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Prove: $\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right) = \arcsin\left(\frac {16}{65}\right)$ This is not a homework question, its from sl loney I'm just practicing. To prove : $$\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right) = \arcsin\left(\frac {16}{65}\right)$$ So I changed all the angles to $\arctan$ which gives: $$\arctan\left(\frac 34\right) - \arctan\left(\frac {12}{5}\right) = \arctan\left(\frac {16}{63}\right)$$ But the problem is after applying formula of $\arctan(X)-\arctan(Y)$ the lhs is negative and not equal to rhs? Is this because I have to add pi somewhere please help.

How exactly did you convert to arctan? Careful: $$\arccos\left(\frac {12}{13}\right) = \arctan\left(\frac {5}{12}\right) \ne \arctan\left(\frac {12}{5}\right)$$ Draw a right triangle with hypotenuse of length 13, adjacent side (from an angle $\alpha$) with length 12 and opposite side with length 5; then $\cos\alpha = 12/13$ and $\tan\alpha = 5/12$. Perhaps easier: take the sine of both sides in the original equation: $$\sin\left(\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right)\right) = \sin\left( \arcsin\left(\frac {16}{65}\right)\right)$$ The RHS is $16/65$ and simplify the LHS with $\sin(a-b)=\sin a \cos b - \cos a \sin b$ to get: $$\frac{3}{5}\frac{12}{13}-\sqrt{1-\frac{9}{25}}\sqrt{1-\frac{144}{169} } = \frac{3}{5}\frac{12}{13}-\frac{4}{5}\frac{5}{13}= \cdots$$

A human way to simplify $ \frac{((\sqrt{a^2 - 1} - a)^2 - 1)^2}{(\sqrt{a^2 - 1} - a)^22 \sqrt{a^2 - 1}} - 2 a $ I end up with simplifying the following fraction when I tried to calculate an integral(*) with the residue theory in complex analysis: $$ \frac{((\sqrt{a^2 - 1} - a)^2 - 1)^2}{(\sqrt{a^2 - 1} - a)^22 \sqrt{a^2 - 1}} - 2 a $$ where $a>1$. With Mathematica, I can quickly get $$ \frac{((\sqrt{a^2 - 1} - a)^2 - 1)^2}{(\sqrt{a^2 - 1} - a)^22 \sqrt{a^2 - 1}} - 2 a=2(\sqrt{a^2 - 1} - a). $$ Would anybody give a calculation for the simplification in a human way? (*)The integral I did is $$ \int_{-\pi}^\pi\frac{\sin^2 t}{a+\cos t}\ dt $$ with $a>1$.

Start with $$\begin{align}\left(\sqrt{a^2-1}-a\right)^2-1&=a^2-1-2a\sqrt{a^2-1}+a^2-1\\ &=\sqrt{a^2-1}\left(2\sqrt{a^2-1}-2a\right)\\ &=2\sqrt{a^2-1}\left(\sqrt{a^2-1}-a\right)\end{align}$$ So you are now down to $$\frac{\left(2\sqrt{a^2-1}\left(\sqrt{a^2-1}-a\right)\right)^2}{\left(\sqrt{a^2-1}-a\right)^2\cdot2\sqrt{a^2-1}}-2a=2\sqrt{a^2-1}-2a$$

Prove that $a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$ for $a, b, c > 0$ Prove for $a, b, c > 0$ that $$a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$$ Could you give me some hints on this? I thought that Jensen's inequality might be of use in this exercise, but I haven't managed to solve this on my own.

Use the Cauchy-Schwarz inequality on the two vectors $(\sqrt a, \sqrt b, \sqrt c)$ and $(\sqrt{a(b+c)}, \sqrt{b(a+c)}, \sqrt{c(a+b)})$ (then take the square root on both sides or not, depending on which version of the CS inequality you use), and lastly note that we have: $$ a(b+c) + b(a + c) + c(a + b) = 2(bc + ac + ab) $$

Function that is second differential continuous Let $f:[0,1]\rightarrow\mathbb{R}$ be a function whose second derivative $f''(x)$ is continuous on $[0,1]$. Suppose that f(0)=f(1)=0 and that $|f''(x)|<1$ for any $x\in [0,1]$. Then $$|f'(\frac{1}{2})|\leq\frac{1}{4}.$$ I tried to use mean value theorem to prove it, but I found no place to use condition of second derivative and I cannot prove this.

A bit more tricky than I thought at first. The idea is easy, the calculation may look complicated. The idea is to find a second order polynomial with second derivate $=1$ which has the same values as $f$ for $x=0 $ and $x= \frac{1}{2}$, and then to show that this function $-f$ is convex, which allows to get an estimate for the derivative at the boundary. First note that the assumptions and the claim are invariant with respect to multiplication by $-1$. Now, since we may look at $-f$ instead of $f$, we may assume $f(\frac{1}{2})\le 0$ We will bound $f^\prime(\frac{1}{2})$ by looking at the comparison function $$ h(x) = \frac{1}{2}x^2 + \left(2f(\frac{1}{2}) -\frac{1}{4}\right)x $$ You'll easily verify $h(0) = 0, \, h(\frac{1}{2}) =f(\frac{1}{2})$ and $h^{\prime \prime}(x) = 1$. So for $$\psi(x) = h(x)-f(x)$$ we have $\psi(0)= 0 = \psi(\frac{1}{2})$ and $ \psi^{\prime \prime}(x)=1- f^{\prime \prime}(x)>0$. This means $\psi$ is strictly convex with zero boundary values at $x= \frac{1}{2}$, which implies $\psi^\prime(\frac{1}{2})>0$ Since $h^\prime(x) = x+2f(\frac{1}{2})-\frac{1}{4}$ we have $h^\prime(\frac{1}{2}) = 2f(\frac{1}{2})+\frac{1}{4}$ and so $$ \begin{eqnarray} 0< \psi^\prime(\frac{1}{2}) = 2f(\frac{1}{2})+\frac{1}{4} -f^\prime (\frac{1}{2}) \end{eqnarray} $$ which implies (using the assumption on the sign of $f$) $$ f^\prime (\frac{1}{2})< 2f(\frac{1}{2})+\frac{1}{4} \le \frac{1}{4}$$ To get a bound $f^\prime (\frac{1}{2})\ge- \frac{1}{4}$ with the same assumption $f(\frac{1}{2})\le 0$ the same trick is applied on the interval $[\frac{1}{2},1]$ using the comparison function $$h(x)=\frac{1}{2}(x-1)^2 - \left( 2f(\frac{1}{2})-\frac{1}{4}\right)(x-1) $$ If you now look at $\phi=h-f$ this is again convex with zero boundary values for $x= \frac{1}{2}$ and $x=1$, so $\phi^\prime(\frac{1}{2})< 0$ this time. The calculations are the same.

Calculate $\int_D \rvert x-y^2 \rvert dx \ dy $ $$\int_D \rvert x-y^2 \rvert dx \ dy $$ $D$ is the shape that is delimited from the lines: $$ y=x \\ y=0 \\ x=1 \\$$ $$D=\{ (x,y) \in \mathbb{R}^2: 0 \le x \le 1 \ , \ 0 \le y \le x \}$$ $$\rvert x-y^2 \rvert=x-y^2 \qquad \forall (x,y) \in D $$ $$\int_0^1 \Big( \int_0^x (x-y^2) \ dy \Big) \ dx= \int_0^1 \Big[_0^x xy-\frac{1}{3} y^3 \Big] \ dx= \int_0^1 \left(x^2-\frac{1}{3} x^3\right) \ dx=\Big[_0^1 \frac{1}{3} x^3-\frac{1}{12} x^4 \Big]=\frac{1}{4}$$ Is it correct?

This is correct, you are correct that the absolute value integrand is equal to x-y^2 for all x and y in your region of integration.

Maximum of $xy+y^2$ subject to right-semicircle $x\ge 0,x^2+y^2\le 1$ Maximum of: $$ xy+y^2 $$ Domain: $$ x \ge 0, x^2+y^2 \le1 $$ I know that the result is: $$ \frac{1}{2}+\frac{1}{\sqrt{2}} $$ for $$ (x,y)=\left(\frac{1}{\sqrt{2(2+\sqrt{2})}},\frac{\sqrt{2+\sqrt{2}}}{2}\right) $$ But I don't know how to get this result. I know that: $$ xy+y^2 \le \frac{1}{2}+y^2 $$ so: $$ xy \le \frac{1}{2} $$ And also: $$ xy+y^2 \le xy+1-x^2 \equiv 1+x(y-x) $$ But I don't know what to do next...

Hint$$x^2+y^2=x^2+(3-2\sqrt{2})y^2+(2\sqrt{2}-2)y^2 $$ Now notice $$x^2+(3-2\sqrt{2})y^2+(2\sqrt{2}-2)y^2 \ge 2(3-2\sqrt{2})^{\frac{1}{2}}xy+(2\sqrt{2}-2)y^2 (\because \text{AM-GM})$$Now note $(\sqrt{2}-1)^2=3-2\sqrt{2}$.

Given $\tan a = -7/24$ in $2$nd quadrant and $\cot b = 3/4$ in $3$rd quadrant find $\sin (a + b)$. Say $\tan a = -7/24$ (second quadrant) and $\cot b = 3/4$ (third quadrant), how would I find $\sin (a + b)$? I figured I could solve for the $\sin/\cos$ of $a$ & $b$, and use the add/sub identities, but I got massive unwieldy numbers outside the range of $\sin$. How ought I to go about this problem? Thanks.

$\tan(a) = -7/24$ Opposite side $= 7$ and adjacent side $= 24$ Pythagorean theorem $\Rightarrow$ hypotenuse $= \sqrt{49+576} = 25$ $\sin(a) = 7/25$ (sin is positive in second quadrant) $\cos(a) = - 24/25$ (cos is negative in second quadrant) $\cot(b) = 3/4 \Rightarrow \tan(b) = 4/3$ Opposite side $= 4$ and adjacent side $= 3$ Pythagorean theorem $\Rightarrow$ hypotenuse $= \sqrt{9+16} = 5$ $\sin(b) = - 4/5$ (sin is negative in third quadrant) $\cos(b) = -3/5$ (cos is negative in third quadrant) $\sin(a+b) = \sin a \cos b + \cos a \sin b$ $\sin(a+b) = 3/5$

Prove that the sum of the squares of two odd integers cannot be the square of an integer. Prove that the sum of the squares of two odd integers cannot be the square of an integer. My method: Assume to the contrary that the sum of the squares of two odd integers can be the square of an integer. Suppose that $x, y, z \in \mathbb{Z}$ such that $x^2 + y^2 = z^2$, and $x$ and $y$ are odd. Let $x = 2m + 1$ and $y = 2n + 1$. Hence, $x^2 + y^2$ = $(2m + 1)^2 + (2n + 1)^2$ $$= 4m^2 + 4m + 1 + 4n^2 + 4n + 1$$ $$= 4(m^2 + n^2) + 4(m + n) + 2$$ $$= 2[2(m^2 + n^2) + 2(m + n) + 1]$$ Since $2(m^2 + n^2) + 2(m + n) + 1$ is odd it shows that the sum of the squares of two odd integers cannot be the square of an integer. This is what I have so far but I think it needs some work.

Let $a=2n+1$, $b=2m+1$. Then $a^2 + b^2=4n^2 + 4n +4m^2 +4m+2$. This is divisible by $2$, a prime number, but not by $4=2^2$. Hence it cannot be the square of an integer.

Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix $$A = \begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{pmatrix}.$$ I am trying to find $e^{At}$. The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, $A$ is not diagonalizable. How does one find the exponential of a non-diagonalizable matrix? My attempt: Write $\begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{pmatrix} = M + N$, with $M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ and $N = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 0 & -4 \\ 0 & 0 & 0 \end{pmatrix}$. We have $N^3 = 0$, and therefore $\forall x > 3$, $N^x = 0$. Thus: $$\begin{aligned} e^{At} &= e^{(M+N)t} = e^{Mt} e^{Nt} \\ &= \begin{pmatrix} e^t & 0 & 0 \\ 0 & e^t & 0 \\ 0 & 0 & e^t \end{pmatrix} \left(I + \begin{pmatrix} 0 & t & 2t \\ 0 & 0 & -4t \\ 0 & 0 & 0 \end{pmatrix}+\begin{pmatrix} 0 & 0 & -2t^2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\right) \\ &= e^t \begin{pmatrix} 1 & t & 2t \\ 0 & 1 & -4t \\ 0 & 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} e^t & te^t & 2t(1-t)e^t \\ 0 & e^t & -4te^t \\ 0 & 0 & e^t \end{pmatrix}. \end{aligned}$$ Is that the right answer?

this is my first answer on this site so if anyone can help to improve the quality of this answer, thanks in advance. That said, let us get to business. * *Compute the Jordan form of this matrix, you can do it by hand or check this link. (or both). *Now, we have the following case: $$ A = S J S^{-1}.$$ You will find $S$ and $S^{-1}$ on the previous link. For the sake of simplicity, $J$ is what actually matters, $$ J = \begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{pmatrix} $$ because: *$e^A = e^{SJS^{-1}} = e^J$ And the matrix $J$ can be written as: $J = \lambda I + N$, where $I$ is the identity matrix and $N$ a nilpotent matrix. *So, $e^J = e^{\lambda I + N} = \mathbf{e^{\lambda} \cdot e^N}$ By simple inspection, we get that: $$ J = \lambda I + N = 1 \cdot \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix} + \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{pmatrix} $$ where you can check that $\lambda =1$ and N is $$ \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{pmatrix} $$ *So, $e^A = e \cdot e^N$ we just apply the definition $ e^N \equiv \sum^{\infty}_{k=0} \frac{1}{k!} N^k$. And, of course, it converges fast: $N^2 \neq 0$ but $N^3=0$. *Finally: $$ e^A = e \cdot \left[ 1 \cdot I + 1 \cdot N^1 + \frac{1}{2} N^2 \right] $$ where $$ N^2 = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{pmatrix} $$ then: $$ \mathbf{ e^A = e \cdot \begin{pmatrix} 1 & 1 & 1/2\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{pmatrix} } $$ Last but not least, $$ e^{At} = e^{A \cdot t} = e^{\lambda \cdot t} \cdot e^{N \cdot t} = e^t \cdot \begin{pmatrix} 1 & t & 1/2 t^2\\ 0 & 1 & t\\ 0 & 0 & 1\\ \end{pmatrix} $$ You replace $N$ by $At$ in the exp definition and that's it.

Calculation of total number of real ordered pairs $(x,y)$ Calculation of total number of real ordered pairs $(x,y)$ in $x^2-4x+2=\sin^2 y$ and $x^2+y^2\leq 3$ $\bf{My\; Try::}$ Given $x^2-4x+2=x^2-4x+4-2=\sin^2 y\Rightarrow (x-2)^2-2=\sin^2 y$ Now Using $0 \leq \sin^2 y\leq 1$. So we get $0\leq (x-2)^2-2\leq 1\Rightarrow 2\leq (x-2)^2\leq 3$ So we get $\sqrt{2}\leq |x-2|\leq \sqrt{3}\Leftrightarrow \bigg(-\sqrt{3}\leq (x-2)\leq \sqrt{3}\bigg)\;\;\cap\;\;\bigg((x-2)\leq -\sqrt{2} \;\cup\; (x-2)\;\geq \sqrt{2}\bigg)$ So we get $$x\in \left[2-\sqrt{3}\;\;,2-\sqrt{2}\right]\;\cup \;\left[2+\sqrt{2}\;,2+\sqrt{3}\right]$$ Now How can I solve it after that, Help me, Thanks

Now How can I solve it after that I don't know how to continue from that. So, let us take another approach. Solving $x^2-4x+2-\sin^2 y=0$ for $x$ gives $$x=2\pm\sqrt{2+\sin^2y}$$ Now $x=2+\sqrt{2+\sin^2y}$ does not satisfy $x^2\le 3$. So, we have $x=2-\sqrt{2+\sin^2y}$. Now $$\left(2-\sqrt{2+\sin^2y}\right)^2+y^2\le 3$$ is equivalent to $$3+\sin^2y+y^2\le 4\sqrt{2+\sin^2y}\tag1$$ By the way, for $-\pi/4\le y\le \pi/4$, we have $$y^2\le \frac{\pi^2}{16},\quad 0\le \sin^2y\le \frac 12$$ and so using $\pi\le 4$ gives $$3+\sin^2y+y^2\le 3+\frac 12+\frac{\pi^2}{16}\le 3+\frac 12+\frac{4^2}{16}=4\times 1.125\le 4\sqrt 2\le 4\sqrt{2+\sin^2y}$$ Hence, if $-\pi/4\le y\le \pi/4$, then $(1)$ holds. It follows from this that there are infinitely many such pairs.

Curious inequality: $(1+a)(1+a+b)\geq\sqrt{27ab}$ I was recently trying to play with mean inequalities and Jensen inequality. My question is, if the following relation holds for any positive real numbers $a$ and $b$ $$(1+a)(1+a+b)\geq\sqrt{27ab}$$ and if it does, then how to prove it. By AM-GM inequality we could obtain $1+a\geq\sqrt{4a}$ and $1+a+b\geq\sqrt[3]{27ab}$, so by putting this together we obtain$$(1+a)(1+a+b)\geq\sqrt{4a}\sqrt[3]{27ab},$$but this is slightly different from what I wanted. It's probably true that for all positive $a$ and $b$ there is $\sqrt{4a}\sqrt[3]{27ab}\geq\sqrt{27ab}$, but could there be equality? Thanks a lot.

Use AM-GM: $$\frac{1}{2}+\frac{1}{2}+a\ge3\sqrt[3]{\frac{a}{4}}\\ \frac{1}{2}+\frac{1}{2}+a+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}\ge6\sqrt[6]{\frac{ab^3}{108}}\\\therefore(1+a)(1+a+b)\ge\left(3\sqrt[3]{\frac{a}{4}}\right)\left(6\sqrt[6]{\frac{ab^3}{108}}\right)=\sqrt{27ab}$$ Equality holds iff $\frac{1}{2}=a=\frac{b}{3}$.

Sum of squares of integers divisible by 3 Suppose that $n$ is a sum of squares of three integers divisible by $3$. Prove that it is also a sum of squares of three integers not divisible by $3$. From the condition, $n=(3a)^2+(3b)^2+(3c)^2=9(a^2+b^2+c^2)$. As long as the three numbers inside are divisible by $3$, we can keep pulling out a factor of $9$, until we get $n=9^k(x^2+y^2+z^2)$. Modulo $3$, squares leave a remainder of either $0$ or $1$. Therefore one or three of $x,y,z$ are divisible by $3$.

Induction on the power of $9$ dividing the number. Begin with any number not divisible by $9,$ although it is allowed to be divisible by $3.$ The hypothesis at this stage is just that this number is the sum of three squares, say $n = a^2 + b^2 + c^2.$ Since this $n$ is not divisible by $9,$ it follows that at least one of $a,b,c$ is not divisible by $3.$ Let us order so that we can demand $c$ not divisible by $3.$ If $a+b+c$ is divisible by $3,$ replace it by $-c$ and use the same name. We now have $$ a + b + c \neq 0 \pmod 3. $$ Induction: let $n = a^2 + b^2 + c^2,$ with $a+b+c \neq 0 \pmod 3.$ Then $$ 9n = (a-2b-2c)^2 + (-2a+b-2c)^2 + (-2a-2b+c)^2, $$ where all three summands are nonzero $\pmod 3.$ We can multiply by another $9,$ then another, and so on. The formula is $- \bar{q} p q $ in the quaternions with rational integer coefficients, where $q=i+j+k$ and $p= ai+bj+ck.$

Calculate $ \lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx}$ My attempt: \begin{align*} \lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx} &= (\frac{\ln1}{0}) \text{ (we apply L'Hopital's rule)} \\ &= \lim_{x \to0}\frac{\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{x^n+x^{n-1}+\dots+1}}{n} \\ &= \lim_{x \to0}\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{n(1+x+\dots+x^n)} \\ &= \frac{1}{n}. \end{align*} Are my steps correct? Thanks.

Besides using L'Hospital's Rule, By the definition of derivative, $\displaystyle\lim \limits_{x\to0}\frac{\ln(1+x+x^2+...+x^n)}{nx}=\lim \limits_{x\to0} \frac{\ln(1+x+x^2+...+x^n)-\ln1}{n(x-0)}=\left.\frac{1}{n}\frac{d}{dx}(ln(1+x+x^2+...+x^n))\right|_{x=0}$ $=\displaystyle\left.\frac{1}{n}\frac{1+2x+...+nx^{n-1}}{1+x+x^2+...+x^n}\right|_{x=0}=\frac{1}{n}$

Find all integral solutions of the equation $x^n+y^n+z^n=2016$ Find all integral solutions of equation $$x^n+y^n+z^n=2016,$$ where $x,y,z,n -$ integers and $n\ge 2$ My work so far: 1) $n=2$ $$x^2+y^2+z^2=2016$$ I used wolframalpha n=2 and I received the answer to the problem (Number of integer solutions: 144) 2) $n=3$ I used wolframalpha n=3 and I not received the answer to the problem How to do without wolframalpha?

