmatrix}.$$ Which gave me $f\left(\frac{1}{x}\right)-f(x)=\frac{2f(x)-f(x)^2}{f(x)-1}$. Now putting the value of $f\left(\frac{1}{x}\right)$ obtained in terms of $f(x)$ in the original determinant I became helpless when it reduced to $f(x)=f(x)$ and I achieved nothing. Please help me out with this problem.thanks.

Your starting idea is indeed great. Using $f(x)$ and $f(1/x)$ both, and adding them up, we have $$ f(x)+f(1/x) = f(x)f(1/x). $$ This gives $$ f(1/x)-1=\frac{1}{f(x)-1}. \ \ \ \ (*) $$ As $x\rightarrow\infty$, we have $f(0)=1$. By assuming that $n=\textrm{deg}(f)$, we have by multiplying $x^n$ both sides, $$ \frac{x^n}{f(x)-1} \ \textrm{ is a polynomial.} $$ Thus, $f(x)-1$ must be a divisor of $x^n$. Then only possibilities that we have $$ f(x) = ax^k + 1. $$ By (*), we have $a^2=1$, giving that $a=\pm 1$. With $f(3)=244$, we are left with $a=1$. Then $f(x) = x^k +1$. Again by $f(3)=244$, we obtain $k=5$. Therefore, $$ f(x)=x^5+1, \ \textrm{this gives } f(2) = 33. $$

Proof that $ \|x+y\|^2 - \|x\|^2 \geq b(1 - 2^{-n})\|y\|^2 + 2^n( \|x+2^{-n}y\|^2 - \|x\|^2), \quad \forall x,y \in E, \forall n \in \mathbb N $ Suppose that $E$ is a Banach space over $\mathbb R$ satisfying the following inequality, for some $b > 0$ $$ \|x+y\|^2 + b\|x-y\|^2 \leq 2\|x\|^2 + 2 \|y\|^2, \quad \forall x,y \in E $$ I'm trying to proof that: $$ \|x+y\|^2 - \|x\|^2 \geq b(1 - 2^{-n})\|y\|^2 + 2^n( \|x+2^{-n}y\|^2 - \|x\|^2), \quad \forall x,y \in E, \forall n \in \mathbb N \quad (I) $$ My attempt: An induction proof over $n$. For $n = 1$, using the hypothesis replacing $x$ and $y$ for $(x+y)/2$ and $x/2$, respectively, we have: $$ \|x + y/2\|^2 + b \|y/2\|^2 \leq 2\|(x+y)/2 \|^2 + 2\|x/2\|^2 = \frac{\|x+y\|^2}{2} + \frac{\|x\|^2}{2} $$ $$ \therefore \quad 2\|x+ 2^{-1}y\|^2 + b2^{-1}\|y\|^2 \leq \|x+y\|^2 + \|x\|^2 $$ $$ \therefore \quad \|x+y\|^2 - \|x\|^2 \geq b2^{-1}\|y\|^2 + 2(\|x+ 2^{-1}y\|^2 - \|x\|^2) $$ which is the inequality $(I)$ for $n = 1$. If I suppose that $(I)$ works for some $n > 1$, I have to proof that $$ \|x+y\|^2 - \|x\|^2 \geq b(1 - 2^{-n-1})\|y\|^2 + 2^{n+1}( \|x+2^{-n-1}y\|^2 - \|x\|^2), \quad \forall x,y \in E \quad (II) $$ However I don't know how to conclude that. Thank you for any help!!

I think I might have solved the problem. Let's re-write (I) as $$ \|x+y\|^2 - \|x\|^2 \geq b(\frac{2^n - 1}{2^n})\|y\|^2 + 2^n\|x+\frac{1}{2^n}y\|^2 - 2^n\|x\|^2 $$ $$ 2^n\|x+y\|^2 - 2^n\|x\|^2 \geq b(2^n - 1)\|y\|^2 + (2^n)^2\|x+\frac{1}{2^n}y\|^2 - (2^n)^2\|x\|^2 $$ $$ 2^n\|x+y\|^2 - 2^n\|x\|^2 \geq b(2^n - 1)\|y\|^2 + \|2^n x+y\|^2 - \|2^n x\|^2 $$ Now, replacing $x$ for $2x$, we have $$ 2^n\|2x+y\|^2 - 2^n\|2x\|^2 \geq b(2^n - 1)\|y\|^2 + \|2^{n+1} x+y\|^2 - \|2^{n+1} x\|^2 $$ $$ 2^{n+1} [2( \|x + y/2\|^2 - \|x\|^2 ) ] \geq b(2^n - 1)\|y\|^2 + (2^{n+1})^2\| x+\frac{1}{2^{n+1}}y\|^2 - (2^{n+1})^2\| x\|^2 $$ $$ 2( \|x + y/2\|^2 - \|x\|^2 ) \geq b\frac{2^n - 1}{2^{n+1}}\|y\|^2 + 2^{n+1}\| x+\frac{1}{2^{n+1}}y\|^2 - 2^{n+1}\| x\|^2 $$ Summing $b\frac{1}{2}\|y\|^2$ in both sides: $$ b\frac{1}{2}\|y\|^2 + 2( \|x + y/2\|^2 - \|x\|^2 ) \geq b(\frac{1}{2}+\frac{2^n - 1}{2^{n+1}})\|y\|^2 + 2^{n+1}(\| x+\frac{1}{2^{n+1}}y\|^2 - \| x\|^2) $$ Applying the inequality for $n = 1$ $$ \|x+y\|^2 - \|x\|^2 \geq b(1-\frac{1}{2^{n+1}})\|y\|^2 + 2^{n+1}(\| x+\frac{1}{2^{n+1}}y\|^2 - \| x\|^2) $$ since $\frac{1}{2}+\frac{2^n - 1}{2^{n+1}} = \frac{1}{2} (1 + 1 - \frac{1}{2^n}) = (1 -\frac{1}{2^{n+1}})$

Explanations about some Mittag-leffler partial fraction expansions Is it possible to show where the following series come from? $$\sum _{k=1}^{\infty } \left(\frac{1}{\pi ^2 k^2}-\frac{2}{(x-2 \pi k)^2}-\frac{2}{(2 \pi k+x)^2}\right)+\left(-\frac{2}{x^2}-\frac{1}{6}\right)=\frac{1}{\cos (z)-1}$$ $$\sum _{k=1}^{\infty } \left(-\frac{2 \pi ^2 k^2+3}{6 \pi ^4 k^4}+\frac{2}{3 (x-2 \pi k)^2}+\frac{4}{(x-2 \pi k)^4}+\frac{2}{3 (2 \pi k+x)^2}+\frac{4}{(2 \pi k+x)^4}\right)+\frac{4}{x^4}+\frac{2}{3 x^2}+\frac{11}{180}=\frac{1}{(\cos (z)-1)^2}$$ $$\sum _{k=1}^{\infty } \left(\frac{8 \pi ^4 k^4+15 \pi ^2 k^2+15}{60 \pi ^6 k^6}-\frac{4}{15 (x-2 \pi k)^2}-\frac{2}{(x-2 \pi k)^4}-\frac{8}{(x-2 \pi k)^6}-\frac{4}{15 (2 \pi k+x)^2}-\frac{2}{(2 \pi k+x)^4}-\frac{8}{(2 \pi k+x)^6}\right)+\left(-\frac{8}{x^6}-\frac{2}{x^4}-\frac{4}{15 x^2}-\frac{191}{7560}\right)=\frac{1}{(\cos (z)-1)^3}$$ sorry for the inconvenient I forget to add a part of the formula.

As I commented earlier, I have a problem with the first expression. So, since Maple said that it is correct, I suppose I am wrong but I would like to know where. Let me consider $$S_1=\sum _{k=1}^{\infty } \frac{1}{\pi ^2 k^2}\qquad S_2=\sum _{k=1}^{\infty }\frac{1}{(x-2 \pi k)^2}\qquad S_3=\sum _{k=1}^{\infty }\frac{1}{(x+2 \pi k)^2}$$ So $$S_1=\frac 16\qquad S_2=\frac{\psi ^{(1)}\left(1-\frac{x}{2 \pi }\right)}{4 \pi ^2}\qquad S_3=\frac{\psi ^{(1)}\left(1+\frac{x}{2 \pi }\right)}{4 \pi ^2}$$ where appears the first derivative of the digamma function.$$S_1-2S_2-2S_3=\frac{1}{6}-\frac{\psi ^{(1)}\left(1-\frac{x}{2 \pi }\right)+\psi ^{(1)}\left(1+\frac{x}{2 \pi }\right)}{2 \pi ^2}=\frac{1}{6}+\frac{2}{x^2}+\frac{1}{\cos (x)-1}$$ the last simplification being obtained using a CAS. Using Taylor around $x=0$, what I find is $$S_1-2S_2-2S_3=-\frac{x^2}{120}-\frac{x^4}{3024}-\frac{x^6}{86400}+O\left(x^{8}\right)$$ while $$\frac{1}{\cos (x)-1}=-\frac{2}{x^2}-\frac{1}{6}-\frac{x^2}{120}-\frac{x^4}{3024}-\frac{x^6}{86400}+O\left(x^{8}\right)$$

What is the value of the nested radical $\sqrt[3]{1+2\sqrt[3]{1+3\sqrt[3]{1+4\sqrt[3]{1+\dots}}}}$? The closed-forms of the first three are well-known, $$x_1=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}\tag1$$ $$x_2=\sqrt[3]{1+\sqrt[3]{1+\sqrt[3]{1+\sqrt[3]{1+\dots}}}}\tag2$$ $$x_3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\dots}}}}\tag3$$ $$x_4=\sqrt[3]{1+2\sqrt[3]{1+3\sqrt[3]{1+4\sqrt[3]{1+\dots}}}}=\;???\tag4$$ with $x_1$ the golden ratio, $x_2$ the plastic constant, and $x_3=3\,$ (by Ramanujan). Questions: * *Trying to generalize $x_3$, what is the value of $x_4$ to a $100$ or more decimal places? (The Inverse Symbolic Calculator may then come in handy to figure out its closed-form, if any.) *What is the Mathematica command to compute $x_4$? P.S. This other post is related but only asks for its closed-form which resulted in speculation in the comments. (A method/code to compute $x_4$, and a verifiable numerical value is more desirable.)

