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Find $\sin^3 a + \cos^3 a$, if $\sin a + \cos a$ is known Given that $\sin \phi +\cos \phi =1.2$, find $\sin^3\phi + \cos^3\phi$. My work so far: (I am replacing $\phi$ with the variable a for this) $\sin^3 a + 3\sin^2 a *\cos a + 3\sin a *\cos^2 a + \cos^3 a = 1.728$. (This comes from cubing the already given statement with 1.2 in it.) $\sin^3 a + 3\sin a * \cos a (\sin a + \cos a) + \cos^3 a = 1.728$ $\sin^3 a + 3\sin a * \cos a (1.2) + \cos^3 a = 1.728$ $\sin^3 a + \cos^3 a = 1.728 - 3\sin a * \cos (a) *(1.2)$ Now I am stuck. What do I do next? Any hints?

Squaring $\sin a + \cos a = b$, $b^2 =\sin^2a+2\sin a \cos a + \cos^2 a = 1+2\sin a \cos a $, so $\sin a \cos a =(b^2-1)/2 $.

Finding value of 1 variable in a 3-variable $2^{nd}$ degree equation The question is: If $a,b,\space (a^2+b^2)/(ab-1)=q$ are positive integers, then prove that $q=5$. Also prove that for $q=5$ there are infinitely many solutions in $\mathbf N$ for $a$ and $b$. I simplified the equation as follows:-$$\frac {a^2+b^2}{ab-1}=q\\\begin{align}\\&=>\frac {2a^2+2b^2}{ab-1}=2q\\&=>\frac{a^2+b^2+2ab+a^2+b^2-2ab}{ab-1}=2q\\&=>(a+b)^2+(a-b)^2=2q(ab-1)\\&=>2(a+b)^2+2(a-b)^2=q(4ab-4)\\&=>2(a+b)^2+2(a-b)^2=q((a+b)^2-(a-b)^2-4)\end{align}$$Substituting $a+b=X$ and $a-b=Y$, we get $$2X^2+2Y^2=q(X^2-Y^2-4)\\\begin{align}&=>(q-2)X^2=(q+2)Y^2+4q\end{align}$$Now using the quadratic residues modulo $5$, I know that $X^2,Y^2\equiv0, \pm1(mod\space 5)$. But using this directly doesn't give the answer. So what to do after this? An answer without the use of co-ordinate geometry would be greatly appreciated as it seems there is a very good resemblance of the equation to a pair of hyperbolas which are symmetric with respect to the line $y=x$ but I don't understand co-ordinate geometry very well.

For such equations: $$\frac{x^2+y^2}{xy-1}=-t^2$$ Using the solutions of the Pell equation. $$p^2-(t^4-4)s^2=1$$ You can write the solution. $$x=-4tps$$ $$y=t(p^2+2t^2ps+(t^4-4)s^2)$$ It all comes down to the Pell equation - as I said. Considering specifically the equation: $$\frac{x^2+y^2}{xy-1}=5$$ Decisions are determined such consistency. Where the next value is determined using the previous one. $$p_2=55p_1+252s_1$$ $$s_2=12p_1+55s_1$$ You start with numbers. $(p_1;s_1) - (55 ; 12)$ Using these numbers, the solution can be written according to a formula. $$y=p^2+2ps+21s^2$$ $$x=3p^2+26ps+63s^2$$ If you use an initial $(p_1 ; s_1) - (1 ; 1)$ Then the solutions are and are determined by formula. $$y=s$$ $$x=\frac{p+5s}{2}$$ As the sequence it is possible to write endlessly. Then the solutions of the equation, too, can be infinite.

Help on proving a trigonometric identity involving cot and half angles Prove: $\cot\frac{x+y}{2}=-\left(\frac{\sin x-\sin y}{\cos x-\cos y}\right)$. My original idea was to do this: $\cot\frac{x+y}{2}$ = $\frac{\cos\frac{x+y}{2}}{\sin\frac{x+y}{2}}$, then substitute in the formulas for $\cos\frac{x+y}{2}$ and $\sin\frac{x+y}{2}$, but that became messy very quickly. Did I have the correct original idea, but overthink it, or is there any easier way? Hints only, please.

Try using $ \displaystyle \sin{x} - \sin{y} = \sin{\frac{2x}{2}} - \sin{\frac{2y}{2}} = 2 \frac{\sin \left( {\frac{x+y}{2} + \frac{x-y}{2}} \right) - \sin \left( {\frac{x+y}{2} - \frac{x-y}{2}} \right)}{2}$ and then the product to sum-formula for sine, i.e. $ \displaystyle \frac {\sin \left({x + y}\right) - \sin \left({x - y}\right)}{2} = \cos{x} \sin{y} $. You can prove it geometrically as well.

What is the coefficient of $x^{18}$ in the expansion of $(x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6})^{4}$? How to approach this type of question in general? * *How to use binomial theorem? *How to use multinomial theorem? *Are there any other combinatorial arguments available to solve this type of question?

We really seek the coefficient of $x^{14}$, factoring out an $x$ from each term in the generating function. Then observe that: $(1 + x + x^{2} + x^{3} + x^{4} + x^{5}) = \frac{1-x^{6}}{1-x}$ Now raise this to the fourth to get: $f(x) = \left(\frac{1-x^{6}}{1-x}\right)^{4}$. We have the identities: $$(1-x^{m})^{n} = \sum_{i=0}^{n} \binom{n}{i} (-1)^{i} x^{mi}$$ And: $$\frac{1}{(1-x)^{n}} = \sum_{i=0}^{\infty} \binom{i + n - 1}{i} x^{i}$$ So we expand out the numerator and denominator, picking terms of $x^{14}$. Note that we are multiplying the numerator expansion by the denominator expansion. $$\binom{14 + 4 - 1}{14}x^{14} - \binom{4}{1} \binom{8 + 4 - 1}{8} x^{14} + \binom{4}{2} \binom{2 + 4 - 1}{2} x^{14}$$

If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$ If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$ I tried Schturm's method for quite some time, and Cauchy Schwarz for numerators because of the given product condition.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$. Hence, our inequality is equivalent to $f(v^2)\geq0$, where $f$ is a linear increasing function. Hence, $f$ gets a minimal value, when $v^2$ gets a minimal value, which happens for equality of two variables or maybe one of them equal to $\frac{1}{2}$. * *$y=x$, $z=\frac{1}{x^2}$, which gives $(x-1)^2(2x^2+2x-1)\geq0$; *$z=\frac{1}{2}$, $y=\frac{2}{x}$, which gives $\frac{1}{2}\geq0$. Done!

How does one expression factor into the other? How does $$(k+1)(k^2+2k)(3k+5)$$ factor into $$(k)(k^2-1)(3k+2) + 12k(k+1)^2$$

Well, $RHS=k(k^2-1)(3k+2) + 12k(k+1)^2 = k(k+1)((k-1)(3k+2)+12(k+1))=k(k+1)(3k^2+2k-3k-2+12k+12)=k(k+1)(3k^2+11k+10)=k(k+1)(k+2)(3k+5)=LHS$

How can I show the two limits How can I show the two limits $$ \displaylines{ \mathop {\lim }\limits_{x \to + \infty } \frac{{x^2 e^{x + \frac{1}{x}} }}{{e^{ - x} \left( {\ln x} \right)^2 \sqrt x }} = \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{x^{\frac{3}{4}} e^{\left( {x + \frac{1}{{2x}}} \right)} }}{{\ln x}}} \right)^2 = + \infty \cr \mathop {\lim }\limits_{x \to 0^ + } \left( {\frac{{x^{\frac{3}{4}} e^{\left( {x + \frac{1}{{2x}}} \right)} }}{{\ln x}}} \right)^2 = + \infty \cr} $$ think you

$$\frac{x^2e^x}{e^{-x} (\ln x)^2\sqrt x} < \frac{x^2e^{x + \frac1x}}{e^{-x} (\ln x)^2\sqrt x} \text{ for large $x$ since } e^{\frac1x} > 1$$ $$$$ $$\lim_{x \to \infty} \frac{x^2e^x}{e^{-x} (\ln x)^2\sqrt x} = e^{2x}\frac{x^{1.5}}{(\ln x)^2} = +\infty$$ $$$$ $$\text{Thus, the right hand side at the top should diverge to $+\infty$ too.}$$

The roots of a certain recursively-defined family of polynomials are all real Let $P_0=1 \,\text{and}\,P_1=x+1$ and we have $$P_{n+2}=P_{n+1}+xP_n\,\,n=0,1,2,...$$ Show that for all $n\in \mathbb{N}$, $P_n(x)$ has no complex root?

Interlacing is a good hint, but let we show a brute-force solution. By setting: $$ f(t) = \sum_{n\geq 0}P_n(x)\frac{t^n}{n!} \tag{1}$$ we have that the recursion translates into the ODE: $$ f''(t) = f'(t) + x\, f(t) \tag{2}$$ whose solutions are given by: $$ f(t) = A \exp\left(t\frac{1+\sqrt{1+4x}}{2}\right) + B\exp\left(t\frac{1-\sqrt{1+4x}}{2}\right)\tag{3}.$$ This gives: $$ P_n(x) = A\left(\frac{1+\sqrt{1+4x}}{2}\right)^n + B\left(\frac{1-\sqrt{1+4x}}{2}\right)^n\tag{4}$$ and by plugging in the initial conditions we have: $$ A=\frac{1+2x+\sqrt{1+4x}}{2\sqrt{1+4x}},\quad B=\frac{-1-2x+\sqrt{1+4x}}{2\sqrt{1+4x}}\tag{5}$$ so $P_n(x)=0$ is equivalent to: $$\left(\frac{1+\sqrt{1+4x}}{1-\sqrt{1+4x}}\right)^n = \frac{1+2x-\sqrt{1+4x}}{1+2x+\sqrt{1+4x}},\tag{6}$$ or, by setting $x=\frac{y^2-1}{4}$, to: $$\left(\frac{y+1}{1-y}\right)^n = \frac{y^2-y+1}{y^2+y+1} \tag{7} $$ or, by setting $y=\frac{z-1}{z+1}$, to: $$ z^n = \frac{3+z^2}{1+3z^2}\tag{8} $$ or to: $$ 3z^{n+2}+z^n-z^2-3 = 0.\tag{9} $$

$\lim_{n \rightarrow \infty} \frac{1-(1-1/n)^4}{1-(1-1/n)^3}$ Find $$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}$$ I can't figure out why the limit is equal to $\dfrac{4}{3}$ because I take the limit of a quotient to be the quotient of their limits. I'm taking that $\lim_{n \rightarrow \infty}1-\left(1-\frac{1}{n}\right)^4 = 0$ and likewise that $\lim_{n \rightarrow \infty}1-\left(1-\frac{1}{n}\right)^3 = 0$, which still gives me that the limit should be 0.

$$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3} \stackrel{\mathscr{L}}{=}\lim_{n \rightarrow \infty} \dfrac{4\left(1-\dfrac{1}{n}\right)^3 \dfrac{1}{n^2}}{3\left(1-\dfrac{1}{n}\right)^2\dfrac{1}{n^2}} =\lim_{n \rightarrow \infty} \dfrac{4\left(1-\dfrac{1}{n}\right)^3}{3\left(1-\dfrac{1}{n}\right)^2} = \color{blue}{\dfrac{4}{3}}$$

Finding determinant of $n \times n$ matrix I need to find a determinant of the matrix: $$ A = \begin{pmatrix} 1 & 2 & 3 & \cdot & \cdot & \cdot & n \\ x & 1 & 2 & 3 & \cdot & \cdot & n-1 \\ x & x & 1 & 2 & 3 & \cdot & n-2 \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ x & x & \cdot & \cdot & x & 1 & 2 \\ x & x & \cdot & \cdot & \cdot & x & 1 \\ \end{pmatrix} $$ We know that $x \in R$ So far I managed to transform it to the form: $$ \begin{pmatrix} 1-x & 1 & 1 & \cdot & \cdot & \cdot & 1 \\ 0 & 1-x & 1 & 1 & \cdot & \cdot & 1 \\ 0 & 0 & 1-x & 1 & 1 & \cdot & 1 \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 0 & 0 & \cdot & \cdot & 0 & 1-x & 1 \\ x & x & \cdot & \cdot & \cdot & x & 1 \\ \end{pmatrix} $$ by the operations: (Let's say $r_i$ is the ith row) $$r_1 = r_1 - r_n,r_2 = r_2-r_n, r_3 = r_3 - r_n, ..., r_{n-1} = r_{n-1} - r_n$$ and then $$r_1 = r_1 - r_2, r_2 = r_2 - r_3, r_3 = r_3 - r_4,...,r_{n-2} = r_{n-2} - r_{n-1}$$ Unfortunately, I have no idea how to eliminate the last row. Any hints?