An approach that can sometimes help to find solutions to this type of equation is to consider the prime factors of the given integer. In this case: $$2016 = 2^5.3^2.7 = 2^5(2^6-1) = 2^5((2.2^5)-1)$$ Hence: $$\mathbf{2016 = 4^5 + 4^5 + (-2)^5}$$ And also (finding a solution of $x^3+y^3+z^3=252$ by trial and error or from this list): $$2016 = 2^3.252 = 2^3(7^3 - 3^3 - 4^3)$$ Hence: $$\mathbf{2016 = 14^3 + (-6)^3 + (-8)^3}$$

Integrate $\frac{\sin^3 \frac{x}{2}}{\cos \frac{x}{2} \sqrt{\cos x+\cos^2 x+\cos^3 x}}$ Evaluate $$\int \frac{\sin^3 \frac{x}{2}}{\cos \frac{x}{2} \sqrt{\cos x+\cos^2 x+\cos^3 x}}dx$$ I saw terms like $1+t+t^2$ in the denominator , so I thought of $t^3-1$ and then converting back into half angle but it doesn't help me. I am unable to simplify it further. Please tell me a better way to start. Thanks.

Let $$\displaystyle I = \int \frac{\sin ^3(x/2)}{\cos(x/2)\sqrt{\cos^3 x+\cos^2x+\cos x}}dx = \frac{1}{2}\int\frac{2\sin^2 \frac{x}{2}\cdot 2\sin \frac{x}{2}\cdot \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}\sqrt{\cos^3 x+\cos^2 x+\cos x}}dx$$ So we get $$\displaystyle I = \frac{1}{2}\int\frac{(1-\cos x)\cdot \sin x}{(1+\cos x)\sqrt{\cos^3 x+\cos^2 x+\cos x}}dx$$ Now Put $\cos x = t\;,$ Then $\sin x dx = -dt$ So Integral $$\displaystyle I = -\frac{1}{2}\int\frac{(1-t)}{(1+t)\sqrt{t^3+t^2+t}}dt = -\frac{1}{2}\int\frac{(1-t^2)}{(1+t)^2\sqrt{t^3+t^2+t}}dt$$ So we get $$\displaystyle I = \frac{1}{2}\int\frac{\left(1-\frac{1}{t^2}\right)}{\left(t+\frac{1}{t}+2\right)\sqrt{t+\frac{1}{t}+1}}dt$$ Now Let $\displaystyle \left(t+\frac{1}{t}+1\right) = u^2\;,$ Then $\left(1-\frac{1}{t^2}\right)dt = 2udu$ So Integral $$\displaystyle I = \frac{1}{2}\int\frac{2u}{u^2+1}\cdot \frac{1}{u}du = \tan^{-1}(u)+\mathcal{C}$$ So we get $$\displaystyle I = \tan^{-1}\sqrt{\left(t+\frac{1}{t}+1\right)}+\mathcal{C}$$ So we get $$\displaystyle \displaystyle I = \tan^{-1}\sqrt{\left(\cos x+\sec x+1\right)}+\mathcal{C}$$

Evaluate $\int\frac{\sqrt{x^2+2x-3}}{x+1}d\,x$ by trig substitution I am preparing for an exam and found this integral in a previous test. Did I do it correctly? My attempt. $$ \int\frac{\sqrt{x^2+2x-3}}{x+1}\,dx $$ Complete the square of $x^2+2x-3$; I changed the integral to $$ \int\frac{\sqrt{(x-1)^2-4}}{x+1}\,dx $$ then set $u=x+1$ to get $$ \int\frac{\sqrt{u^2-4}}{u}\,dx $$ Using the triangle, $2\sec\theta=u$ and $du=2\sec\theta\tan\theta d\theta$ $$ \int\frac{\sqrt{(2\sec\theta)^2-2^2}}{2\sec\theta}2\sec\theta \tan\theta\, d\theta $$ This I simplified to $$ 2\int\tan^2\theta\, d\theta = 2\int\sec^2\theta-1\,d\theta =2[\tan\theta-\theta]+C $$ Back substitute $$ \theta=\tan^{-1}\frac{\sqrt{u^2-4}}{2} $$ and $$ \tan\theta=\frac{\sqrt{u^2-4}}{2} $$ Back substitute $u=x+1$ $$ \int\frac{\sqrt{x^2+2x-3}}{x+1}\,dx= \sqrt{(x-1)^2-4}-\tan^{-1}{\sqrt{(x-1)^2-4}}+C $$

You've made a mistake. I hope you can find it using my answer $$\int\frac{\sqrt{x^2+2x-3}}{x+1}\space\text{d}x=\int\frac{\sqrt{(x+1)^2-4}}{x+1}\space\text{d}x=$$ Substitute $u=x+1$ and $\text{d}u=\text{d}x$: $$\int\frac{\sqrt{u^2-4}}{u}\space\text{d}u=$$ Substitute $u=2\sec(s)$ and $\text{d}u=2\tan(s)\sec(s)\space\text{d}s$. We get that $\sqrt{u^2-4}=\sqrt{4\sec^2(s)-4}=2\tan(s)$ and $s=\text{arcsec}\left(\frac{u}{2}\right)$: $$2\int\tan^2(s)\space\text{d}s=2\int\left[\sec^2(s)-1\right]\space\text{d}s=2\left[\int\sec^2(s)\space\text{d}s-\int1\space\text{d}s\right]$$ Notice now, the integral of $\sec^2(s)$ is equal to $\tan(s)$ and the integral of $1$ is just $s$.

Evaluation of $\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}$ Evaluation of $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} = $$ $\bf{My\; Try::}$ I have solved Using Direct formula:: $$\sin \frac{\pi}{n}\cdot \sin \frac{2\pi}{n}\cdot......\sin \frac{(n-1)\pi}{n} = \frac{n}{2^{n-1}}$$ Now Put $n=7\;,$ We get $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}\cdot \sin \frac{4\pi}{7}\cdot \sin \frac{5\pi}{7}\cdot \sin \frac{6\pi}{7}=\frac{7}{2^{7-1}}$$ So $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} =\frac{\sqrt{7}}{8}$$ Now my question is how can we solve it without using Direct Formula, Help me Thanks

Using $2\sin a\sin b=\cos(a-b)-\cos(a+b)$ and $2\sin a\cos b=\sin(a+b)+\sin(a-b)$, write $$\sin \frac{\pi}7\cdot\sin \frac{2\pi}7\cdot \sin \frac{3\pi}7 = \frac12\left(\cos\frac{\pi}7-\cos\frac{3\pi}7\right)\sin\frac{3\pi}7=\frac14\left(\sin\frac{4\pi}7+\sin\frac{2\pi}7-\sin\frac{\pi}7\right)\\=\frac14\left(\sin\frac{2\pi}7+\sin\frac{4\pi}7+\sin\frac{8\pi}7\right)$$ Then have a look at this question: Trigo Problem : Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$

Prove ths sum of $\small\sqrt{x^2-2x+16}+\sqrt{y^2-14y+64}+\sqrt{x^2-16x+y^2-14y+\frac{7}{4}xy+64}\ge 11$ Let $x,y\in R$.show that $$\color{crimson}{f(x,y)=\sqrt{x^2-2x+16}+\sqrt{y^2-14y+64} + \sqrt{x^2-16x+y^2-14y+\frac{7}{4}xy+64} \ge 11}$$ Everything I tried has failed so far.use Computer found this inequality $\color{blue}=$ iff only if $\color{blue}{x=2,y=6}$ Here is one thing I tried, but obviously didn't work. $$f(x,y)=\sqrt{(x-1)^2+15}+\sqrt{(y-7)^2+15}+\sqrt{(x-8)^2+(y-7)^2+\dfrac{7}{4}xy-49}$$ Thanks in advance

For convenience, we make the translation $x=2+a$ and $y=6+b$, so that the equality case is $a=b=0$. Then the expression to bound is: $$\sqrt{(a+1)^2+15}+\sqrt{(b-1)^2+15}+\sqrt{\frac{7}{8}(a+b)^2+\frac{1}{8}(a-6)^2+\frac{1}{8}(b+6)^2} $$ Now recall the following form of Cauchy-Schwarz for $n$ nonnegative variables $x_1, \cdots, x_n$: $$\sqrt{n(x_1+\cdots+x_n)}=\sqrt{(1+\cdots+1)(x_1+\cdots+x_n)}\ge \sqrt{x_1}+\cdots+\sqrt{x_n}$$ with equality iff $x_1=\cdots=x_n$. We use this three times, keeping in mind the equality case $a=b=0$: $$\sqrt{(a+1)^2+15}=\frac{1}{4}\sqrt{16((a+1)^2+15)}\ge \frac{1}{4}(|a+1|+15)$$ $$\sqrt{(b-1)^2+15}=\frac{1}{4}\sqrt{16((b-1)^2+15)}\ge \frac{1}{4}(|b-1|+15)$$ $$\sqrt{\frac{7}{8}(a+b)^2+\frac{1}{8}(a-6)^2+\frac{1}{8}(b+6)^2}\ge \frac{1}{4}\sqrt{2((a-6)^2+(b+6)^2)}\ge \frac{1}{4}(|a-6|+|b+6|)$$ Now since $|a-6|+|a+1|\ge 7$ and $|b+6|+|b-1|\ge 7$ by the triangle inequality, the expression must be at least $\frac{15}{2}+\frac{7}{2}=11$, as required.

If $a$ and $b$ are roots of $x^4+x^3-1=0$, $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$. I have to prove that: If $a$ and $b$ are two roots of $x^4+x^3-1=0$, then $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$. I tried this : $a$ and $b$ are root of $x^4+x^3-1=0$ means : $\begin{cases} a^4+a^3-1=0\\ b^4+b^3-1=0 \end{cases}$ which gives us : $(ab)^4+(ab)^3=a^3+b^3+a^4+b^4+a^4b^3-a^3b^4-1$ can you help me carry on ? or propose another solution ? thanks in advance

Let $a,b,c,d$ be the roots of $x^4+x^3-1=0$. By Vieta's formula, $$a+b+c+d=-1\quad\Rightarrow\quad c+d=-1-(a+b)\tag1$$ $$abcd=-1\quad\Rightarrow \quad cd=-\frac{1}{ab}\tag2$$ Since we have $$a^4+a^3=1\quad\text{and}\quad b^4+b^3=1$$ we can have $$1=(a^4+a^3)(b^4+b^3)$$ $$(ab)^4+(ab)^3(a+b+1)=1,$$ i.e. $$a+b=\frac{1-(ab)^4}{(ab)^3}-1\tag3$$ Similarly, $$(cd)^4+(cd)^3(c+d+1)=1$$ Using $(1)(2)$, this can be written as $$\left(-\frac{1}{ab}\right)^4+\left(-\frac{1}{ab}\right)^3(-1-(a+b)+1)=1,$$ i.e. $$a+b=\left(1-\frac{1}{(ab)^4}\right)(ab)^3\tag4$$ From $(3)(4)$, letting $ab=x$, we have $$\frac{1-x^4}{x^3}-1=\left(1-\frac{1}{x^4}\right)x^3$$ to get $$x^6+x^4+x^3-x^2-1=0.$$

Find the sum of the infinite series $\sum n(n+1)/n!$ How do find the sum of the series till infinity? $$ \frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+\cdots$$ I know that it gets reduced to $$\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}$$ But I don't know how to proceed further.

You could also consider that $$A(x)=\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}x^n=\sum\limits_{n=0}^∞ \frac{n(n+1)}{n!}x^n=\sum\limits_{n=0}^∞ \frac{n(n-1)+2n}{n!}x^n$$ $$A(x)=\sum\limits_{n=0}^∞ \frac{n(n-1)}{n!}x^n+2\sum\limits_{n=0}^∞ \frac{n}{n!}x^n=x^2\sum\limits_{n=0}^∞ \frac{n(n-1)}{n!}x^{n-2}+2x\sum\limits_{n=0}^∞ \frac{n}{n!}x^{n-1}$$ $$A(x)=x^2 \left(\sum\limits_{n=0}^∞ \frac{x^n}{n!} \right)''+2x\left(\sum\limits_{n=0}^∞ \frac{x^n}{n!} \right)'=x^2e^x+2x e^x=x(x+2)e^x$$ Now, compute $A(1)$.

Another of $\frac{1^2}{1^2}+\frac{1-2^2+3^2-4^2}{1+2^2+3^2+4^2}+\cdots=\frac{\pi}{4}$ type expressable in cube? Gregory and Leibniz formula (1) $$-\sum_{m=1}^{\infty}\frac{(-1)^m}{2m-1}=\frac{\pi}{4}$$ We found another series equivalent to (1) This is expressed in term of square numbers $$-\sum_{m}^{\infty}\frac{\sum_{n=1}^{3m-2}(-1)^nn^2 }{\sum_{n=1}^{3m-2}(+1)^nn^2}=\frac{\pi}{4}$$ $$\frac{1^2}{1^2}+\frac{1-2^2+3^2-4^2}{1+2^2+3^2+4^2}+\frac{1-2^2+3^2-4^2+5^2-6^2+7^2}{1^2+2^2+3^2+4^2+5^2+6^2+7^2}+\cdots=\frac{\pi}{4}$$ Is there another series equivalent to (1) but expressable in term of cube numbers?

A proof of mahdi's result: $$ \sum_{k=1}^{n}k^3 = \frac{n^2(n+1)^2}{4}\tag{1}$$ $$ \sum_{k=1}^{2n}(-1)^{k+1} k^3 = -n^2(4n+3),\qquad \sum_{k=1}^{2n-1}(-1)^{k+1} k^3 = n^2(4n-3)\tag{2}$$ lead to: $$\begin{eqnarray*}\sum_{n\geq 1}\frac{1^3-2^3+\ldots}{1^3+2^3+\ldots+n^3} &=& \sum_{m\geq 1}\frac{-(4m+3)}{(2m+1)^2}+\sum_{m\geq 1}\frac{(4m-3)}{(2m-1)^2}\\&=&3+\sum_{m\geq 1}\frac{(4m-3)-(4m-1)}{(2n-1)^2}\\&=&3-2\sum_{m\geq 1}\frac{1}{(2n-1)^2}=\color{red}{3-\frac{\pi^2}{4}}.\tag{3}\end{eqnarray*}$$

Recurrent sequence limit Let $a_n$ be a sequence defined: $a_1=3; a_{n+1}=a_n^2-2$ We must find the limit: $$\lim_{n\to\infty}\frac{a_n}{a_1a_2...a_{n-1}}$$ My attempt The sequence is increasing and does not have an upper bound. Let $b_n=\frac{a_n}{a_1a_2...a_{n-1}},n\geq2$. This sequence is decreasing(we have $b_{n+1}-b_{n}<0$. How can I find this limit?

since take $a_{1}=x+\dfrac{1}{x},x>1$,then $$a_{2}=x^2+\dfrac{1}{x^2},a_{3}=x^4+\dfrac{1}{x^4}\cdots,a_{n}=x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}$$ so we have $$a_{1}a_{2}\cdots a_{n}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}\right)\cdots\left(x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}\right)=\dfrac{x^{2^n}-\dfrac{1}{x^{2^n}}}{x-\dfrac{1}{x}}$$ so $$\dfrac{a_{n+1}}{a_{1}a_{2}\cdots a_{n}}=\left(x-\frac{1}{x}\right)\dfrac{x^{2^n}+\dfrac{1}{x^{2^n}}}{x^{2^n}-\dfrac{1}{x^{2^n}}}\to \dfrac{x^2-1}{x},n\to+\infty$$ and $$\left(x-\dfrac{1}{x}\right)^2+4=\left(x+\dfrac{1}{x}\right)^2=9\Longrightarrow x-\dfrac{1}{x}=\sqrt{5}$$ another approach $$a^2_{n+1}-4=a^4_{n}-4a^2_{n}=a^2_{n}(a^2_{n}-4)=a^2_{n}a^2_{n-1}(a^2_{n-1}-4)=\cdots=a^2_{n}a^2_{n-1}\cdots a^2_{1}(a^2_{1}-4)$$ so we have $$\dfrac{a_{n+1}}{a_{1}a_{2}\cdots a_{n}}=\sqrt{a^2_{1}-2}\cdot \dfrac{a_{n+1}}{\sqrt{a^2_{n+1}-4}}\to\sqrt{a^2_{1}-4}=\sqrt{5},a_{n+1}\to+\infty$$ since $a_{n+1}\to +\infty,n\to +\infty$

Minimize $a^5+b^5+c^5+d^5+e^5 = p^4+q^4+r^4+s^4 = x^3+y^3+z^3 = m^2 + n^2$ with distinct positive integers Find the minimum value of the following: $$a^5+b^5+c^5+d^5+e^5 = p^4+q^4+r^4+s^4 = x^3+y^3+z^3 = m^2 + n^2$$ where all numbers are different/distinct positive integers. I know the answer (see below), but want to confirm the same. Is there any way to prove following conjecture? Conjecture. There is always unique way to write down $\sum_{i=1}^{n} a_{i}^n$ for any arbitrary value of $n$ such that it gives same value for all values of $n$. Answer is given below, spoiler alert: $$1^5+2^5+4^5+7^5+9^5 = 3^4+6^4+10^4+16^4 = 17^3+20^3+40^3 = 88^2 + 263^2 = 76913$$

the smallest is 76913 squares of: 263 88 76913 cubes of: 40 17 20 76913 fourth powers of: 16 3 6 10 76913 fifth powers of: 9 1 2 4 7 76913 the second smallest is 1560402 squares of: 1239 159 1560402 cubes of: 101 45 76 1560402 fourth powers of: 35 5 12 14 1560402 fifth powers of: 17 1 6 8 10 1560402 Here is another one. It may or may not be the third smallest. 2091473 squares of: 1367 472 2091473 cubes of: 122 10 65 2091473 fourth powers of: 32 14 21 30 2091473 fifth powers of: 17 4 8 10 14 2091473

Cardano's method returns incorrect answer for $x = u + v$ I'm trying to use Cardano's method to solve this equation: $$x^3+6x=20 \tag{1}$$ As described on Wikipedia, I let $x = u + v$ and expand in $(1)$: $$(u+v)^3+6(u+v)=20$$ $$u^3 + v^3 + (3uv+6)(u+v)-20=0 \tag{2}$$ I then let $3uv + 6 = 0$ and substitute in $(2)$: $$u^3 + v^3 - 20 = 0$$ $$u^3 + v^3 = 20 \tag{3}$$ I also express $uv$ as a product of cubics: $$3uv + 6 = 0$$ $$uv = -2$$ $$u^3v^3 = -8$$ $$-4u^3v^3 = 32 \tag{4}$$ At this point, Wikipedia says "the combination of these two equations [$(3)$ and $(4)$] leads to a quadratic equation" which I think I can also be achieved by squaring $(3)$ and adding $(4)$ to both sides: $$u^6 + 2u^3v^3 + v^6 = 400$$ $$u^6 - 2u^3v^3 + v^6 = 432$$ $$(u^3 - v^3)^2 = 432$$ $$u^3 - v^3 = \pm 12\sqrt{3} \tag{5}$$ I then get $u$ by adding $(3)$ and $(5)$: $$2u^3 = 20 + 12\sqrt{3} \textrm{ or } 20 - 12\sqrt{3}$$ $$u = \sqrt[3]{10 + 6\sqrt{3}} \textrm{ or } \sqrt[3]{10 - 6\sqrt{3}}$$ and $v$ by subtracting $(3)$ and $(5)$: $$2v^3 = 20 - 12\sqrt{3} \textrm{ or } 20 + 12\sqrt{3}$$ $$v = \sqrt[3]{10 - 6\sqrt{3}} \textrm{ or } \sqrt[3]{10 + 6\sqrt{3}}$$ I finally get $x$ by adding $u$ and $v$: $$x = \sqrt[3]{10 + 6\sqrt{3}} + \sqrt[3]{10 - 6\sqrt{3}}$$ $$\textrm{ or } \sqrt[3]{10 - 6\sqrt{3}} + \sqrt[3]{10 + 6\sqrt{3}}$$ $$\textrm{ or } 2\sqrt[3]{10 + 6\sqrt{3}}$$ $$\textrm{ or } 2\sqrt[3]{10 - 6\sqrt{3}}$$ I know there's a real solution, so that only leaves $x = 2\sqrt[3]{10 + 6\sqrt{3}}$ which equals approximately $195$ instead of $20$ in the original equation. I can only find the correct real solution by using $x = u - v$ instead of $x = u + v$: $$x = \sqrt[3]{10 + 6\sqrt{3}} - \sqrt[3]{6\sqrt{3} - 10}$$ So, am I misusing Cardano's method somehow?

You forgot the condition $u^3v^3=-8$. The correct solution is the first you enunciated. That said, your way of solving the system of equations $\; \begin{cases}u^3+v^3=20\\u^3v^3=-8\end{cases}\;$ is over complicated. Just use what any high-school student knows to solve the problem of finding two numbers the sum $s$ and product $p$ of which are given: they're roots of the quadratic equation (if any): $$t^2-st+p=t^2-20t-8=0$$ The reduced discriminant is $\Delta'=100+8$, hence the roots $$u^3,v^3=10\pm\sqrt{108}=10\pm6\sqrt 3.$$ You must take both roots, because the problem involves the condition on the product is $8$ – i.e. it implies the qudratic equation, but is not equivalent to it. Thus $$x=u+v=\sqrt[3]{10-6\sqrt 3}+\sqrt[3]{10+6\sqrt 3}.$$ Edit: As pointed out by @André Nicolas $10\pm6\sqrt 3=(1\pm\sqrt3)^3$, so that the root is equal to $2$.