To answer this question we need to find the system that is error resilient. We can write the equation as $$y(x)=x\sqrt[3]{1+(x+1)\sqrt[3]{1+(x+2)\sqrt[3]{1+...}}}$$ from where we have $$y(x)=x\sqrt[3]{1+y(x+1)}$$ or $$y(x-1)=(x-1)\sqrt[3]{1+y(x)}$$ Now we need to estimate how this function behaves and we can easily see that $$y(x) \sim x^{\frac{3}{2}}$$ because $$(x^{\frac{3}{2}})^3\sim(x-1)\sqrt[3]{1+x^{\frac{3}{2}}}$$ From here we have an algorithm. Take large $N$ start with $y(N)=N^{\frac{3}{2}}$ and go backwards using $$y(k-1)=(k-1)\sqrt[3]{1+y(k)}$$ With, for example, $N=20$ we have $y(1)=1.70221913267155$ $N=50$ we have $y(1)=1.70221913269546$ already fixing 14 digits which is obvious from $N=100$ $y(1)=1.70221913269546$ It is not difficult to estimate the error terms. If we have missed initial value by $\Delta x$ the error term exponentially diminishes until we reach $y(1)$. Extracting Mathematica or any other command or language from this is rather trivial.

mimimum value of expression $a^2+b^2$ If $a,b$ are two non zero real numbers and $ab(a^2-b^2) = a^2+b^2,$ Then $\min(a^2+b^2)$ $\bf{My\; Try::}$ We can write it as $$ab=\frac{a^2+b^2}{a^2-b^2}\Rightarrow a^2b^2=\frac{(a^2+b^2)^2}{(a^2+b^2)^2-4a^2b^2}$$ Now Put $a^2+b^2=u$ and $a^2b^2=v,$ Then expression convert into $$v=\frac{u^2}{u^2-4v}\Rightarrow 4v^2-u^2v+u^2=0$$ For real roots, $\bf{Discriminant \geq 0}$ $$u^4-16u^2\geq 0\Rightarrow u^2(u^2-16)\geq 0$$ So we get $$u^2\geq 16\Rightarrow u\geq 4\Rightarrow x^2+y^2\geq 4,$$ My question is can we solve it any other way (without Trigonometric substution), If yes , Then plz explain here, Thanks

If we let $z = a + b i$ then we get $$ab = \frac{ z^2 - \bar{z}^2}{4i}$$ $$a^2 - b^2 = \frac{ z^2 + \bar{z}^2}{2}$$ $$a^2 + b^2 = z \bar{z}$$ Thus your original equation can be rewritten as $$\frac{Im(z^4)}{4} = \frac{z^4 - \bar{z}^4}{8 i} = \frac{(z^2 - \bar{z}^2)(z^2 + \bar{z}^2)}{8i} = z \bar{z}$$ We have $\frac{\|z\|^4}{4} \geq \frac{Im(z^4)}{4} = \|z\|^2$, so $\|z\|^2 \geq 4$, as desired. This is achieved exactly when $z^4$ is purely imaginary.

$ \int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx$ $$I=\int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx$$ I can't think of a substitution to solve this problem, by parts won't work here. Can anyone tell how should I solve this problem?

Let $$I = \int\frac{2+\sqrt{x}}{(x+\sqrt{x}+1)^2}dx = \int \frac{2+\sqrt{x}}{x^2\left(1+x^{-\frac{1}{2}}+x^{-1}\right)^2}dx$$ So $$I = \int\frac{2x^{-2}+x^{-\frac{3}{2}}}{\left(1+x^{-\frac{1}{2}}+x^{-1}\right)^2}dx$$ Put $\left(1+x^{-\frac{1}{2}}+x^{-1}\right) = t\;,$ Then $\displaystyle \left(-\frac{1}{2}x^{-\frac{3}{2}}-x^{-2}\right)dx = dt\Rightarrow \left(2x^{-2}+x^{-\frac{3}{2}}\right)dx = -2dt$ So $$I = -2\int\frac{1}{t^2}dt = \frac{2}{t}+\mathcal{C} = \frac{2}{1+x^{-\frac{1}{2}}+x^{-1}}+\mathcal{C}=\frac{2x}{x+\sqrt{x}+1}+\mathcal{C}$$ So $$I = 2\left[\frac{(x+\sqrt{x}+1)-(\sqrt{x}+1)}{x+\sqrt{x}+1}\right]+\mathcal{C} = -\frac{2(\sqrt{x}+1)}{x+\sqrt{x}+1}+\mathcal{C'}$$

Use congruence's to find the reminder when $2^{50}$ and $41^{65}$ are divided by $7$ Use congruence's to find the reminder when $2^{50}$ and $41^{65}$ are divided by 7 $2^{50}$ $50=(7)^2+1$ $2^{50}=2^{7\cdot7+1}$ and I'm not sure where to go from here?

Note that, $$\begin{align} & 2^3\equiv 1 \pmod7 \\ \implies & (2^3)^{16}\equiv 1^{16} \pmod7 \\ \implies & 2^{48}\equiv 1 \pmod7 \\ \implies & 2^{48}\cdot 2^2\equiv 1\cdot 2^2 \pmod7 \\ \implies & \color{blue}{2^{50}\equiv 4 \pmod7}\end{align}$$ Also note that $$\begin{align} & 41\equiv -1\pmod7 \\ \implies & 41^{65}\equiv (-1)^{65}\equiv -1\pmod7 \\ \implies & \color{blue}{41^{65}\equiv 6\pmod7}\end{align}$$

How to prove that $(a\cos\alpha)^n + (b\sin\alpha)^n = p^n$ under the following conditions? How to prove that $(a\cos\alpha)^n + (b\sin\alpha)^n = p^n$ when then line $x\cos\alpha + y\sin\alpha = p$ touches the curve $$\left (\frac{x}{a} \right )^\frac{n}{n-1} + \left (\frac{y}{b} \right )^\frac{n}{n-1}=1$$ What I've tried: I've equated the derivative of the given line with the general derivative of the given curve but couldn't proceed to any meaningful step thereafter.

At the tangent point (x, y), the normal of the two curves are parallel, so we can get $$\{\cos \alpha, \sin \alpha \}//\{\frac{x^\frac{1}{n-1}}{a^\frac{n}{n-1}}, \frac{x^\frac{1}{n-1}}{a^\frac{n}{n-1}}\} $$ so we get the equation: $$\frac{x^\frac{1}{n-1}}{a^\frac{n}{n-1}\cos \alpha}=\frac{x^\frac{1}{n-1}}{a^\frac{n}{n-1}\sin \alpha}=k$$ Hence: $$x=k^{n-1}a^n(\cos\alpha)^{n-1}$$ $$y=k^{n-1}b^n(\sin\alpha)^{n-1}$$ Take x and y into $x\cos\alpha + y\sin\alpha = p$ we get: $$k^{n-1}((a\cos\alpha)^n+(b\sin\alpha)^n)=p$$ Take x and y into $\left (\frac{x}{a} \right )^\frac{n}{n-1} + \left (\frac{y}{b} \right )^\frac{n}{n-1}=1$ we get: $$k^n((a\cos\alpha)^n+(b\sin\alpha)^n)=1$$ Therefore:$$k=1/p$$ Take K into $k^n(a^n\cos\alpha^n+b^n\sin\alpha^n)=1$ get the result: $$(a\cos\alpha)^n + (b\sin\alpha)^n = p^n$$

How would you calculate the surface of the part of the paraboloid $z=x^2+y^2$ with $1 \le z \le 4$? Do you calculate it like done below? can you calculate it in another way? $z=x^2+y^2$ Let $x=\sqrt{z}\cos\theta$ $y=\sqrt{z}\sin\theta$ $z=z$ where $\theta\in[0,2\pi]$ and $z\in[1,4]$ $ \dfrac{\partial{x}}{\partial{\theta}}=-\sqrt{z}\sin\theta$ $ \dfrac{\partial{y}}{\partial{\theta}}=\sqrt{z}\cos\theta$ $ \dfrac{\partial{z}}{\partial{\theta}}=0$ $ \dfrac{\partial{x}}{\partial{z}}=\dfrac{1}{2\sqrt{z}}\cos\theta$ $ \dfrac{\partial{y}}{\partial{z}}=\dfrac{1}{2\sqrt{z}}\sin\theta$ $ \dfrac{\partial{z}}{\partial{z}}=1$ $\begin{Vmatrix}\hat{i} & \hat{j} & \hat{k} \\ -\sqrt{z}\sin\theta &\sqrt{z}\cos\theta & 0 \\ \dfrac{1}{2\sqrt{z}}\cos\theta & \dfrac{1}{2\sqrt{z}}\sin\theta & 1\end{Vmatrix}$ $=\sqrt{z}\cos\theta\hat{i}+\sqrt{z}\sin\theta\hat{j}+\dfrac{1}{2}(-\sin^2 \theta-\cos^2 \theta)\hat{k}$ $=\left<\sqrt{z}\cos\theta,\sqrt{z}\sin\theta,-\dfrac{1}{2}\right>$ And its magnitude is given by $\sqrt{z\cos^2\theta+z\sin^2\theta+\dfrac{1}{4}}=\sqrt{z+\dfrac{1}{4}}=\dfrac{1}{2}\sqrt{4z+1}$ Surface $=\displaystyle \int \int_S dS$ $=\displaystyle \int_1^4 \int_0^{2\pi} \dfrac{1}{2}\sqrt{4z+1} \,d\theta\,dz$ $=\bigg|2\pi\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}(4z+1)^{\frac{3}{2}}\dfrac{1}{4}\bigg|_1^4$ $=\dfrac{\pi}{6}(17\sqrt{17}-5\sqrt{5})$

Your result is correct. By using polar coordinates, we obtain $$\iint_S dS=\int_{\rho=1}^2 \int_0^{2\pi} \sqrt{1+f_x^2+f_y^2} \,(d\theta\,\rho d\rho)=2\pi\int_{\rho=1}^2 \sqrt{1+4\rho^2} \,\rho d\rho\\ =\dfrac{\pi}{6}\left[(1+4\rho^2)^{3/2}\right]_{\rho=1}^2 =\dfrac{\pi}{6}(17\sqrt{17}-5\sqrt{5})$$ where $f(x,y)=x^2+y^2$, $x=\rho\cos\theta$ and $y=\rho\sin\theta$. P.S. We have that $f_x(x,y)=2x$, $f_y(x,y)=2y$ and therefore $$\sqrt{1+f_x^2+f_y^2}=\sqrt{1+4(x^2+y^2)}=\sqrt{1+4\rho^2}$$ For details take a look here: https://en.wikipedia.org/wiki/Surface_integral

Find positive integer $x,y$ such that $7^{x}-3^{y}=4$ Find all positive integers $x,y$ such that $7^{x}-3^{y}=4$. It is the problem I think it can be solve using theory of congruency. But I can't process somebody please help me . Thank you

Let us go down the rabbit hole. Assume that there is a solution with $ x, y > 1 $, and rearrange to find $$ 7(7^{x-1} - 1) = 3(3^{y-1} - 1) $$ Note that $ 7^{x-1} - 1 $ is divisible by $ 3 $ exactly once (since $ x > 1 $): the contradiction will arise from this. Reducing modulo $ 7 $ we find that $ 3^{y-1} \equiv 1 $, and since the order of $ 3 $ modulo $ 7 $ is $ 6 $, we find that $ y-1 $ is divisible by $ 6 $, hence $ 3^{y-1} - 1 $ is divisible by $ 3^6 - 1 = 2^3 \times 7 \times 13 $. Now, reducing modulo $ 13 $ we find that $ 7^{x-1} \equiv 1 $, and since the order of $ 7 $ modulo $ 13 $ is $ 12 $, we find that $ x-1 $ is divisible by $ 12 $. As above, this implies that $ 7^{x-1} - 1 $ is divisible by $ 7^{12} - 1 $, which is divisible by $ 9 $. This is the desired contradiction, hence there are no solutions with $ x, y > 1 $. For an outline of the method I used here, see Will Jagy's answers to this related question.

Is there a series approximation in terms of $n$ for the sum of the harmonic progression : $\sum_{k=0}^{n}\frac{1}{1+ak}$? When $a=1$ the sum is given by $ H_{n} $ and we have : $$H_{n}=log(n)+γ+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4} \hspace{0.5cm}.\hspace{.1cm}.\hspace{.1cm}. $$ Does any representation of a similar type exist for arbitrary positive $a$ ?