Multiply the last row by $\frac{1-x}{x}$; this means that the determinant you want will be the determinant of the changed matrix times $-\frac{x}{x-1}$. Now subtract $r_1$ from $r_n$ leaving $$r_n = (0, -x, -x, -x, \cdots, -x, \frac{(x-1)^2 - x^2}{x}) $$ where I have intentionally written $$ \frac{1-x}{x} -1 = \frac{-2x+1}{x} = \frac{(x-1)^2 - x^2}{x} $$ Now we have 0 in the last row in columns 1 through 1. For each remaining column $j$ up to column $n-1$, multiply the last row by $\frac{x-1}{x}$ (giving another power of $\frac{x}{x-1}$ in the factor before the changed matrix), at which point you can eliminate the $(1-x)$ in column $j$ of the last row by adding the $j$-th row. When you do this, the last element of the $n$-th row changes to $$ \frac{(x-1)^{j}- x^{j}}{x^{j-1} } \frac{x-1}{x} - 1 = \frac{(x-1)^{j+1}-x^{j}(x-1) -x^{j}}{x^{j}} = \frac{(x-1)^{j+1}-x^{j+1}}{x^{j}} $$ and the process repeats for the mext $j$ In the end, the final term in $A_{nn}$ involves $$ \frac{(x-1)^n -x^n}{x^{n-1}}$$ and a lot of cancelation with the accumlated factors happens, leaving the answeer $$ (-1)^n \left( (x-1)^n - x^n \right)$$

Evaluate $ \int^\infty_0\int^\infty_0 x^a y^{1-a} (1+x)^{-b-1}(1+y)^{-b-1} \exp(-c\frac{x}{y})dxdy $ Evaluate $$ \int^\infty_0\int^\infty_0 x^a y^{1-a} (1+x)^{-b-1}(1+y)^{-b-1} \exp(-c\frac{x}{y})dxdy $$ under the condition $a>1$, $b>0$, $c>0$. Note that none of $a$, $b$ and $c$ is integer. Mathematica found the following form, but I prefer more compact expression for the faster numerical evaluation. $$ \bigg(2 c^2 \Gamma (a-2) \Gamma (b-2) \Gamma (b-1) \Gamma (b+1) \Gamma (-a+b+1) \, _2F_2(3,b+1;3-a,3-b;-c) \nonumber\\ ~~~ -\pi \csc (\pi a) \Big(\pi c^b \Gamma (b+1) \Gamma (2 b-1) (\cot (\pi (a-b))+\cot (\pi b)) \, _2F_2(b+1,2 b-1;b-1,-a+b+1;-c) \nonumber\\ ~~~~~ ~~~~~ +(a-1) a c^a \Gamma (b-1) \Gamma(b-a) \Gamma (-a+b+1) \Gamma (a+b-1) \, _2F_2(a+1,a+b-1;a-1,a-b+1;-c)\Big)\Bigg) \nonumber\\ ~~~~~ ~~~~~ \bigg/\Big(c^a{\Gamma (b-1) \Gamma (b+1)^2 \Gamma (-a+b+1)}\Big) $$

$\int_0^\infty\int_0^\infty x^ay^{1-a}(1+x)^{-b-1}(1+y)^{-b-1}e^{-c\frac{x}{y}}~dx~dy$ $=\Gamma(a+1)\int_0^\infty y^{1-a}(1+y)^{-b-1}U\left(a+1,a-b+1,\dfrac{c}{y}\right)dy$ (according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Integral_representations) $=\dfrac{\Gamma(a+1)\Gamma(b-a)}{\Gamma(b+1)}\int_0^\infty y^{1-a}(1+y)^{-b-1}{_1F_1}\left(a+1,a-b+1,\dfrac{c}{y}\right)dy+c^{b-a}\Gamma(a-b)\int_0^\infty y^{1-b}(1+y)^{-b-1}{_1F_1}\left(b+1,b-a+1,\dfrac{c}{y}\right)dy$ (according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Kummer.27s_equation) $=\dfrac{\Gamma(a+1)\Gamma(b-a)}{\Gamma(b+1)}\int_\infty^0\left(\dfrac{1}{y}\right)^{1-a}\left(1+\dfrac{1}{y}\right)^{-b-1}{_1F_1}(a+1,a-b+1,cy)~d\left(\dfrac{1}{y}\right)+c^{b-a}\Gamma(a-b)\int_\infty^0\left(\dfrac{1}{y}\right)^{1-b}\left(1+\dfrac{1}{y}\right)^{-b-1}{_1F_1}(b+1,b-a+1,cy)~d\left(\dfrac{1}{y}\right)$ $=\dfrac{\Gamma(a+1)\Gamma(b-a)}{\Gamma(b+1)}\int_0^\infty y^{a+b-2}(y+1)^{-b-1}{_1F_1}(a+1,a-b+1,cy)~dy+c^{b-a}\Gamma(a-b)\int_0^\infty y^{2b-2}(y+1)^{-b-1}{_1F_1}(b+1,b-a+1,cy)~dy$ $=\dfrac{\Gamma(a+1)\Gamma(b-a)}{\Gamma(b+1)}\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(a+1)_nc^ny^{a+b+n-2}(y+1)^{-b-1}}{(a-b+1)_nn!}dy+c^{b-a}\Gamma(a-b)\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(b+1)_nc^ny^{2b+n-2}(y+1)^{-b-1}}{(b-a+1)_nn!}dy$ $=\dfrac{\Gamma(a+1)\Gamma(b-a)}{\Gamma(b+1)}\sum\limits_{n=0}^\infty\dfrac{(a+1)_nc^nB(a+b+n-1,2-a-n)}{(a-b+1)_nn!}+c^{b-a}\Gamma(a-b)\sum\limits_{n=0}^\infty\dfrac{(b+1)_nc^nB(2b+n-1,2-b-n)}{(b-a+1)_nn!}$ (according to http://en.wikipedia.org/wiki/Beta_function#Properties) $=\sum\limits_{n=0}^\infty\dfrac{\Gamma(a+1)\Gamma(b-a)(a+1)_n\Gamma(a+b+n-1)\Gamma(2-a-n)c^n}{\Gamma(b+1)\Gamma(b+1)(a-b+1)_nn!}+\sum\limits_{n=0}^\infty\dfrac{\Gamma(a-b)(b+1)_n\Gamma(2b+n-1)\Gamma(2-b-n)c^{b-a+n}}{\Gamma(b+1)(b-a+1)_nn!}$ $=\sum\limits_{n=0}^\infty\dfrac{\Gamma(a+1)\Gamma(2-a)\Gamma(b-a)\Gamma(a+b-1)(a+1)_n(a+b-1)_n(-1)^nc^n}{(\Gamma(b+1))^2(a-1)_n(a-b+1)_nn!}+\sum\limits_{n=0}^\infty\dfrac{\Gamma(a-b)\Gamma(2-b)\Gamma(2b-1)(b+1)_n(2b-1)_n(-1)^nc^{b-a+n}}{\Gamma(b+1)(b-1)_n(b-a+1)_nn!}$ (according to http://en.wikipedia.org/wiki/Pochhammer_symbol#Properties) $=\dfrac{\Gamma(a+1)\Gamma(2-a)\Gamma(b-a)\Gamma(a+b-1)}{(\Gamma(b+1))^2}{_2F_2}(a+1,a+b-1;a-1,a-b+1;-c)+\dfrac{\Gamma(a-b)\Gamma(2-b)\Gamma(2b-1)c^{b-a}}{\Gamma(b+1)}{_2F_2}(b+1,2b-1;b-1,b-a+1;-c)$

Number theory: prove that if $a,b,c$ odd then $2\gcd(a,b,c) = \gcd(a+b,b+c, c+a)$ Please help! Am lost with the following: Prove that if $a,b,c$ are odd integers, then $2 \gcd(a,b,c) = \gcd( a+b, b+c, c+a)$ Thanks a lot!!

Let $d=\gcd(a,b,c)$. Then solve $d=ax+by+cz$. Use that $2a=(a+b)+(a+c)-(b+c)$, $2b=(a+b)+(b+c)-(a+c)$ and $2c=(a+c)+(b+c)-(a+b)$. Then $$2d=2ax+2by+2cz = (a+b)(x+y-z) + (a+c)(x-y+z) + (b+c)(y+z-x)$$ So we have a solution to: $$2d = (a+b)X+(a+c)Y + (b+c)Z$$ So we know $\gcd(a+b,a+c,b+c)\mid 2\gcd(a,b,c)$. The other direction is easier (and there, you use that $a,b,c$ are odd.)

why is $\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}}$=-3? Exercise taken from here: https://mooculus.osu.edu/textbook/mooculus.pdf (page 42, "Exercises for Section 2.2", exercise 4). Why is $\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}}$=-3*? I always find 3 as the solution. I tried two approaches: Approach 1: $$ \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} = \lim_{x\to -\infty} \frac{3x+7}{x} = \lim_{x\to -\infty} \frac{3x}{x}+\lim_{x\to -\infty} \frac{7}{x}\\ = \lim_{x\to -\infty} 3+\lim_{x\to -\infty} \frac{7}{x} = 3 + 0 = 3 $$ Approach 2: $$ \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} = \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} \times \frac{\frac{1}{x}}{\frac{1}{x}} = \lim_{x\to -\infty} \frac{3+\frac{7}{x}}{\frac{\sqrt{x^2}}{x}}\\ = \lim_{x\to -\infty} \frac{3+\frac{7}{x}}{\frac{\sqrt{x^2}}{\sqrt{x^2}}} = \lim_{x\to -\infty} \frac{3+\frac{7}{x}}{\sqrt{\frac{x^2}{x^2}}}\\ = \frac{\lim_{x\to -\infty}3+\lim_{x\to -\infty}\frac{7}{x}}{\lim_{x\to -\infty}\sqrt{1}}\\ = \frac{3 + 0}{1} = 3 $$ * -3 is given as the answer by the textbook (cf. page 247) as well as wolfram|alpha EDIT: I just re-read page 40 of the textbook and realized that I made a mistake in my approach 2. Instead of multiplying with $\frac{1}{x}$ I should have multiplied with $\frac{-1}{x}$ which is positive because as $x\to -\infty$, x is a negative number. It thus reads: $$ \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} = \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} \times \frac{\frac{-1}{x}}{\frac{-1}{x}} = \lim_{x\to -\infty} \frac{-3+\frac{-7}{x}}{\frac{\sqrt{x^2}}{-x}}\\ = \lim_{x\to -\infty} \frac{-3+\frac{-7}{x}}{\frac{\sqrt{x^2}}{\sqrt{x^2}}} = \lim_{x\to -\infty} \frac{-3+\frac{-7}{x}}{\sqrt{\frac{x^2}{x^2}}}\\ = \frac{\lim_{x\to -\infty}-3+\lim_{x\to -\infty}\frac{-7}{x}}{\lim_{x\to -\infty}\sqrt{1}}\\ = \frac{-3 + 0}{1} = -3 $$

For $x \to -\infty$, we have that $\sqrt{x^2} = |x| = -x$.