Factor out (m+1) in the following so that the final answer is $\frac{(2m+1) (m+2) (m+1)} {6}$ Question: $\frac{m (m+1) (2m+1) + 6(m+1)^2}{6}$=$\frac{(2m+3)(m+2)(m+1)}{6}$ I must multiply by 6 on both sides and expand the brackets and collect like terms. I'm I correct? Edit notes: The original problem was posed as: $$ \frac{(m+1) (2m+1) + 6(m+1)^2}{6}= \frac{(2m+3)+(m+2)(m+1)}{6} $$ which did not match the title. The edit provides the corrected question.

The description as given is correct. Multiplying both sides by $6$ yields $$(2m+1) (m+1) + 6(m+1)^2= (2m+3)+(m+2)(m+1)$$ Now what remains is to determine the value of both sides. Let $L_{m}$ be the left and $R_{m}$ be the right. \begin{align} L_{m} &= (2m+1)(m+1)m + 6(m+1)^2 \\ &= (2m^2 + 3m + 1)m + 6(m^2 + 2m +1) \\ &= 2 m^3 + 9 m^2 + 13 m + 6. \end{align} \begin{align} R_{m} &= (2m+3)(m+2)(m+1) \\ &= 2 m^3 + 9m^2 + 13 m + 6. \end{align} This demonstrates that $L_{m} = R_{m}$, as required.

Power Diophantine equation: $(a+1)^n=a^{n+2}+(2a+1)^{n-1}$ How to solve following power Diophantine equation in positive integers with $n>1$:$$(a+1)^n=a^{n+2}+(2a+1)^{n-1}$$

If the equation holds, then $(a+1)^n>(2a+1)^{n-1}$ and $(a+1)^n>a^{n+2}$. Multiplying gives $(a+1)^{2n} > (2a+1)^{n-1}a^{n+2}$. This can be rewritten as $(a^2+2a+1)^n > (2a^2+a)^{n-1} a^3$. However, if $a\geq 3$, then $a^2+2a+1 < 2a^2+a$ and $a^2+2a+1 <a^3$, which gives a contradiction. Hence $a=1$ or $a=2$. The first gives $2^n=1+3^{n-1}$. With induction, one may prove that the right hand side is larger for $n>2$. This only gives the solution $(1,2)$. The second gives $3^n=2^{n+2}+5^{n-1}$. With induction, one may prove that the right hand side is larger for $n>3$. One can also check that there is no equality for $n=1,2,3$.

Evaluation of Definite Integral Evaluation of $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x\sin 2x \sin 3x}{x}dx$ $\bf{My\;Try::}$ Let $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x\sin 2x \sin 3x}{x}dx = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{2\sin 3x\cos x\sin 2x}{x}dx$ So we get $$I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{(\sin4x+\sin 2x)\sin 2x}{x}dx$$ So $$I=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\frac{\cos2x-\cos6x+1-\cos 4x}{x}dx$$ Now How can i solve after that , Help me Thanks

Hint 1: $$ \int f(x) = F(X) + C \Longrightarrow \int f(ax+b) = \frac{1}{a} \cdot F(ax+b) + C$$ Hint 2: $$ \int \left(f(x) + g(x)\right) = \int f(x) + \int g(x) $$ Hint 3: $$ \int \frac{\cos x}{x} = Ci(x) + C$$ First step: (use hint 2) $$ 4I=\int_{0}^{\frac{\pi}{2}}\frac{\cos2x-\cos6x+1-\cos 4x}{x}dx =\\ =2\int_{0}^{\frac{\pi}{2}} \frac{1}{x} + 2\int_{0}^{\frac{\pi}{2}}\frac{\cos 2x -1}{2x} - 6 \int_{0}^{\frac{\pi}{2}} \frac{\cos 6x -1}{6x} - 4\int_{0}^{\frac{\pi}{2}} \frac{\cos 4x -1}{4x} $$ Now let $f(x) := \frac{\cos x - 1}{x}$ so $$ 4I = 2\int_{0}^{\frac{\pi}{2}} \frac{1}{x} + 2 \int_{0}^{\frac{\pi}{2}} f(2x) - 6 \int_{0}^{\frac{\pi}{2}} f(6x) - 4 \int_{0}^{\frac{\pi}{2}} f(4x) $$ Edit after comment: You can also use Taylor expansion of $\cos x$. But then you will have series as result (without $Ci$ function).

Mean-value Theorem $f(x)=\sqrt{x+2}; [4,6]$ Verify that the hypothesis of the mean-value theorem is satisfied for the given function on the indicated interval. Then find a suitable value for $c$ that satisfies the conclusion of the mean-value theorem. $$f(x)=\sqrt{x+2}; [4,6]$$ So, $$f'(x) = {1 \over 2} (x+2)^{-{1\over 2}}$$ $f(x)$ is differentiable for all x. Now, $$f'(c) = {f(b)-f(a)\over b- a} \\ = {f(6) - f(4)\over 6-4} \\ = {2\sqrt{2} - \sqrt{6}\over2}$$ Since $f'(c) = {2\sqrt{2} - \sqrt{6}\over2}$, $${1\over 2}(c+2)^{-{1\over 2}} = {2\sqrt{2} - \sqrt{6}\over2}$$ Then after, this will give me $c$ right? I just want to know if I did it right so far. I tried simplifying the last equation, but it wasn't right. Please let me know if there is anything wrong in my steps, if not could anyone help me solve for $c$ at the end? Thank you.

You did everything correctly, let's solve for $c$ together. You have $$ \frac{1}{2} (c+2)^{-1/2} = \frac{a}{2}\\ (c+2)^{-1/2} = a \\ \frac{1}{\sqrt{c+2}} = a \\ c+2 = \frac{1}{a^2} $$ so $$ \begin{split} c &= \frac{1}{a^2} - 2 = \frac{1}{\left(2 \sqrt{2} - \sqrt{6}\right)^2} - 2 \\ &= \frac{1}{8 + 6 - 4\sqrt{2}\sqrt{6}} - 2 \\ &= \frac{1}{14 - 8\sqrt{3}} - 2 \\ &= \frac{14 + 8 \sqrt{3}}{14^2 - 3 \cdot 8^2} - 2 \\ &= \frac{14 + 8 \sqrt{3}}{4} - 2 \\ &= \frac{7}{2} +2\sqrt{3} - 2 \\ &= \frac{3}{2} +2\sqrt{3} \end{split} $$

find the maximum of the function $f(x)=a+b\sqrt{2}\sin{x}+c\sin{2x}$ let $a,b,c\in R$,and such $a^2+b^2+c^2=100$, find the maximum value and minimum value of the function $$f(x)=a+b\sqrt{2}\sin{x}+c\sin{2x},0<x<\dfrac{\pi}{2}$$ Use Cauchy-Schwarz inequality?

Use Cauchy-Schwarz inequality: $$\left(a+b\sqrt{2}\sin{x}+c\sin{2x}\right)^2\le (a^2+b^2+c^2)(1+2\sin^2x+\sin^22x)$$ $$\left(a+b\sqrt{2}\sin{x}+c\sin{2x}\right)^2\le 100\cdot(1+2\sin^2x+\sin^22x)$$ $$|a+b\sqrt{2}\sin{x}+c\sin{2x}|\le 10\cdot\sqrt{1+2\sin^2x+\sin^22x}$$ $$1\le1+2\sin^2x+\sin^22x\le \frac{13}{4}$$ Then $$-5\sqrt{13}\le a+b\sqrt{2}\sin{x}+c\sin{2x}\le 5\sqrt{13}$$

Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$ Is there an easy to solve the problem? The way I did it is to find the value of $a$ from the second expression and then use it to find the value of the first expression. I believe there must be an simple and elegant approach to tackle the problem. Any help is appreciated. Find the value of $$\frac{a^2}{a^4+a^2+1}$$ if $$\frac{a}{a^2+a+1}=\frac{1}{6}$$

From the first equation (inverted), $$\frac{a^2+a+1}a=6$$ or $$\frac{a^2+1}a=5.$$ Then squaring, $$\frac{a^4+2a^2+1}{a^2}=25$$ or $$\frac{a^4+a^2+1}{a^2}=24.$$

Geometrical Description of $ \arg\left(\frac{z+1+i}{z-1-i} \right) = \pm \frac{\pi}{2} $ The question is in an Argand Diagram, $P$ is a point represented by the complex number. Give a geometrical description of the locus of $P$ as $z$ satisfies the equation: $$ \arg\left(\frac{z+1+i}{z-1-i} \right) = \pm \frac{\pi}{2} $$ What I have done: Consider $$\left(\frac{z+1+i}{z-1-i} \right)$$ Let $z=x+iy$ $$\left(\frac{x+iy+1+i}{x+iy-1-i} \right) \Leftrightarrow \left(\frac{(x+1)+i(y+1)}{(x-1)+i(y-1)} \right) $$ $$ \Leftrightarrow \left(\frac{(x+1)+i(y+1)}{(x-1)+i(y-1)} \right) \cdot \left(\frac{(x-1)-i(y-1)}{(x-1)-i(y-1)} \right) $$ $$ \Leftrightarrow \left( \frac{ (x+1)(x-1) -i(y+1)(x+1) + i(y+1)(x-1) - i^2(y+1)(y-1)}{(x-1)^2 -i^2(y-1)^2} \right)$$ $$ \Leftrightarrow \left (\frac{ x^2 -1+y^2-1 +i(xy-y+x-1)-i(xy+y+x+1)}{(x-1)^2 +(y-1)^2} \right) $$ $$ \Leftrightarrow \left (\frac{ x^2 -1+y^2-1 +i(xy-y+x-1)-i(xy+y+x+1)}{(x-1)^2 +(y-1)^2} \right) $$ $$ \Leftrightarrow \left (\frac{ (x^2 +y^2-2) +i(-2-2y)}{(x-1)^2 +(y-1)^2} \right) $$ $$ \Leftrightarrow \left( \frac{x^2+y^2-2}{(x-1)^2 +(y-1)^2} \right) + i \left( \frac{-2-2y}{(x-1)^2 +(y-1)^2} \right) $$ As the $\arg = \pm \frac{\pi}{2}$ , which means locus of $P$ is purely imaginary hence $\Re(z)=0$ $$\Rightarrow \frac{x^2+y^2-2}{(x-1)^2 +(y-1)^2} = 0 $$ $$ \therefore x^2 + y^2 = 2 $$ So the Locus of $P$ should be a circle with centre $(0,0)$ and radius $\sqrt{2}$ However my answer key states that the answer is $|z|=2$ which is a circle with centre $(0,0)$ and radius $2$ so where exactly did I go wrong?

Let the points $z_A = -1-i$ and $z_B=1+i$. Then we look for the locus of all points Pwith $z_P=z$ such that $\vert\arg \frac{z-z_A}{A-z_B} \vert = \pi$ (in other words, the angle $\angle APB=\pi$). This is the circle in the complex plane with diameter $AB$, since we know that the angle under which the segment $AB$ is seen from a point $P$ on the circle with the same diameter is $\pi$.

$\sum_{n=0}^{\infty}\frac{a^2}{(1+a^2)^n}$ converges for all $a\in \mathbb{R}$ $$\sum_{n=0}^{\infty}\frac{a^2}{(1+a^2)^n}$$ Can I just see this series as a geometric series? Since $c = \frac{1}{1+a^2}<1$, we can see this as the geometric series: $$\sum_{n=0}^{\infty}bc^n = \sum_{n=0}^{\infty}a^2\left(\frac{1}{1+a^2}\right)^n$$ that converges because $c<1$. The sum of this series is: $$S_n = b(c^0+c^1+c^2+\cdots c^n)\rightarrow cS_n = b(c^1+c^2 + c^3 + \cdots + c^{n}+c^{n+1})\rightarrow $$$$cS_n - S_n = b(c^{n+1}-1)\rightarrow S_n(c-1) = b(c^{n+1}-1)\rightarrow S_n = b\frac{c^{n+1}-1}{c-1}$$ $$S = \lim S_n = b\frac{1}{1-c}$$ So $$\sum_{n=0}^{\infty}\frac{a^2}{(1+a^2)^n} = b\frac{1}{1-c} = a^2\frac{1}{1-\frac{1}{1+a^2}} =$$

Let $a=0$. Then the series obviously converges to $0$. Now suppose that $a\ne 0$. Then our series is the geometric series $$a^2+a^2r+a^2r^2+a^2r^3+\cdots,$$ where $r=\frac{1}{1+a^2}\lt 1$. Since $|r|\lt 1$, the series converges. It is probably by now a familiar fact that when $|r|\lt 1$ the series $1+r+r^2+r^3+\cdots$ converges to $\frac{1}{1-r}$. So our series converges to $$\frac{a^2}{1-\frac{1}{1+a^2}}.$$ This simplifies to $1+a^2$. Remark: Your calculation included an almost correct proof that the series that you called $1+c+c^2+c^3+\cdots$ converges to $\frac{1}{1-c}$. There was a little glitch. At one stage you had $\frac{c^{n+1}-1}{c-1}$. Since $c^{n+1}$ has limit $0$, the sum that you called $S_n$ converges to $\frac{-1}{c-1}$, which I prefer to call $\frac{1}{1-c}$. I think that part of your calculation is not necessary, since the convergence of $1+c+c^2+\cdots$ to $\frac{1}{1-c}$ if $|c|\lt 1$ can probably by now be viewed as a standard fact.

Solution of $4 \cos x(\cos 2x+\cos 3x)+1=0$ Find the solution of the equation: $$4 \cos x(\cos 2x+\cos 3x)+1=0$$ Applying trigonometric identity leads to $$\cos (x) \cos \bigg(\frac{x}{2} \bigg) \cos \bigg(\frac{5x}{2} \bigg)=-\frac{1}{8}$$ But I can't understand what to do from here. Could some suggest how to proceed from here?

Thinking about the answer, we might notice that if $\theta=\frac{2\pi k}9$, then $\cos9\theta=1$. We can write this as $$\begin{align}\cos9\theta-1&=4(4\cos^3\theta-3\cos\theta)^3-3(4\cos^3\theta-3\cos\theta)-1\\ &=(16\cos^4\theta+8\cos^3\theta-12\cos^2\theta-4\cos\theta+1)^2(\cos\theta-1)=0\end{align}$$ From this we can see that the solutions were all the solutions to $\cos9\theta=1$ except for $\cos\theta=1$. If $\theta=\frac{\pm2\pi}3$, then $\cos\theta=-\frac12$, and if $\theta=\frac{2\pi(3k\pm1)}9$, then $\cos3\theta=-\frac12$ so all the cases in the big factor are taken into account by $(2\cos\theta+1)(2\cos3\theta+1)=0$. Indeed we can go back to the original equation and find that $4\cos\theta\cos2\theta=2(\cos(1+2)\theta+\cos(1-2)\theta)=2\cos3\theta+2\cos\theta$, so it reads $$\begin{align}4\cos x(\cos2x+\cos3x)+1&=2\cos3x+2\cos x+4\cos3x\cos x+1\\ &=(2\cos3x+1)(2\cos x+1)=0\end{align}$$ Thus if we could have seen through this at the outset, we could have made quick work of the problem.

A curious approximation to $\cos (\alpha/3)$ The following curious approximation $\cos\left ( \frac{\alpha}{3} \right ) \approx \frac{1}{2}\sqrt{\frac{2\cos\alpha}{\sqrt{\cos\alpha+3}}+3}$ is accurate for an angle $\alpha$ between $0^\circ$ and $120^\circ$ In fact, for $\alpha = 90^\circ$, the result is exact. How can we derive it?

Close but not exact even with regard to the linear term. With $x$ and $y$ as per Paramanand Singh's answer, if $x=1-\delta$, then, ignoring $o(\delta)$ throughout, $$\begin{align} y&=(1-\delta)[4(1-2\delta)-3]\\ &=(1-\delta)(4-8\delta-3)\\ &=(1-\delta)(1-8\delta)\\ &=1-9\delta. \end{align}$$ By Paramanand Singh's equation (2), $$\begin{align} x&\approx \frac12\sqrt{\frac{2(1-9\delta)}{\sqrt{1-9\delta+3}}+3}\\ &\approx \frac12\sqrt{\frac{2(1-9\delta)}{\sqrt{4-9\delta}}+3}\\ &\approx \frac12\sqrt{\frac{1-9\delta}{\sqrt{1-9\delta/4}}+3}\\ &\approx \frac12\sqrt{\frac{1-72\delta/8}{1-9\delta/8}+3}\\ &\approx \frac12\sqrt{1-63\delta/8+3}\\ &\approx \frac12\sqrt{4-63\delta/8}\\ &\approx \sqrt{1-63\delta/32}\\ &\approx 1-63\delta/64. \end{align}$$

Proving $\int_{0}^{\infty}\frac{4x^2}{(x^4+2x^2+2)^2}dx\stackrel?=\frac{\pi}{4}\sqrt{5\sqrt2-7}$ $$\int_{0}^{\infty}\frac{4x^2}{(x^4+2x^2+2)^2}dx=\frac{\pi}{15}$$ $$\int_{0}^{\infty}\frac{4x^2}{[(1+(1+x^2)^2]^2}dx=\frac{\pi}{15}$$ $u=\tan(z)$ $\rightarrow$ $du=\sec^2(z)$ $u$ $\rightarrow \infty$, $\tan(z)=\frac{\pi}{2}$ $u$ $\rightarrow 0$, $\tan(z)=0$ $$\int_{0}^{\pi \over 2}\frac{4\tan^2(z)}{[(1+(1+\tan^2(z))^2]^2}\frac{du}{\sec^2(z)}=\frac{\pi}{15}$$ $1+\tan^2(z)=\sec^2(z)$ $$\int_{0}^{\pi \over 2}\frac{4\sin^2(z)}{[(1+\sec^4(z)]^2}dx=\frac{\pi}{15}$$ $$\int_{0}^{\pi \over 2}\frac{4\sin^2(z)}{[(1+i\sec^2(z))(1-i\sec^2(z))]^2}dx=\frac{\pi}{15}$$ Hopeless! Any suggestion? Try again $$\int_{0}^{\pi \over 2}\frac{4\sin^2(z)}{[(1+\sec^4(z)]^2}dx=\frac{\pi}{15}$$ $$\int_{0}^{\infty}\frac{\sin^2(2z)\cos^6(z)}{(1+\cos^4(z))^2}dx=\frac{\pi}{15}$$ $\sin^2(2z)=\frac{1-\cos(4z)}{2}$ No more I gave up. Any hints?

Hint. One may write $$ \begin{align} \int_{0}^{\infty}\frac{4x^2}{(x^4+2x^2+2)^2}dx&=4\int_0^{\infty}\frac1{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2}\:\frac{dx}{x^2} \\\\&=2\sqrt{2}\int_0^{\infty}\frac1{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2}\:dx \quad (x \to \sqrt{2}/x) \\\\&=\frac1{\sqrt{2}}\int_{-\infty}^{\infty}\frac{\left(1+\frac{\sqrt{2}}{x^2}\right)}{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2}\:dx \\\\&=\frac1{\sqrt{2}}\int_{-\infty}^{\infty}\frac{d\left(x-\frac{\sqrt{2}}x\right)}{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2} \\\\&=\frac1{\sqrt{2}}\int_{-\infty}^{\infty}\frac{du}{(u^2+2+2\sqrt{2})^2} \\\\&=\color{red}{\frac14 \sqrt{5\sqrt{2}-7} \:\pi} \\\\ \end{align} $$ where we have made $u:=\sqrt{2+2\sqrt{2}}\:\sinh v$ to get the last step.

How to verify $(1+\frac{1}{n})^2(1-\frac{1}{n^2})^{n-1}\geq \exp(\frac{1}{n})$ How to verify this inequality? Assuming that $n\in \mathbb{N}^+$, we have: $$\left(1+\frac{1}{n}\right)^2\left(1-\frac{1}{n^2}\right)^{n-1}\geq \exp\left(\frac{1}{n}\right).$$

Consider $$A_n=\left(1+\frac{1}{n}\right)^2\times\left(1-\frac{1}{n^2}\right)^{n-1}$$ Take logarithms $$\log(A_n)=2\log\left(1+\frac{1}{n}\right)+(n-1)\log\left(1-\frac{1}{n^2}\right)$$ Now, use the Taylor series $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$ and replace $x$ by $\frac 1n$ in the first term and by $-\frac 1{n^2}$ in the second term. This should give $$\log(A_n)=\frac{1}{n}+\frac{1}{6 n^3}+\frac{1}{15 n^5}+O\left(\frac{1}{n^6}\right) >\frac 1n$$ You can check, using limits, that $A_1=4 > e$. Edit Taking more terms, the expansion of $\log(A_n)$ does not seem to contain any negative coefficient (I did stop at $O\left(\frac{1}{n^{1000}}\right)$, using, for sure, a CAS for this last part). A closer look at the coefficients reveals that $$\log(A_n)=\sum_{k=1}^\infty \frac{1}{k (2 k-1)}\frac 1 {n^{2 k-1}}$$

Show that $(a,b)+(c,d) = (a+c,b+d)$ Let $a<b$ and $c<d$ be real numbers. Show that $(a,b)+(c,d) = (a+c,b+d)$. I don't understand the question. Since $(a,b)$ and $(c,d)$ are intervals, what does it mean to add them?