As Felix Marin answered,$$\sum_{k = 1}^{n}{1 \over 1 + ak}=\frac{H_{n+\frac{1}{a}}-H_{\frac{1}{a}}}{a}$$ Now, using the asymptotics $$H_m=\gamma +\log \left({m}\right)+\frac{1}{2 m}-\frac{1}{12 m^2}+\frac{1}{120 m^4}+O\left(\frac{1}{m^5}\right)$$ $$H_{n+\frac{1}{a}}=\gamma +\log \left({n+\frac{1}{a}}\right)+\frac{1}{2 (n+\frac{1}{a})}-\frac{1}{12 (n+\frac{1}{a})^2}+\frac{1}{120(n+\frac{1}{a} )^4}+O\left(\frac{1}{n^5}\right)$$ Now, developing each term as Taylor series $$\log \left({n+\frac{1}{a}}\right)=\log \left({n}\right)+\frac{1}{a n}-\frac{1}{2 a^2 n^2}+\frac{1}{3 a^3 n^3}-\frac{1}{4 a^4 n^4}+O\left(\frac{1}{n^5}\right)$$ $$\frac{1}{ (n+\frac{1}{a})}=\frac{1}{n}-\frac{1}{a n^2}+\frac{1}{a^2 n^3}-\frac{1}{a^3 n^4}+O\left(\frac{1}{n^5}\right)$$ $$\frac{1}{ (n+\frac{1}{a})^2}=\frac{1}{n^2}-\frac{2}{a n^3}+\frac{3}{a^2 n^4}+O\left(\frac{1}{n^5}\right)$$ $$\frac{1}{ (n+\frac{1}{a})^4}=\frac{1}{n^4}+O\left(\frac{1}{n^5}\right) $$and replacing, you should get $$\sum_{k = 1}^{n}{1 \over 1 + ak}=\frac{\gamma-H_{\frac{1}{a}} }{a}+\frac{\log \left({n}\right) }{a}+\frac{a+2}{2 a^2 }\frac 1{n}-\frac{a^2+6 a+6}{12 a^3 }\frac 1{n^2}+\frac{a^2+3 a+2}{6 a^4 }\frac 1{n^3}+\frac{a^4-30 a^2-60 a-30}{120 a^5 }\frac 1{n^4}+O\left(\frac{1}{n^5}\right)$$ Take care that, for $a=1$ $$\sum_{k = 1}^{n}{1 \over 1 + k}=H_{n+1}-1=\gamma -1+\log \left({n}\right)+\frac{3}{2 n}-\frac{13}{12 n^2}+\frac{1}{n^3}-\frac{119}{120 n^4}+O\left(\frac{1}{n^5}\right)$$ and not $H_n$ (for which you gave the correct expansion) because of the shift of the index.

Find the values of $\alpha $ satisfying the equation(determinant) Find the values of $\alpha $ satisfying the equation $$\begin{vmatrix} (1+\alpha)^2 & (1+2\alpha)^2 & (1+3\alpha)^2\\ (2+\alpha)^2& (2+2\alpha)^2 & (2+3\alpha)^2\\ (3+\alpha)^2& (3+2\alpha)^2 & (3+3\alpha)^2 \end{vmatrix}=-648\alpha $$ I used $$R_3 \rightarrow R_3- R_2 \qquad R_2 \rightarrow R_2- R_1$$ $$\begin{vmatrix} (1+\alpha)^2 & (1+2\alpha)^2 & (1+3\alpha)^2\\ 3+2\alpha& 3+4\alpha & 3+6\alpha\\ 5+2\alpha& 5+4\alpha & 5+6\alpha \end{vmatrix}$$ Then $$R_3 \rightarrow R_3- R_2$$ $$\begin{vmatrix} (1+\alpha)^2 & (1+2\alpha)^2 & (1+3\alpha)^2\\ 3+2\alpha& 3+4\alpha & 3+6\alpha\\ 2& 2 & 2 \end{vmatrix}$$ Now applying column will make zero but the question will become too long . This is contest question and so it should not be that long .

Hint write it as a product of two determinants after taking $\alpha,\alpha^2$ common from one of the determinants to get $\alpha=\pm 9$ or to continue your method use $R_1\to R_1-R_2$

Why does $\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$? In the paper "A Parametric Texture Model based on Joint Statistics of Complex Wavelet Coefficients", the authors use this equation for the angular part of the filter in polar coordinates: $$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right)$$ My friend and I have tested many values of $k > 1$, and in each case this summation is equal to $$\frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$ The paper asserts this as well. We are interested in having an analytic explanation of this equality, if it really holds. How can we derive this algebraically? TL;DR Is this true, and if so, why? $$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$

Suppose we seek to verify that $$\sum_{k=0}^n \cos^{2n}\left(x+\frac{k\pi}{n+1}\right) = \frac{n+1}{2^{2n}} {2n\choose n}.$$ The LHS is $$\sum_{k=0}^n \cos^{2n}\left(x+\frac{k\times 2\pi}{2n+2}\right).$$ Observe also that $$\sum_{k=0}^n \cos^{2n}\left(x+\frac{(k+n+1)\times 2\pi}{2n+2}\right) \\ = \sum_{k=0}^n \cos^{2n}\left(x+\pi+\frac{k\times 2\pi}{2n+2}\right) = \sum_{k=0}^n \cos^{2n}\left(x+\frac{k\times 2\pi}{2n+2}\right)$$ because the cosine is raised to an even power. Therefore the LHS is in fact $$\frac{1}{2} \sum_{k=0}^{2n+1} \cos^{2n}\left(x+\frac{k\times 2\pi}{2n+2}\right).$$ Hence we need to prove that $$\frac{1}{2} \sum_{k=0}^{2n+1} \left(\exp\left(ix+k\times\frac{2\pi i}{2n+2}\right) + \exp\left(-ix-k\times\frac{2\pi i}{2n+2}\right)\right)^{2n} \\ = (n+1)\times {2n\choose n}.$$ Introducing $$f(z) = \left(\exp(ix)z+\exp(-ix)/z\right)^{2n} \frac{(2n+2)z^{2n+1}}{z^{2n+2}-1}$$ We have that the sum is $$\frac{1}{2} \sum_{k=0}^{2n+1} \mathrm{Res}_{z=\exp(2\pi ik/(2n+2))} f(z).$$ The other potential poles are at $z=0$ and at $z=\infty$ and the residues must sum to zero. For the candidate pole at zero we write $$f(z) = \left(\exp(ix)z^2+\exp(-ix)\right)^{2n} \frac{(2n+2)z}{z^{2n+2}-1}$$ and we see that it vanishes. Therefore the target sum is given by $$-\frac{1}{2} \mathrm{Res}_{z=\infty} f(z) \\ = \frac{1}{2} \mathrm{Res}_{z=0} \frac{1}{z^2} \left(\exp(ix)/z+\exp(-ix)z\right)^{2n} \frac{1}{z^{2n+1}} \frac{2n+2}{1/z^{2n+2}-1} \\ = \frac{1}{2} \mathrm{Res}_{z=0} \frac{1}{z} \left(\exp(ix)/z+\exp(-ix)z\right)^{2n} \frac{2n+2}{1-z^{2n+2}} \\ = \frac{1}{2} \mathrm{Res}_{z=0} \frac{1}{z^{2n+1}} \left(\exp(ix)+\exp(-ix)z^2\right)^{2n} \frac{2n+2}{1-z^{2n+2}}.$$ This is $$(n+1) [z^{2n}] \left(\exp(ix)+\exp(-ix)z^2\right)^{2n} \frac{1}{1-z^{2n+2}}.$$ Now we have $$\frac{1}{1-z^{2n+2}} = 1 + z^{2n+2} + z^{4n+4} + \cdots$$ and only the first term contributes, leaving $$(n+1) [z^{2n}] \left(\exp(ix)+\exp(-ix)z^2\right)^{2n} \\ = (n+1) \times {2n\choose n} \exp(ixn)\exp(-ixn) = (n+1) \times {2n\choose n}.$$ This is the claim. Remark. Inspired by the work at this MSE link.

Prime modulo maximum Prove that the remainder of division of a positive number $n$ by a prime $p \le n$ is maximized when $p$ is the smallest prime larger than $\frac{n}{2}.$ It is easy to see that for any number of the form $\frac{n}{2}+k$ where $k \gt 0$, if $k$ is increased remainder will decrease. How to prove that for any number $p \le \frac{n}{2}$, I cannot obtain remainder more than what can be obtained from $\frac{n}{2}+k$ (first prime number large than half of n).

This does not hold true. Consider $n=14$, the smallest prime larger than $\frac{14}{2}=7$ is $11$, but the maximum remainder is attained for prime $5 \lt 7$: $$14 \bmod 5 = 4 \;\;\gt\;\; 14 \bmod 11 = 3$$ [ EDIT ]    The following shetches the proof to the related question asked in a comment below. The remainder of the division of a positive number $n$ by a positive $p \le n$ (not necessarily a prime) is maximized when $p = \lfloor \frac{n}{2}\rfloor + 1$. First, let $p = \lfloor \frac{n}{2}\rfloor + k$ where $k \ge 1$. Then $2 p \gt n$ so the quotient of the division must be $1$, and the remainder will be $n - p = n - \lfloor \frac{n}{2}\rfloor - k$. This is obviously maximized when $k = 1$ in which case it is $r_{max} = n - \lfloor \frac{n}{2}\rfloor - 1$. Now, take $p \le \lfloor \frac{n}{2}\rfloor$. Then the remainder will be by definition a number $r \lt p \le \lfloor \frac{n}{2}\rfloor$. Since $r$ is an integer, it follows that $r \le \lfloor \frac{n}{2}\rfloor - 1 \le n - \lfloor \frac{n}{2}\rfloor - 1 = r_{max}$. The equality could only be attained when $2 \lfloor \frac{n}{2}\rfloor = n$ but in that case $n$ would need to be even, $p = \lfloor \frac{n}{2} \rfloor$ would be a divisor of $n$ and the remainder would be $0$. This proves the strict inequality $r \lt r_{max}$.

show that $X/(X+Y) $ has a cauchy distribution if $X $ and $X+Y $ are standard normal A random variable $X$ has a cauchy distribution with parameters $a$ and $b$ is the density of $X$ is $f(x\mid a,b)=\dfrac{1}{\pi b}\dfrac{1}{1+(\frac{x-a}{b})^2}$ where $-\infty <x< \infty $, $-\infty <a <\infty$, $b>0$ Suppose $X$ and $Y$ are independent standard normal random variables then show that $X/(X+Y)$ has a cauchy distribution. Since $X$ and $Y$ are independent standard normal random variables then the joint pdf for $(X,Y)$ is $$f_{X,Y}(x,y)=\frac{1}{2\pi}e^{-x^2/2}e^{-y^2/2}$$ I used the Jacobian method to try and find $f_{U,V}$ so I let $U=X+Y$ and $V=X/(X+Y)$ so $x=uv$ and $y=u-uv$ and then $J=u$. So $f_{U,V}(u,v)=\dfrac{u}{2\pi}e^{-(uv)^2/2}e^{-(u-uv)^2/2}=\dfrac{u}{2\pi}e^{-(uv)^2/2}e^{-(u^2-2u^2v + u^2v^2)/2}=\dfrac{u}{2\pi}e^{-(u^2v^2)/2}e^{(-u^2/2)+(u^2v) - (u^2v^2/2)}= \dfrac{u}{2\pi}e^{-(u^2v^2)-(u^2/2)+(u^2v)} = \dfrac{u}{2\pi}e^{-u^2(v^2-v+(1/2))} $ Therefore $$f_{U,V}(u,v)= \dfrac{u}{2\pi}e^{-u^2(v^2-v+(1/2))} ; -\infty<u,v<\infty$$ I now want to find the $V$ marginal density and that should have a cauchy distribution. $f_V(v)=2\int_0^{\infty} \dfrac{u}{2\pi}e^{-u^2(v^2-v+(1/2))}du=\int_0^{\infty} \dfrac{u}{\pi}e^{-u^2(v^2-v+(1/2))}du$. If I let $s=u^2$ then $\dfrac{ds}{2u}=du$ and then integral then becomes $ \int_0^{\infty} \dfrac{1}{2\pi}e^{-s(v^2-v+(1/2))}ds = \dfrac{1}{2\pi}\bigg( \dfrac{1}{v^2-v+(1/2)} \bigg)$. I think I might have made a mistake somewhere because I cant see how I can rearrange this to show that $f_V$ is a cauchy distribution.