Considering $ (1+i)^n - (1 - i)^n $, Complex Analysis I have been working on problems from Complex Analysis by Ahlfors, and I got stuck in the following problem: Evaluate: $$ (1 + i)^n - (1-i)^n $$ I have just "reduced" to: $$ (1 + i)^n - (1-i)^n = \sum_{k=0} ^n i^k(1 - (-1)^k) $$ by using expansion of each term. Thanks.

There are a number of spiffy techniques one could use on this problem, but Ahlfors doesn't get to conjugation and modulus until 1.3 and geometry of the complex plane until Section 2 (of Chapter 1), while this is still in 1.1. [I dug up my copy of the third edition to see how much was discussed to that point.] abel shows one approach in using the binomial theorem that lies within the (rather) limited means available. Given what the author covers in this section, this is another possibility: $$ ( 1 \ + \ i )^n \ + \ ( 1 \ - \ i )^n \ \ = \ \ ( 1 \ + \ i )^n \ \left[ \ 1 \ + \ \frac{( 1 \ - \ i )^n}{( 1 \ + \ i )^n} \ \right] $$ $$ = \ \ ( 1 \ + \ i )^n \ \left[ \ 1 \ + \ \left(\frac{ 1 \ - \ i }{ 1 \ + \ i } \right)^n \ \right] \ \ = \ \ ( 1 \ + \ i )^n \ \left[ \ 1 \ + \ \left(\frac{ [ \ 1 \ - \ i \ ] \ [ \ 1 \ - \ i \ ] }{ [ \ 1 \ + \ i \ ] [ \ 1 \ - \ i \ ] } \right)^n \ \right] $$ [the conjugate is being applied as shown in that section, but Ahlfors hasn't called it that yet] $$ = \ \ ( 1 \ + \ i )^n \ \left[ \ 1 \ + \ \left(\frac{ 1 \ - \ 2i \ - \ 1 }{ 2 } \right)^n \ \right] \ \ = \ \ ( 1 \ + \ i )^n \ [ \ 1 \ + \ ( - i) ^n \ ] \ \ . $$ The binomial theorem can now be applied to the first factor: $$ ( 1 \ + \ i )^n \ + \ ( 1 \ - \ i )^n $$ $$ = \ \ \left( \ 1 \ + \ \left( \begin{array}{c} n \\ 1 \end{array} \right) i \ + \ \left( \begin{array}{c} n \\ 2 \end{array} \right) i^2 \ + \ \ldots \ + \ \left( \begin{array}{c} n \\ n-1 \end{array} \right) i^{n-1} \ + \ i^n \ \right) \ [ \ 1 \ + \ ( - i) ^n \ ] \ \ . $$ [The "typoed" version of the problem that David Cardozo originally posted is analogous: $$ ( 1 \ + \ i )^n \ - \ ( 1 \ - \ i )^n $$ $$ = \ \ \left( \ 1 \ + \ \left( \begin{array}{c} n \\ 1 \end{array} \right) i \ + \ \left( \begin{array}{c} n \\ 2 \end{array} \right) i^2 \ + \ \ldots \ + \ \left( \begin{array}{c} n \\ n-1 \end{array} \right) i^{n-1} \ + \ i^n \ \right) \ [ \ 1 \ - \ ( - i) ^n \ ] \ \ . \ \ ] $$ $ \ \ $ Presumably, we'd like to consolidate this a bit. Because of that $ \ (-i)^n \ $ term in the second factor, that factor has a cycle of period 4. We see that this product is zero for $ \ n \ = \ 4m \ + \ 2 \ $ , with integer $ \ m \ \ge \ 0 \ $ . [These will be "out-of-phase" with abel's expressions, since I am using Ahlfors' version of the problem.] For the other cases, we will write the first factor as $$ \left[ \ 1 \ - \ \left( \begin{array}{c} n \\ 2 \end{array} \right) \ + \ \left( \begin{array}{c} n \\ 4 \end{array} \right) \ + \ \text{etc.} \ \right] \ \ + \ \ i \ \left[ \ \left( \begin{array}{c} n \\ 1 \end{array} \right) \ - \ \left( \begin{array}{c} n \\ 3 \end{array} \right) \ + \ \left( \begin{array}{c} n \\ 5 \end{array} \right) \ - \ \text{etc.} \ \right] \ \ . $$ For $ \ n \ = \ 4m \ $ , the factor $ \ [ \ 1 \ + \ ( - i) ^n \ ] \ = \ 2 \ $ and the imaginary part of the binomial series is zero, owing to the symmetry of the binomial coefficients. The real part also simplifies due to this symmetry, so we have $$ ( 1 \ + \ i )^{4m} \ + \ ( 1 \ - \ i )^{4m} \ \ = \ \ \left[ \ 2 \cdot 1 \ - \ 2 \ \left( \begin{array}{c} 4m \\ 2 \end{array} \right) \ + \ 2 \ \left( \begin{array}{c} 4m \\ 4 \end{array} \right) \ - \ \ldots \ + \ \left( \begin{array}{c} 4m \\ 2m \end{array} \right) \ \right] \cdot \ 2 \ \ . $$ The remaining cases are somewhat more complicated to work out: for $ \ n \ = \ 4m \ + \ 1 \ $ and $ \ n \ = \ 4m \ + \ 3 \ $ , respectively, we obtain $$ \left( \ \left[ \ 1 \ - \ \left( \begin{array}{c} n \\ 2 \end{array} \right) \ + \ \left( \begin{array}{c} n \\ 4 \end{array} \right) \ \ldots \ \pm \ n \ \right] \ \ + \ \ i \ \left[ \ n \ - \ \left( \begin{array}{c} n \\ 3 \end{array} \right) \ + \ \left( \begin{array}{c} n \\ 5 \end{array} \right) \ \ldots \ \pm \ 1 \ \right] \ \right) \ \cdot \ ( 1 \ \mp \ i) \ \ . $$ For either of these cases, since $ \ n \ $ is odd, the number of binomial coefficients is even. So the real part has the symmetry in which the first half of the terms are identical to the second half of them; also, the symmetry among the coefficients produces a sum which is always a power of 2 . In the imaginary part, we do not get a simple alternation of signs (which we know gives a sum of zero for the binomial coefficients), but the "double-alternating" signs proves to have the same effect; the result is that the imaginary part is zero for these cases as well. Hence, the expression $ \ ( 1 \ + \ i )^n \ + \ ( 1 \ - \ i )^n \ $ is always real; by analogous reasoning, the expression $ \ ( 1 \ + \ i )^n \ - \ ( 1 \ - \ i )^n \ $ is always pure imaginary. We find the sequences (including the values abel presents) for $ \ 0 \ \le \ n \ \le \ 9 \ $ $$ \ ( 1 \ + \ i )^n \ + \ ( 1 \ - \ i )^n \ \ : \ \ 2 \ , \ 2 \ , \ 0 \ , \ -4 \ , \ -8 \ , \ -8 \ , \ 0 \ , \ 16 \ , \ 32 \ , \ 32 \ \ \text{and} $$ $$ \ ( 1 \ + \ i )^n \ - \ ( 1 \ - \ i )^n \ \ : \ \ 0 \ , \ 2i \ , \ 4i \ , \ 4i \ , \ 0 \ , \ -8i \ , \ -16i \ , \ -16i \ , \ 0 \ , \ 32i \ \ . $$ [Incidentally, these results indicate the interesting identities$ ^* $ $$ ( 1 \ + \ i )^{4m} \ + \ ( 1 \ - \ i )^{4m} \ \ = \ \ ( 1 \ + \ i )^{4m+1} \ + \ ( 1 \ - \ i )^{4m+1} \ \ \text{and} $$ $$ ( 1 \ + \ i )^{4m+2} \ - \ ( 1 \ - \ i )^{4m+2} \ \ = \ \ ( 1 \ + \ i )^{4m+3} \ - \ ( 1 \ - \ i )^{4m+3} \ \ ] $$ $ ^* $ with unintentional alliteration on the theme of $ \ i \ $ ... $ \ \ $ To be sure, this is a cumbersome description of the result, but it is a consequence of using Cartesian coordinates for the description of the complex values. Once you reach Section 2 and the use of polar coordinates, you will have the far more compact expressions $$ \ ( 1 \ + \ i )^n \ + \ ( 1 \ - \ i )^n \ \ = \ \ 2^{(n+2)/2} \ \cos\left( \frac{n \pi}{4} \right) \ \ \text{and} $$ $$( 1 \ + \ i )^n \ - \ ( 1 \ - \ i )^n \ \ = \ \ 2^{(n+2)/2} \ \sin\left( \frac{n \pi}{4} \right) \ i \ \ . $$ (The exponential factor simply grows by a factor of $ \ \sqrt{2} \ $ at each successive stage, but its product with the trigonometric factors create the complications in the sequences above. The trigonometric factors also immediately explain the "out-of-phase" behavior between the two versions of the expression we have been evaluating. These products follow from the methods being described by dustin and RikOsuave. )

If $G$ is a group whereby $(a\cdot b)^{i} =a^i\cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show $G$ is abelian. If $G$ is a group in which $(a\cdot b)^{i} =a^i\cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show that $G$ is abelian. Proof: Let $x$ be the smallest of the 3 consecutive integers. Then, we have $(1)(a\cdot b)^{x} =a^x\cdot b^x$, $(2)(a\cdot b)^{x+1} =a^{x+1}\cdot b^{x+1}$ and $(3)(a\cdot b)^{x+2} =a^{x+2}\cdot b^{x+2}$. Using $(2)$ and multiplying $a^{-1}$ on the left and $b^{-1}$ on the right, we get $baba...ba = a^{x}b^{x}(4)$ whereby there are $x$ number of $a$ and $x$ number of $b$ on both sides. Using $(1)$ and multiplying both sides on the right by $ab$, we have $\overbrace{abab...ab}^{(x+1)ab} = (a^{x}b^{x})ab (5)$. Substitute $(4)$ into $(5)$, we get $\overbrace{ab...ab}^{(x+1)ab} =\overbrace{ba...ba}^{(x)ba}ab(6)$. Using $(3)$ and multiplying $a^{-1}$ on the left and $b^{-1}$ on the right, we get $\overbrace{ba...ba}^{(x+1)ba} = a^{x+1}b^{x+1} = \overbrace{abab...ab}^{(x+1)ab}(7)$. Combining $(6)$ and $(7)$, and multiply $a^{-1}b^{-1}...a^{-1}b^{-1}$ on the left, we get $ab = ba$. Hence $G$ is abelian.