In terms of sets and set notation: $(a,b)$ = all the points of R that are between a and b exclusively =$\{x\in \mathbb R| a < x < y\}$ If $A$ and $B$ are sets, than we say $A + B =\{x+y|x \in A; y \in B \}$. So the statement $(a,b)+(c,d) = (a+b,c+d)$ means that $(a,b) + (c,d)=\{x+y|a <x <b;c <y<d\}$ is the same as $(a+b, c+d) = \{z|a+b <z <c+d\}$ Proof: 1) if $z=x+y \in (a,b) + (c,d)$, that is, $a <x <b;c <y <d$ then $a+c <x+y <b+d$ so $(a,b)+(c,d) \subset (a+c,b+d) $. That was easy. 2) if $z \not \in (a+b,c+d)$ a) if $z \le a + c$. Let $x$ be any number such that $a < x < b$. Then if $z =x+y$ for some $y$ it follows that $y = z-x < a+c-a =c$ so $z \not \in (a,b) +(c,d)$. b) Likewise if $z \ge b + d$. Let $y$ be any number such that $c < y < d$. Then if $z =x+y$ for some $x$ it follows that $x = z-y > b+d-d =b$ so $z \not \in (a,b) +(c,d)$. So $z \not \in (a+b,c+d)$ implies $z \not \in (a,b)+(c,d)$. So $(a,b)+(c,d) \subset (a+b,c+d)$ 1 and 2 together mean $(a+b,c+d)=(a,b)+(c+d)$.

I'm stuck in a logarithm question: $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ If $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ so $x + 2y= ?$ I've tried this far, and I'm stuck $$\begin{align}4^{y+3x}&= 64 \\ 4^{y+3x} &= 4^3 \\ y+3x &= 3 \end{align}$$ $$\begin{align}\log_x (x+12)- 3 \log_x 4 &= -1 \\ \log_x (x+12)- \log_x 4^3 &= -1 \\ \log_x(x+12)- \log_x 64 &= -1 \end{align}$$ then I substituted $4^{y+3x} = 64$ $\log_x (x+12) - \log_x 4^{y+3x} = -1$ I don't know what should I do next. any ideas?

You're right up to $y+3x=3$. Now consider the other statement $\log_x(x+12)-3\log_x 4=-1$ $\log_x{x+12 \over 64 }=-1$ ${x+12 \over 64 }={1 \over x}$

Let $n \in \mathbb{N}$. Proving that $13$ divides $(4^{2n+1} + 3^{n+2})$ Let $n \in \mathbb{N}$. Prove that $13 \mid (4^{2n+1} + 3^{n+2} ). $ Attempt: I wanted to show that $(4^{2n+1} + 3^{n+2} ) \mod 13 = 0. $ For the first term, I have $4^{2n+1} \mod 13 = (4^{2n} \cdot 4) \mod 13 = \bigg( ( 4^{2n} \mod 13) \cdot ( 4 \mod 13 ) \bigg) \mod 13. $ But still I don't know how to simplify the first term in the large bracket. Any help/suggestions?

By the binomial theorem, $$ 4^{2n+1} + 3^{n+2} =4\cdot 16^n+9\cdot 3^n =4\cdot (13+3)^n+9\cdot 3^n =4(13a+3^n)+9\cdot 3^n =13(4a+3^n) $$

$\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$ when $\alpha + \beta + \gamma = \pi$ Assume: $\alpha + \beta + \gamma = \pi$ (Say, angles of a triangle) Prove: $\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$ There is already a solution on Math-SE, however I want to avoid using the sum-to-product identity because technically the book I go by hasn't covered it yet. So, is there a way to prove it with identities only as advanced as $\sin\frac{\alpha}{2}$? Edit: Just giving a hint will probably be adequate (i.e. what identity I should manipulate).

You may go the other way around: $$ \cos\frac{\gamma}{2}=\cos\frac{\pi-\alpha-\beta}{2}= \sin\frac{\alpha+\beta}{2}= \sin\frac{\alpha}{2}\cos\frac{\beta}{2}+ \cos\frac{\alpha}{2}\sin\frac{\beta}{2} $$ so the right hand side becomes $$ 4\cos\frac{\alpha}{2}\cos\frac{\beta}{2} \sin\frac{\alpha}{2}\cos\frac{\beta}{2}+ 4\cos\frac{\alpha}{2}\cos\frac{\beta}{2} \cos\frac{\alpha}{2}\sin\frac{\beta}{2} $$ Recalling the duplication formula for the sine we get $$ 2\sin\alpha\cos^2\frac{\beta}{2}+2\sin\beta\cos^2\frac{\alpha}{2} $$ and we can recall $$ 2\cos^2\frac{\delta}{2}=1+\cos\delta $$ to get $$ \sin\alpha+\sin\alpha\cos\beta+\sin\beta+\sin\beta\cos\alpha = \sin\alpha+\sin\beta+\sin(\alpha+\beta)= \sin\alpha+\sin\beta+\sin\gamma $$

Finding all pairs of integers that satisfy a bilinear Diophantine equation The problem asks to "find all pairs of integers $(x,y)$ that satisfy the equation $xy - 2x + 7y = 49$. So far, I've got \begin{align} xy - 2x + 7y &= 49 \\ x\left(y - 2\right) + 7 &= 49 \\ y &\leq 49 \end{align} I can't get any further. Any help?

hint: $xy+7y = 2x+49 \implies (x+7)y = 2x+49 \implies y = \dfrac{2x+49}{x+7}= 2 + \dfrac{35}{x+7}\implies (x+7) \mid 35\implies x+7 = \pm 1, \pm 5, \pm 7, \pm 35$

Condition on $a$ for $(x^2+x)^2+a(x^2+x)+4=0$ Find the set of values of $a$ if $$(x^2+x)^2+a(x^2+x)+4=0$$ has $(i)$ All four real and distinct roots $(ii)$ Four roots in which only two roots are real and distinct. $(iii)$ All four imaginary roots $(iv)$ Four real roots in which only two are equal. Now if I set $x^2+x=t$ then even if $t^2+at+4=0$ has real roots in is not necessary that $(x^2+x)^2+a(x^2+x)+4=0$ will have real roots too. So how to derive the condition on a? Could someone give me some direction?

Alternatively, graphing it: Let $a=y$, then $(x^2+x)^2+y(x^2+x)+4=0$ $\displaystyle y=\frac{1}{4}-\left( x+\frac{1}{2} \right)^{2}+ \frac{4}{\frac{1}{4}-\left( x+\frac{1}{2} \right)^{2}}$

proof that diagonals of a quadrilateral are perpendicular if $AB^2+CD^2=BC^2+AD^2$ proof that diagonals of a quadrilateral are perpendicular if $AB^2+CD^2=BC^2+AD^2$. My Attempt:we know that if diagonals of a quadrilateral are perpendicular then we have $AB^2+CD^2=BC^2+AD^2$.But have to proof opposite of it?

Assume the intersection of the to diagonals is $O$. Let $|OA|=a,|OB|=b,|OC|=c,|OD|=d$. Assume $\angle AOB=\gamma$. Then $$|AB|^2=a^2+b^2-2ab \cos\gamma,$$ $$|CD|^2=c^2+d^2-2cd \cos\gamma,$$ $$|BC|^2=b^2+c^2-2bc \cos(\pi-\gamma)=b^2+c^2+2bc \cos\gamma,$$ $$|AB|^2=a^2+d^2-2ad \cos(\pi-\gamma)=a^2+d^2+2ad \cos\gamma.$$ From the condition, we have $-2ab \cos\gamma-2cd \cos\gamma=2bc \cos\gamma+2ad \cos\gamma$. Thus, $(a+c)(b+d)\cos\gamma=0\Rightarrow \cos\gamma=0\Rightarrow \gamma=\frac{\pi}{2}$. Thus, $AC\perp BD$.

Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but that it has two different irreducible factors in $\mathbb{R}[X]$ Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but it has two different irreducible factors in $\mathbb{R}[X]$. I've tried to use the cyclotomic polynomial as: $$X^5-1=(X-1)(X^4+X^3+X^2+X+1)$$ So I have that my polynomial is $$\frac{X^5-1}{X-1}$$ and now i have to prove that is irreducible. The lineal change of variables are ok*(I don't know why) so I substitute $X$ by $X+1$ then I have: $$\frac{(X+1)^5-1}{X}=\frac{X^5+5X^4+10X^3+10X^2+5X}{X}=X^4+5X^3+10X^2+10X+5$$ And now we can apply the Eisenstein criterion with p=5. So my polynomial is irreducible in $\mathbb{Q}$ Now let's prove that it has two different irreducible factors in $\mathbb{R}$ I've tryed this way: $X^4+X^3+X^2+X+1=(X^2+AX+B)(X^2+CX+D)$ and solve the system. But solve the system is quite difficult. Is there another way?

let $$P(x)=x^4+x^3+x^2+x^1+1$$ We know if $x=\frac{a}{b}$ is root of $P(x)$ then $b|1\,$ , $\,a|1$. In the other words $a=\pm 1 $ and $b=\pm 1 $ but $P(1)=5$ and $P(-1)=1$, thus we let $$P(x)=(x^2+ax+b)(x^2+cx+d)$$ as a result \begin{align} & bd=1 \\ & ad+bc=1 \\ & b+d+ac=1 \\ & a+c=1 \\ \end{align} This system has not solution in $Q$ because $$d(ad+bc)=d\times\,1\to\,ad^2+c=d$$ On the other hand $\,c=1-a$ thus $$ad^2+1-a=d\to\,a(d^2-1)=d-1$$ This implies $d=1$ or $ad+a=1$. If $d=1$ then $\left\{\begin{matrix} a+c=1 \\ ac=-1 \\ \end{matrix}\right.$ that this system has not rational roots . If $\,ad+a=1\,$ then $a=\frac{1}{d+1}=\frac{b}{b+1}$ as a result $$b+d+ac=1\to b+\frac{1}{b}+\frac{b}{b+1}\left(1-\frac{b}{b+1}\right)=1$$ we have $$\frac{(b+1)^2}{b}+\frac{b}{(b+1)^2}=-1$$ This equation has not solution in $\mathbb{R}$

When a loop with inverse property is commutative Question How to prove that a loop $L$ with inverse property and $x^3=e$ for all $x$ is commutative iff $(x y)^2=x^2 y^2$ for all $x,y$? Definitions: A loop is a quasigroup with identity $e$. $L$ has the inverse property if every element has a two sided inverse and $x^{-1}(xy) = y = (yx)x^{-1}$ for all $x,y \in L$. Edit: I just figured one half of the proof: Suppose $L$ is commutative. From $x^3 = e$, we have $x^2 = x^{-1}$. Then $(xy)^2 = (xy)^{-1}$. In an IP loop, $(xy)^{-1} = y^{-1} x^{-1}$. So that by using commutativity, $(xy)^2 = y^{-1} x^{-1} = x^{-1} y^{-1} = x^2 y^2$. Edit2: The other side is easy now: Suppose $(xy)^2 = x^2y^2$. Also, we have $(xy)^2 = y^{-1} x^{-1}.$ Hence $x^{-1} y^{-1} = y^{-1}x^{-1}$. Because $L = L^{-1}$, we get that $L$ is commutative.

Note that $x^3 = e$ means that $x^2 = x^{-1}$. Similarly, this means that $(xy)^2 = (xy)^{-1}, y^2 = y^{-1}$. We now write our condition as $$ (xy)^{-1} = x^{-1}y^{-1} $$ This implies that $(x^{-1}y^{-1})^{-1} = xy$. We now apply the inverse property repeatedly: $$ x^{-1} = (x^{-1}y^{-1})y\\ (x^{-1}y^{-1})^{-1}x^{-1} = (x^{-1}y^{-1})^{-1}((x^{-1}y^{-1})y) = y\\ (x^{-1}y^{-1})^{-1} = ((x^{-1}y^{-1})^{-1}x^{-1})x = yx $$ Combining we get $xy = (x^{-1}y^{-1})^{-1} = yx$, as desired.

Prove that as $PP'$ varies,the circle generates the surface $(x^2+y^2+z^2)(\frac{x^2}{a^2}+\frac{y^2}{b^2})=x^2+y^2.$ $POP'$ is a variable diameter of the ellipse $z=0,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ and a circle is described in the plane $PP'ZZ'$ on $PP'$ as diameter.Prove that as $PP'$ varies,the circle generates the surface $(x^2+y^2+z^2)(\frac{x^2}{a^2}+\frac{y^2}{b^2})=x^2+y^2.$ My Attempt Let the circle be described on the plane $PP'ZZ'$.Let $P$ be $(a\cos\theta,b\sin\theta)$ and $P'$ be $(-a\cos\theta,-b\sin\theta)$. Then the equation of the circle be $(x-a\cos\theta)(x+a\cos\theta)+(y+b\sin\theta)(y-b\sin\theta)=0,z=0$ $x^2-a^2\cos^2\theta+y^2-b^2\sin^2\theta=0,z=0$ I am stuck here.Please help.

The circle has two endpoints $(-X,-Y)$ and $(X,Y)$ with $\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}=1$ and is perpendicular to the $z$-plane. Let $(x,y,z)$ be a point on this circle. Then we have: $x^2+y^2+z^2=X^2+Y^2$ and $y/x=Y/X$. So let $Y/y=X/x=c$. We get: $c^2\left (\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} \right)=1$ as well as: $x^2+y^2+z^2=c^2(x^2+y^2)$. Multiplying these two equations gives the desired equation.

Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Using the Trigonometric Addition Formulae, \begin{align} \tan 2x & = \frac{2\tan x}{1-\tan ^2 x} \\ \Rightarrow \frac{2\tan x}{1-\tan ^2 x}+\tan x & = 0 \\ \ 2\tan x+\tan x(1-\tan ^2 x) & = 0 \\ 2+1-\tan ^2 x & = 0 \\ \tan ^2 x & = 3 \end{align} This is as far as I can get, and when I look at the Mark Scheme no other Trignometric Identities have been used. Thanks

By the double angle formula we get $$\tan(2x)+\tan(x)=\frac{2\tan(x)}{1-\tan^2(x)}+\tan(x)=\frac{3-\tan^2(x)}{1-\tan^2(x)}\tan(x),$$ so that $\tan(x)=0$ is certainly a solution. But $\tan(x)=\pm\sqrt3$ as well, so that the initial claim is false.

Sum $1+(1+x)^2+(1+x+x^2)^2+\cdots+(1+x+x^2+\cdots+x^{N-2})^2 $ Is there a way to find the sum the following series: $$1+(1+x)^2+(1+x+x^2)^2+\cdots+(1+x+x^2+\cdots+x^{N-2})^2 \text{ ?}$$ Any ideas ? Perhaps someone knows already the result.. Thank you in advance for your time.

$$ 1+(1+x)^2+(1+x+x^2)^2+\cdots+(1+x+x^2+...+x^{N-2})^2 = \sum_{i=0}^{N-2}(1+x+\cdots + x^i)^2$$ Let $x<1$, though same can be repeated for $x>1$ - we do not consider $x=1$ since the answer is straightforward in this case $$\sum_{i=0}^{N-2}(1+x+\cdots + x^i)^2 = \sum_{i=0}^{N-2}\left(\frac{1-x^{i+1}}{1-x}\right)^2 = \frac{1}{(1-x)^2}\sum_{i=0}^{N-2}(1+(x^{i+1})^2-2x^{i+1}).$$ Thus, we have $$\sum_{i=0}^{N-2}(1+x+\cdots + x^i)^2 = \frac{1}{(1-x)^2}(N-1+\sum_{i=0}^{N-2}(x^{i+1})^2-2\sum_{i=0}^{N-2}x^{i+1}).$$ Thus, the answer is $$\frac{1}{(1-x)^2}\left(N-1+\frac{1-x^{(2(N-1)})}{1-x^2}-2\frac{1-x^{N-1}}{1-x}\right)$$ For $x=1$, this is sum of $i^2$ from 1 to $N-1$, which is $(N-1)(N)(2N-1)/6$. For $x>1$, it will be the same expression as for $x<1$.

Prove that $\frac{dy}{dx} = -\frac1{(1+x)^2}$ for given that $x\sqrt{1+y} + y\sqrt{1+x} = 0$ $$x\sqrt{1+y} + y\sqrt{1+x} = 0$$ Please tell me where I went wrong. Why I am not getting correct answer ?

There is nothing wrong. Put the value of $y$ to get your result. However, a simpler approach: $$x\sqrt{1+y} + y\sqrt{1+x} = 0$$ $$x\sqrt{1+y} = - y\sqrt{1+x}$$ Squarring both sides, we get $$x^2(1+y) = y^2(1+x)$$ $$x^2(1+y) - y^2(1+x)=0$$ $$(x-y)(x+y+xy)=0$$ So either $x-y=0$ or $x+y+xy=0$. Now if, $x-y=0$, then we have $x=y$. But this does not satisfy the equation $x\sqrt{1+y} + y\sqrt{1+x} = 0$ for all $x,y \in \mathbb{R}$. Hence required solution is $x+y+xy=0$. Thus we can write that $$1+\frac{dy}{dx}+y+x\frac{dy}{dx}=0$$ $$\frac{dy}{dx}=\frac{-y-1}{1+x}=\frac{\frac{x}{1+x}-1}{1+x}=-\frac{1}{(1+x)^2}$$

Finding the second derivative of $f(x) = \frac{4x}{x^2-4}$. What is the second derivative of $$f(x) = \frac{4x}{x^2-4}?$$ I have tried to use the quotient rule but I can't seem to get the answer.

Avoiding the quotient rule, just for an option: $$\begin{align} \ln(f(x)) &=\ln(4)+\ln(x)-\ln(x+2)-\ln(x-2)\\ \frac{f'(x)}{f(x)} &=x^{-1}-(x+2)^{-1}-(x-2)^{-1}\\ f'(x) &=f(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)\\ f''(x) &=f(x)\left(-x^{-2}+(x+2)^{-2}+(x-2)^{-2}\right)+f'(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)\\ &=f(x)\left(-x^{-2}+(x+2)^{-2}+(x-2)^{-2}\right)+f(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)^2\\ &=f(x)\left[-x^{-2}+(x+2)^{-2}+(x-2)^{-2}+\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)^2\right]\\ &=\frac{4x}{x^2-4}\left[-x^{-2}+(x+2)^{-2}+(x-2)^{-2}+\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)^2\right]\\ \end{align}$$

Prove that $ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $ Prove: $$ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $$ Hypothesis: $$ F(x) = 1+2q+3q^2+...+xq^{x-1} = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} $$ Proof: $$ P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2} $$ $$ P2 | \frac{1-(x+1)q^x+xq^{x+1}+[(x+1)(1-q)^2]q^x}{(1-q)^2} = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2} $$ $$ P3| \frac{x\color{red}{q^{x+1}}+[-(x+1)]\color{red}{q^x}+1+[(x+1)(1-q)^2]\color{red}{q^x}}{(1-q)^2} = \frac{x\color{red}{q^{x+2}}-(x+2)\color{red}{q^{x+1}}+1}{(1-q)^2} | $$ Here I just reorganize both sides of the equation, so LHS is explicity an expression with a degree of x+1, while the degree of RHS is x+2. Both LHS' $\color{red}{q^x}$ are added next. $$P4| \frac{xq^{x+1}+[-(x+1)+(x+1)(<1^2q^0+\binom{2}{1}1q-1^0q^2>)]q^x+1}{(1-q)^2}=\frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $$ $$P5 | \frac{xq^{x+1}+[2xq-xq^2+2q-q^2]q^x+1}{(1-q)^2} = \frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $$ I get stuck at this point. I don't know if i'm approaching the problem the right way. So, any help would be appreciated. Thanks in advance.

$$\begin{align} S&=1+2q+3q^2+\qquad\cdots\qquad \qquad+nq^{n-1}\\ qS&=\qquad q+2q^2+3q^3+\cdots +\quad(n-1)q^{n-1}+nq^n \\ \text{Subtracting,}&\\ (1-q)S&=1+\;\ q \ +\ q^2 +\ q^3+\cdots \qquad \qquad +q^{n-1}-nq^n\\ &=\frac {\;\ 1-q^n}{1-q}-nq^n\\ S&=\frac{1-q^n-nq^n(1-q)}{(1-q)^2}\\ &=\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2}\qquad\blacksquare\end{align}$$

What is the sum of the first $17$ terms of an arithmetic sequence if $a_9=35$? What is the sum of the first $17$ terms of an arithmetic sequence if $a_9=35$? This is what I did: $a_9=a_1+8d=35$ $S_{17}=\frac{17}{2}(a_1+a_{17})=\frac{17}{2}(a_1+a_1+16d)=\frac{17}{2}(2a_1+16d)=\frac{17}{2}\cdot 70= 595$ This solution is correct, however I don't understand the solution given in the book: $a_1+a_2+...+a_{17}=17a_9=17\cdot 35=595$ How did they get $a_1+a_2+...+a_{17}=17a_9$?