You had almost reached the end. It suffices to transform: $$\left(\dfrac{1}{2\pi}\right)\dfrac{1}{v^2-v+\frac{1}{2}}=\left(\dfrac{1}{2\pi}\right)\dfrac{1}{(v-\frac{1}{2})^2+(\frac{1}{2})^2}=\left(\dfrac{1}{2\pi}\right)\dfrac{4}{1+\left(\frac{v-\frac{1}{2}}{\frac{1}{2}}\right)^2}$$ giving $$\left(\dfrac{1}{\pi \frac{1}{2}}\right)\dfrac{1}{1+\left(\frac{v-\frac{1}{2}}{\frac{1}{2}}\right)^2}.$$ Thus $a=b=\frac{1}{2}.$

How can the surd $\sqrt{2-\sqrt{3}}$ be expressed? I was wondering how $\sqrt{2-\sqrt{3}}$ could be expressed in terms of $\frac{\sqrt{3}-1}{\sqrt{2}}$. I did try to solve both the expressions separately but none of them seemed to match. I would appreciate it if someone could also mention the procedure

Theorem: Given a nested radical of the form $\sqrt{X\pm Y}$, it can be rewritten into the form $$\sqrt{\frac {X+\sqrt{X^2-Y^2}}{2}}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}{2}}\tag{1}$$ Where $X>Y$. Therefore, we have $X=2,Y=\sqrt{3}$ because $2>\sqrt{3}$. So plugging that into $(1)$ gives us $$\sqrt{\frac {2+\sqrt{4-3}}{2}}-\sqrt{\frac {2-\sqrt{4-3}}{2}}\tag{2}$$ Simplifying $(2)$ gives us $$\sqrt{\frac {2+1}{2}}-\sqrt{\frac {2-1}{2}}\implies \sqrt{\frac 32}-\sqrt{\frac 12}$$ $$\therefore\sqrt{2-\sqrt{3}}=\frac {\sqrt{3}-1}{\sqrt{2}}$$ Alternatively, one can rewrite it as a sum of two surds, and simplify from there. Specifically, let $\sqrt{2-\sqrt3}$ equal $\sqrt d-\sqrt e$. Squaring, we get\begin{align*} & 2-\sqrt3=d+e-2\sqrt{de}\\ & \therefore\begin{cases}d+e=2\\de=\frac 34\end{cases}\end{align*} With solving for $d$ and $e$ gives the simplification.

Number of integer triangles with sides 4 or less Consider triangles having integer sides such that no side is greater than 4 units. How many such triangles are possible? I suspect a relation to the following question: How many ways can $r$ things be taken from $n$ with repetition and without regard to order?

Without loss of generality, assume that $a \leq b \leq c < a+b$. This leaves us with very few choices. 1) All three could be equal. That gives us four choices. 2) $a=1$. Then, $b+1 > c$, so $b=c$ must happen, this gives three choices. 3) $a=2$. Then, $b \leq c < b+2$, so $b=2$,$c=3$ and $b=3, c=3,4$ , and $b=c=4$ are the possibilities. 4) $a=3$, then $b=3,c=4$ and $b=c=4$ are the only possibilities. Hence, the total is $4+3+4+2 = 13$ possibilities.

Prove that $S = \{a^2 + b^2: a,b \in \Bbb N\}$ is closed under multiplication. Is it true? Can you prove or disprove this? $S = \{a^2 + b^2: a,b \in \Bbb N\}$ is closed under multiplication.

Suppose that $x=a^2+b^2$ and $y=c^2+d^2$. Then $$xy=\det \begin{pmatrix} a&b\\-b&a \end{pmatrix} \det\begin{pmatrix} c&d\\-d&c \end{pmatrix} =\det \begin{pmatrix} a&b\\-b&a \end{pmatrix} \begin{pmatrix} c&d\\-d&c \end{pmatrix} =\det\left( \begin{array}{cc} a c-b d & b c+a d \\ -b c-a d & a c-b d \\ \end{array} \right)=(ac-bd)^2+(bc+ad)^2$$

Question in complex numbers from GRE This is a question motivated from GRE subgect test exam. if f(x) over the real number has the complex numbers $2+i$ and $1-i$ as roots,then f(x) could be: a) $x^4+6x^3+10$ b) $x^4+7x^2+10$ c) $x^3-x^2+4x+1$ d) $x^3+5x^2+4x+1$ e) $x^4-6x^3+15x^2-18x+10$ What I thought at first was to calculate $(x-2-i)(x-1+i)$ and find the polynomial that is divisible by it.Unfortunately it turns out that $(x-2-i)(x-1+i)$ is complex polynomial which makes thing harder to calculate and since this is a multiple choice question with very limited time there must be an easier way. Then I thought maybe it will be easy if I write the complex numbers in polar form and check explicitly if they are the roots.But I don't think that's a very efficient way as well. Then I noticed that the question ends with "$f(x)$ could be" which may suggest that there is a way eliminate the wrong choices, however I have no idea what to eliminate or not. Does anyone have any ideas? Thanks in advance

1. $(x-(2-i))(x-(2+i))$ $x^2-x(2+i)-x(2-i)+(2-i)(2+i)$ $x^2-2x-xi-2x+xi+(4-2i+2i+1)$ $x^2-4x+5$ 2. $(x-(1-i))(x-(1+i))$ $x^2-x(1+i)-x(1-i)+(1-i)(1+i)$ $x^2-x-xi-x+xi+(1+i-i+1)$ $x^2-2x+2$ 3. $(x^2-4x+5)(x^2-2x+2)$ $x^4-2x^3+2x^2-4x^3+8x^2-8x+5x^2-10x+10$ $x^4-6x^3+15x^2-18x+10$

Prove that $(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$ Let $a,b,c,d>0$. Prove that $$(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$$ I don't know how to begin to solve this problem

We can assume that $a+b+c+d=2$. Then the inequality becomes $$ (1-d)(1-c)(1-b)(1-a)\le\left(1-\tfrac{c+d}2\right)\left(1-\tfrac{a+d}2\right)\left(1-\tfrac{a+b}2\right)\left(1-\tfrac{c+b}2\right)\tag{1} $$ If any of $a$, $b$, $c$, or $d$ is greater than $1$, then the left side is negative and the inequality is trivial. So we can assume $0\le a,b,c,d\le1$. Subsituting $a\mapsto1-a$, $b\mapsto1-b$, $c\mapsto1-c$, and $d\mapsto1-d$ shows that the inequality is equivalent to $$ \begin{align} abcd &\le\left(\frac{c+d}2\right)\left(\frac{a+d}2\right)\left(\frac{a+b}2\right)\left(\frac{c+b}2\right)\\ &=\left(\frac{ac-bd}4\right)^2+\left(\frac1a+\frac1b+\frac1c+\frac1d\right)\frac{abcd}8\tag{2} \end{align} $$ Since $\frac1x$ is convex for $x\gt0$, Jensen's Inequality says that $$ \begin{align} \frac14\left(\frac1a+\frac1b+\frac1c+\frac1d\right)&\ge\frac1{\frac14(a+b+c+d)}\\ &=2\tag{3} \end{align} $$ $(3)$ shows that $(2)$ is true, which in turn shows that $(1)$ is true.

Calculating $\sum_{n=1}^∞ \frac{1}{(2 n-1)^2+(2 n+1)^2}$ using fourier series of $\sin x$ I have to calculate $\frac{1}{1^2+3^2}+\frac{1}{3^2+5^2}+\frac{1}{5^2+7^2}+...$ using half range Fourier series $f(x)=\sin x$ which is: $f(x)=\frac{2}{\pi}-\frac{2}{\pi}\sum_{n=2}^\infty{\frac{1+\cos n\pi}{n^2-1}\cos nx}$ I have no idea how to proceed. I'll appreciate if someone guide me.

A different approach. Since $\frac{1}{(2n-1)^2+(2n+1)^2}=\frac{1}{8n^2+2}$ we have: $$\begin{eqnarray*} \sum_{n\geq 1}\frac{1}{(2n-1)^2+(2n+1)^2}&=&\frac{1}{8}\sum_{n\geq 1}\frac{1}{n^2+\frac{1}{4}}\\&=&\frac{1}{8}\int_{0}^{+\infty}\sum_{n\geq 1}\frac{\sin(nx)}{n}e^{-x/2}\,dx\\&=&\frac{1}{8(1-e^{-\pi})}\int_{0}^{2\pi}\frac{\pi-x}{2}e^{-x/2}\,dx\\&=&\color{red}{\frac{1}{8}\left(-2+\pi\coth\frac{\pi}{2}\right)}.\end{eqnarray*} $$ We exploited $\int_{0}^{+\infty}\sin(ax)e^{-bx}\,dx = \frac{a}{a^2+b^2}$ and the fact that $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ is the Fourier series of a sawtooth wave.

Verify the triple angle formula Verify the triple angle formula $$\tan(3x) = \frac{3 \tan(x) − \tan^3(x)}{1 − 3 \tan^2(x)}$$ I have tried simplifying the right side by the following $$\tan(3x) = \frac{\tan(x)(3 − \tan^2(x))}{1 − 3 \tan^2(x)}$$ but then I am getting stuck trying to verify the equation

$$\tan 3x =\tan(x+2x)= \frac{\tan x + \tan 2x}{1 − \tan x \tan 2x}=$$ $$=\frac{\tan x + \frac{2\tan x}{1-\tan^2x}}{1 − \tan x \frac{2\tan x}{1-\tan^2x}}=\frac{3\tan x- \tan ^3x}{1-3\tan^2x}$$

Limit of the two variable function $f(x,y).$ How to show that $$\lim_{(x,y)\to(0,0)}\frac{x^{2}y^{2}}{\sqrt{x^{2}+y^{2}}(x^{4}+y^{2})}=0$$ I tried with different paths as $x=0,y=0, y=x$ its comes to zero but i have no general idea. Please help. Thanks to lot.

You have \begin{align} \left| \frac{x^{2}y^{2}}{\sqrt{x^{2}+y^{2}}(x^{4}+y^{2})}\right| &=\frac{x^{2}y^{2}}{\sqrt{x^{2}+y^{2}}(x^{4}+y^{2})} \leq \frac{x^{2}y^{2}}{\sqrt{x^{2}+y^{2}}\,(y^{2})}\\ \ \\ &=\frac{x^2}{\sqrt{x^2+y^2}}\leq\frac{x^2}{\sqrt{x^2}}=|x|\\ \ \\ &\leq\sqrt{x^2+y^2} \end{align}

Does the sum $\sum_{n \geq 1} \frac{2^n\operatorname{mod} n}{n^2}$ converge? I am somewhat a noob, and I don't recall my math preparation from college. I know that the sum $\displaystyle \sum_{n\geq 1}\frac{1}{n}$ is divergent and my question is if the sum$$\sum \limits _{n\geq 1}\frac{2^n\mod n}{n^2}$$converges. I think is not but I do not know how to prove that! Thanks!