Use $\backslash$overbrace{below}^{above}, as in (right click and select to see LaTeX commands): $$ \overbrace{a\ldots a}^{27} $$ The proof looks fine to me.

Which arrangement produces the largest number? I learnt that the power tower $2\uparrow3\uparrow4\uparrow...\uparrow n$ is larger than any power tower with a different order of the numbers $2,3,4,...,n$. Is this also true for conway-chains and for bowers array notation ? Are $$2\rightarrow 3\rightarrow 4\rightarrow...\rightarrow n$$ and {$2,3,4,...,n$} also larger than any other number of this form with different order of the numbers ?

No for both. For example, for $n = 4$ we have $2 \rightarrow 3 \rightarrow 4 = 2 \rightarrow 4 \rightarrow 3 = 2 \uparrow \uparrow 65536$, whereas $3 \rightarrow 2 \rightarrow 4 = 3 \uparrow \uparrow 3^{27}$. We can show by induction that $2 \rightarrow 3 \rightarrow n < 3 \rightarrow 2 \rightarrow n$, as $$ 2 \rightarrow 3 \rightarrow 1 = 8 < 9 = 3 \rightarrow 2 \rightarrow 1 $$ and assuming $2 \rightarrow 3 \rightarrow n = 2 \rightarrow 4 \rightarrow (n-1) < 3 \rightarrow 2 \rightarrow n = 3 \rightarrow 3 \rightarrow (n-1)$, we have $$ 2 \rightarrow 3 \rightarrow (n+1) = 2 \rightarrow (2 \rightarrow 2 \rightarrow (n+1)) \rightarrow n = 2 \rightarrow 4 \rightarrow n $$ $$ = 2 \rightarrow (2 \rightarrow 4 \rightarrow (n-1)) \rightarrow (n-1) < 2 \rightarrow (3 \rightarrow 3 \rightarrow (n-1)) \rightarrow (n-1) $$ $$ < 3 \rightarrow (3 \rightarrow 3 \rightarrow (n-1)) \rightarrow (n-1) = 3 \rightarrow 3 \rightarrow n = 3\rightarrow 2 \rightarrow (n+1). $$ It follows that $2 \rightarrow 3 \rightarrow X < 3 \rightarrow 2 \rightarrow X$ for any chain $X$, since when you evaluate the chains you will get the same expressions, except one will have chains starting with $3 \rightarrow 2$ and one will have chains starting with $2 \rightarrow 3$. Whenever the expressions reduce a chain to a 3-chain, we will have $3 \rightarrow 2 \rightarrow n$ evaluate to a higher value than $2 \rightarrow 3 \rightarrow n$, so in the end $3 \rightarrow 2 \rightarrow X$ will evaluate to a higher value than $2 \rightarrow 3 \rightarrow X$. In particular, $2 \rightarrow 3 \rightarrow \cdots \rightarrow n < 3 \rightarrow 2 \rightarrow \cdots \rightarrow n$. For Bowers arrays the situation is more extreme. Any expression of the form {2,2,...} will evaluate to the number 4, as applying the evaluation rules will either keep the first two entries the same, or will replace the array with {2, {2,1,...},...} = {2,2,...}. So the array will eventually evaluate to {2,2} = 4. Further, any array {2,b,c,d,...} with four or more entries (not counting trailing 1's) will evaluate to 4 as well, as {2,b,c,d,...} will evaluate to {2,b',c-1,d,...} and then {2,b'',c-2,d,...} eventually reaching {2,n,1,d,...}. This evaluates to {2,2,{2,n-1,1,d,...},d-1,...}, which we have already determined equals 4. In particular, {2,3,4,...,n} will evaluate to 4 for n > 4, whereas {3,2,4,...,n} will grow extremely fast.

Show that $\frac{(n-a)^2}{n}$ can be written as $1-\left(\frac{n}{a}\right)^2\cdot\frac{n}{(n/a)^2}$ \left(\frac{n}{a}\right)^2\cdot\frac{n}{(n/a)^2}$. I have got so far to $(a^2/n)-2a+n$ But I can not see how to proceed. Can anyone help please?

$$\frac{(n-a)^2}{n} =(n-a)^2\left(\frac{1}{n}\right) =(n-a)^2\left(\frac{1}{n}\right)\frac{a^2}{a^2} =\left(\frac{n-a}{a}\right)^2 \frac{a^2}{n}= \left(\frac{n}{a}-1\right)^2 \frac{a^2}{n} = \left(\frac{n}{a}-1\right)^2 \frac{n^2}{n^2} \frac{a^2}{n} = (-1)^2\left(1-\frac{n}{a}\right)^2 \frac{n}{(\frac{n}{a})^2}=\left(1-\frac{n}{a}\right)^2 \frac{n}{(\frac{n}{a})^2}$$ The real trick here is multiplying by $1=\frac{a^2}{a^2}=\frac{n^2}{n^2}$.

Summation of an infinite series The sum is as follows: $$ \sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1}\\ $$ This is how I started: $$ = \frac{1}{6}\sum_{n=1}^{\infty} n \left ( \frac{5}{6} \right )^{n-1} \\ = \frac{1}{5}\sum_{n=1}^{\infty} n \left ( \frac{5}{6} \right )^{n}\\\\ = \frac{1}{5}S\\ S = \frac{5}{6} + 2\left (\frac{5}{6}\right)^2 + 3\left (\frac{5}{6}\right)^3 + ... $$ I don't know how to group these in to partial sums and get the result. I also tried considering it as a finite sum (sum from 1 to n) and applying the limit, but that it didn't get me anywhere! PS: I am not looking for the calculus method. I tried to do it directly in the form of the accepted answer, $$ \textrm{if} \ x= \frac{5}{6},\\ S = x + 2x^2 + 3x^3 + ...\\ Sx = x^2 + 2x^3 + 3x^4 + ...\\ S(1-x) = x + x^2 + x^3 + ...\\ \textrm{for x < 1},\ \ \sum_{n=1}^{\infty}x^n = -\frac{x}{x-1}\ (\textrm{I looked up this eqn})\\ S = \frac{x}{(1-x)^2}\\ \therefore S = 30\\ \textrm{Hence the sum} \sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1} = \frac{30}{5} = 6 $$

hint: differentiate the identity $$\sum_{k=0}^{\infty} x^k = \frac{1}{1-x} $$

Integral $\int \frac{x+2}{x^3-x} dx$ I need to solve this integral but I get stuck, let me show what I did: $$\int \frac{x+2}{x^3-x} dx$$ then: $$\int \frac{x}{x^3-x} + \int \frac{2}{x^3-x}$$ $$\int \frac{x}{x(x^2-1)} + 2\int \frac{1}{x^3-x}$$ $$\int \frac{1}{x^2-1} + 2\int \frac{1}{x^3-x}$$ now I need to resolve one integral at the time so: $$\int \frac{1}{x^2-1}$$ with x = t I have: $$\int \frac{1}{t^2-1}$$ Now I have no idea about how to procede with this...any help?

The Heaviside cover-up method for solving partial fraction decompositions deserves to be more widely known. We want to find $A,B,C$ in this equation: $$\frac{x+2}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$$ To find $A$, multiply by $x$: $$\frac{x+2}{(x-1)(x+1)}=A+\frac{Bx}{x-1}+\frac{Cx}{x+1}$$ Now put $x=0$ to get $\dfrac{2}{-1}=A$. It looks like a swindle, because the starting equation is not valid when $x=0$. But if this bothers you, you can make it rigorous by taking the limit as $x \to 0$ in the second equation. To find $B$, multiply by $x-1$ and put $x=1$ to get $\dfrac{3}{2}=B$. To find $C$, multiply by $x+1$ and put $x=-1$ to get $\dfrac{1}{2}=C$. (Things are not quite so simple if the denominator has a repeated root, but it's still doable. See the link for details.)

Does $\int_0^\infty \sin(x^{2/3}) dx$ converges? My Try: We substitute $y = x^{2/3}$. Therefore, $x = y^{3/2}$ and $\frac{dx}{dy} = \frac{2}{3}\frac{dy}{y^{1/3}}$ Hence, the integral after substitution is: $$ \frac{3}{2} \int_0^\infty \sin(y)\sqrt{y} dy$$ Let's look at: $$\int_0^\infty \left|\sin(y)\sqrt{y} \right| dy = \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| \sqrt{y} dy \ge \sum_{n=0}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| dy \\= \sum_{n=1}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\sqrt{\sin(y)^2}$$

$\sin x^{2/3}$ remains above $1/2$ for $x$ between $[(2n+1/6)\pi]^{3/2}$ and $[2n+5/6]^{3/2}$, so the integral rises by more than $\left([2n+5/6]^{3/2}-[2n+1/6]^{3/2}\right)\pi^{3/2}/2$ during that time. $$[2n+5/6]^{3/2}-[2n+1/6]^{3/2}=\frac{[2n+5/6]^3-[2n+1/6]^3}{[2n+5/6]^{3/2}+[2n+1/6]^{3/2}}\\ >\frac{8n^2}{2[2n+1]^{3/2}}$$ That increases as a function of $n$, so the integral does not converge.

(Infinite) Nested radical equation, how to get the right solution? I've been tasked with coming up with exam questions for a high school math contest to be hosted at my university. I offer the following equation, $$\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}=2$$ and ask for the solution for $x$. Here's what I attempted so far. The first utilizes some pattern recognition, but it gives me two solutions (only one of which is correct). $$\begin{align*} \sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}&=2\\ \sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}&=4-x\\ \sqrt{x+\sqrt{x-\cdots}}&=x-(4-x)^2\\ 2&=x-(4-x)^2&\text{(from line 1)}\\ (x-6)(x-3)&=0 \end{align*}$$ $x=6$ is the extraneous solution. Where did I go wrong, and how can I fix this? I know there's a closed form for non-alternating nested radicals $\sqrt{n+\sqrt{n+\cdots}}$ and $\sqrt{n-\sqrt{n-\cdots}}$, but I can't seem to find anything on alternating signs.

You know that: $$\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}=A = 2$$ and hence $$A = \sqrt{x+\sqrt{x-A}} = 2$$ or equivalently $$\sqrt{x+\sqrt{x-2}} = 2$$ Clearly, $\sqrt{x-2}$ is well defined when $$x \geq 2. ~~~(1)$$ Then, squaring both side, you get: $$x+ \sqrt{x-2} = 4 \Rightarrow \sqrt{x-2} = 4-x ~~~(2).$$ Since $\sqrt{x-2} \geq 0$, then also $4-x \geq0$, and hence $$x \leq 4. ~~~(3)$$ Joining conditions $(1)$ and $(3)$, one obtain the existence set for $x$: $$x \geq 2 \wedge x \leq 4 \Rightarrow 2\leq x \leq 4. ~~~(4)$$ Going back to $(3)$, we can square both side and we get: $$x-2 = (4-x)^2 \Rightarrow x-2=16+x^2-8x \Rightarrow x^2-9x+18=0 \Rightarrow $$ $$\Rightarrow (x-3)(x-6) = 0. ~~~(5)$$ The solution of $(5)$ are $x_1 = 3$ and $x_2 = 6$, but according to $(4)$, only $x_1 = 3$ is feasible.