Dang, @gt6989b is right: \begin{align} a_1 + a_{17} &= a_1 + (a_1 + 16 d) = 2 a_1 + 16 d = 2(a_1 + 8 d) = 2 a_9 \\ a_2 + a_{16} &= (a_1 + d) + (a_1 + 15 d) = 2 a_1 + 16 d = 2 a_9 \\ & \vdots \\ a_8 + a_{10} &= (a_1 + 7d) + (a_1 + 9 d) = 2 a_1 + 16 d = 2 a_9 \end{align} so $$ \sum_{i=1}^{17} a_i = 8 \cdot 2 a_9 + a_9 = 17 a_9 $$

$\int_0^\infty \frac{ x^{1/3}}{(x+a)(x+b)} dx $ $$\int_0^\infty \frac{ x^{1/3}}{(x+a)(x+b)} dx$$ where $a>b>0$ What shall I do? I have diffucty when I meet multi value function.

Let us assume $A^3=a, B^3=b$, for simplicity. Now make a substitution $x=t^3$ which will transform the integral like this $$I=\int_{0}^\infty \frac {3t^3dt}{(t^3+A^3)(t^3+B^3)}$$ Now break this into partial fractions like this $$I=3\int_{0}^\infty [\frac {1}{(t^3+B^3)}-\frac {A^3}{B^3-A^3}(\frac {1}{t^3+A^3}-\frac {1}{t^3+B^3})]$$ Now, by using this $$I_1=\int_{0}^\infty \frac{dt}{t^3+A^3}=\frac {2\pi}{3^{\frac{3}{2}}A^2}$$ (You can compute this integral very easily. Just put $x=At$ and $t=\frac {1}{p}$ and after making these two substitutions, add both integral, you will see that a quadratic is left in denominator. To remove that put $p-\frac {1}{2}=\lambda$. After making this substitution, you will easily get the value of above integral.) Now using $I_1$, value of the $I$ is equal to $$I=\frac {2\pi}{3^{1/2}}[\frac{1}{B^2}-\frac {A^3}{B^3-A^3}(\frac{1}{A^2}-\frac{1}{B^2})]$$ which after simplication equal to $$I=\frac{2\pi}{\sqrt3 \times(B^2+A^2+AB))}$$ where $A^3=a, B^3=b$

Find stationary points of the function $f(x,y) = (y^2-x^4)(x^2+y^2-20)$ I have problem in finding some of the stationary points of the function above. I proceeded in this way: the gradient of the function is: $$ \nabla f = \left( xy^2-3x^5-2x^3y^2+40x^3 ; x^2y+2y^3-x^4y-20y \right) $$ So in order to find the stationary points, I must resolve the system: $$ \begin{cases} xy^2-3x^5-2x^3y^2+40x^3 = 0 \\ x^2y+2y^3-x^4y-20y = 0 \end{cases} $$ So far I've found the points: $$ (0,0) \qquad \left(\pm 2 \sqrt{10 \over 3} , 0 \right) \qquad (0, \pm \sqrt{10}) $$ But, I'm still blocked when I've to found the points deriving by the system: $$ \begin{cases} 2x^6 + 3x^4 +x^2 -20 = 0 \\ y^2 = \frac{1}{2} \left( x^4 - x^2 + 20 \right) \end{cases} $$ Which I don't know how to solve. Can someone help me ? Thanks.

WA gets $$ \DeclareMathOperator{grad}{grad} \grad((x^2+y^2-20) (y^2-x^4)) = (-6 x^5-4 x^3 (y^2-20)+2 x y^2, 2 y (-x^4+x^2+2 y^2-20)) $$ (link) and nine real critical points (link).

A sum of squared binomial coefficients I've been wondering how to work out the compact form of the following. $$\sum^{50}_{k=1}\binom{101}{2k+1}^{2}$$

$$\begin{align}\sum_{k=0}^m \binom {2m+1}{2k+1}^2 &=\sum_{k=0}^m \binom {2m+1}{2k+1}\binom {2m+1}{2m-2k} \color{lightgrey}{=\sum_{j=0}^m\binom {2m+1}{2(m-j)+1}\binom {2m+1}{2j}\quad \scriptsize (j=m-k)}\\ &=\frac 12 \sum_{k=0}^m \binom {2m+1}{2k}\binom {2m+1}{2(m-k)+1}+\binom {2m+1}{2k+1}\binom{2m+1}{2m-2k}\\ &=\frac 12 \sum_{i=0}^{2m+1}\binom {2m+1}i\binom {2m+1}{2m+1-i}\\ &=\frac 12 \binom {4m+2}{2m+1}\\ \sum_{k=1}^m \binom {2m+1}{2k+1}^2&=\frac 12 \binom {4m+2}{2m+1}-\binom {2m+1}1^2\\ &=\frac 12 \binom {4m+2}{2m+1}-(2m+1)^2 \end{align}$$ Put $m=50$: $$\sum_{k=1}^{50}\binom {101}{2k+1}^2=\color{red}{\frac 12 \binom {202}{101}-101^2}\qquad\blacksquare$$

What is the value of $\frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b}$ if $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 1$? If $$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 1$$ then find the values of $$\frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b}.$$ How can I solve it? Please help me. Thank you in advance.

HINT: $$\dfrac{a^2}{b+c}+a=\dfrac{a(a+b+c)}{b+c}$$ $$\sum_{\text{cyc}}\left(\dfrac{a^2}{b+c}+a\right)=(a+b+c)\sum_{\text{cyc}}\dfrac a{b+c}$$

Limit of the minimum value of an integral Let $$f(a)=\frac{1}{2}\int_{0}^{1}|ax^n-1|dx+\frac{1}{2}$$ Here $n$ is a natural number. Let $b_n$ be the minimum value of $f(a)$ for $a>1$. Evaluate $$\lim_{m \to \infty}b_mb_{m+1}\ldots b_{2m}$$ Some starters please. Thanks.

$$\begin{eqnarray*}f(a) = \frac{a}{2}\int_{0}^{1}\left| x^n-\frac{1}{a}\right|\,dx+\frac{1}{2}&=&\frac{1}{2}+\frac{a}{2}\int_{0}^{1}(x^n-1/a)\,dx+a\int_{0}^{\frac{1}{\sqrt[n]{a}}}\left(\frac{1}{a}-x^n\right)\,dx\\&=&\frac{1}{2}+\frac{a}{2n+2}-\frac{1}{2}+\frac{1}{\sqrt[n]{a}}-\frac{1}{(n+1)\sqrt[n]{a}}\\&=&\frac{a}{2n+2}+\frac{n}{(n+1)\sqrt[n]{a}}\end{eqnarray*}$$ attains its minimum at $a=2^{\frac{n}{n+1}}$: $$ b_n = 2^{-\frac{1}{n+1}}.$$ Then consider that: $$ \sum_{k=m}^{2m}\frac{1}{k+1}\stackrel{m\to +\infty}{\longrightarrow}\log 2.$$

Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let $$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$ $$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$ Prove that $I=J={\pi \over 2\sqrt3}$ Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$ $x=\infty \rightarrow u={\pi\over 2}$, $x=0\rightarrow u=0$ Rewrite $(1)$ as $$I=\int_{0}^{\infty}{1\over (1+x^2)^2-x^2}dx$$ then $$\int_{0}^{\pi/2}{\sec^2{u}\over \sec^4{u}-\tan^2{u}}du\tag3$$ Simplified to $$I=\int_{0}^{\pi/2}{1\over \sec^2{u}-\sin^2{u}}du\tag4$$ Then to $$I=2\int_{0}^{\pi/2}{1+\cos{2u}\over (2+\sin{2u})(2-\sin{2u})}du\tag5$$ Any hints on what to do next? Re-edit (Hint from Marco) $${1\over x^8+x^4+1}={1\over 2}\left({x^2+1\over x^4+x^2+1}-{x^2-1\over x^4-x^2+1}\right)$$ $$M=\int_{0}^{\infty}{x^2+1\over x^4+x^2+1}dx=\int_{0}^{\infty}{x^2\over x^4+x^2+1}dx+\int_{0}^{\infty}{1\over x^4+x^2+1}dx={\pi\over \sqrt3}$$ $$N=\int_{0}^{\infty}{x^2-1\over x^4-x^2+1}dx=0$$ $$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx={1\over 2}\left({\pi\over \sqrt3}-0\right)={\pi\over 2\sqrt3}.$$

$$ \begin{aligned} I & =\int_0^{\infty} \frac{\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} d x \\ & =\frac{1}{2} \int_0^{\infty} \frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+1} d x \\ & =\frac{1}{2}\left[\int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+3}-\int_0^{\infty} \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2-1}\right] \\ & =\frac{1}{2}\left[\frac{1}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)\right]_0^{\infty}-\frac{1}{2}\ln \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|_0^{\infty}\right]\\ & =\frac{\pi}{2 \sqrt{3}} \end{aligned} $$ $$ \begin{aligned} & J=\int_0^{\infty} \frac{1}{x^8+x^4+1} d x=\frac{1}{2}\left(\underbrace{\int_0^{\infty} \frac{x^2+1}{x^4+x^2+1}}_K d x-\underbrace{\int_0^{\infty} \frac{x^2-1}{x^4-x^2+1} d x}_L\right) \\ & K=\int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+3} d x=\int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+1} =\frac{1}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)\right]_0^{\infty}=\frac{\pi}{\sqrt{3}} \end{aligned} $$ $$ \begin{aligned} L & =\int_0^{\infty} \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}-1} d x =\int_0^{\infty} \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2-3} =\frac{1}{2 \sqrt{3}}\left[\ln \left(\frac{x+\frac{1}{x}-\sqrt{3}}{x+\frac{1}{x}+\sqrt{3}}\right)\right]_0^{\infty} =0 \end{aligned} $$ Hence we can conclude that$$\boxed{J=I =\frac{\pi}{2 \sqrt{3}}} $$

How do I find the Integral of $\sqrt{r^2-x^2}$? How can I find the integral of the following function using polar coordinates ? $$f(x)=\sqrt{r^2-x^2}$$ Thanks!

$\displaystyle\int \sqrt{r^2-x^2}dx$ Let be $\;x=r.\sin\alpha$ or $\quad x=r.\cos\alpha$, Let be $\;x=r.\sin\alpha$, and $\quad dx=r.\cos \alpha \;d\alpha$ Integral be, $\displaystyle\int \sqrt{r^2-x^2}dx=\displaystyle\int r.\sqrt{1-\sin^2\alpha}\;.r.\cos \alpha \;d\alpha=\displaystyle\int r^2.\cos^2\alpha\; d\alpha$ And use this equation,$\quad \cos2\alpha=2\cos^2\alpha-1\longrightarrow \cos^2\alpha=\dfrac{\cos 2\alpha+1}{2}$ Then, $=\displaystyle\int r^2.\cos^2\alpha\; d\alpha=\displaystyle\int r^2.\dfrac{\cos 2\alpha+1}{2}\; d\alpha=\dfrac{r^2}{2}\left[\dfrac{\sin2\alpha}{2}+\alpha\right]+C$ From here , $\boxed{\;x=r.\sin\alpha}$ Answer'll be, $\boxed{\boxed{\displaystyle\int \sqrt{r^2-x^2}dx=\dfrac{r^2}{2}\left[\dfrac{\sin2\alpha}{2}+\alpha\right]+C=\dfrac{r^2}{2}\left[\dfrac{2x\sqrt{r^2-x^2}}{r^2}+\arcsin\left(\dfrac{x}{r}\right)\right]}}$

$5^{th}$ degree polynomial expression $p(x)$ is a $5$ degree polynomial such that $p(1)=1,p(2)=1,p(3)=2,p(4)=3,p(5)=5,p(6)=8,$ then $p(7)$ $\bf{My\; Try::}$ Here We can not write the given polynomial as $p(x)=x$ and $p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ for a very complex system of equation, plz hel me how can i solve that question, Thanks

hint : write the polynomial in this form $$f(x)= a(x-1)(x-2)(x-3)(x-4)(x-5)+b(x-1)(x-2)(x-3)(x-4)(x-6) +c(x-1)(x-2)(x-3)(x-5)(x-6)+d(x-1)(x-2)(x-4)(x-5)(x-6)+e(x-1)(x-3)(x-4)(x-5)(x-6)+f(x-2)(x-3)(x-4)(x-5)(x-6)$$ now finding constants are easy

How to solve asymptotic expansion: $\sqrt{1-2x+x^2+o(x^3)}$ Determinate the best asymptotic expansion for $x \to 0$ for: $$\sqrt{1-2x+x^2+o(x^3)}$$ How should I procede? In other exercise I never had the $o(x^3)$ in the equation but was the maximum order to consider.

You have the following asymptotic expansion : $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+o(x^3)$$ So : $$\sqrt{1-2x+x^2+o(x^3)}=1+\frac{-2x+x^2+o(x^3)}{2}-\frac{(-2x+x^2+o(x^3))^2}{8}+\frac{(-2x+x^2+o(x^3))^3}{16}+o((-2x+x^2+o(x^3))^3)\\=1+\frac{-2x+x^2}{2}-\frac{4x^2-4x^3+x^4}{8}+\frac{-8x^3+12x^4-6x^5+x^8}{16}+o(x^3)\\=1-x+\frac{x^2}{2}-\frac{x^2}{2}+\frac{x^3}{2}-\frac{x^3}{2}+o(x^3)=1-x+o(x^3)$$

Matrix with orthonormal base I have the two following given vectors: $\vec{v_{1} }=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ $\vec{v_{2} }=\begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix} $ I have to calculate matrix $B$ so that these vectors in $\mathbb{R}^{3}$ construct an orthonormal basis. The solution is: $$B=\begin{pmatrix} 0 & -\frac{\sqrt{2} }{2} & \frac{\sqrt{2} }{2} \\ 1 & 0 & 0 \\ 0 & -\frac{\sqrt{2} }{2} & -\frac{\sqrt{2} }{2} \end{pmatrix}$$ I really don't have any idea how to get this matrix. I'm also confused because I only have 2 vectors.

Maybe these calculations would help you. We need to find vertor $\vec{v}_3$ such that $\vec{v}_3\perp\vec{v}_1$ and $\vec{v}_3\perp \vec{v}_2$, i.e. $$ \begin{cases} (\vec{v}_1, \vec{v}_3) = 0, \\ (\vec{v}_2, \vec{v}_3) = 0. \end{cases} $$ Here $(\vec{x},\vec{y})$ is a scalar product of vectors $\vec{x}$ and $\vec{y}$. If we denote $\vec{v}_3$ as $(x_1,x_2,x_3)^T$ we get the system $$ \begin{cases} 0\cdot x_1 + 1\cdot x_2 + 0\cdot x_3 = 0, \\ 3\cdot x_1 +0\cdot x_2 - 3\cdot x _3 = 0 \end{cases} \iff \begin{cases} x_2 = 0, \\ 3x_1 - 3x_3 = 0 \end{cases}\iff \begin{cases} x_2 = 0, \\ x_1 = x_3. \end{cases} $$ So vector $\vec{v}_3$ is depends on one parameter $x$ and has form $(x,0,x)^T$. Then we need to normalize this system, i.e. calculate vectors $\vec{u}_i = \dfrac{\vec{v}_i}{||\vec{v}_i||}$. We get $$ \vec{u}_1 = \frac{1}{\sqrt{1^2}} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}; $$ $$ \vec{u}_2 = \frac{1}{\sqrt{3^2 + (-3)^2}} \begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{2}}{2} \\ 0 \\ -\frac{\sqrt{2}}{2} \end{pmatrix}; $$ $$ \vec{u}_3 = \frac{1}{\sqrt{x^2 + x^2}} \begin{pmatrix} x \\ 0 \\ x \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{2}}{2} \\ 0 \\ \frac{\sqrt{2}}{2} \end{pmatrix}. $$ One may see that system of vectors $(\vec{u}_1,\vec{u}_2,\vec{u}_3)$ is orthonormal.

If $\sin(\pi \cos\theta) = \cos(\pi\sin\theta)$, then show ........ If $\sin(\pi\cos\theta) = \cos(\pi\sin\theta)$, then show that $\sin2\theta = \pm 3/4$. I can do it simply by equating $\pi - \pi\cos\theta$ to $\pi\sin\theta$, but that would be technically wrong as those angles could be in different quadrants. So how to solve?

We first rewrite $$ \cos(\pi/2 - \pi \cos \theta) = \cos(\pi \sin \theta) $$ (cofunction identity). We notice that $\cos(x)$ is a periodic function with period $2\pi$, so we need a period offset term to be sure that we find all solutions. We also need to account for the fact that $\cos(x)$ is symmetric, so: $$ \cos(\pi/2 - \pi \cos \theta) = \cos(\pm \pi \sin \theta + 2 \pi k) \\ \pi/2 - \pi \cos \theta = \pm \pi \sin \theta + 2 \pi k $$ Then we do basic algebra $$ 1/2 - \cos \theta = \pm \sin \theta + 2k \\ \cos \theta \pm \sin \theta = 1/2 - 2k $$ We'll get back to this. Trigonometric trick time! We can do a neat little trick to all functions of the form $a \cos \theta + b \sin \theta$. We rewrite: \begin{align*} a \cos \theta + b \sin \theta &= \sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}} \cos \theta + \frac{b}{\sqrt{a^2+b^2}} \sin \theta\right) \\ &=: \sqrt{a^2+b^2} (a' \cos \theta + b' \sin \theta) \\ &=: \sqrt{a^2+b^2} (\cos \phi \cos \theta + \sin \phi \sin \theta) \\ &=: \sqrt{a^2+b^2} \cos (\theta - \phi) \end{align*} where we can find $\phi=\arctan(b'/a') + \pi n = \arctan(b/a) +\pi n$ (note the $\pi n$ because, like $\sin(x)$ and $\cos(x)$, $\tan(x)$ is also a periodic function, so we have to account for all possible inverse values). End trigonometric trick We apply the trick to get $$ \cos \theta \pm \sin \theta = 1/2 - 2k \\ \sqrt{1^2+1^2} \cos (\theta - \arctan (\pm1/1)) = 1/2 - 2k \\ \sqrt 2 \cos (\theta \mp \pi/4) = 1/2 - 2k $$ (You can verify that the rewritten forms do indeed evaluate to the original.) We continue: $$ \cos (\theta \mp \pi/4) = 1/(2\sqrt{2}) + \sqrt{2} k = \sqrt{2} (1/4 - k) $$ Notice that $-1 \le \cos(x) \le 1$, so only $k = 0$ is valid. Hence we get $$ \cos (\theta \mp \pi/4) = \sqrt{2}/4 $$ Next, we notice \begin{align*} \sin 2\theta &= \cos (\pi/2 - 2 \theta) \\ &= \cos (2 \theta - \pi/2) \\ &= \cos(2 (\theta - \pi/4)) \\ &= 2 \cos^2 (\theta - \pi/4) - 1 \\ &= 2 (\sqrt{2}/4)^2 - 1 \\ &= 1/4 - 1 \\ &= -3/4 \end{align*} We notice the other solution \begin{align*} \sin 2\theta &= -\cos (\pi/2 - 2 \theta) \\ &= \cos (2 \theta - \pi/2) \\ &= -\cos (2 \theta - \pi/2 + \pi) \\ &= -\cos (2 \theta + \pi/2) \\ &= -\cos(2 (\theta + \pi/4)) \\ &= -(2 \cos^2 (\theta + \pi/4) - 1) \\ &= -(2 (\sqrt{2}/4)^2 - 1) \\ &= -(1/4 - 1) \\ &= 3/4 \end{align*}

If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$then.. If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$ then find the value of $f(0) + f'(0) + f''(0)$. I tried differentiating the given. But it is getting too long and complicated. So there must be a way to simplify $f(x)$. What is it?

we can simplify the fraction as $$\frac{2\cos3x\cos2x+5\cos3x}{2\cos^23x-1+6[2\cos3x\cos x]+9\cos2x+10}$$ $$=\frac{(2\cos2x+5)\cos3x}{2\cos^23x+12\cos3x\cos x+18\cos^2x}$$ $$=\frac{(2\cos2x+5)\cos3x}{2(\cos3x+3\cos x)^2}$$ $$=\frac{[2(2c^2-1)+5](4c^3-3c)}{2(4c^3)^2}$$ $$=\frac{(4c^2+3)(4c^2-3)}{32c^5}$$ $$=\frac 12\sec x-\frac{9}{32}\sec^5x$$ Now you can differentiate this twice quite easily

Determining a basis for a space of polynomials. Let $V = \mathbb R[x]_{\le 3}$ I have the space of polynomials $U_2 = \{ p = a_0 + a_1x + a_2x^2 + a_3x^3 \in V \mid a_1 - a_2 + a_3 = 0, a_0 = a_1 \}$ I am asked to find a basis, so I proceed by noticing that in $U_2$: $$a_0 + a_1x + a_2x^2 + a_3x^3 = a_0 (1+x-x^3) + a_2 (x^2 +x^3)$$ So I figure that a basis is $(1+x-x^3,x^2 + x^3 )$ In $\mathbb R^4$ the vectors would be $(1 10 -1), (001 1)$ but the solution gives the vectors $(1110), (-1-1 0 1)$. What am I doing wrong? Edit: corrected an error pointed out in the comments.