In this answer, we prove that $$ \sum_{n=1}^{\infty} \frac{2^n \text{ mod } n}{n^2} = \infty. \tag{*} $$ Idea. The intuition on $\text{(*)}$ comes from the belief that the sequence $(2^n \text{ mod } n)/n$ is equidistributed on $[0, 1]$, which is quite well supported by numerical computation. Proving this seems quite daunting, though, so we focus on a particular subsequence which is still good to give a diverging lower bound of $\text{(*)}$. To be precise, we focus on the indices of the form $n = 5p$ for some prime $p$ and prove that the corresponding sum is comparable to the harmonic series for primes $\sum_p \frac{1}{p}$, which also diverges. Proof. To this end, we consider the sequence $(a_k : k \geq 0)$ in $[0, 1)$ defined by $$ a_k = \left\{ \frac{2^{5p_k}}{5p_k} \right\},$$ where $\{ x \} = x - \lfloor x \rfloor$ is the fractional part of $x$ and $p_k$ is the $k$-th prime number. Now focusing only on the index $n = 5p_k$ for some $k$, we can bound the sum $\text{(*)}$ below by $$ \sum_{n=1}^{\infty} \frac{2^n \text{ mod } n}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n}\left\{ \frac{2^n}{n} \right\} \geq \sum_{k=1}^{\infty} \frac{a_k}{5p_k}. $$ So it suffices to prove that this bound diverges. First, for any prime $p$ we have $$ 2^{5p} \equiv 2^5 \equiv 32 \pmod{p}. $$ This allows us to write $2^{5p} = mp + 32$ for some non-negative $m$. Next, notice that any prime $p$ other than $2$ and $5$ are either of the form $p = 4k+1$ or of the form $p = 4k+3$. Depending on which class $p$ falls in, we find that $$ 2^{5p} \equiv 2^p \equiv \begin{cases} 2, & \text{if } p =4k+1 \\ 3, & \text{if } p =4k+3 \end{cases} \pmod{5}. $$ What this tells about $m$ is as follows: $$ m \equiv \begin{cases} 0, & \text{if } p =4k+1 \\ p^{-1}, & \text{if } p =4k+3 \end{cases} \pmod{5}. $$ (Here, $p^{-1}$ is the multiplicative inverse of $p$ modulo $5$.) From this, for $p_k > 32$ we have the following estimate: $$ a_k \geq \frac{1}{5} \quad \text{if } p_k \equiv 3 \pmod{4}. $$ Consequently, by the PNT for arithmetic progression, $$ \frac{a_1 + \cdots + a_n}{n} \geq \frac{1}{5} \frac{\pi_{4,3}(p_n) + \mathcal{O}(1)}{\pi(p_n)} \xrightarrow[n\to\infty]{} \frac{1}{10}. $$ (The $\mathcal{O}(1)$-term appears by discarding terms with $p_k \leq 32$.) Finally, let $s_n = a_1 + \cdots + a_n$. Then by summation by parts, as $N \to \infty$ we have \begin{align*} \sum_{k=1}^{N} \frac{a_k}{5p_k} &= \frac{1}{5} \bigg( \frac{s_N}{p_N} + \sum_{k=1}^{N-1} \left( \frac{1}{p_k} - \frac{1}{p_{k+1}} \right) s_k \bigg) \\ &\geq \frac{1}{5} \sum_{k=1}^{N-1} \left( \frac{1}{p_k} - \frac{1}{p_{k+1}} \right) \frac{k}{11} + \mathcal{O}(1) \\ &\geq \frac{1}{55} \sum_{k=1}^{N} \frac{1}{p_k} + \mathcal{O}(1). \end{align*} Taking $N \to \infty$, this series diverges by the harmonic series for primes. Therefore the claim $\text{(*)}$ follows. //// Elaborating this argument, we find that $$ a_k \equiv \frac{2^{5p_k}}{5p_k} \equiv \tilde{a}_k + \frac{32}{5p_k} \pmod{1} $$ where $\tilde{a}_k$ satisfies $$ \tilde{a}_k = \begin{cases} 0, & \text{if } p_k \equiv 1, 9, 13, 17 \pmod{20} \\ 1/5, & \text{if } p_k \equiv 11 \pmod{20} \\ 2/5, & \text{if } p_k \equiv 3 \pmod{20} \\ 3/5, & \text{if } p_k \equiv 7 \pmod{20} \\ 4/5, & \text{if } p_k \equiv 19 \pmod{20} \end{cases}. $$ Thus by the PNT for arithmetic progression again, we have the following convergence in distribution: $$ \frac{1}{n} \sum_{k=1}^{n} \delta_{a_k} \xrightarrow{d} \frac{1}{2}\delta_{0} + \frac{1}{8}\sum_{j=1}^{4} \delta_{j/5} \quad \text{as } n \to \infty$$ The following numerical simulation using first 1,000,000 terms of $(a_k)$ clearly demonstrates this behavior:

$4x^2y′′-8x^2y′+(4x^2+1)y=0$ solve by Frobenius Method. I would like to ask if someone can explain to me how can we solve the following DE using this method. $4x^2y′′-8x^2y′+(4x^2+1)y=0$

Given differential equation $4x^2y^{\prime\prime}-8x^2y^{\prime}+(4x^2+1)y=0$. $P(x)=-2$ and $Q(x)=\frac{1+4x^2}{4x^2}$, since $Q(x)$ is not analytic at $x=0$, we say $x=0$ is a singular point of this differential equation . Singular points are futher classified into regular singular and irregular singular points. Since $xP(x)$ and $x^2Q(x)$ are analytic at $x=0$, we say $x=0$ is a regular singular point. So we can assume that $y=\displaystyle\sum_{n=0}^{\infty} a_n x^{n+m}$ for real number $m$ which we will find out. Then $y^{\prime}=\displaystyle\sum_{n=0}^{\infty} a_n (n+m)x^{n+m-1}$ and $y^{\prime\prime}=\displaystyle\sum_{n=0}^{\infty} a_n(n+m)(n+m-1) x^{n+m-2}$. We will substitute this in the given differential equation. We get, $$(4a_0m(m-1)+a_0)x^m+(4a_1m(m+1)-8a_0m+a_1)x^{m+1}+\sum_{n=0}^{\infty}\left[ 4a_n+4a_{n+2}(n+m+1)(n+m+2)-8a_{n+1}(n+m+1)+a_n+2\right]x^{n+m+2}=0$$ $m=\frac{1}{2}$, $a_1=a_0$ and $a_{n+2}(4(n+m+2)(n+m+1)+1)-8a_{n+1}(n+m+1)=4a_n=0$ Put $m=\frac{1}{2}$ , the recussion relation is $a_{n+2} = \dfrac{1}{(n+2)^2}\left((2n+3)a_{n+1}-a_n\right)$. Now by induction you can prove that $a_n = \dfrac{a_0}{n!}$. Therefore $y_1 = \sqrt{x}\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!} = \sqrt{x}e^x$. Since the roots of the Indicial equation equation are equal, we cannot find the second Inependent solution using this method. It can be solved as $y_2 = y_1\displaystyle\int \dfrac{1}{y_1^2}e^{-\int p(x)dx}dx$. So $y_2 = \sqrt{x}e^x\displaystyle\int \dfrac{1}{xe^{2x}}e^{2x}dx = \sqrt{x}e^x\ln{|x|}$. Thefore our final answer wil be $y=c_1y_1+c_2y_2 = c_1\sqrt{x}e^x+c_2\sqrt{x}e^x\ln{|x|}$. Where $c_1$ and $c_2$ are arbitrary constants. Hope this helps :)

What is $\lim\limits_{n\to\infty}\frac {1}{n^2}\sum\limits_{k=0}^{n}\ln\binom{n}{k} $? It was originally asked on another website but nobody has been able to prove the numerical result. The attempts usually go by Stirling's approximation or try to use the Silverman-Toeplitz theorem.

By Stolz Cezaro $$\lim\limits_{n\to\infty}\frac {1}{n^2}\sum\limits_{k=0}^{n}\ln\binom{n}{k}=\lim\limits_{n\to\infty}\frac {1}{2n+1} \left(\sum\limits_{k=0}^{n+1}\ln\binom{n+1}{k}- \sum\limits_{k=0}^{n}\ln\binom{n}{k} \right)\\ =\lim\limits_{n\to\infty}\frac {1}{2n+1} \sum\limits_{k=0}^{n}\left(\ln\binom{n+1}{k}- \ln\binom{n}{k} \right)\\ =\lim\limits_{n\to\infty}\frac {1}{2n+1} \sum\limits_{k=0}^{n}\ln\left(\frac{(n+1)!}{k!(n+1-k)!} \frac{k!(n-k)!}{n!} \right)\\ =\lim\limits_{n\to\infty}\frac { \sum\limits_{k=0}^{n}\ln\left(\frac{(n+1)}{(n+1-k)} \right)}{2n+1}\\ =\lim\limits_{n\to\infty}\frac { \ln\left(\prod\limits_{k=0}^{n}\frac{(n+1)}{(n+1-k)} \right)}{2n+1}\\ =\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+1)^n}{(n+1)!} \right)}{2n+1}\\ $$ Applying stolz Cezaro again, we get $$ =\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n+1}}{(n+2)!} \right)-\ln\left(\frac{(n+1)^n}{(n+1)!} \right)}{2} \\ =\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n+1}}{(n+2)!} \frac{(n+1)!}{(n+1)^n} \right)}{2} \\ =\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n}} {(n+1)^n} \right)}{2}\\ =\lim\limits_{n\to\infty}\frac { \ln\left(1+\frac{1}{n+1} \right)^n}{2}\\ =\frac{\ln(e)}{2}=\frac{1}{2} $$

Diophantine equation $\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n$ Let $a,b,c$ and $n$ be natural numbers and $\gcd(a,b,c)=\gcd(\gcd(a,b),c)=1$. Does it possible to find all tuples $(a,b,c,n)$ such that: $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n?$$

First note that we can assume $a,b,c \in \mathbb{Q}$, without loss of generality. A rational solution can then be scaled to integers. Combining into a single fraction gives the quadratic in $a$ \begin{equation*} a^2+\frac{b^2-nbc+c^2}{b+c}a+bc=0 \end{equation*} and, for rational solutions, the discriminant must be a rational square. Thus, the quartic \begin{equation*} D^2=b^4-2(n+2)b^3c+(n^2-6)b^2c^2-2(n+2)bc^3+c^4 \end{equation*} must have rational solutions. This quartic is birationally equivalent to the elliptic curve \begin{equation*} V^2=U^3+(n^2-12)U^2+16(n+3)U \end{equation*} with \begin{equation*} \frac{b}{c}=\frac{V+(n+2)U}{2(U-4(n+3))} \end{equation*} The elliptic curve is singular when $n=-2,-3,6$. $n=6$, for example, corresponds to $a=b=c=K$ as a solution. If $n \ne -2,-3,6$, the curve has $5$ finite torsion points at $(0,0)$, $(4,\pm 4(n+2))$ and $(4n+12,\pm 4(n+2)(n+3))$ none of which give a solution. For $n=7$, there are a further $6$ torsion points which lead to the solutions $(a,b,c)=(1,1,2)$ and $(a,b,c)=(1,2,2)$. Thus, if $n \ne -3,-2,6,7$, we need the elliptic curve to have rank greater than zero to find a solution. Computations using the Birch and Swinnerton-Dyer conjecture suggest the first few solutions are with $n=8,11,12,15,\ldots$. For example, the $n=15$ curve has a generator $(-36,468)$ which gives the solution $a=2, b=3, c=15$. As $n$ gets larger, the size of solutions generally increases.

Probability that a number is divisible by 11 The digits $1, 2, \cdots, 9$ are written in random order to form a nine digit number. Then, the probability that the number is divisible by $11$ is $\ldots$ I know the condition for divisibility by $11$ but I couldn't guess how to apply it here. Please help me in this regard. Thanks.