Four different positive integers a, b, c, and d are such that $a^2 + b^2 = c^2 + d^2$ Four different positive integers $a, b, c$, and $d$ are such that $a^2 + b^2 = c^2 + d^2$ What is the smallest possible value of $abcd$? I just need a few hints, nothing else. How should I begin? Number theory?

The number of representations of a positive integer as a sum of two squares depends on the number of prime divisors of the form $4k+1$ (see Cox, Primes of the form $a^2+k b^2$). If we take the first two primes of such a form, $5$ and $13$, we have that $5\cdot 13$ can be represented as: $$ 65 = 1^2+8^2 = 4^2+7^2 $$ so we have a solution with $abcd = 224$. You can complete the proof by exhaustive search (it is quite easy to check that the first $64$ positive integers do not have a double representation in terms of positive integers, and we just have to check till $n=225$ or so to find the minimum $abcd$), or proving that the number of representations of $n$ as a sum of two squares is given by the number of divisors of $n$ of the form $4k+1$ minus the number of divisors of the form $4k+3$.

Can the cube of every perfect number be written as the sum of three cubes? I found an amazing conjecture: the cube of every perfect number can be written as the sum of three positive cubes. The equation is $$x^3+y^3+z^3=\sigma^3$$ where $\sigma$ is a perfect number (well it holds good for the first three perfect numbers that is): $${ 3 }^{ 3 }+{ 4 }^{ 3 }+{ 5 }^{ 3 }={ 6 }^{ 3 }$$ $${ 21 }^{ 3 }+{ 18 }^{ 3 }+{ 19 }^{ 3 }={ 28 }^{ 3 }$$ $${ 495 }^{ 3 }+{ 82 }^{ 3 }+{ 57 }^{ 3 }={ 496 }^{ 3 }$$ Is this what I am proposing that the cube of any perfect number can be expressed in terms of the sum of three positive cubes true? If it is then can we prove it?

This is more of a comment as opposed to an answer There is a formula for finding the values of $a, b, c, d$ in the following equation: $$a^3 + b^3 + c^3 = d^3$$ Where $$\forall x, y\in \mathbb{Z}, \ \begin{align} a &= 3x^2 + 5y(x - y), \ b = 2\big(2x(x - y) + 3y^2\big) \\ c &= 5x(x - y) - 3y^2, \ d = 2\big(3x^2 - 2y(x + y)\big) \end{align}$$ Therefore if we prove this conjecture, we prove that every perfect number $d$ must be even! I also would like to expand this theorem as well with a theorem of mine: If $$\forall\{x, y, z\}\subset \mathbb{N}, \ 6^3 + (2x)^3 + (2y - 1)^3 = z^3$$ Then $$z \equiv \pm 1 \pmod 6$$ Where $z$ is a prime number. If $z$ is not a prime number, then $z\equiv3\pmod 6$ Examples: $$\begin{align} 6^3 + 8^3 + 1^3 &= 9^3 \qquad \ \ \ \ \mid9 &\equiv 3 \pmod 6 \\ 6^3 + 32^3 + 33^3 &= 41^3 \qquad \ \ \mid41 &\equiv 5 \pmod 6 \\ 6^3 + 180^3 + 127^3 &= 199^3 \qquad \mid199 &\equiv 1 \pmod 6 \\ 6^3 + 216^3 + 179^3 &= 251^3 \qquad \mid251 &\equiv 5 \pmod 6 \\ 6^3 + 718^3 + 479^3 &= 783^3 \qquad \mid783 &\equiv 3 \pmod 6 \\ 6^3 + 768^3 + 121^3 &= 769^3 \qquad \mid769 &\equiv 1 \pmod 6 \end{align}$$

Use integration by substitution I'm trying to evaluate integrals using substitution. I had $$\int (x+1)(3x+1)^9 dx$$ My solution: Let $u=3x+1$ then $du/dx=3$ $$u=3x+1 \implies 3x=u-1 \implies x=\frac{1}{3}(u-1) \implies x+1=\frac{1}{3}(u+2) $$ Now I get $$\frac{1}{3} \int (x+1)(3x+1)^9 (3 \,dx) = \frac{1}{3} \int \frac{1}{3}(u+2)u^9 du = \frac{1}{9} \int (u+2)u^9 du \\ = \frac{1}{9} \int (u^{10}+2u^9)\,du = \frac{1}{9}\left( \frac{u^{11}}{11}+\frac{2u^{10}}{10} \right) + C$$ But then I get to this one $$\int (x^2+2)(x-1)^7 dx$$ and the $x^2$ in brackets is throwing me off. I put $u=x-1\implies x=u+1,$ hence $x^2+2 =(u+1)^2 +2 = u^2+3$. So $$ \int(x^2+2)(x-1)^7\, dx = \int (u+1)u^7 du = \int (u+u^7) du = \frac{u^2}{2}+\frac{u^8}{8} + C $$ Is this correct or have I completely missed the point?

$\frac{19683 x^{11}}{11}+\frac{39366 x^{10}}{5}+15309 x^9+17496 x^8+13122 x^7+6804 x^6+\frac{12474 x^5}{5}+648 x^4+117 x^3+14 x^2+x$

Evaluating $\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx$ How do I evaluate the definite integral $$\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx ?$$ I used trig substitution, and then a u substitution for $\sec\theta$. I tried doing it and got an answer of: $-\sqrt{125}+12\sqrt{5}-16$, but apparently its wrong. Can someone help check my error?

A way to compute this is as follows: \begin{align*} \int_0^1 \frac {x^3}{\sqrt {4+x^2}}\mathrm d x &=\int_0^1\frac{4x+x^3-4x}{\sqrt{4+x^2}}\mathrm d x\\ &=\int_0^1x\sqrt{4+x^2}\mathrm d x -2\int_0^1\frac{2x}{\sqrt{4+x^2}}\mathrm d x\\ &=\left.\frac{1}{3}(4+x^2)^{3/2}\right|_0^1-4\left.\left(4+x^2\right)^{1/2}\right|_0^1\\ &=\frac{5\sqrt{5}-8}{3}-4\left(\sqrt{5}-2\right)\\ &=\boxed{\color{blue}{\dfrac{16-7\sqrt{5}}{3}}} \end{align*}

Chinese Remainder Theorem for non prime-numbers. Let's say I want to find x such that x leaves remainder 2 when divided by 3 and x leaves remainder 3 when divided by 5. x % 3 = 2 x % 5 = 3 We break down the problem to: x % 3 = 1 x % 5 = 0 Therefore, 5k % 3 = 1 2k % 3 = 1 k = 2 10, when remainder = 1 20, when remainder = 2 Now x % 3 = 0 and x % 5 = 1 3k % 5 = 1 k = 2 6, when remainder = 1 18, when remainder = 3 Therefore, final solution is 20 + 18 = 38. 38 is a solution LCM of 3,5 = 15. 38 - 15 - 15 = 8. 8 is the least number, that is the solution. But now if I have x % 7 = 3 x % 4 = 2 How do I solve the question ?

Exactly the same way. The equation $x\equiv 3 \mod 7$ tells you that $x=3+7y$ . Plugging this into the second equation gives you $3+7y\equiv 2 \mod 4$, that is $-y\equiv -1 \mod 4$, so $y=1+4z$, and $x=10+28z$, i.e. $x\equiv 10 \mod 28$.

How to prove this: $a^4+b^4+2 \ge 4ab$? How to prove this: $a^4+b^4+2 \ge 4ab$? $a$ and $b$ are reals.

Another possible method is generation of a function and calculating its minimum: $$f(a,b)=a^4+b^4+2-4 a b$$ $$D_a f(a,b)=4 a^3-4 b = 0$$ $$ a^3=b$$ $$ f(a)=a^4+a^{12}+2-4a^4=a^{12}+2-3a^4$$ $$D_a f(a)=12a^{11}-12a^3=0$$ $$a^3 (a^8-1)=0$$ Minimum must be at one of: $$f(0,0)=2$$ $$f(1,1)=0$$ $$f(-1,-1)=0$$ so $$f(a,b)=a^4+b^4+2-4 a b\geq0$$

Number theory: show that ${ 1^2, 2^2, 3^2,... , m^2}$ cannot be a complete residue system Is this an acceptable answer? Question: show that ${ 1^2, 2^2, 3^2,... , m^2}$ cannot be a complete residue system. Since the above has $m$ elements, one must show it cannot be a complete residue system modulo $m$. Consider the complete residue system modulo m: {1,2,3,... ,$m$}. Now between the two the first element $1^2$ and $1$ matches. But what about the second last element $ (m-1)^2$. Here we have $(m-1)^2$ = $m^2 -2m +1$ which is clearly congruent to the first element, $1$, modulo m. So there exist two elements congruent modulo $m$ to each other in ${ 1^2, 2^2, 3^2,... , m^2}$ so it cannot be a complete residue system.

Let's start with an example. Below is a congruence table modulo &11&. \begin{array}{rrr} k & k^2 & k^2 \pmod{11} \\ \hline 0 & 0 & 0\\ 1 & 1 & 1\\ 2 & 4 & 4\\ 3 & 9 & 9\\ 4 & 16 & 5\\ 5 & 25 & 3\\ 6 & 36 & 3\\ 7 & 49 & 5\\ 8 & 64 & 9\\ 9 & 81 & 4\\ 10 & 100 & 1\\ \hline \end{array} Notice that the last column does not contain all eleven of the integers from $0$ to $10$ because, of the eleven numbers, there are repeated numbers: $k^2 \equiv (11-k)^2 \pmod{11}$. This is true for all moduli, N. $$(N-k)^2 \equiv N^2 - 2kN + k^2 \equiv k^2 \pmod N$$

How to solve this ordinary differential equation? I am just trying to find general solution $$\frac{dy}{dx} = 1 + \sqrt{1 - xy}$$