These are both bases. Your basis is $\{1+x-x^3, x^2+x^3\}$; the solution gives the basis $\{1+x+x^2, -1-x+x^3\}$. But each of these is expressible in terms of the other: \begin{gather*} 1+x+x^2 = (1+x-x^3) + (x^2+x^3),\quad -1-x+x^3 = -(1+x-x^3) \\ 1+x-x^3 = -(-1-x+x^3),\quad x^2+x^3 = (1+x+x^2) + (-1-x+x^3). \end{gather*}

How can I solve this inequality? Have a nice day, how can I solve this inequality? $$a<b<-1$$ $$ |ax - b| \le |bx-a|$$ what is the solution set for this inequality

There may be more efficient ways to do these but for clarity I like to break absolute values into cases: Case 1: $ax -b \ge 0$ and $bx -a \ge 0$. [This implies $ax \ge b\implies x \le b/a$ and likewise $x \le a/b$ so $x \le \min(a/b,b/a) = b/a < 1$. Let's keep in mind $b/a < 1 < a/b$] Then $|ax - b| \le |bx -a| \implies ax - b \le bx - a$ So $(a-b)x \le (b-a)$; $a < b$ so $(a - b) < 0$ so $x \ge (b-a)/(a-b) = -1$ So $-1 \le x \le b/a < 1$. OR Case 2: $ax -b <0$ and $bx-a < 0$. [This implies $ax < b\implies x > b/a$ and $x > a/b$ so $x > \max(a/b,b/a) =a/b > 1$] then $b - ax \le a - bx$ so $(b - a) \le (a - b)x$ so $(b-a)/(a-b) \ge x$ so $x \le -1$ which is a contradiction. OR Case 3: $ax -b < 0$ and $bx - a \ge 0$ [which implies $ax < b\implies x >b/a > 1$ and $bx \ge a\implies x \le a/b \le 1$. So $b/a < x \le a/b$. Remember $b/a < 1 < a/b$] So $b - ax \le bx - a$ so $(b+a) \le (a + b)x$. Now (a+b) < 0. So $(b+a)/(a+b) = 1 \ge x$ So $b/a < x \le 1$. OR Case 4: $ax - b \ge 0$ and $bx - a < 0$ [which implies $ax \ge b \implies x\le b/a$ and $x > a/b$ so $a/b < x \le b/a$ which is a contradiction.] So we know $-1 \le x \le b/a$ OR $b/a < x \le -1$ so combining the two results we know $-1 \le x \le 1$. But there are probably more efficient ways to do this.

I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$ I want to show that $$\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$$ Expand $(x^4-x+\pi)^2=x^4-2x^3+2x^2-2x\pi+\pi{x^2}+\pi^2$ Let see (substitution of $y=x^2$) $$\int_{-\infty}^{\infty}{x\over (x^4-x^2+1)^2}dx={1\over 2}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$ Substituion of $y=x^3$ $$\int_{-\infty}^{\infty}{x^3\over (x^4-x^2+1)^2}dx={1\over 4}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$ As for $\int_{-\infty}^{\infty}{x^2\over (x^4-x^2+1)^2}dx$ and $\int_{-\infty}^{\infty}{x^4\over (x^4-x^2+1)^2}dx$ are difficult to find a suitable substitution. This is the point where I am shrugged with to find a suitable substitution To lead me to a particular standard integral. Need some help, thank. standard integral of the form $$\int{1\over (ax^2+bx+c)^2}dx={2ax+b\over (4ac-b^2)(ax^2+bx+c)}+{2a\over 4ac-b^2}\int{1\over ax^2+bx+c}dx$$ And $$\int{1\over ax^2+bx+c}dx={2\over \sqrt{4ac-b^2}}\tan^{-1}{2ax+b\over \sqrt{4ac-b^2}}$$

Elaborating user @Dr. MV's answer, we have \begin{equation} \int_0^\infty\frac{1}{a^2x^4+bx^2+c^2}\ dx=\frac{c\pi}{2a\sqrt{b+2ac}} \end{equation} Putting $a=1$, $b=a$, and $c^2=b$, then \begin{equation} I(a,b)=\int_0^\infty\frac{1}{x^4+ax^2+b}\ dx=\frac{\pi}{2}\sqrt{\frac{b}{a+2\sqrt{b}}} \end{equation} Hence \begin{equation} \frac{\partial}{\partial a}I(a,b)=\int_0^\infty\frac{x^2}{\left(x^4+ax^2+b\right)^2}\ dx=-\frac{\pi}{2b}\sqrt{\left(\!\frac{b}{a+2\sqrt{b}}\!\right)^3} \end{equation} and \begin{equation} \frac{\partial}{\partial b}I(a,b)=\int_0^\infty\frac{1}{\left(x^4+ax^2+b\right)^2}\ dx=\frac{\pi(a+\sqrt{b})}{2\sqrt{b\left(a+2\sqrt{b}\right)^3}} \end{equation}

prove inequation a,b,c,d $\in \mathbb{R}$ $a,b,c,d \gt 0$ and $ c^2 +d^2=(a^2 +b^2)^3$ prove that $$ \frac{a^3}{c} + \frac{b^3}{d} \ge 1$$ If I rewrite the inequation like $ \frac{a^3}{c} + \frac{b^3}{d} \ge \frac{c^2 +d^2}{(a^2 +b^2)^3}$ and manage to simplfy it brings me nowhere. I try with Cauchy-Schwarz Inequality but still can not solve it. I would very much appreciate it if anyone could help me

Using Titu's Lemma, we have $$ \dfrac{a^3}{c} + \dfrac{b^3}{d} \ge \dfrac{(a^2+b^2)^2}{ac+bd}$$ So, we are left to prove that $$ (a^2+b^2)^2 \geq (ac+bd)\tag{1} $$ Using Cauchy-Schwarz inequality, we have $$(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2\tag{2}$$ Using the given proposition, $$c^2+d^2 =(a^2+b^2)^3$$ in $(2)$ and the fact that $a,b,c,d \in \mathbb{R^+}$, we get $(1)$, which was required to be proved.

Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is: Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer. I'm stuck at the basis step. If I started with $1$. I get the right hand side is $18$ which is clearly not even close. It says prove shouldn't it be always true?

Helping out with the problem. I'm stuck at the basis step. $p(n)$: $3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \dfrac{3(5^{n+1} -1)}{4}$. Where $n \in \{0, 1, 2, \dots \}$. We can rewrite the predicate. $p(n)$: $\sum_0^n3\cdot5^n = \dfrac{3(5^{n+1} -1)}{4}$ Base case: Here you need to start at 0 because we only "induct" upwards. $p(0): \sum_0^03\cdot5^n = \dfrac{3(5^{n+1} -1)}{4}$ $\implies 3 = \dfrac{3(5^1 -1)}{4} = 3 \checkmark$ Now it's up to you to show that $p(n) \implies p(n+1)$ and interpret the results.

Let $P(x)=ax^{2014}-bx^{2015}+1$ and $Q(x)=x^2-2x+1$ such that $Q(x)|P(x)$, find $a+b$ Let $P(x)=ax^{2014}-bx^{2015}+1$ and $Q(x)=x^2-2x+1$ be the polynomials where $a$ and $b$ are real numbers. If polynomial $P$ is divisible by $Q$, what is the value of $a+b$. This is what I have tried so far: Since $Q(x)|P(x)$ we have $P(1)=0$, therefore $a-b+1=0$. Problem is because we cannot obtain system of equations, because polynomial $Q(x)$ has double root at $x=1$. From equation $a-b+1$ we cannot find $a+b$, so how to find out value of $a+b$ or $a^2-b^2$?

Since $Q$ divides $P$, we have that $x-1$ divides $P$ twice. Hence $1$ is a root of $P$ and of its derivative $P'(x)=2014ax^{2013}$. So $a-b^{2015}+1=0$ and $2014a=0$. Now we can solve to get $a=0$, but in that case $Q$ cannot divide $P$, since $P$ has degree $0$. If we actually have $P(x)=ax^{2015}-bx^{2014}+1$ then its derivative is $2015ax^{2014}-2014bx^{2013}$ and we get instead the system $a-b+1=0, \, 2015a-2014b=0.$

Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. So I have to prove $$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$ I rearranged it $$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$ My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas? Thanks

Because $$\sum_{cyc}\left(\frac{1}{a^2}-\frac{1}{ab}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a}-\frac{1}{b}\right)^2\geq0.$$

How to solve $x<\frac{1}{x+2}$ Need some help with: $$x<\frac{1}{x+2}$$ This is what I have done: $$Domain: x\neq-2$$ $$x(x+2)<1$$ $$x^2+2x-1<0$$ $$x_{1,2} = \frac{-2\pm\sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{-2\pm2\sqrt{2}}{2}$$ What about now?

Multiply the inequality by $\;(x+2)^2>0\;$ ( obviously, $\;x\neq-2\;$) : $$x(x+2)^2<x+2\iff x^3+4x^2+3x-2<0\iff$$ $$\iff (x+2)(x+1-\sqrt2)(x+1+\sqrt2)<0\iff \color{red}{x<-1-\sqrt 2}\;\;\text{or}\;\color{red}{-2<x<-1+\sqrt2}$$

Prove that $\sum_{n=1}^{\infty }\frac{B_{2n}}{(2n-1)!}=\frac{1}{2}-\frac{1}{(e-1)^2}$ Prove that $$\sum_{n=1}^{\infty }\frac{B_{2n}}{(2n-1)!}=\frac{1}{2}-\frac{1}{(e-1)^2}$$ My idea is to find the Taylor series of $\frac{1}{(e^x-1)^2}$, but it seems not useful. Any helps, thanks

An alternative approach is to use the integral representation $$ B_{2n} = (-1)^{n}4n \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2 \pi x}-1} \, dx.$$ Specifically, $$ \begin{align}\sum_{n=1}^{\infty} \frac{B_{2n}}{(2n-1)!} &= \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} (-1)^{n-1} 4n \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2 \pi x}-1} \, dx \\ &= 4 \int_{0}^{\infty} \frac{1}{e^{2 \pi x}-1} \sum_{n=1}^{\infty} \frac{n (-1)^{n-1} x^{2n-1}}{(2n-1)!} \, dx. \end{align}$$ But notice that $$ \begin{align} \sin (x) + x \cos(x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n-1}}{(2n-1)!} + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{2n-1}}{(2n-2)!} \\ &= \sum_{n=1}^{\infty} \frac{[1+(2n-1)](-1)^{n-1} x^{2n-1}}{(2n-1)!} \\ &= \sum_{n=1}^{\infty} \frac{2n(-1)^{n-1} x^{2n-1}}{(2n-1)!}. \end{align}$$ So using the fact that $$\int_{0}^{\infty} \frac{\sin ax}{e^{2 \pi x}-1} \, dx = \frac{1}{4} \, \coth \left(\frac{a}{2} \right) - \frac{1}{2a}, $$ we get $$ \begin{align} \sum_{n=1}^{\infty} \frac{B_{2n}}{(2n-1)!} &= 2 \left[ \int_{0}^{\infty} \frac{\sin x}{e^{2 \pi x}-1} \,dx + \int_{0}^{\infty} \frac{x \cos x}{e^{2 \pi x}-1} \, dx \right] \\ &= 2 \left[\frac{1}{4} \, \coth \left(\frac{1}{2} \right) - \frac{1}{2} + \frac{\mathrm{d}}{\mathrm{d}a} \left(\frac{1}{4} \, \coth \left(\frac{a}{2} \right) - \frac{1}{2a}\right)\Bigg|_{a=1} \right] \\ &= 2 \left[\frac{1}{4} \, \coth \left(\frac{1}{2} \right) - \frac{1}{2} + \frac{1}{2} - \frac{1}{8} \, \text{csch}^{2} \left(\frac{1}{2} \right) \right] \\ &= \frac{1}{2} \, \coth \left(\frac{1}{2} \right) - \frac{1}{4} \, \text{csch}^{2} \left(\frac{1}{2} \right) \\&= \frac{1}{2} \frac{e+1}{e-1} - \frac{1}{4} \frac{4e}{(e-1)^{2}} \\ &= \frac{(e^{2}-1) -2e}{2(e-1)^{2}} \\ &= \frac{(e-1)^{2}-2}{2(e-1)^{2}} \\ &= \frac{1}{2} - \frac{1}{(e-1)^{2}}.\end{align}$$

Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$ \sin 2x = \sin x; \\ 0 \le x < 2 \pi $$ My method: $$ \sin 2x - \sin x = 0 $$ I apply the formula $$ \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$ So: $$ 2\sin\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) = 0 $$ $$ \sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right) = 0 $$ Here one of the factors has to be $0$, $$ \sin x = 0 \ \Rightarrow \ x = 0 \ or \ x = \pi $$ $$ \sin\left(\frac{x}{2}'\right) = 0 \ \Rightarrow \ x = 0 ;\ x \text{ can't be } \pi \text{ because of its range} $$ $$ \cos x = 0 \ \Rightarrow \ x = \frac{\pi}{2} \text{ or } \ x = \frac{3\pi}{2} $$ $$ \cos\left(\frac{3x}{2}\right) = 0 => x = \frac{\pi}{3} \text{ or } x = \pi $$ So the solutions are : $$ 0, \pi, \frac{\pi}{3} $$ I have seen other methods to solve this, so please don't post them. I'm really interested what's wrong with this one.

$\sin A=0\implies A=n\pi$ $\cos B=0\implies B=(2m+1)\pi/2$ $m,n$ are arbitrary integers

If $x\in \left(0,\frac{\pi}{4}\right)$ then $\frac{\cos x}{(\sin^2 x)(\cos x-\sin x)}>8$ If $\displaystyle x\in \left(0,\frac{\pi}{4}\right)\;,$ Then prove that $\displaystyle \frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$ $\bf{My\; Try::}$ Let $$f(x) = \frac{\cos x}{\sin^2 x(\cos x-\sin x)}=\frac{\sec^2 x}{\tan^2 x(1-\tan x)} = \frac{1+\tan^2 x}{\tan^2(1-\tan x)}$$ Now Put $\tan x= t \in (0,1)\;,$ Then $$h(t) = \frac{1+t^2}{t^2(1-t)}\;\;, 0<t<1 $$ where $h(t)=f(x)$. Now How can i solve it after that, Help Required ,Thanks

We are to prove that $$\frac{\cos x}{\sin x (\cos x - \sin x)}> 8 \sin x $$ By Cauchy-Schwarz Inequality, since all quantities involved are positive $$\bf{LHS = }\frac{1}{\cos x} + \frac{1}{\cos x-\sin x} \ge \frac{4}{\sin x + \cos x - \sin x} = \frac{4}{\sin x}$$ For the given range of x, we have $1> 2 \sin^2 x$ So $$\frac{4}{\sin x} > \frac{4}{\sin x} \times 2 \sin^2 x = 8 \sin x$$ and we are done

Need a solution to this Integration problem How to evaluate:$\displaystyle\int_{0}^{r}\frac{x^4}{(x^2+y^2)^{\frac{3}{2}}}dx$ I have tried substituting $x =y\tan\ A$, but failed.

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Note that $\ds{I \equiv \int_{0}^{r}{x^{4} \over \pars{x^{2} + y^{2}}^{3/2}}\,\dd x = y^{2}\ \overbrace{\int_{0}^{R}{x^{4} \over \pars{x^{2} + 1}^{3/2}} \,\dd x}^{\ds{\equiv\ \,\mathcal{I}}}\ \,,\qquad R \equiv {r \over \verts{y}}}$. \begin{align} \,\mathcal{I} & = \int_{0}^{R}{x^{4} \over \pars{x^{2} + 1}^{3/2}}\,\dd x = \int_{0}^{R}{\bracks{\pars{x^{2} + 1} - 1}^{2}\over \pars{x^{2} + 1}^{3/2}} \,\dd x \\[4mm] & = \int_{0}^{R}\pars{x^{2} + 1}^{1/2}\,\,\dd x - 2\int_{0}^{R}\pars{x^{2} + 1}^{-1/2}\,\dd x + \int_{0}^{R}\pars{x^{2} + 1}^{-3/2}\,\,\dd x \end{align} With $\ds{x \equiv \tan\pars{t}}$: \begin{align} &\int\bracks{\sec\pars{t} - 2\cos\pars{t} + \cos^{3}\pars{t}}\,\dd t = \ln\pars{\sec\pars{t} + \tan\pars{t}} - \int\bracks{1 + \sin^{2}\pars{t}}\cos\pars{t}\,\dd t \\[4mm] = &\ \ln\pars{\sec\pars{t} + \tan\pars{t}} - \sin\pars{t} - {1 \over 3}\,\sin^{3}\pars{t} \\[4mm] & = \ln\pars{\root{x^{2} + 1} + x} - {x \over \root{x^{2} + 1}} - {1 \over 3}\,{x^{3} \over \pars{x^{2} + 1}^{3/2}} \end{align} \begin{align} \color{#f00}{\,\mathcal{I}} & = \color{#f00}{\ln\pars{\root{R^{2} + 1} + R} - {R \over \root{R^{2} + 1}} - {1 \over 3}\,{R^{3} \over \pars{R^{2} + 1}^{3/2}}} \end{align} \begin{align} \color{#f00}{I} & = \color{#f00}{y^{2}\ln\pars{\root{r^{2} + y^{2}} + r} - y^{2}\ln\pars{\verts{y}} - {y^{2}r \over \root{r^{2} + y^{2}}} - {1 \over 3}\,{y^{2}r^{3} \over \pars{r^{2} + y^{2}}^{3/2}}} \end{align}

How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$ How to find ?$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$$ I tried using the substitution $x^2=z$.But that did not help much.

By setting $x^2=z$ we are left with: $$ \frac{1}{2}\int\frac{z-1}{z^2\sqrt{2z^2-2z+1}}\,dz=C+\frac{\sqrt{2z^2-2z+1}}{2z}.$$

Finding the coefficient of $x^{50}$ in $\frac{(x-3)}{(x^2-3x+2)}$ First, the given answer is: $$-2 + (\frac{1}{2})^{51}$$ I have tried solving the problem as such: $$[x^{50}]\frac{(x-3)}{(x^2-3x+2)} = [x^{50}]\frac{2}{x-1} + [x^{50}]\frac{-1}{x-2}$$ $$ = 2[x^{50}](x-1)^{-1} - [x^{50}](x-2)^{-1}$$ $$=2\binom{-1}{50}-\binom{-1}{50} = \binom{-1}{50} = \binom{50}{50} = 1$$ which is different from the correct answer. Can anyone tell me what I'm doing wrong here? Edit: As Did mentions in the comments, $$2[x^{50}](x-1)^{-1} = -2\binom{-1}{50} = -2 \neq 2\binom{-1}{50}$$ Also, $$- [x^{50}](x-2)^{-1} = [x^{50}]\frac{1}{2-x} = \frac{1}{2}[x^{50}]\frac{1}{1-\frac{x}{2}}$$ $$ = \frac{1}{2}[x^{50}](1-\frac{x}{2})^{-1} = \frac{1}{2}\binom{-1}{50}(\frac{-1}{2})^{50}$$ $$=(\frac{1}{2})^{51}$$ Both added together gives the correct answer, which is the correct solution using the original method.

The beginning looks good, but I do not see how you justify the last line. I would use the geometric series instead: $$\begin{align*}\frac{x-3}{x^2-3x+2} &= \frac{2}{x-1} - \frac{1}{x-2}\\ &= -2\frac{1}{1-x} + \frac{1}{2}\frac{1}{1-\frac 12 x} \\ & = -2 \sum_{n=0}^\infty x^n + \frac{1}{2}\sum_{n=0}^\infty \frac{x^n}{2^n} \end{align*}$$ From that we can easily see that $$[x^{50}]\frac{x-3}{x^2-3x+2} = -2+\frac{1}{2^{51}}$$

What is the range of $λ$? Suppose $a, b, c$ are the sides of a triangle and no two of them are equal. Let $λ ∈ IR$. If the roots of the equation $x^ 2 + 2(a + b + c)x + 3λ(ab + bc + ca) = 0$ are real, then what is the range of $λ$? I got that $$λ ≤\frac{ (a + b + c)^ 2} {3(ab + bc + ca)}$$ After that what to do?