The rule of divisibility by $11$ is as follows: The difference between the sum of digits at odd places and the sum of the digits at even places should be $0$ or a multiple of $11$. We also know that the sum of all the digits will be $45$ as $1 + 2 + ... + 9 = 45$. Let $x$ denote sum of digits at even position s and $y$ denote sum of digits at odd places, or vice versa. Case 1 (difference is $0$): $$x + y = 45$$ $$x - y = 0$$ Thus, $2x = 45$, or $x = 22.5$ which cannot be obtained. Case 2 (difference is $11$): $$x + y = 45$$ $$x - y = 11$$ Thus, $2x = 56$, or $x = 28$ and $y = 17$. This is a valid possibility. Case 3 (difference is $22$): $$x + y = 45$$ $$x - y = 22$$ Thus, $2x = 67$, or $x = 33.5$, which cannot be obtained. As you can see, the difference between the sum of the digits at odd places and the sum of the digits at even places must be $11$. Now, imagine that there are $9$ placeholders (representing the $9$ digits of the $9$-digit number). Either the sum of the digits at odd places ($5$ odd places) should be $28$, or the sum of the digits at even places ($4$ even places) should be $28$. We write down the possibilities: $2$ ways to express $28$ as a sum of $4$ numbers between $1$ and $9$. $9$ ways to express $28$ as a sum of $5$ numbers between $1$ and $9$. In the first case, there are $4!$ ways of arranging the $4$ numbers (that add up to $28$) and $5!$ ways of arranging the $5$ other numbers (that add up to $17$). Hence, no. of ways$ = 2 * 4! * 5!$ In the second case, there are $5!$ ways of arranging the $5$ numbers ( that add up to $28$) and $4!$ ways of arranging the $4$ other numbers (that add up to $17$). Hence, no. of ways$ = 9 * 5! * 4!$ Total favourable possibilities$$= 2 * 4! * 5! + 9 * 5! * 4!$$ $$= 4! * 5! * (2 + 9)$$ $$= 4! * 5! * 11$$ Also, total no. of ways of arranging $9$ numbers to form a $9$-digit number = $9!$ Hence, probability$=P= (4! * 5! * 11)/9!$ $$= 11/126$$

A better way to evaluate a certain determinant Question Statement:- Evaluate the determinant: $$\begin{vmatrix} 1^2 & 2^2 & 3^2 \\ 2^2 & 3^2 & 4^2 \\ 3^2 & 4^2 & 5^2 \\ \end{vmatrix}$$ My Solution:- $$ \begin{align} \begin{vmatrix} 1^2 & 2^2 & 3^2 \\ 2^2 & 3^2 & 4^2 \\ 3^2 & 4^2 & 5^2 \\ \end{vmatrix} &= (1^2\times2^2\times3^2)\begin{vmatrix} 1 & 1 & 1 \\ 2^2 & \left(\dfrac{3}{2}\right)^2 & \left(\dfrac{4}{3}\right)^2 \\ 3^2 & \left(\dfrac{4}{2}\right)^2 & \left(\dfrac{5}{3}\right)^2 \\ \end{vmatrix}&\left[\begin{array}{11}C_1\rightarrow\dfrac{C_1}{1} \\ C_2\rightarrow\dfrac{C_2}{2^2}\\ C_3\rightarrow\dfrac{C_3}{3^2}\end{array}\right]\\ &= (1^2\times2^2\times3^2)\begin{vmatrix} 1 & 0 & 0 \\ 2^2 & \left(\dfrac{3}{2}\right)^2-2^2 & \left(\dfrac{4}{3}\right)^2-2^2 \\ 3^2 & 2^2-3^2 & \left(\dfrac{5}{3}\right)^2-3^2 \\ \end{vmatrix} &\left[\begin{array}{11}C_2\rightarrow C_2-C_1 \\ C_3\rightarrow C_3-C_1\end{array}\right]\\ &= (1^2\times2^2\times3^2) \begin{vmatrix} 1 & 0 & 0 \\ 2^2 & 2^2-\left(\dfrac{3}{2}\right)^2 & 2^2-\left(\dfrac{4}{3}\right)^2 \\ 3^2 & 3^2-2^2 & 3^2-\left(\dfrac{5}{3}\right)^2 \\ \end{vmatrix}\\ &=(1^2\times2^2\times3^2) \begin{vmatrix} 1 & 0 & 0 \\ 2^2 & \dfrac{7}{4} & \dfrac{20}{9} \\ 3^2 & 5 & \dfrac{56}{9} \\ \end{vmatrix}\\ &=(1^2\times2^2) \begin{vmatrix} 1 & 0 & 0 \\ 2^2 & \dfrac{7}{4} & 20 \\ 3^2 & 5 & 56 \\ \end{vmatrix}\\ &=4\times(-2)\\ &=-8 \end{align} $$ As you can see, my solution is a not a very promising one. If I encounter such questions again, so would you please suggest a better method which doesn't include this ridiculous amount of calculations.

Using the rule of Sarrus, the computation is really not too long, and we get in general for all $n\ge 1$, $$ \det \begin{pmatrix} n^2 & (n+1)^2 & (n+2)^2\cr (n+1)^2& (n+2)^2 & (n+3)^2\cr (n+2)^2& (n+3)^2 & (n+4)^2\end{pmatrix}=-8. $$

How to evaluate $\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ I have a stuck on the problem of L'Hospital's Rule, $\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ which is in I.F. $\frac{0}{0}$ If we use the rule, we will have $\lim\limits_{x\to 0} \frac{\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}}{\sec^2x-\cos x}$. So, I think that I approach this problem in the wrong way. Have you guy any idea?

Alternatively, one may use standard Taylor expansions, as $x \to 0$, $$ \begin{align} \sin x&=x-\frac{x^3}{6}+o(x^4) \\\tan x&=x+\frac{x^3}{3}+o(x^4) \\\arctan x&=x-\frac{x^3}{3}+o(x^4) \\\arcsin x&=x+\frac{x^3}{6}+o(x^4) \end{align} $$ giving, as $x \to 0$, $$ \frac{\arctan x - \arcsin x}{\tan x - \sin x}= \frac{-\frac{x^3}{2}+o(x^4)}{\frac{x^3}{2}+o(x^4)}=-1+o(x) \to -1. $$

System of quadratic equations with parameter I'd appreciate your help with this problem: $p - a^2 = b$ $p - b^2 = c$ $p - c^2 = d$ $p - d^2 = a$ Where $a$, $b$, $c$, $d$ are real numbers, $p$ is a real parameter lower or equal to 1 and greater or equal to 0. Thank you a lot. I am capable of solving the problem for positive numbers, but that's it. :(

* *Subtracting two adjacent equations, $$ \left \{ \begin{array}{ccc} (a-b)(a+b) &=& c-b \\ (b-c)(b+c) &=& d-c \\ (c-d)(c+d) &=& a-d \\ (d-a)(d+a) &=& b-a \end{array} \right.$$ Note that $$a=b \iff b=c \iff c=d \iff a=d$$ Now $$a^2+a-p=0$$ $$a=b=c=d=\frac{-1\pm \sqrt{1+4p}}{2}$$ * *Subtracting alternate equations, $$ \left \{ \begin{array}{ccc} (a-c)(a+c) &=& d-b \\ (b-d)(b+d) &=& a-c \end{array} \right.$$ Note that $$a=c \iff b=d$$ $$ \left \{ \begin{array}{ccc} p-a^2 &=& b \\ p-b^2 &=& a \end{array} \right.$$ Now $$p-(p-a^2)^2=a$$ $$p^2-(2a^2+1)p+(a^4+a)=0$$ $$(a^2+a-p)(a^2-a+1-p)=0$$ Since $a^2+a-p=0$ reproduces the previous case, we only need to solve $$a^2-a+1-p=0$$ $$a=c=\frac{1\pm \sqrt{4p-3}}{2}$$ $$b=d=\frac{1\mp \sqrt{4p-3}}{2}$$ * *For distinct $a$, $b$, $c$ and $d$, that'll be a solution of $12$-th order polynomial equations in terms of $p$. P.S. $a,b,c,d$ are critical points of the iteration $u_{n+1}=p-u_{n}^2$ of period $4$.

Show that $f(g(x)) = x$ and $g(f(x))= x$, with $f(x) = x^e \bmod n$ and $g(x) = x^d \bmod n$ I want to solve the following problem: Let $d$ and $e$, both natural numbers, be each others inverses modulo $\varphi(n)$, where $n = p\cdot q$ is a product of two different prime numbers $p$ and $q$. Let $M = \{0,1,2,\dots,(n-1)\}$ be the set of nonnegative numbers smaller than $n$. Define two functions $f: M \rightarrow M$ and $g: M \rightarrow M$ as \begin{align*} f(x) = x^e \bmod n \quad \mbox{and}\quad g(x) = x^d \bmod n \end{align*} Show that $f(g(x)) = x$ and $g(f(x))= x$ for all $x \in M$. I understand that $f(x)$ and $g(x)$ will always produce numbers between 0 and $n$, since $x$ is smaller than $n$. In that respect, $f(x) = g(x)$ no matter what $e$ and $d$ we choose. But I don't understand why $f(g(x)) = x$ and $g(f(x))= x$.

I think I figured it out. First I have to prove that $x^{k\varphi(n) + 1} \equiv x \pmod{n}$, even when I don't know if $\gcd(x,n) = 1$. We look at the system \begin{align} \begin{cases} y \equiv x \pmod p \\ y \equiv x \pmod q \end{cases} \end{align} Since $q$ and $p$ are two different prime numbers, they are relatively prime to eachother. Then we have \begin{align*} &\varphi(n) = \varphi(p)\cdot \varphi(q) \end{align*} and so, by Eulers theorem, \begin{align*} &x^{k\varphi(n) + 1} = (x^{\varphi(p)})^{k\varphi(q)} \cdot x \equiv 1^{k\varphi(q)} x \equiv x \pmod{p}\\ &x^{k\varphi(n) + 1} = (x^{\varphi(q)})^{k\varphi(p)} \cdot x \equiv 1^{k\varphi(p)} x \equiv x \pmod{q} \end{align*} Thus, a solutions to the set of congruences above is \begin{align*} y = x^{k\varphi(n) + 1} \end{align*} By the Chinese Remainder Theorem, this solution is unique modulo $p\cdot q =n$. Thus, \begin{align*} x^{k\varphi(n) + 1} \equiv x \pmod{n} \end{align*} Then, I can apply the solution as proposed by marwalix, namely \begin{align*} &f(g(x)) = x^{ed} = x^{k\varphi(n) + 1} \equiv x \pmod{n}\\ &g(f(x)) = x^{de} = x^{k\varphi(n) + 1} \equiv x \pmod{n} \end{align*}

trouble with double/iterated integration of $\int^1_0[\int^1_0v(u+v^2)^4du]dv$ I have: For $\int^1_0v(u+v^2)^4du$: u substitution (using x instead since there's a u in there already) with $x=(u+v^2), dx/du=1$ $v\int^1_0x^4=v\frac{1}{5}x^5=v\frac{(u+v^2)^5}{5}-\frac{v(0+v^2)^5}{5}|^1_0=\frac{(v+v^3)^5}{5}-\frac{v^{15}}{5}$ then $ \frac{1}{5}[\int^1_0(v+v^3)^5dv-\int^1_0v^{15}dv]$ used substitution again with $x= (v+v^3),dx/du=(1+3v^2)$ for the first one $\int^1_0\frac{x^5dx}{(1+3v^2)}->\frac{1}{1+3v^2}\int^1_0x^5dx=\frac{1}{1+3v^2}\frac{1}{6}x^6|^1_0->\frac{1}{1+3v^2}\frac{1}{6}(v+v^3)^6|^1_0$ $= \frac{1}{24}*1-0$ for the second: $\int^1_0 v^{15}=\frac{1}{16}v^{16}|^1_0=\frac{1}{16}$ $\frac{1}{5}[\frac{1}{24}-\frac{1}{16}] = -\frac{1}{240}$ The answer is supposed to be 31/30.