Let $u=-\sqrt{1-xy}$ , Then $y=\dfrac{1-u^2}{x}$ $\dfrac{dy}{dx}=\dfrac{u^2-1}{x^2}-\dfrac{2u}{x}\dfrac{du}{dx}$ $\therefore\dfrac{u^2-1}{x^2}-\dfrac{2u}{x}\dfrac{du}{dx}=1-u$ $\dfrac{2u}{x}\dfrac{du}{dx}=u-1+\dfrac{u^2-1}{x^2}$ Approach $1$: $\dfrac{2u}{x}\dfrac{du}{dx}=\dfrac{(u-1)x^2+(u+1)(u-1)}{x^2}$ $(x^2+u+1)\dfrac{dx}{du}=\dfrac{2ux}{u-1}$ Let $v=x^2$ , Then $\dfrac{dv}{du}=2x\dfrac{dx}{du}$ $\therefore\dfrac{x^2+u+1}{2x}\dfrac{dv}{du}=\dfrac{2ux}{u-1}$ $(x^2+u+1)\dfrac{dv}{du}=\dfrac{4ux^2}{u-1}$ $(v+u+1)\dfrac{dv}{du}=\dfrac{4uv}{u-1}$ This belongs to an Abel equation of the second kind. Let $w=v+u+1$ , Then $v=w-u-1$ $\dfrac{dv}{du}=\dfrac{dw}{du}-1$ $\therefore w\left(\dfrac{dw}{du}-1\right)=\dfrac{4u(w-u-1)}{u-1}$ $w\dfrac{dw}{du}-w=\dfrac{4uw}{u-1}-\dfrac{4u(u+1)}{u-1}$ $w\dfrac{dw}{du}=\dfrac{(5u-1)w}{u-1}-\dfrac{4u(u+1)}{u-1}$ Approach $2$: $\dfrac{2u}{x}\dfrac{du}{dx}=u-1+\dfrac{u^2-1}{x^2}$ $u\dfrac{du}{dx}=\dfrac{u^2}{2x}+\dfrac{xu}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$ Let $u=\sqrt{x}v$ , Then $\dfrac{du}{dx}=\sqrt{x}\dfrac{dv}{dx}+\dfrac{v}{2\sqrt{x}}$ $\therefore\sqrt{x}v\left(\sqrt{x}\dfrac{dv}{dx}+\dfrac{v}{2\sqrt{x}}\right)=\dfrac{xv^2}{2x}+\dfrac{x\sqrt{x}v}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$ $xv\dfrac{dv}{dx}+\dfrac{v^2}{2}=\dfrac{v^2}{2}+\dfrac{x\sqrt{x}v}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$ $xv\dfrac{dv}{dx}=\dfrac{x\sqrt{x}v}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$ $v\dfrac{dv}{dx}=\dfrac{\sqrt{x}v}{2}-\dfrac{1}{2}-\dfrac{1}{2x^2}$ Let $t=\dfrac{x^\frac{3}{2}}{3}$ , Then $\dfrac{dv}{dx}=\dfrac{dv}{dt}\dfrac{dt}{dx}=\dfrac{\sqrt{x}}{2}\dfrac{dv}{dt}$ $\therefore\dfrac{\sqrt{x}v}{2}\dfrac{dv}{dt}=\dfrac{\sqrt{x}v}{2}-\dfrac{1}{2}-\dfrac{1}{2x^2}$ $v\dfrac{dv}{dt}=v-\dfrac{1}{\sqrt{x}}-\dfrac{1}{x^\frac{5}{2}}$ $v\dfrac{dv}{dt}=v-\dfrac{1}{\sqrt[3]{3t}}-\dfrac{1}{(3t)^\frac{5}{3}}$ This belongs to an Abel equation of the second kind in the canonical form. Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf

$ A_n $ =$[\frac{n}{n+1},\frac{n+1}{n+2}] $ be closed subsets find $\bigcup_{n=1}^\infty A_n $ Let $ A_n =\frac{n}{n+1},\frac{n+1}{n+2}] $, $n=1,2,3...$ be closed subsets of real line R. Then $\bigcup_{n=1}^\infty A_n $ is * *(1/2,1) *[1/2,1) *(1/2,1] *[1/2,1] My attempt : think it could be [1/2,1) since $\lim \frac{n+1}{n+2}$=1 . I am not sureit is just a guess.any hints plz.

Hint: Since $f(x)=\frac{x}{x+1}$ is monotonically increasing, we have that for any $0\le x\lt1$, $$ n=\left\lfloor\frac{x}{1-x}\right\rfloor\iff n\le\frac{x}{1-x}\lt n+1\iff\frac{n}{n+1}\le x\lt\frac{n+1}{n+2} $$ Thus, for any $\frac12\le x\lt1$, if $n=\left\lfloor\frac{x}{1-x}\right\rfloor$, then $n\ge1$ and $x\in\left[\frac{n}{n+1},\frac{n+1}{n+2}\right]$. Furthermore, note that for all $n\ge0$, $\frac{n+1}{n+2}\lt1$.

Differential calculus: integrate $\frac{1}{x \log^3 (x)}$ I would like a step by step description of how to integrate $$\frac{1}{x \log^3 (x)}$$ * *I know that the answer is - $\frac{1}{2\log^2(x)}$ *and that the integral of $\frac{1}{\log^2(x)}$ is $\frac{1}{\log(x)}$ *and that the integral of $\frac{1}{\log(x)}$ is $\log(\log(x))$ But I don't see how these are obtained.

For this problem:$$\int\frac{1}{x\log^3(x)}dx\tag{1}$$we can use the substitution:$$u=\log^3(x)\tag{2}$$$$\therefore \log(x)=u^{\frac{1}{3}}\tag{3}$$which leads to:$$du=3\log^2(x)\times\frac{1}{x}dx$$$$=3(u^{\frac{1}{3}})^2\times\frac{1}{x}dx=\frac{3u^{\frac{2}{3}}}{x}dx$$$$\therefore \frac{dx}{x}=\frac{du}{3u^{\frac{2}{3}}}\tag{4}$$Now we substitute this into (1) to get:$$\int\frac{1}{x\log^3(x)}dx=\int\frac{1}{3u^{\frac{2}{3}}\times u}du=\frac{1}{3}\int\frac{1}{u^{\frac{5}{3}}}du$$$$=\frac{1}{3}\int u^{-\frac{5}{3}}du=\frac{1}{3}\left(\frac{u^{-\frac{2}{3}}}{-\frac{2}{3}}\right)=-\frac{1}{2}\times\frac{1}{u^{\frac{2}{3}}}=-\frac{1}{2\log^2(x)}$$

Number of digits $d$ in $d^k$ The title says it all really. For example, how many occurences of $6$ are there in $6^k$? It starts $6, 36, 216, \dots$ so $1, 1, 1,\dots$ The question can now be generalized into any digit or group of digits.

Here is a general formula for base $10$ and $d\in[1,10-1]$: $$\sum\limits_{n=1}^{k}1-\left\lceil\left(\frac{\left\lfloor\frac{d^k}{10^{n-1}}\right\rfloor-10\left\lfloor\frac{d^k}{10^n}\right\rfloor-d}{\left\lfloor\frac{d^k}{10^{n-1}}\right\rfloor-10\left\lfloor\frac{d^k}{10^n}\right\rfloor+d}\right)^2\right\rceil$$ Here is a general formula for base $B$ and $d\in[1,B-1]$: $$\sum\limits_{n=1}^{k}1-\left\lceil\left(\frac{\left\lfloor\frac{d^k}{B^{n-1}}\right\rfloor-B\left\lfloor\frac{d^k}{B^n}\right\rfloor-d}{\left\lfloor\frac{d^k}{B^{n-1}}\right\rfloor-B\left\lfloor\frac{d^k}{B^n}\right\rfloor+d}\right)^2\right\rceil$$

Show that the series $\sum_{k=0} ^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$ Show that the series $\sum_{k=0} ^\infty (-1)^k \dfrac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$ I tried using the ratio test but I got that it equals 1, so it is inconclusive. How can I show that it converges, and that it converges to arctan x? I think I'm trying to show that the Legrange error term is 0 but I'm not sure how to proceed with it - the nth derivatives of arctan x keep changing so...

Modifying Jack D'Aurizio's answer, since $\sum_{k=0}^n x^k = \dfrac{1-x^{n+1}}{1-x}$, putting $-x^2$ for $x$ we get $\sum_{k=0}^n (-1)^kx^{2k} = \dfrac{1-(-1)^{n+1}x^{2n+2}}{1+x^2} = \dfrac1{1+x^2}+\dfrac{(-1)^{n}x^{2n+2}}{1+x^2} $ or $\dfrac1{1+x^2} =\sum_{k=0}^n (-1)^kx^{2k} -\dfrac{(-1)^{n}x^{2n+2}}{1+x^2} $. Integrating from $0$ to $y$, $\begin{array}\\ \arctan(y) &=\int_0^y \dfrac{dx}{1+x^2}\\ &=\int_0^y \left(\sum_{k=0}^n (-1)^{k-1}x^{2k}-(-1)^{n}\dfrac{x^{2n+2}}{1+x^2}\right)dx\\ &=\sum_{k=0}^n (-1)^{k-1}\int_0^y x^{2k}dx-(-1)^{n}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\ &=\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}-(-1)^{n}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\ \end{array} $ so $\arctan(y)-\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1} =(-1)^{n+1}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2} $ so that $\begin{array}\\ |\arctan(y)-\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}| &\le\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\ &\le\int_0^y x^{2n+2}dx\\ &=\dfrac{x^{2n+3}}{2n+3}\\ &\to 0 \qquad\text{as } n \to \infty\\ \end{array} $

Let $x,y,z>0,xyz=1$. Prove that $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\ge \frac34$ Let $x,y,z>0$ and $xyz=1$. Prove that $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{y^3}{(1+x)(1+z)}+\dfrac{z^3}{(1+x)(1+y)}\ge \dfrac34$ My attempt: Since it is given that $xyz=1$, I tried substituting $x=\dfrac{a}{b},y=\dfrac{b}{c},z=\dfrac{c}{a}$. But the expansion looked really ugly and I didn't think I could make out anything out of it. So, I made another attempt, if each element was greater than $\dfrac{1}{4}$, we could have a solution, so, treating that way, I get $4x^3\ge 1+x+y+xy, 4y^3\ge 1+x+z+xz, 4z^3\ge 1+x+y+xy$. Using AM-GM I get an equality. So, please help. Thank you.

As another approach, you could prove it as follows. Due to Hölder's inequality, we have: $$ \left(\sum_{cyc} \frac{x^3}{(1+y)(1+z)}\right)\cdot\left(\sum_{cyc} (1+y)\right)\cdot\left(\sum_{cyc} (1+z)\right)\ge(x+y+z)^3\iff \sum_{cyc} \frac{x^3}{(1+y)(1+z)}≥\frac{(x+y+z)^3}{\left(\sum_{cyc} (1+y)\right)\cdot\left(\sum_{cyc} (1+z)\right)}=\frac{(x+y+z)^3}{\left(3+x+y+z\right)^2} $$ So it remains to prove that: $$ \frac{(x+y+z)^3}{\left(3+x+y+z\right)^2}≥\frac34 $$ Setting $x+y+z=a$, this is equivalent to: $$ \frac{a^3}{\left(3+a\right)^2}≥\frac34\iff 4a^3≥27+18a+3a^2\iff 4a^3-3a^2-18a-27≥0 $$ We have equality at $a=3$, so we can factor out $a-3$: $$ (a-3)\left(4a^2+9a+9\right)≥0 $$ But since $$ a=x+y+z≥3(xyz)^{\frac{1}{3}}=3\iff a-3≥0 $$ due to AM-GM, this is always true, so we're done.