For a triangle with sides $a,b,c$ by triangle inequality, we have $$|a-b|<c$$ Squaring both sides we get, $$(a-b)^2<c^2\tag{1}$$ Similarly, $$(b-c)^2<a^2\tag{2}$$ And $$(c-a)^2<b^2\tag{3}$$ Adding $(1),(2)$ and $(3)$, we get $$a^2+b^2+c^2 <2(ab+bc+ca) \Longleftrightarrow (a+b+c)^2 <4(ab+bc+ca)\tag{4}$$ From $(4)$, we get $\lambda <\dfrac{4}{3}$ To see why the upper bound is the supremum consider a degenerate triangle with sides $(a,a,c)$,(where $c \rightarrow 0$) and permutations. For example, when $(a,b,c)=(1,1, 0.01)$, then $\lambda \leq \dfrac{4.0401}{3.0600} \approx 1.321 $

Series with digammas (Inspired by a comment in answer https://math.stackexchange.com/a/699264/442.) corrected Let $\Psi(x) = \Gamma'(x)/\Gamma(x)$ be the digamma function. Show $$ \sum_{n=1}^\infty (-1)^n\left(\Psi\left(\frac{n+1}{2}\right) -\Psi\left(\frac{n}{2}\right)\right) = -1 $$ As noted, it agrees to many decimals. But care may be required since the convergence is only conditional. added Both solutions are good. But could use explanation for exchange of summation. Note that $$ 2\sum_{n=1}^\infty\sum_{m=0}^\infty \left(\frac{(-1)^n}{2m+n} -\frac{(-1)^n}{2m+n+1}\right) = 2\sum_{m=0}^\infty\sum_{n=1}^\infty \left(\frac{(-1)^n}{2m+n} -\frac{(-1)^n}{2m+n+1}\right) = -1 $$ is correct. But "Fibini" justification fails, since $$ \sum_{m=0}^\infty\sum_{n=1}^\infty \left|\frac{(-1)^n}{2m+n} -\frac{(-1)^n}{2m+n+1}\right| = \infty $$ Similarly in Random Variable's solution, the exchange $\sum_{n=1}^\infty \int_0^1 = \int_0^1 \sum_{n=1}^\infty$, although correct, cannot be justified by Fubini.

Using the integral representation $$\psi(s+1) = -\gamma +\int_{0}^{1} \frac{1-x^{s}}{1-x} \, dx ,$$ we get $$ \begin{align} \sum_{n=1}^{\infty} (-1)^{n} \left(\psi \left(\frac{n}{2} \right)- \psi \left(\frac{n+1}{2}\right) \right) &= \sum_{n=1}^{\infty} (-1)^{n} \int_{0}^{1} \frac{x^{(n+1)/2-1} - x^{n/2-1}}{1-x} \, dx \\ &= \int_{0}^{1} \frac{1}{1-x} \sum_{n=1}^{\infty} (-1)^{n} \left(x^{n/2-1/2} - x^{n/2-1} \right) \, dx \\ &= \int_{0}^{1} \frac{1}{1-x} \left(- \frac{1}{1+\sqrt{x}}+ \frac{1}{(1+\sqrt{x})\sqrt{x}} \right) \, dx \\ &= 2 \int_{0}^{1} \frac{u}{1-u^{2}} \left(- \frac{1}{1+u} + \frac{1}{(1+u)u} \right) \, du \\&= \int_{0}^{1} \frac{1-u}{(1-u^{2})(1+u)} \, du\\ &= 2 \int_{0}^{1} \frac{1}{(1+u)^{2}} \, du \\ &=1. \end{align}$$ EDIT: Daniel Fischer was kind enough to point out to me that we can use the dominated convergence theorem to justify interchanging the order of integration and summation. Specifically, we may write $$ \begin{align} \sum_{n=1}^{\infty} (-1)^{n} \int_{0}^{1} \frac{x^{(n+1)/2-1} - x^{n/2-1}}{1-x} \, dx &= \lim_{N \to \infty}\sum_{n=1}^{N} (-1)^{n} \int_{0}^{1} \frac{x^{(n+1)/2-1} - x^{n/2-1}}{1-x} \, dx \\ &=\lim_{N \to \infty}\sum_{n=1}^{N} (-1)^{n+1} \int_{0}^{1} \frac{x^{n/2-1}} {1+\sqrt{x}} \, dx \\ &= \lim_{N \to \infty} \int_{0}^{1} \sum_{n=1}^{N} (-1)^{n+1} \frac{x^{n/2}} {x(1+\sqrt{x})} \, dx. \end{align}$$ But for $x \in [0,1]$, $$\left|\sum_{n=1}^{N} (-1)^{n+1} \frac{x^{n/2}} {x(1+\sqrt{x})} \right| \le \frac{\sqrt{x}}{x(1+\sqrt{x})},$$ which is integrable on $[0,1]$. This, combined with the fact that $$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{n/2}} {x(1+\sqrt{x})} $$ converges pointwise on $[0, 1)$, allows us to move the limit inside the integral.

How to turn the reflection about $y=x$ into a rotation. If we reflect $(x,y)$ about $y=x$ then we get $(y,x)$. And because $x^2+y^2=y^2+x^2$ this can also be represented by a rotation. Using this we get: $$(x,y)•(y,x)=2xy=(x^2+y^2)\cos (\theta)$$ Hence $\theta=\arccos (\frac{2xy}{x^2+y^2})$ So using complex numbers we rotate $(x,y)$ clockwise/counterclockwise an angle of $\arccos (\frac{2xy}{x^2+y^2})$ depending on which side of $y=x$ the point is, i.e. Depending on wether or not $y > x$. My question : How can we know the angle is $\arccos (\frac{2xy}{x^2+y^2})$ without using the already known result that $(x,y)$ rotated about $x=y$ is $(y,x)$?

If $\theta$ is the angle between the x-axis and the line from 0 to $(x,y)$, then $\theta = \arccos(\frac{y}{\sqrt{x^2+y^2}})$. We reflect about a line with angle $\frac{\pi}{4}$, so the angle between $(x,y)$ and the line $x=y$ is $\frac{\pi}{4} - \theta$. The angle between $(x,y)$ and the reflected point will be double the previous angle, or $\frac{\pi}{2} -2 \theta$. Compute the cosine of $\frac{\pi}{2} - 2\theta$ and you will get $\frac{2xy}{x^2+y^2}$.

A rectangle with perimeter of 100 has area at least of 500, within what bounds must the length of the rectangle lie? Problem The problem states that there is a rectangle that has a perimeter of $100$ and an area of at least $500$ and it asks for the bounds of the length which can be given in interval notation or in the <> (greater than or less than) signs My steps and thought process So I set some few equations 1)$2x+2y=100$ which becomes $x+y=50$ 2)$xy \geq 500$ 3)I then made $y = 50-x$ so that I can substitute it into equation #2 to get: $50x-x^2 \geq 500$ which eventually got me $0 \geq x^2-50x+500$ and this is where I got stuck.

You are correct that $x + y = 50$ and that $xy \geq 500$. We can solve the inequality by completing the square. \begin{align*} xy & \geq 500\\ x(50 - x) & \geq 500\\ 50x - x^2 & \geq 500\\ 0 & \geq x^2 - 50x + 500\\ 0 & \geq (x^2 - 50x) + 500\\ 0 & \geq (x^2 - 50x + 625) - 625 + 500\\ 0 & \geq (x - 25)^2 - 125\\ 125 & \geq (x - 25)^2\\ \sqrt{125} & \geq |x - 25|\\ 5\sqrt{5} & \geq |x - 25| \end{align*} Hence, \begin{align*} -5\sqrt{5} & \leq x - 25 \leq 5\sqrt{5}\\ 25 - 5\sqrt{5} & \leq x \leq 25 + 5\sqrt{5} \end{align*} Alternatively, note that since the perimeter of the rectangle is $100$ units, the average side length is $100/4 = 25$ units. Hence, we can express the lengths of adjacent sides as $25 + k$ and $25 - k$. Hence, the area is $$A(k) = (25 + k)(25 - k) = 625 - k^2$$ The requirement that the area must be at least $500$ square units means \begin{align*} 625 - k^2 & \geq 500\\ 125 & \geq k^2\\ 5\sqrt{5} & \geq |k|\\ 5\sqrt{5} & \geq k \geq -5\sqrt{5} \end{align*} Thus, for the area of the rectangle to be at least $500$ square units, the length of the longer side of the rectangle must be at most $25 + 5\sqrt{5}$ units and the length of the shorter side of the rectangle must be at least $25 - 5\sqrt{5}$ units. Also, note that the maximum area of $625$ square units occurs when $k = 0$, that is, when the rectangle is a square.

Is $77!$ divisible by $77^7$? Can $77!$ be divided by $77^7$? Attempt: Yes, because $77=11\times 7$ and $77^7=11^7\times 7^7$ so all I need is that the prime factorization of $77!$ contains $\color{green}{11^7}\times\color{blue} {7^7}$ and it does. $$77!=77\times...\times66\times...\times55\times...\times44\times...\times33\times...\times22\times...\times11\times...$$ and all this $\uparrow$ numbers are multiples of $11$ and there are at least $7$ so $77!$ contains for sure $\color{green}{11^7}$ And $77!$ also contains $\color{blue} {7^7}:$ $$...\times7\times...\times14\times...\times21\times...\times28\times...\times35\times...42\times...49\times...=77!$$ I have a feeling that my professor is looking for other solution.

If $p$ is a prime number, the largest number $n$ such that $p^n \mid N!$ is $\displaystyle n = \sum_{i=1}^\infty \left \lfloor \dfrac{N}{p^i}\right \rfloor$. Note that this is really a finite series since, from some point on, all of the $\left \lfloor \dfrac{N}{p^i}\right \rfloor$ are going to be $0$. There is also a shortcut to computing $\left \lfloor \dfrac{N}{p^{i+1}}\right \rfloor$ because it can be shown that $$\left \lfloor \dfrac{N}{p^{i+1}}\right \rfloor = \left \lfloor \dfrac{\left \lfloor \dfrac{N} {p^i} \right \rfloor}{p}\right \rfloor$$ For $77!$, we get $\qquad \left \lfloor \dfrac{77}{11}\right \rfloor = 7$ $\qquad \left \lfloor \dfrac{7}{11}\right \rfloor = 0$ So $11^7 \mid 77!$ and $11^8 \not \mid 77!$ Since $7 < 11$, it follows immediatley that $7^7 \mid 77!$. But we can also compute $\qquad \left \lfloor \dfrac{77}{7}\right \rfloor = 11$ $\qquad \left \lfloor \dfrac{11}{7}\right \rfloor = 1$ $\qquad \left \lfloor \dfrac{1}{7}\right \rfloor = 0$ So $7^{12} \mid 77!$ and $7^{13} \not \mid 77!$ It follows that $77^7 = 7^{7} 11^7 \mid 77!$. Added 3/9/2018 The numbers are small enough that we can show this directly Multiples of powers of $7$ between $1$ and $77$ \begin{array}{|r|ccccccccccc|} \hline \text{multiple} & 7 & 14 & 21 & 28 & 35 & 42 & 49 & 56 & 63 & 70 & 77 \\ \hline \text{power} & 7 & 7 & 7 & 7 & 7 & 7 & 7^2 & 7 & 7 & 7 & 7 \\ \hline \end{array} So $7^{12} \mid 77!$. Multiples of powers of $11$ between $1$ and $77$ \begin{array}{|r|ccccccc|} \hline \text{multiple} & 11 & 22 & 33 & 44 & 55 & 66 & 77\\ \hline \text{power} & 11 & 11 & 11 & 11 & 11 & 11 & 11 \\ \hline \end{array} So $11^7 \mid 77!$. Hence $77^7 \mid 7^{12}11^7 \mid 77!$.

How to find the union&intersection of two lines by their equations? I will try to be as clear as possible concerning my confusion, and I will use some examples(several ones). Case number 1. Assume two equations(in cartesian form) of two planes. $2x+2y-5z+2=0$ and $x-y+z=0$ Now,we need to find their vectors. For the first on, we get: {1,-1,0}, {0,5/2,1} and {1,0,2/5}. For the second equation, we get: {1,1,0}, {0,1,1},and {1,0,-1}. Now, I have a hard time understanding how I have to figure out where is their intersection and their union? Case 2: Assume the one of the previous planes $x-y+z=0$ and the line $x-y=0$. How do I find the intersection and union of these two?

First of all: $Ax+By+Cz+D=0$ is plane equation. Case 1: Intersection of to planes is line. To find equation of that line you have to solve system of equations: $$ 2x+2y-5z+2=0\\ x-y+z=0 \Rightarrow x=y-z \\ $$ If we substitute second equation into first we got $$ 2(y-z)+2y-5z+2=0 \Rightarrow 4y-7z+2=0 \Rightarrow y=\frac{7z-2}{4} $$ and for x $$ x=\frac{7z-2}{4}-z=\frac{3z-2}{4} $$ Now we have parametric equation of the intersection $$ x=-\frac{1}{2}+\frac{3}{4}z\\ y=-\frac{1}{2}+\frac{7}{4}z\\ z=0+1z $$ It's line equation which can be written in another form $$ \frac{x+\frac{1}{2}}{\frac{3}{4}}=\frac{y+\frac{1}{2}}{\frac{7}{4}}=\frac{z}{1} $$ If we multiply denominators by 4 we get shorter equation $$ \frac{x+\frac{1}{2}}{3}=\frac{y+\frac{1}{2}}{7}=\frac{z}{4} $$ Case 2: $x-y=0$ is not line equation - it's plane equation $x-y+0z=0$. So this case is also intersection of two planes - like case 1. Solution: $$ x-y+z=0 \Rightarrow x=y-z\\ x-y=0 \Rightarrow y-z-y=0 \Rightarrow z=0 $$ Using $x=y-z$ we have $x=y$. So, intersection is a line whose equation is $$ x=0+1y\\ y=0+1y\\ z=0+0y $$ or in another form $$ \frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{0} $$

Double integration over a general region $\iint x^2 +2y$ bound by $y=x$ $y=x^3$ $x \geq 0$ this is either a type I or type II since the bounds are already nicely given for a type I, I integrated it as a type I: Finding the bounds: $x^3=x \to x^3-x=0 \to x(x^{2}-1)= 0 \to x=0, x=\pm1$ Since $-1\lt 0$ my bounds for $x$ are $[0,1]$ and since $x \gt x^3$, $x$ is my upper bound for dy. $\int_{0}^{1} \int_{x^3}^{x} x^2+2y$ $dydx$ $\int_{x^3}^{x}$ $x^2+y^2$ $\Big\vert_{x}^{x3}$dy $=$ $(2x^2) - (-x^2+x^6)$ $\int_{0}^{1} 3x^2+x^6 dx=$ $x^{3}+\frac{1}{7}x^7 \Big\vert_{0}^{1} \to 1 -\frac{1}{7}$ $=\frac{6}{7}$

Note that $x^3\lt x$ in the interval $(0,1)$. (A picture always helps in this kind of problem.) So $y$ travels from $y=x^3$ to $y=x$. One can see without checking details that the answer $-\frac{4}{21}$ cannot be right. Your integrand is positive in the region, so the answer must be positive.

Proving that $2^{2a+1}+2^a+1$ is not a perfect square given $a\ge5$ I am attempting to solve the following problem: Prove that $2^{2a+1}+2^a+1$ is not a perfect square for every integer $a\ge5$. I found that the expression is a perfect square for $a=0$ and $4$. But until now I cannot coherently prove that there are no other values of $a$ such that the expression is a perfect square. Any help would be very much apreciated.

I will assume that $a \ge 1$ and show that the only solution to $2^{2a+1}+2^a+1 = n^2$ is $a=4, n=23$. This is very non-elegant but I think that it is correct. I just kept charging forward, hoping that the cases would terminate. Fortunately, it seems that they have. If $2^{2a+1}+2^a+1 = n^2$, then $2^{2a+1}+2^a = n^2-1$ or $2^a(2^{a+1}+1) = (n+1)(n-1)$. $n$ must be odd, so let $n = 2^uv+1$ where $v $ is odd and $u \ge 1$. Then $2^a(2^{a+1}+1) = (2^uv+1+1)(2^uv+1-1) = 2^u v(2^u v+2) = 2^{u+1} v(2^{u-1} v+1) $. If $u \ge 2$, then $a = u+1$ and $2^{a+1}+1 =v(2^{u-1} v+1) $ or $2^{u+2}+1 =v(2^{u-1} v+1) =v^22^{u-1} +v $. If $v \ge 3$, the right side is too large, so $v = 1$. But this can not hold, so $u = 1$. Therefore $2^a(2^{a+1}+1) = 2^{2} v( v+1) $ so that $a \ge 3$. Let $v = 2^rs-1$ where $s$ is odd and $r \ge 1$. Then $2^{a-2}(2^{a+1}+1) = v( 2^rs) $ so $a-2 = r$ and $2^{a+1}+1 = vs \implies 2^{r+3}+1 = vs = (2^rs-1)s = 2^rs^2-s $. Therefore $s+1 =2^rs^2-2^{r+3} =2^r(s^2-8) \ge 2(s^2-8) \implies 2s^2-s \le 17$ so $s = 1$ or $3$. If $s = 1$, then $2^{r+3}+1 =2^r-1 $ which can not be. If $s = 3$ then $2^{r+3}+1 =9\cdot 2^r-3 \implies 4 =9\cdot 2^r-2^{r+3} =2^r \implies r = 2, v = 11, a = 4$ and $2^9+2^4+1 =512+16+1 =529 =23^2 $.

Finding the basis and dimension of a subspace of the vector space of 2 by 2 matrices I am trying to find the dimension and basis for the subspace spanned by: $$ \begin{bmatrix} 1&-5\\ -4&2 \end{bmatrix}, \begin{bmatrix} 1&1\\ -1&5 \end{bmatrix}, \begin{bmatrix} 2&-4\\ -5&7 \end{bmatrix}, \begin{bmatrix} 1&-7\\ -5&1 \end{bmatrix} $$ in the vector space $M_{2,2}$. I don't really care about the answer, I am just hoping for an efficient algorithm for solving problems like this for matrices. I am not sure how to account for interdependence within the matrices. My instinct as of now is to find the maximum restriction imposed by the matrices. It is clear that the $1$ in position $a_{1,1}$ in each matrix will allow me to get any number in that position, so one vector in the basis will be: $$ \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix} $$ But depending on which of the matrices I scale, I have restrictions on the other entries. So I don't think I can include that matrix in my basis. It just occurred as I was writing this that I could maybe just think about these as $4$ by $1$ vectors and proceed as usual. Is there any danger in doing so?

Inputs $$ \alpha = \left( \begin{array}{rr} 1 & -5 \\ -4 & 2 \end{array} \right), \qquad \beta = \left( \begin{array}{rr} 1 & 1 \\ -1 & 5 \end{array} \right), \qquad \gamma = \left( \begin{array}{rr} 2 & -4 \\ -5 & 7 \end{array} \right), \qquad \delta = \left( \begin{array}{rr} 1 & -7 \\ -5 & 1 \end{array} \right) $$ Find the basis for these matrices. Matrix of row vectors As noted by @Bernard, compose a matrix of row vectors. Flatten the matrices in this manner $$ \left( \begin{array}{rr} 1 & -5 \\ -4 & 2 \end{array} \right) \quad \Rightarrow \quad \left( \begin{array}{crrc} 1 & -5 & -4 & 2 \end{array} \right) $$ Compose the matrix $$ \mathbf{A} = \left( \begin{array}{crrr} 1 & -5 & -4 & 2 \\\hline 1 & 1 & -1 & 5 \\\hline 2 & -4 & -5 & 7 \\\hline 1 & -7 & -5 & 1 \end{array} \right) $$ Row reduction Column 1 $$ \left( \begin{array}{rccc} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ -2 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ \end{array} \right) % \left( \begin{array}{crrc} 1 & -5 & -4 & 2 \\ 1 & 1 & -1 & 5 \\ 2 & -4 & -5 & 7 \\ 1 & -7 & -5 & 1 \\ \end{array} \right) % = % \left( \begin{array}{crrr} \boxed{1} & -5 & -4 & 2 \\ 0 & 6 & 3 & 3 \\ 0 & 6 & 3 & 3 \\ 0 & -2 & -1 & -1 \\ \end{array} \right) % $$ Column 2 $$ \left( \begin{array}{rccc} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ -2 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ \end{array} \right) % \left( \begin{array}{crrr} \boxed{1} & -5 & -4 & 2 \\ 0 & 6 & 3 & 3 \\ 0 & 6 & 3 & 3 \\ 0 & -2 & -1 & -1 \\ \end{array} \right) % = % \left( \begin{array}{cccc} \boxed{1} & 0 & -\frac{3}{2} & \frac{9}{2} \\ 0 & \boxed{1} & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) % $$ The fundamental rows are marked by the unit pivots. Solution The basis is $$ \mathcal{B} = \left\{ \alpha, \, \beta \right\} = \left\{ \left( \begin{array}{rr} 1 & -5 \\ -4 & 2 \end{array} \right), \ \left( \begin{array}{rr} 1 & 1 \\ -1 & 5 \end{array} \right) \right\} $$

How many integers $\leq N$ are divisible by $2,3$ but not divisible by their powers? How many integers in the range $\leq N$ are divisible by both $2$ and $3$ but are not divisible by whole powers $>1$ of $2$ and $3$ i.e. not divisible by $2^2,3^2, 2^3,3^3, \ldots ?$ I hope by using the inclusion–exclusion principle one may derive such a formula and part of the formula has a form $$ N-\left[\frac{N}{2} \right]+\left[\frac{N}{2^2} \right]-\left[\frac{N}{2^3} \right]+\cdots -\left[\frac{N}{3} \right]+\left[\frac{N}{3^2} \right]-\left[\frac{N}{3^3} \right]+\cdots+\left[\frac{N}{2 \cdot 3} \right]+\text{some terms like as $\pm \left[\frac{N}{2^i \cdot 3^j} \right]$} $$ Question. What is the exact sign for a term $ \left[\frac{N}{2^i \cdot 3^j} \right]$?