An alternate approach would be to change the order of integration. This will greatly reduce the difficulty of the integral. As the limits of the inner integral do not depend upon $v$ this trivially becomes: $$\int^1_0[\int^1_0v(u+v^2)^4\ dv]\ du$$ $$=\int^1_0\left[\frac{1}{10}\bigg((u+v^2)^5\bigg)_0^1\right]\ du$$ $$=\int^1_0\frac{1}{10}\left[(u+1)^5-(u+0)^5\right]\ du$$ $$=\frac{1}{10}\int^1_0(u+1)^5-u^{5}\ du$$ $$=\frac{1}{10}\bigg(\frac{1}{6}(u+1)^6-\frac{1}{6}u^6\bigg)_0^1$$ $$=\frac{1}{10}\bigg(\frac{1}{6}(1+1)^6-\frac{1}{6}1^6-\left(\frac{1}{6}(1+0)^6-\frac{1}{6}0^6\right)\bigg)_0^1$$ $$=\frac{31}{30}$$

$\sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}} = \frac{1+\sqrt{5}}{2} = \phi$, is this a coincidence? I was playing around with square roots today when I "discovered" this. $\sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}} = x$ $\sqrt{1 + x} = x$ $1 + x = x^2$ Which, via the quadratic formula, leads me to the golden ratio. Is there any significance to this or is it just a random coincidence?

If you play around a little more, you will also notice that: $$ \frac{1+\sqrt{5}}{2} = 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+ \ldots} } } } $$ Which simplifies to $x = 1+ \frac 1x \implies x^2=x+1$. It's no coincidence. I mean to say, it comes directly from the equation itself. Just to give you another example: The equation $x^2 = 4+x$ is satisfied by the fraction $\frac{1+\sqrt{17}}{2}$. Now, we can use the same logic to extend this fellow: $$ x = \sqrt{4 + x} = \sqrt{4 + \sqrt{4 + \sqrt{4 + \sqrt{4 + \ldots}}}} $$ While at the same time, this also expands as a continuous fraction, namely: $$ x = 1 + \frac{4}{x} = 1 + \frac{4}{1 + \frac{4}{1 + \frac{4}{1 + \frac{4}{1 + \ldots} } } } $$ You see, it's not a coincidence, yet it's wonderful. The question arises: Can we do this with other quadratic polynomials? Take for example, $ax^2+bx+c=0$. Then $ax^2 = -bx-c$ and $x^2 = -\frac{b}{a}x -\frac{c}{a}$. This will expand now in an interesting way: $$ x = \sqrt{-\frac{c}{a}-\frac{b}{a}x} = \sqrt{-\frac{c}{a}-\frac{b}{a}\sqrt{-\frac{c}{a}-\frac{b}{a}\sqrt{-\frac{c}{a}-\frac{b}{a} \ldots}}} $$ And as a continuous fraction: $$ x = -\frac{b}{a} - \frac{c}{ax} =-\frac{b}{a} - \frac{c}{a(-\frac{b}{a} - \frac{c}{a(-\frac{b}{a} - \frac{c}{a \ldots} )}) } $$ That is your license to play around. Please do so. Also, see what you get if $ax^3+bx^2+cx+d=0$, and if you can find something interesting here do comment.

Are these derivatives correct (the respective functions involve a square root, fraction and expansion)? Question: Use rules of differentiation to answer the following. There is no need to simplify your answer. * *If $y = 3x{\sqrt x}$, find $\frac{dy}{dx}$ My working: $y = 3x^{1+\frac{1}{2}}$ $y = 3x^{\frac{3}{2}}$ $\frac{dy}{dx} = 3 \left(\frac{3}{2}x^{\frac{3}{2}-1}\right)$ $\frac{dy}{dx} = 3 \left(\frac{3}{2}x^{\frac{1}{2}}\right)$ $\frac{dy}{dx} = \frac{9}{2}x^{\frac{1}{2}}$ (differentiating $x^n$) *If $f(x) = \frac{2x-1}{5x}$, find $f'(x)$ My working: $f(x) = \frac{2x-1}{5x}$ $f'(x) = \frac{2(1x^{1-1})-1}{5(1x^{1-1})}$ $f(x) = \frac{1}{5}$ (differentiating $x^n$) *If $y = (2-3x)^2$, find $\frac{dy}{dx}$ My working: $y = 9x^2-12x+4$ $f'(x) = 9(2x^{2-1})-12(1x^{1-1})+0$ $f'(x) = 18x-12$ (differentiating $x^n$ and differentiating a constant)

Your work for Question 1 is fine. For Question 2, use the Quotient Rule (you seem to have assumed that the derivative of a fraction is the derivative of the numerator over the derivative of the denominator, which isn't the case): \begin{align} f' \left ( x \right ) & = \frac{5x \cdot 2 - \left ( 2x - 1 \right ) \cdot 5}{\left ( 5x \right )^2} = \frac{1}{5x^2} \end{align} In the comments you mentioned that your course permits neither the use of the Product Rule nor of the Quotient Rule. In that case, you can split the numerator into two fractions (this is probably a simpler method anyway, but it wasn't the first way that came to mind): \begin{align} f' \left ( x \right ) & = \frac{d}{dx} \left ( \frac{2 x}{5 x} - \frac{1}{5x} \right ) = \frac{d}{dx} \left ( \frac{2}{5} - \frac{1}{5x} \right ) = \frac{d}{dx} \left ( - \frac{1}{5x} \right ) = - \frac{1}{5} \left ( - x^{-2} \right ) = \frac{1}{5x^2} \end{align} For Question 3, you could have also used the Chain rule, although what you've done is fine.

What is the number that, when divided by $3$, $5$, $7$, leaves remainders of $2$, $3$, $2$, respectively? What is the number? The LCM of the divisors is 105. I think this has something to do with the Chinese remainer theorem, but I am not sure how to apply this knowledge.

We can solve in a simple way without the Chinese Remainder Theorem. Let $n = 3x + 2 = 5y + 3 = 7z + 2$. Then, $5y + 3 \equiv 3x + 2 \equiv (\mod 3) \implies y \equiv 1(\mod 3)y$ Hence let $y = 3k + 1$ which gives $n = 5y + 3 = 15k + 8$. Again $n = 15k + 8 \equiv 7z + 2 (\mod 7) \implies k \equiv 1 (\mod 7)$ Hence let $k = 7m + 1$. This gives $n = 15k + 8 = 105m + 23$ Thus in general the number $n$ is of the form $n = 105m + 23$. The smallest such number is when $m = 0$, we get $n = 23$.

Calculate the sum of the series $\sum_{n=1}^\infty \frac 1{n(n+1)^2}$ I have already proven that $$\int_0^1 \ln(1-x)\ln (x)\mathrm d x=\sum_{n=1}^\infty \frac 1{n(n+1)^2}.$$ So now I want so calculate $$S:=\sum_{n=1}^\infty \frac 1{n(n+1)^2}.$$ I think I should use the fact that $$\sum_{n=1}^\infty \frac 1{n^2}=\frac{\pi^2}6$$ and I know using Wolfram Alpha that $S=2-\frac{\pi^2}6$. So finally I will get $$\int_0^1 \ln(1-x)\ln (x)\mathrm d x=2-\frac{\pi^2}6.$$ But how to calculate $S$ ?

we have that $$\sum_{n=1}^{m }\frac{1}{n}- \frac{1}{n+1}=\frac{m}{m+1}=1-\frac{1}{m+1}$$ the sum is $1$when the $m\rightarrow \infty$ $$\sum_{n=1}^{\infty }\frac{1}{n(n+1)^2} = \sum_{n=1}^{\infty }(\frac{1}{n}- \frac{1}{n+1}) -\sum_{n=1}^{\infty } \frac{1}{(n+1)^2}=1 -\sum_{n=2}^{\infty } \frac{1}{n^2}$$ $$=1-(\frac{\pi^2}{6}-1)=2-\frac{\pi^2}{6}$$

$x + y = 7 $and $x^3 - y^3 = 37$, find $xy$ $x+y=7$ $x^3-y^3 = 37$ Find $xy$ I have tackled this one but I am stuck, when I get to this point: $$(x-y)((x+2)^2 -xy) = 37$$ I would be glad if you could give me any suggestions.

Since $(x^3-y^3)^2$ is symmetric in $x$ and $y$, you can express it as a polynomial in terms of $x+y$ and $xy$ : $(x^3-y^3)^2 = x^6 - 2x^3y^3 + y^6 = (x+y)^6 - (6x^5y+15x^4y^2+22x^3y^3+15x^2y^4+6xy^5) \\ = (x+y)^6 - 6xy(x+y)^4 + (9x^4y^2+14x^3y^3+9x^2y^4) \\ = (x+y)^6 - 6xy(x+y)^4 + 9(xy)^2(x+y)^2 - 4(xy)^3$ Plugging the values $x+y = 7$ and $x^3-y^3 = 37$ gives you the equation $4(xy)^3 - 441(xy)^2 +14406(xy)-116280 = 0$ The only rational root of this cubic polynomial is $12$, so this factors as $(xy-12)(4(xy)^2-393(xy)+9690) = 0$ And now you can check that the second factor doesn't have real roots. Finally, getting a value for $xy$ (here, $12$) determines the values of $x$ and $y$ completely, because $x = (x+y)/2 + (x^3-y^3)/(2((x+y)^2-xy))= 7/2 + 37/(2.(7^2-12)) = 4$ and then $y = (x+y)-x = 7-4 = 3$

Prove that $a^2$ when divided by $5$ cannot have a remainder of $3$. For any natural number $a$, prove that $a^2$ when divided by $5$ cannot have a remainder of $3$. (Hint: What are the possible values of the remainder when $a$ is divided by $5$? Using the hint, I found that the possible values of the remainder are $1, 2, 3, 4$, and $0$, and $3$ is only a remainder when the last digit of $a$ is $3$ or $8$. But I'm not sure how to explain this in a proof, and I don't know how to extend it to $a^2$. Any help would be much appreciated!

if $a \equiv 0 \mod 5$, then $a^2 \equiv 0 \mod 5$. if $a \equiv 1 \mod 5$, then $a^2 \equiv 1 \mod 5$. if $a \equiv 2 \mod 5$, then $a^2 \equiv 4 \mod 5$. if $a \equiv -2 \mod 5$, then $a^2 \equiv 4 \mod 5$. if $a \equiv -1 \mod 5$, then $a^2 \equiv 1 \mod 5$.