Prove $\int_t^{\infty} e^{-x^2/2}\,dx > e^{-t^2/2}\left(\frac{1}{t} - \frac{1}{t^3}\right)$ How to formally prove the following inequality - $$\int_t^{\infty} e^{-x^2/2}\,dx > e^{-t^2/2}\left(\frac{1}{t} - \frac{1}{t^3}\right)$$

For a better lower-bound you may use the following proof by @robjohn: $$x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t \le \int_x^\infty e^{-t^2/2}\,t\,\mathrm{d}t =e^{-x^2/2}$$ Integrate both sides of the preceding: $$ \begin{align} \int_s^\infty e^{-x^2/2}\,\mathrm{d}x &\ge\int_s^\infty x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t\,\mathrm{d}x\\ &=\int_s^\infty\int_s^txe^{-t^2/2}\,\mathrm{d}x\,\mathrm{d}t\\ &=\int_s^\infty\frac12(t^2-s^2)e^{-t^2/2}\,\mathrm{d}t\\ \left(1+\frac12s^2\right)\int_s^\infty e^{-x^2/2}\,\mathrm{d}x &\ge\frac12\int_s^\infty t^2e^{-t^2/2}\,\mathrm{d}t\\ &=-\frac12\int_s^\infty t\,\mathrm{d}e^{-t^2/2}\\ &=\frac12se^{-s^2/2}+\frac12\int_s^\infty e^{-t^2/2}\,\mathrm{d}t\\ \left(s+\frac1s\right)\int_s^\infty e^{-x^2/2}\,\mathrm{d}x &\ge e^{-s^2/2} \end{align} $$ and note that $\displaystyle \left(s+\frac{1}{s}\right)^{-1} = \frac{s}{1+s^2} > \frac{1}{s} - \frac{1}{s^3}$ for $s > 0$, which is equivalent to $\displaystyle s^4 > s^4 - 1$. Expalanation for integartion by parts: \begin{align*} &\int_x^{\infty} e^{-t^2/2} \mathrm dt\\ =& \int_x^{\infty} \color{red}{\frac{1}{t}} .\color{blue}{te^{-t^2/2}} \mathrm dt\\ =& \left[\color{red}{\frac{1}{t}} .\int\color{blue}{te^{-t^2/2}}\,dt\right]_x^{\infty} - \int_x^{\infty} \left( \color{red}{\frac{1}{t}} \right )' \left (\int\color{blue}{te^{-t^2/2}}\,dt\right)\mathrm dt\\ =& \frac{e^{-x^2/2}}{x} - \int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \mathrm dt. \end{align*} Now, for the second integral that we obtained in the previous line we employ similar idea as done above: \begin{align*} &\int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \mathrm dt\\ =& \int_x^{\infty} \color{red}{\frac{1}{t^3}} .\color{blue}{te^{-t^2/2}} \mathrm dt\\ =& \left[\color{red}{\frac{1}{t^3}} .\int\color{blue}{te^{-t^2/2}}\,dt\right]_x^{\infty} - \int_x^{\infty} \left( \color{red}{\frac{1}{t^3}} \right )' \left (\int\color{blue}{te^{-t^2/2}}\,dt\right)\mathrm dt\\ =& \frac{e^{-x^2/2}}{x^3} -3 \int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \mathrm dt. \end{align*} Combining these together we have $$\int_x^{\infty} e^{-t^2/2} = e^{-x^2/2}\left(\frac{1}{x} - \frac{1}{x^3}\right)+3 \int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \mathrm dt$$ Since, the integrand $\dfrac{e^{-t^2/2}}{t^4}$ is always positive, we get the desired inequality: $$\int_x^{\infty} e^{-t^2/2} > e^{-x^2/2}\left(\frac{1}{x} - \frac{1}{x^3}\right)$$

Sum $\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$ I want to evaluate the sum $$\large\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}.$$ I did partial fraction decomposition to get $$\frac{1}{2^n}\left(\frac{-1}{n^2+2n+2}+\frac{4}{n^2-2n+2}+1\right)$$ I am absolutely stuck after this.

Note that $$\dfrac{n^4+3n^2+10n+10}{2^n(n^4+4)}=\dfrac{1}{2^n}+\dfrac{3n^2+10n+6}{2^n[(n^2+2)^2-(2n)^2]}$$ Then let's find constants $A,B$ suct that $$\dfrac{3n^2+10n+6}{(n^2+2n+2)(n^2-2n+2)}=\dfrac{A(n+1)+B}{(n+1)^2+1}-4\Big[\dfrac{A(n-1)+B}{(n-1)^2+1}\Big]$$ to obtain the form $$f(n+1)-f(n-1).$$ For $n=-1,$ we have $-\dfrac{1}{5}=B+4\Big(\dfrac{2A-B}{5}\Big)\iff8A+B=-1.$ For $n=+1,$ we have $\dfrac{19}{5}=\Big(\dfrac{2A+B}{5}\Big)-4B\iff2A-19B=19.$ By solving these equations, $$A=0,\,\,\,\,\,B=-1$$ Now $$\dfrac{n^4+3n^2+10n+10}{2^n(n^4+4)}=\dfrac{1}{2^n}-\dfrac{1}{2^n((n+1)^2+1)}+\dfrac{1}{2^{n-2}((n-1)^2+1)}$$ Can you continue from here? Good Luck.

Derivative of $\frac { y }{ x } +\frac { x }{ y } =2y$ with respect to $x$ $$\frac { y }{ x } +\frac { x }{ y } =2y$$ Steps I took: $$\frac { d }{ dx } \left[yx^{ -1 }1+xy^{ -1 }\right]=\frac { d }{ dx } [2y]$$ $$\frac { dy }{ dx } \left(\frac { 1 }{ x } \right)+(y)\left(-\frac { 1 }{ x^{ 2 } } \right)+(1)\left(\frac { 1 }{ y } \right)+(x)\left(-\frac { 1 }{ y^{ 2 } } \right)\frac { dy }{ dx } =(2)\frac { dy }{ dx } $$ $$-\frac { y }{ x^{ 2 } } +\frac { 1 }{ y } =(2)\frac { dy }{ dx } -\left(\frac { 1 }{ x } \right)\frac { dy }{ dx } +\left(\frac { x }{ y^{ 2 } } \right)\frac { dy }{ dx } $$ $$-\frac { y }{ x^{ 2 } } +\frac { 1 }{ y } =\left(2-\frac { 1 }{ x } +\frac { x }{ y^{ 2 } } \right)\frac { dy }{ dx } $$ $$\frac { -\frac { y }{ x^{ 2 } } +\frac { 1 }{ y } }{ \left(2-\frac { 1 }{ x } +\frac { x }{ y^{ 2 } } \right) } =\frac { dy }{ dx } $$ At this point I get stuck because once I simplify the result of the last step I took, the answer is not what it should be. I think that I am making a careless mistake somewhere but I cannot seem to find it. Hints only, please. The direct answer does nothing for me. Actual answer: $$\frac { d y}{ dx } =\frac { y(y^{ 2 }-x^{ 2 }) }{ x(y^{ 2 }-x^{ 2 }-2xy^{ 2 }) } $$

$\text{ Assuming your y' is correct... } \\ \text{ then we should get rid of the compound fractions.. } \\ y'=\frac{\frac{-y}{x^2}+\frac{1}{y}}{2-\frac{1}{x}+\frac{x}{y^2}} \\ \text{ now we need to multiply top and bottom by } \\ x^2y^2 \text{ this is the lcm of the bottoms of the mini-fractions } \\ y'=\frac{-y(y^2)+1(x^2)(y)}{2x^2y^2-xy^2+x(x^2)} \\ y'=\frac{-y^3+x^2y}{2x^2y^2-xy^2+x^3}$

In how many ways can you distribute 100 lemons between Dana, Sara and Lena so that Lena will get more lemons than Dana? Assume Dana has 0 lemons, so Lena must have 1 lemon. Now all i need to distribute is $$x_1 + x_2 = 99 \text{ // because Lena already has 1 and Dana has 0}$$ The answer to above is 100. Now assume Dana has 1 Lemon. So Lena must have 2 Lemons and now all I need to distribute is $$x_1+x_2 = 97 \text{ // because Lena has 2 and Dana has 1.}$$ the answer again to above is 98. and it goes so on until 2. so i think the answer is : $2 + 4 + 6 + ... + 98 + 100$ which is 2550. that was the answer i wrote in my exam... so, am I right?

Alternatively, let $D, S$ and $L$ be the values, and let $L=D+1+L_0$. Then you want non-negative integer solutions to $1+D+S+L_0=100$, or $2D+S+L_0=99$. You are doing $$\sum_{D=0}^{49} \sum_{S=0}^{99-2D} 1=\sum_{D=0}^{49} (100-2D)$$ which is a correct way to count this value. A generating function solution would be to write it as seeking the coefficient of $99$ in the power series: $$(1+x+x^2\cdots)^2(1+x^2+x^4\cdots) = \frac{1}{(1-x)^3(1+x)}$$ Then using partial fractions: $$\frac{1}{(1-x)^3(1+x)} = \frac{a}{(1-x)^3} +\frac{b}{(1-x)^2} + \frac{c}{1-x} + \frac{d}{1+x}$$ Then you can get an exact formula for any number of lemons. Wolfram Alpha gives the values: $$a=\frac{1}{2}, b=\frac{1}{4},c=\frac{1}{8},d=\frac{1}{8}$$ Then the number of ways to distribute $N$ lemon is the coeficient of $x^{N-1}$ which is: $$\frac{1}{2}\binom{N+1}{2} + \frac{1}{4}\binom{N}{1} + \frac{1}8\left(1+(-1)^{N-1}\right)=\left\lceil\frac{(N+2)N}{4}\right\rceil$$ When $N$ is even, it is exactly $$\frac{N(N+2)}{4}=2\frac{\frac N2\left(\frac N2+1\right)}{2} = 2+4+6+\cdots+ N.$$ When $N$ is odd, we get $$\left(\frac{N+1}2\right)^2 = 1+3+5+\cdots + N$$ Again we get $2550$ when $N=100$.

$\alpha = \frac{1}{2}(1+\sqrt{-19})$ I have asked a similar question here before, which was about ring theory, but it is slightly different today and very trivial. $\alpha = \frac{1}{2}(1+\sqrt{-19})$ Here, $\alpha$ is a root of $\alpha^2 - \alpha + 5$ and $\alpha^2 = \alpha - 5$, but I can't seem to understand this. Could anyone please explain how $\alpha$ is a root of $\alpha^2 - \alpha + 5$?

$\alpha^2-\alpha+5=\frac{1}{4}(1+\sqrt{-19})^2-\frac{1}{2}(1+\sqrt{-19})+5=\frac{1}{4}(1+2\sqrt{-19}+(-19))-\frac{1}{4}(2+2\sqrt{-19})+\frac{1}{4}\cdot 20=\frac{1}{4}(1+2\sqrt{-19}+(-19)-2-2\sqrt{-19}+20)=0$

Proving $\frac{n^n}{e^{n-1}} 2$. I am trying to prove $$\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^{n}} \text{ for all }n > 2.$$ Here is the original source (Problem 1B, on page 12 of PDF) Can this be proved by induction? The base step $n=3$ is proved: $\frac {27}{e^2} < 6 < \frac{256}{e^3}$ (since $e^2 > 5$ and $e^3 < 27$, respectively). I can assume the case for $n=k$ is true: $\frac{k^k}{e^{k-1}}<k!<\frac{(k+1)^{k+1}}{e^{k}}$. For $n=k+1$, I am having trouble: \begin{align} (k+1)!&=(k+1)k!\\&>(k+1)\frac{k^k}{e^{k-1}}\\&=e(k+1)\frac{k^k}{e^{k}} \end{align} Now, by graphing on a calculator, I found it true that $ek^k >(k+1)^k$ (which would complete the proof for the left inequality), but is there some way to prove this relation? And for the other side of the inequality, I am also having some trouble: \begin{align} (k+1)!&=(k+1)k!\\&<(k+1)\frac{(k+1)^{k+1}}{e^{k}}\\&=\frac{(k+1)^{k+2}}{e^k}\\&<\frac{(k+2)^{k+2}}{e^k}. \end{align} I can't seem to obtain the $e^{k+1}$ in the denominator, needed to complete the induction proof.