The rules permit all numbers divisible by $6$, but excluding those also divisble by $4$ or $9$. This is given by: $$\lfloor\frac{N}{6}\rfloor-\lfloor\frac{N}{12}\rfloor-\lfloor\frac{N}{18}\rfloor+\lfloor\frac{N}{36}\rfloor$$ Firstly - Start by enumerating number of numbers divisible by 6. Next term: Remove numbers divisible by 6 and 4. These are all numbers divisible by 12 since the intersection of prime factors is $3\times2^2$ Next term: Remove numbers divisible by 6 and 9. These are all numbers divisible by 18 since the intersection of prime factors is $2\times3^2$ Next term: Add back in the multiples of 36 since these are the numbers we have deducted twice; divisible by 6, 4, and 9. Intersection of prime factors is $2^2\times3^2$.

$\lfloor x\rfloor \cdot \lfloor x^2\rfloor = \lfloor x^3\rfloor$ means that $x$ is close to an integer Suppose $x>30$ is a number satisfying $\lfloor x\rfloor \cdot \lfloor x^2\rfloor = \lfloor x^3\rfloor$. Prove that $\{x\}<\frac{1}{2700}$, where $\{x\}$ is the fractional part of $x$. My heuristic is that $x$ needs to be "small": i.e. as close to $30$ as possible to get close to the upper bound on $\{x\}$, but I'm not sure how to make this a proof.

Let $\lfloor x \rfloor =y$ and $\{x\}=b$ Then $\lfloor x\rfloor \cdot \lfloor x^2\rfloor = \lfloor x^3\rfloor =y\lfloor y^2+2by+b^2 \rfloor= \lfloor y^3+3y^2b+3yb^2+b^3\rfloor$ One way this can happen is that $b$ is small enough that all the terms including $b$ are less than $1$, which makes both sides $y^3$. This requires $3y^2b \lt 1$, which gives $b \lt \frac 1{2700}$ as required. Now you have to argue that if $2by+b^2 \ge 1$ the right side will be too large.

A closed form for $1^{2}-2^{2}+3^{2}-4^{2}+ \cdots + (-1)^{n-1}n^{2}$ Please look at this expression: $$1^{2}-2^{2}+3^{2}-4^{2} + \cdots + (-1)^{n-1} n^{2}$$ I found this expression in a math book. It asks us to find a general formula for calculate it with $n$. The formula that book suggests is this: $$-\frac{1}{2}\times (-1)^{n} \times n(n+1)$$ Would you mind explaining to me how we get this formula?

We wish to show that $$ 1^{2}-2^{2}+3^{2}-4^{2} + \dotsb + (-1)^{n-1} n^{2}= (-1)^{n+1}\frac{n(n+1)}{2}\tag{1} $$ To do so, induct on $n$. The base case $n=1$ is simple to verify. Now, suppose that $(1)$ holds. Then \begin{align*} 1^{2}-2^{2}+3^{2}-4^{2} + \dotsb + (-1)^{n} (n+1)^{2} &= (-1)^{n+1}\frac{n(n+1)}{2}+(-1)^{n} (n+1)^{2} \\ &= (-1)^n\left\{ -\frac{1}{2}\,n^2-\frac{1}{2}\,n+n^2+2\,n+1 \right\} \\ &= (-1)^{n+2}\left\{ \frac{1}{2}\,n^2+\frac{3}{2}\,n+1 \right\} \\ &= (-1)^{n+2}\frac{1}{2}\left\{ n^2+3\,n+2 \right\} \\ &= (-1)^{n+2}\frac{(n+1)(n+2)}{2}\\ \end{align*} This closes the induction.

Prove That If $(a + b)^2 + (b + c)^2 + (c + d)^2 = 4(ab + bc + cd)$ Then $a=b=c=d$ If the following equation holds $$(a + b)^2 + (b + c)^2 + (c + d)^2 = 4(ab + bc + cd)$$ Prove that $a$,$b$,$c$,$d$ are all the same. What I did is I let $a$,$b$,$c$,$d$ all equal one number. Then I substituted and expanded. I'm sort of proud of my self (first proof I done). I'm wondering, is there another way? (I'm teaching my self maths and I'm only a humble precalc student)

Consider the following steps $$\begin{align} (a + b)^2 + (b + c)^2 + (c + d)^2 &= 4(ab + bc + cd) \\ \left[ (a + b)^2-4ab \right] + \left[ (b + c)^2-4bc \right] + \left[ (c + d)^2-4cd \right] &=0 \\ \left[ a^2+b^2+2ab-4ab \right] + \left[ b^2+c^2+2bc-4bc \right] + \left[ c^2+d^2+2cd-4cd \right] &=0 \\ \left[ a^2+b^2-2ab \right] + \left[ b^2+c^2-4bc \right] + \left[ c^2+d^2-4cd \right] &=0 \\ (a-b)^2 + (b-c)^2 + (c-d)^2 &=0 \end{align}$$ and the sum of three positive numbers is zero if and only if they are all zero. So you will get $$a=b=c=d$$

Proof of the square root inequality $2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}$ I stumbled on the following inequality: For all $n\geq 1,$ $$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}.$$ However I cannot find the proof of this anywhere. Any ideas how to proceed? Edit: I posted a follow-up question about generalizations of this inequality here: Square root inequality revisited

\begin{align*} 2\sqrt{n+1}-2\sqrt{n} &= 2\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})} \\ &= 2\frac{1}{(\sqrt{n+1}+\sqrt{n})} \\ &< \frac{2}{2\sqrt{n}} \text{ since } \sqrt{n+1} > \sqrt{n}\\ &=\frac{1}{\sqrt{n}} \end{align*} Similar proof for the other inequality.

Consider the function $f(x) = x^2 + 4/x^2$ a) Find$f ^\prime(x)$ b) Find the values of $x$ at which the tangent to the curve is horizontal. So far I have this... a) $f^\prime(x) = 2x + (0)(x^2)-(4)\dfrac{2x}{(x^2)^2}$ $= 2x - \dfrac{8x}{x^4}$ $= \dfrac{2x^5 - 8x}{x^4}$ $= \dfrac{2(x^4 - 4)}{x^3}$ I believe I derived this correctly. But i am not sure how to do part b). I know the horizontal slope $= 0$ but when i solve it i get $2(x^4 - 4)$ and don't know how to go from there. `

Your derivative is correct. You could have saved yourself some work by using the power rule. \begin{align*} f(x) & = x^2 + \frac{4}{x^2}\\ & = x^2 + 4x^{-2} \end{align*} Using the power rule yields \begin{align*} f'(x) & = 2x^1 - 2 \cdot 4x^{-3}\\ & = 2x - 8x^{-3}\\ & = 2x - \frac{8}{x^3} \end{align*} which is equivalent to your expression $$f'(x) = \frac{2(x^4 - 4)}{x^3}$$ To find the values of $x$ at which the tangent line is horizontal, set $f'(x) = 0$, which yields \begin{align*} \frac{2(x^4 - 4)}{x^3} & = 0\\ 2(x^4 - 4) & = 0\\ x^4 - 4 & = 0\\ (x^2)^2 - 2^2 & = 0\\ (x^2 + 2)(x^2 - 2) & = 0\\ \end{align*} Setting each factor equal to zero yields \begin{align*} x^2 + 2 & = 0 & x^2 - 2 & = 0\\ x^2 & = -2 & x^2 & = 2\\ x & = \pm i\sqrt{2} & x & = \pm\sqrt{2} \end{align*} Since there can only be a tangent at real values of $x$, we conclude that the only horizontal tangents of the graph occur at $x = \pm\sqrt{2}$.

Find the values of $b$ for which the equation $2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$ has only one solution Find the values of 'b' for which the equation $$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$$ has only one solution. =$$-2/2\log_{5}(bx+28)=-\log_5(12-4x-x^2)$$ My try: After removing the logarithmic terms I get the quadratic $x^2+x(b+4)+16=0$ Putting discriminant equal to $0$ I get $b={4,-12}$ But $-12$ cannot be a solution as it makes $12-4x-x^2$ negative so I get $b=4$ as the only solution. But the answer given is $(-\infty,-14]\cup{4}\cup[14/3,\infty)$.I've no idea how.Help me please.

You have $$x^2+(4+b)x+16=0\tag1$$ This is correct. However, note that when we solve $$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$$ we have to have $$bx+28\gt 0\quad\text{and}\quad 12-4x-x^2\gt 0,$$ i.e. $$bx\gt -28\quad\text{and}\quad -6\lt x\lt 2\tag2$$ Now, from $(1)$, we have to have $(4+b)^2-4\cdot 16\geqslant 0\iff b\leqslant -12\quad\text{or}\quad b\geqslant 4$. Case 1 : $b\lt -14$ $$(2)\iff -6\lt x\lt -\frac{28}{b}$$ Let $f(x)=x^2+(4+b)x+16$. Then, since the equation has only one solution, we have to have $$f(-6)f\left(-\frac{28}{b}\right)\lt 0\iff b\lt -14$$ So, in this case, $b\lt -14$. Case 2 : $-14\leqslant b\leqslant -12$ or $4\leqslant b\lt \frac{14}{3}$ $$(2)\iff -6\lt x\lt 2$$ $b=4$ is sufficient, and $b=-12$ is not sufficient. For $b\not=4,-12$, $$f(-6)f(2)\lt 0\iff b\lt -14\quad\text{or}\quad b\gt \frac{14}{3}$$ So, in this case, $b=4$. Case 3 : $b\geqslant \frac{14}{3}$ $$(2)\iff -\frac{28}{b}\lt x\lt 2$$ $b=\frac{14}{3}$ is sufficient. For $b\gt\frac{14}{3}$, $$f\left(-\frac{28}{b}\right)f(2)\lt 0\iff b\gt \frac{14}{3}$$ So, in this case, $b\geqslant 14/3$. Therefore, the answer is $$\color{red}{(-\infty,-14)\cup{4}\cup\bigg[\frac{14}{3},\infty\bigg)}$$

What are the constraints on $\alpha$ so that $AX=B$ has a solution? I found the following problem and I'm a little confused. Consider $$A= \left( \begin{array}{ccc} 3 & 2 & -1 & 5 \\ 1 & -1 & 2 & 2\\ 0 & 5 & 7 & \alpha \end{array} \right)$$ and $$B= \left( \begin{array}{ccc} 0 & 3 \\ 0 & -1 \\ 0 & 6 \end{array} \right)$$ What are the constraints on $\alpha$ so that the matrix equation $AX=B$ has solution? Since neither $A$ nor $B$ are square, I can't get their inverses. Is the problem wrong?

Ignoring the fourth column, notice that $$\begin{pmatrix} 3 & 2 & -1 \\ 1 & -1 & 2\\ 0 & 5 & 7 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}$$ and $$\begin{pmatrix} 3 & 2 & -1 \\ 1 & -1 & 2\\ 0 & 5 & 7 \end{pmatrix} \begin{pmatrix} \frac15 \\ \frac65 \\ 0 \end{pmatrix}=\begin{pmatrix}3 \\ -1 \\ 6 \end{pmatrix}.$$ Can you comment on whether $\alpha$ influence existence of solutions? Are you able to construct an $X$ for the original problem? Remark: In general, you might like to perform row operations on the system of equations. Tips: If $A_1$ is a matrix that consists of some columns of $A$ and $A_1$ is non-singular, the solution to $AX=B$ always exist. Do you see why?

Evaluate $\int\sin^{7}x\cos^4{x}\,dx$ $$\int \sin^{7}x\cos^4{x}\,dx$$ \begin{align*} \int \sin^{7}x\cos^4{x}\,dx&= \int(\sin^{2}x)^3 \cos^4{x}\sin x \,dx\\ &=\int(1-\cos^{2}x)^{3}\cos^4{x}\sin x\,dx,\quad u=\cos x, du=-\sin x\,dx\\ &=-\int(1-u^{2})^3u^4{x}\,du\\ &=-\int (1-3u^2+3u^4-u^6)u^4\,du\\ &=u^4-3u^6+3u^8-u^{10}\\ &=\frac{u^5}{5}-\frac{3u^7}{7}+\frac{3u^9}{9}-\frac{u^{11}}{11}+c \end{align*} So we get: $$\frac{\cos^5x}{5}-\frac{3\cos^7x}{7}+\frac{3\cos^{9}x}{9}-\frac{\cos^{11}x}{11}+c$$ Where did I got it wrong?

is it correct? No, it isn't. You have errors in the following part : $$=-\int(1-u^{2})^3u^4{x}du=-\int (1-3u^2+3u^4-u^6)u^4du$$ $$=u^4-3u^6+3u^8-u^{10}=\frac{u^5}{5}-\frac{3u^7}{7}+\frac{3u^9}{9}-\frac{u^{11}}{11}+c$$ They should be $$-\int(1-u^{2})^3u^4du=-\int (1-3u^2+3u^4-u^6)u^4du$$ $$=\int \left(-u^4+3u^6-3u^8+u^{10}\right)du=-\frac{u^5}{5}+\frac{3u^7}{7}-\frac{3u^9}{9}+\frac{u^{11}}{11}+c$$

How to solve this Sturm Liouville problem? $\dfrac{d^2\phi}{dx^2} + (\lambda - x^4)\phi = 0$ Would really appreciate a solution or a significant hint because I could find anything that's helpful in my textbook. Thanks!

Hint: Let $\phi=e^{ax^3}y$ , Then $\dfrac{d\phi}{dx}=e^{ax^3}\dfrac{dy}{dx}+3ax^2e^{ax^3}y$ $\dfrac{d^2\phi}{dx^2}=e^{ax^3}\dfrac{d^2y}{dx^2}+3ax^2e^{ax^3}\dfrac{dy}{dx}+3ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y=e^{ax^3}\dfrac{d^2y}{dx^2}+6ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y$ $\therefore e^{ax^3}\dfrac{d^2y}{dx^2}+6ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y+(\lambda-x^4)e^{ax^3}y=0$ $\dfrac{d^2y}{dx^2}+6ax^2\dfrac{dy}{dx}+((9a^2-1)x^4+6ax+\lambda)y=0$ Choose $9a^2-1=0$ , i.e. $a=\dfrac{1}{3}$ , the ODE becomes $\dfrac{d^2y}{dx^2}+2x^2\dfrac{dy}{dx}+(2x+\lambda)y=0$ Let $t=bx$ , Then $b^2\dfrac{d^2y}{dt^2}+\dfrac{2t^2}{b}\dfrac{dy}{dt}+\left(\dfrac{2t}{b}+\lambda\right)y=0$ $\dfrac{d^2y}{dt^2}+\dfrac{2t^2}{b^3}\dfrac{dy}{dt}+\left(\dfrac{2t}{b^3}+\dfrac{\lambda}{b^2}\right)y=0$ Choose $b^3=2$ , i.e. $b=\sqrt[3]2$ , the ODE becomes $\dfrac{d^2y}{dt^2}+t^2\dfrac{dy}{dt}+\left(t+\dfrac{\lambda}{\sqrt[3]4}\right)y=0$ Which relates to Heun's Triconfluent Equation.

Squeeze fractions with $a^n+b^n=c^n+d^n$ Let $0<x<y$ be real numbers. For which positive integers $n$ do there always exist positive integers $a,b,c,d$ such that $$x<\frac ab<\frac cd<y$$ and $a^n+b^n=c^n+d^n$? For $n=1$ this is true. Pick any $a,b$ such that $x<\frac ab<y$ -- this always exists by the density of the rationals. Since $\frac{a}{b}=\frac{ka}{kb}$ for any positive integer $k$, it suffices to choose $c=ka+1$ and $d=kb-1$. Since $\lim_{k\rightarrow\infty}\frac{ka+1}{kb-1}=\frac{a}{b}$, there exists $k$ such that $\frac{ka+1}{kb-1}<y$.

Partial answer I: if $x < 1 < y$, then we can find $a,b$ with $\frac{a}{b}, \frac{b}{a}$ arbitrarily close to one, satisfying the requirements for any $n$. Then it can be seen that that it suffices to prove the result for $y<1$ or $1<x$, since we have symmetry about 1 by inversion: $$x < \frac{a}{b} < \frac{c}{d} < y < 1 \iff 1 < \frac{1}{y} < \frac{d}{c} < \frac{b}{a} < \frac{1}{x}$$ Partial answer II: it can always be done for $n=2$. First suppose we can find $a, b, c, d$ with $x < \frac{a}{b} < \frac{c}{d} < y$ s.t. $a^2 + b^2 = j^2$ and $c^2 + d^2 = k^2$ for some integers $j, k$. Then we would have $(ka)^2 + (kb)^2 = (jc)^2 + (jd)^2$, and hence we would be done. So it suffices to prove that $S = \{ \frac{a}{b} \ | \ a, b\neq0 \in \mathbb{Z}, \exists c \in \mathbb{Z} \ \text{s.t.} \ a^2 + b^2 = c^2 \}$ is dense in $(0, 1)$. To see that this is the case, consult the identity giving Pythagorean triples: $$(m^2-n^2)^2 + (2mn)^2 = (m^2+n^2)^2;$$ $$v(m,n) := \frac{m^2-n^2}{2mn} = \frac{m}{2n} - \frac{n}{2m}$$ Given some small positive $\varepsilon$, we can take $n$ s.t. $\frac{1}{n} < \varepsilon$. Observe that the derivative of $v$ with respect to $m$ is always positive but strictly decreasing, going to $\frac{1}{2n}$ from above as $m$ goes to infinity. Thus for any positive $k$ we can infer that: $$\frac{1}{2n} < v(n+k+1,n)-v(n+k,n) \leq v(n+1,n)-v(n,n) = \frac{2n+1}{2n+2} \cdot \frac{1}{n} < \varepsilon$$ Since $v(n,n) = 0$, it follows that any value in $(0,1)$ is within $\varepsilon$ of a value in $\{ v(m, n) \ | \ m, n \in \mathbb{Z}, m > n \} \subset S$, so we are done.

Solve $\sec (x) + \tan (x) = 4$ $$\sec{x}+\tan{x}=4$$ Find $x$ for $0<x<2\pi$. Eventually I get $$\cos x=\frac{8}{17}$$ $$x=61.9^{\circ}$$ The answer I obtained is the only answer, another respective value of $x$ in $4$-th quadrant does not solve the equation, how does this happen? I have been facing the same problem every time I solved this kind of trigonometric equation.

Using $t$-formula Let $\displaystyle t=\tan \frac{x}{2}$, then $\displaystyle \cos x=\frac{1-t^2}{1+t^2}$ and $\displaystyle \tan x=\frac{2t}{1-t^2}$. Now \begin{align*} \frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2} &=4 \\ \frac{(1+t)^{2}}{1-t^2} &= 4 \\ \frac{1+t}{1-t} &= 4 \quad \quad (t\neq -1) \\ t &= \frac{3}{5} \\ \tan \frac{x}{2} &= \frac{3}{5} \\ x &=2\left( n\pi +\tan^{-1} \frac{3}{5} \right) \\ x &= 2\tan^{-1} \frac{3}{5} \quad \quad (0<x<2\pi) \end{align*}

How do I find the terms of an expansion using combinatorial reasoning? From my textbook: The expansion of $(x + y)^3$ can be found using combinatorial reasoning instead of multiplying the three terms out. When $(x + y)^3 = (x + y)(x + y)(x + y)$ is expanded, all products of a term in the first sum, a term in the second sum, and a term in the third sum are added. Terms of the form $x^3$, $x^2y$, $xy^2$, and $y^3$ arise. What does the bolded part mean? I found that if you find the possible combinations of $x$ and $y$, you can get $xxx = x^3$, $xxy = x^2y$, $xyy = xy^2$, $yyy = y^3$. Is this what it means?

Expanding $(x+y)(x+y)(x+y)$ amounts to adding up all the ways you can pick three factors to multiply together. For example, you could pick an $x$ from the first $(x+y)$, a $y$ from the second $(x+y)$, and another $x$ from the third $(x+y)$ to get $xyx=x^2 y$. You are right, the only possible products we can get are $x^3$, $x^2 y$, $xy^2$, and $y^3$. However, we do need to count how many ways to get each factor. For example there is only one way to get $x^3$ (pick $x$ from each $(x+y)$), but there are three ways to get $x^2 y$: $xxy$, $xyx$, and $yxx$. One way to count this is to realize that there are $3$ ways to choose which $(x+y)$ contributes a $y$ [and the rest will be $x$s]. Similar reasoning for $xy^2$ and $y^3$ shows that the expansion is $x^3 + 3x^2 y + 3 xy^2 + y^3$. In general, if you have $(x+y)^n$, the number of ways to obtain a product of the form $x^k y^{n-k}$ is the number of ways to choose $k$ of the $(x+y)$ factors from which to select an $x$. There are $\binom{n}{k}$ ways to make this choice. This proves the binomial theorem $(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k n^{n-k}$.