Prove: $ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $ Prove $$ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $$ where $n \in \mathbb{N}$ and $x$ not a multiple of $\frac{ \pi }{2^k} $ for any $k \in \mathbb{N}$. My try. If $n=2$, we have $$\begin{align} \frac{1}{\sin 2x} + \frac{ 1}{\sin 4 x} &= \frac{1}{\sin 2x} + \frac{1}{2 \sin 2x \cos 2x } \\[6pt] &= \frac{ 2 \cos 2x + 1 }{2 \sin 2x \cos 2x} \\[6pt] &= \frac{2 \cos^2 x - 2 \sin^2 x + \cos^2 x + \sin^2 x}{2 \sin 2x \cos 2x} \\[6pt] &= \frac{3 \cos^2 x - \sin^2 x}{2 \sin 2x \cos 2x} \end{align}$$ but here I got stuck. I am on the right track? My goal is to ultimately use induction.

$$\csc2A+\cot2A=\dfrac{1+\cos2A}{\sin2A}=\cot A$$ $$\iff\csc2A=\cot A-\cot 2A$$ Put $2A=2x,4x,8x,\cdots,2^nx$ and add See also: Telescoping series

Let f be an arbitrary, twice differentiable function for which $f''\neq0$ Let f be an arbitrary, twice differentiable function for which $f''\neq0$ . The function $u(x,y)=f(x^2+axy+y^2)$ satifsfies equation $U_{xx}-U_{yy}=0$ then the constant a is. $f_{xx}(x^2+axy+y^2)(2x+ay)^2+2f_x(x^2+axy+y^2)-f_{yy}(x^2+axy+y^2)(2y+xa)^2-2f_y(x^2+axy+y^2)=0$

By the chain rule, $$U_{xx} = 2f'(x^2+axy+y^2) + (2x+ay)^2f''(x^2+axy+y^2)$$ and $$U_{yy} = 2f'(x^2+axy+y^2) + (2y+ax)^2f''(x^2+axy+y^2).$$ Since $U_{xx}-U{yy}=0$, and $f''(x) \neq 0, \forall x \in D_f$, we have $$((2x+ay)^2-(2y+ax)^2)f''(x^2+axy+y^2) = 0 \Rightarrow 4x^2+4axy+a^2y^2 - 4y^2 -4axy - a^2x^2 = 0.$$ In particular, taking $x=1$ and $y=0$, we have $$4-a^2= 0$$ therefore the answer, if it exists, must be $2$ or $-2$ but checking on the original equation, we have that both $a = 2$ and $a = -2$ are solutions.

real values of $a$ for which the range of function $ f(x) = \frac{x-1}{1-x^2-a}$ does not contain value from $\left[-1,1\right]$ All real values of $a$ for which the range of function $\displaystyle f(x) = \frac{x-1}{1-x^2-a}$ does not contain any value from $\left[-1,1\right]$ $\bf{My\; Try::}$ Let $\displaystyle y = \frac{x-1}{1-x^2-a}\Rightarrow y-x^2y-ay=x-1$ So $$x^2y+x+y(1-a)-1=0$$ Now for real values of $y<-1\cup y>1$ equation has real roots So $$1-4y^2(1-a)\geq 0\Rightarrow 1\geq 4y^2(1-a)\Rightarrow y^2\leq \frac{1}{4(1-a)}\;, a\neq 1$$ Now how can i solve it after that, Help required, Thanks

Given $$f(x)=\dfrac{x-1}{1-x^2-a}$$ for some $a$ notice that if for $x\ne0$ you multiply both the numerator and denominator by $\dfrac{1}{x}$ you obtain $$ f(x)=\dfrac{1-\frac{1}{x}}{\frac{1}{x}-x-\frac{a}{x}}\text{ for }x\ne0 $$ Now notice what happens to the value of $f(x)$ for large values of $x$. Regardless of the size of $a$ one may make both $\frac{1}{x}$ and $\frac{a}{x}$ as small as one pleases, so small that they are negligible. Thus for sufficiently large positive or negative values of $x$ the graph of $f(x)$ approaches the graph of $y=-\frac{1}{x}$ which has the $x$ axis as a horizontal asymptote. Thus, the $x$-axis is also the horizontal asymptote of $f(x)$. So regardless of the value of the variable $a$, the range of $f$ must always contain numbers in the interval $[-1,1]$. Once you get to calculus there is an even easier way to show that the graph of $f$ approaches the $x$-axis as $x$ approaches either $+\infty$ or $-\infty$.

How many #8 digits can you build? How many different 8-digit numbers can be formed using two 1s, two 2s, two 3s, and two 4s such that no two adjacent digits are the same? So we have $0 -> 9 = 10$ options for the digits but with the constraints we have something different. Cases: $1$ _ _ $2$ _ _ $3$ _ _ $4$ _ _ We can do recursion in the form $A(N), B(N),.. D(N)$ as to how many sequences can be built from the $1, 2, .. 4$ for $N$ letters. We have $T(n) = A(N-1) + B(N-1) + C(N-1) + D(N-1)$ We are after $T(8)$. We have that for instance $A(n-1) = B(n-2) + C(n-2) + D(n-2) = [A(n-3) + C(N-3) + D(n-3)] + [A(n-3) + B(n-3) + D(n-3)] + [A(n-3) + B(n-3) + C(n-3)] = 3A(n-3) + 2B(n-3) + 2C(n-3) + 2D(n-3)$. But this gets long fast.

Use Inclusion-exclusion principle. There are $\frac{8!}{2^4}$ 8-digits numbers that can be formed using two 1s, two 2s, two 3s, and two 4s. i) How many of these 8-digits numbers have two adjacent 1s? There are $7\cdot \frac{6!}{2^3}$ such numbers. ii) How many of these 8-digits numbers have two adjacent 1s and two adjacent 2s in this order? There are $15\cdot \frac{4!}{2^2}$ such numbers. iii) How many of these 8-digits numbers have two adjacent 1s, two adjacent 2s and two adjacent 3s in this order? There are $10\cdot \frac{2!}{2^1}$ such numbers. iv) How many of these 8-digits numbers have two adjacent 1s, two adjacent 2s, two adjacent 3s and two adjacent 4s in this order? They is only $1$ such number. Hence the final result is (why?): $$\frac{8!}{2^4}-4\cdot7\cdot \frac{6!}{2^3} +(4\cdot 3)\cdot 15\cdot \frac{4!}{2^2} -(4\cdot 3\cdot 2)\cdot 10\cdot \frac{2!}{2^1} +(4!)\cdot 1\cdot \frac{0!}{2^0}=864.$$

Proof that $\prod_{k=1}^{n-1}(1+\frac1k)^k = \frac{n^n}{n!}$ for all $n \in \Bbb N \ge 2$ I've tried to prove this for a while now, but I can't get it: $\prod_{k=1}^{n-1}(1+\frac1k)^k = \frac{n^n}{n!}$ for all $n \in \Bbb N \ge 2$ Solution: $\prod_{k=1}^{(n+1)-1}(1+\frac1k)^k=\frac{(n+1)^{(n+1)}}{(n+1)!}$ $\left(1+\frac1n\right)^n\cdot\prod_{k=1}^{n-1}\left(\left(1+\frac1k\right)^k\right)= \frac{(n+1)^n\cdot(n+1)}{n!\cdot(n+1)}$ $\frac{n^n}{n!}\left(1+\frac1n\right)^n= \frac{(n+1)^n}{n!}$ $n^n(1+\frac{1}{n})^n=(n+1)^n$ $(n+1)^n=(n+1)^n$

The following telescopic product in disguise: $$ \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) = n \tag{1}$$ leads to $$ \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k = \frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}}=\frac{n^n}{n\cdot(n-1)\cdot\ldots\cdot 1}=\frac{n^n}{n!}.\tag{2}$$ In the opposite direction, we may notice that $\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k = \frac{n^n}{n!}$ holds for $n=1$ and $$\prod_{k=1}^{n}\left(1+\frac{1}{k}\right)^k/\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k=\frac{(n+1)^n}{n^n}=\frac{(n+1)^{n+1}}{(n+1)!}/\frac{n^n}{n!}.\tag{3}$$

What is $(-1)^{\frac{1}{7}} + (-1)^{\frac{3}{7}} + (-1)^{\frac{5}{7}} + (-1)^{\frac{9}{7}} + (-1)^{\frac{11}{7}}+ (-1)^{\frac{13}{7}}$? The question is as given in the title. According to WolframAlpha, the answer is 1, but I am curious as to how one gets that. I tried simplifying the above into $$6(-1)^{\frac{1}{7}}$$ however that does not give 1.

We have $$x^7+1=(x+1)(x^6-x^5+x^4-x^3+x^2-x+1).$$ Let $x=(-1)^{1/7}$ and get the result.

Parametrization of the intersection between a sphere and a plane I can't find a way to get the parametric equation $\gamma(t)=(x(t),y(t),z(t))$ of a curve that is the intersection of a sphere and a plane (not parallel to any coordinate planes). That is $$\begin{cases} x^2+y^2+z^2=r^2 \\ ax+by+cz=d \end{cases}$$ I don't know how to move the variables in order to get something easily parameterizable.\ Can anyone let me know the main steps to get the parameterization of this type of curve? Example $$\begin{cases} x^2+y^2+z^2=1 \\ x+y+z=0 \end{cases}$$ Answer: $\gamma(t)=(\frac{\sqrt{2}}{2}cost +\frac{\sqrt{6}}{6}sint,-\frac{\sqrt{2}}{2}cost +\frac{\sqrt{6}}{6}sint,\frac{\sqrt{6}}{3}sint) , \,\,\,\, t \in [0,2\pi]$

Just to present a vectorial approach to the problem Consider the plane equation written as: $$ \frac{{a\,x + b\,y + c\,z}}{{\sqrt {a^{\,2} + b^{\,2} + c^{\,2} } }} = \frac{d}{{\sqrt {a^{\,2} + b^{\,2} + c^{\,2} } }} $$ That means that the points ${\bf p} = \left( {x,y,z} \right)$ on the plane shall project onto the unit vector ${\bf n} = \left( {a,b,c} \right)/\sqrt {a^{\,2} + b^{\,2} + c^{\,2} } $ at a constant distance $\delta = d/\sqrt {a^{\,2} + b^{\,2} + c^{\,2} } $, i.e. $$ {\bf p} \cdot {\bf n} = \delta $$ so that the plane is distant $\delta$ from the origin. At the same time we shall have $$ \left| {\,{\bf p}\,} \right| = r $$ So with reference to the sketch, we can put $$ {\bf p} = \delta \,{\bf n} + \sqrt {r^{\,2} - \delta ^{\,2} } \,{\bf t} = \delta \,{\bf n} + \rho \,{\bf t} $$ where ${\bf t}$ is a generic unit vector parallel to the plane, that is normal to ${\bf n}$. We can express ${\bf t}$ by taking two unit vectors ${\bf u}$ and ${\bf v}$, normal to ${\bf n}$ and to each other, and then putting $$ {\bf t} = \cos \theta \,{\bf u} + \sin \theta \,{\bf v} $$ To determine ${\bf u}$ and ${\bf v}$ we can take (if, e.g. $c \ne 0$) $$ {\bf u} = \left( {0, - c,b} \right)/\sqrt {b^{\,2} + c^{\,2} } \quad \quad {\bf v} = {\bf n} \times {\bf u} $$ example with $r=4 \; a=1 \; b=2 \; c=3 \; d=7$ we get $$ {\bf n} = \,\sqrt {14} /14\,\left( {\begin{array}{*{20}c} 1 \\ 2 \\ 3 \\ \end{array}} \right)\;\quad \delta = \sqrt {14} /2\quad \rho = 5\sqrt 2 /2 $$ $$ {\bf u} = \sqrt {13} /13\;\left( {\begin{array}{*{20}c} 1 \\ 2 \\ 3 \\ \end{array}} \right)\quad {\bf v} = \sqrt {182} /182\;\left( {\begin{array}{*{20}c} {13} \\ { - 2} \\ { - 3} \\ \end{array}} \right) $$ and thus $$ \begin{array}{l} {\bf p} = \delta \,{\bf n} + \rho \,\left( {\cos \theta \,{\bf u} + \sin \theta \,{\bf v}} \right)\quad \Rightarrow \\ \Rightarrow \quad \left( {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right) = \frac{1}{2}\,\left( {\begin{array}{*{20}c} 1 \\ 2 \\ 3 \\ \end{array}} \right) + \frac{{5\sqrt {364} }}{{364}}\,\left( {\sqrt {14} \cos \theta \;\left( {\begin{array}{*{20}c} 1 \\ 2 \\ 3 \\ \end{array}} \right) + \sin \theta \,\;\left( {\begin{array}{*{20}c} {13} \\ { - 2} \\ { - 3} \\ \end{array}} \right)} \right) \\ \end{array} $$ and you can verify that: $$ {\bf p} \cdot {\bf n} = \delta \,\quad {\bf p} \cdot {\bf p} = r^{\,2} $$