Proof: We will prove the inequality by induction. Since $e^2 > 5$ and $e^3 < 27$, we have $$\frac {27}{e^2} < 6 < \frac{256}{e^3}.$$ Thus, the statement for $n=3$ is true. The base step is complete. For the induction step, we assume the statement is true for $n=k$. That is, assume $$\frac{k^k}{e^{k-1}}<k!<\frac{(k+1)^{k+1}}{e^{k}}.$$ We want to prove that the statement is true for $n=k+1$. It is straightforward to see for all $k > 2$ that $\left(1+\frac 1k \right)^k < e < \left(1+\frac 1k \right)^{k+1}$; this algebraically implies $$\left(\frac k{k+1} \right)^{k+1} < \frac 1e < \left(\frac k{k+1} \right)^k. \tag{$*$}$$ A separate induction proof for the left inequality of $(*)$ establishes $\left(\frac{k+1}{k+2} \right)^{k+2}<\frac 1e$. We now have \begin{align} (k+1)! &= (k+1)k! \\ &< (k+1) \frac{(k+1)^{k+1}}{e^k} \\ &= \frac{(k+1)^{k+2}}{e^k} \\ &= \frac{(k+1)^{k+2}}{e^k} \left( \frac{k+2}{k+2} \right)^{k+2} \\ &= \frac{(k+2)^{k+2}}{e^k} \left( \frac{k+1}{k+2} \right)^{k+2} \\ &< \frac{(k+2)^{k+2}}{e^k} \frac 1e \\ &= \frac{(k+2)^{k+2}}{e^{k+1}} \end{align} and \begin{align} (k+1)! &= (k+1)k! \\ &> (k+1) \frac{k^k}{e^{k-1}} \\ &= (k+1) \frac{k^k}{e^{k-1}} \left(\frac{k+1}{k+1} \right)^k \\ &= \frac{(k+1)^{k+1}}{e^{k-1}} \left( \frac k{k+1} \right)^k \\ &> \frac{(k+1)^{k+1}}{e^{k-1}} \frac 1e \\ &= \frac{(k+1)^{k+1}}{e^k}. \end{align} We have established that the statement $$\frac{(k+1)^{k+1}}{e^k}<(k+1)!<\frac{(k+2)^{k+2}}{e^{k+1}}$$ for $n=k+1$ is true. This completes the proof.

How are first digits of $\pi$ found? Since Pi or $\pi$ is an irrational number, its digits do not repeat. And there is no way to actually find out the digits of $\pi$ ($\frac{22}{7}$ is just a rough estimate but it's not accurate). I am talking about accurate digits by either multiplication or division or any other operation on numbers. Then how are the first digits of $\pi$ found - 3.1415926535897932384626433832795028841971693993... In fact, more than 100,000 digits of $\pi$ are found (sources - 100,000 digits of $\pi$) How is that possible? If these digits of $\pi$ are found, then it must be possible to compute $\pi$ with some operations. (I am aware of breaking of circle into infinite pieces method but that doesn't give accurate results.) How are these digits of $\pi$ found accurately? Can it be possible for a square root of some number to be equal to $\pi$?

In the 18th century, Leonard Euler discovered an elegant formula: $$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\dots$$ The more terms you add, the more accurate the calculation of $π$ gets. $$\frac{π^4}{90}=\frac{1}{1^4}=1.000$$ then $π=3.080$ $$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}=1.0625$$ then $π=3.080$ $$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}=1.0748$$ then $π=3.136$ $$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}=1.0788$$ then $π=3.139$ $$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}=1.0804$$ then $π=3.140$ etc This is a slow way to calculate them, since after the 100th term, $π$ is $3.141592$, but it calculates them nontheless.

Find The Equation This is the question: Find the equations of the tangent lines to the curve $y = x − \frac 1x + 1$ that are parallel to the line $x − 2y = 3$. There are two answers: 1) smaller y-intercept; 2) larger y-intercept The work: The slope of the line is (1/2). $y' = ((x + 1) - (x - 1))/(x + 1)^2 = 2/(x + 1)^2 = 1/2$ => $(x + 1)^2 = 4$ => $(x + 1) = 2$ and $ x + 1 = -2$ => $x = 1$ and $x = -3$ At $x = 1$, $y = (x - 1)/(x + 1) = 0$ The equation of the tangent is $y/(x - 1) = (1/2)$ => $2y = x - 1$ => $x - 2y - 1 = 0$ At $x = -3$,$ y = (x - 1)/(x + 1) = -4/-2 = 2$ The equation of the tangent is $(y - 2)/(x + 3) = (1/2)$ => $2y - 4 = x + 3$ =>$ x - 2y + 7 = 0$

there are no points on the graph of $y = x - \frac1x + 1$ has a tangent that is parallel to $x - 2y = 3.$ i will explain why. you will get a better idea of the problem if you can draw the graph of $y = x - \frac1x + 1$ either on your calculator or by hand. the shape of the graph is called a hyperbola, similar looking to the rectangular hyperbola $y = \frac1x$ but here the asymptotes $x = 0$ and $y = x+1$ are not orthogonal. you will two branches one in the left half plane and the other in the right half plane. if you at the right branch, you can see that the slope starts very large positive fro $x$ very close to zero and positive, then gradually decreases to match the slope $1$ of the asymptote. that is the smallest slope on this graph. but the line you have $x - 2y = 3$ has slope $\frac12$ that is smaller than $1,$ the least slope on the graph. here is how you see with calculus. taking the derivative of $y = x - \frac1x + 1$ you find $$\frac{dy}{dx} = 1 + \frac1{x^2} \ge 1 \text{ for all} x. $$

$\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$, where numbers $a$ and $b$ are rational If $a$ and $b$ are rational numbers such that $\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$ Then what is the value of $a$? The answer is $-1$. $$\sqrt{4 - 2\sqrt{3}} = a + b\sqrt{3}$$ $$4 - 2\sqrt{3} = 2^2 - 2\sqrt{3}$$ Let $u =2$ hence, $$\sqrt{u^2 - \sqrt{3}u} = a + b\sqrt{3}$$ $$u^2 - \sqrt{3}u = u(u - \sqrt{3})$$ $$a + b\sqrt{3} = \sqrt{u}\sqrt{u - \sqrt{3}}$$ What should I do?

Using formula $$\sqrt{a\pm\sqrt{b}}=\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}$$ you will get $$\sqrt{4-2\sqrt3}=\sqrt{4-\sqrt{12}}=\sqrt{\dfrac{4+\sqrt{16-12}}{2}}-\sqrt{\dfrac{4-\sqrt{16-12}}{2}}=-1+\sqrt3$$ So, $a=-1$ and $b=1$.

how to compute this limits given these conditions. if $f(1)=1$ and $f'(x)=\frac{1}{x^2+[f(x)]^2}$ then compute $\lim\limits_{x\to+\infty}f(x)$ i tried to write it was $$\frac{dy}{dx}=\frac{1}{x^2+y^2}\\ (x^2+y^2)\frac{dy}{dx}=1\\ (x^2+y^2)dy=dx$$ by the help $$\begin{align} f(x)&\le1+\int_1^x\frac{dt}{1+t^2}\\ &\le1+\arctan t\bigg|_1^x\\ &\le1+\arctan x-\arctan 1\\ &\le1+\arctan x-\frac{\pi}{4} \end{align}$$ so $$\begin{align} \lim\limits_{x\to+\infty}f(x)&\le\lim\limits_{x\to+\infty}1+\arctan x-\frac{\pi}{4}\\ &\le1+\frac{\pi}{2}-\frac{\pi}{4}\\ &\le1+\frac{\pi}{4}=\frac{4+\pi}{4} \end{align}$$

Since $f'(x)>0 $ and $f(1)=1$ then we have $$ \frac{1}{x^2+f(x)^2}\leq \frac{1}{x^2+1}. $$ Also we will have $$ \int_{1}^{x}f'(t)dt = \int_{1}^{x}\frac{dt}{t^2+f(t)^2} \leq \int_{1}^{x}\frac{dt}{t^2+1} $$ $$ \implies f(x) \leq 1+ \int_{1}^{x}\frac{dt}{t^2+1} . $$ Try to finish the problem.

Question in relation to completing the square In description of "completing the square" at http://www.purplemath.com/modules/sqrquad.htm the following is given : I'm having difficulty understanding the third part of the transformation. Where is $ -\frac{1}{4}$ derived from $-\frac{1}{2}$ ? Why is $ -\frac {1}{4}$ squared to obtain $ \frac {1}{16}$ ?

Let's begin with the equation $$x^2 - \frac{1}{2}x = \frac{5}{4}$$ What the author wants to do is to create a perfect square on the left hand side. That is, the author wants to transform the expression on the left hand side into the form $(a + b)^2 = a^2 + 2ab + b^2$. Assume that $$a^2 + 2ab = x^2 - \frac{1}{2}x$$ If we let $a = x$, then we obtain \begin{align*} x^2 + 2bx & = x^2 - \frac{1}{2}x\\ 2bx & = -\frac{1}{2}x\\ \end{align*} Since the equation $2bx = -\frac{1}{2}x$ is an algebraic identity that holds for each real number $x$, it holds when $x = 1$. Thus, \begin{align*} 2b & = -\frac{1}{2}\\ b & = -\frac{1}{4} \end{align*} Therefore, \begin{align*} a^2 + 2ab + b^2 & = x^2 + 2\left(-\frac{1}{4}\right)x + \left(-\frac{1}{4}\right)^2\\ & = x^2 - \frac{1}{2}x + \frac{1}{16} \end{align*} If we add $1/16$ to the left hand side of the equation, we must add $1/16$ to the right hand side of the equation to balance it, which yields $$x^2 - \frac{1}{2}x + \frac{1}{16} = \frac{5}{4} + \frac{1}{16}$$ By construction, the expression on the left hand side is the perfect square $(x - \frac{1}{4})^2$, so we obtain $$\left(x - \frac{1}{4}\right)^2 = \frac{21}{16}$$ We can now solve the quadratic equation by taking square roots, which is why we wanted to transform the left hand side into a perfect square.

Prove $\lim\limits_{n \to \infty} \sup \left ( \frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!)} \right ) ^ {\frac 1 n} = \frac {e^2} 4$ This is a problem in Heuer (2009) "Lerbuch der Analysis Teil 1" on page 366. I assume that the proof should use $e = \sum\limits_{k = 0}^{\infty} \frac 1 {k!}$, but I cannot come further.

If $a_n=\frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!}$. Compute the limit $$\begin{align}\lim \frac{a_{n+1}}{a_n}&=\lim \frac{\frac{(2n + 1)^{2n + 1}}{2^{2n+2} (2n+2)!}}{\frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!}}\\&=\lim\frac{(2n+1)(2n+1)}{4(2n+1)(2n+2)}\left(\frac{2n+1}{2n-1}\right)^{2n-1}\\&=\frac{1}{4}\lim\left[\left(1+\frac{2}{2n-1}\right)^{\frac{2n-1}{2}}\right]^2\\&=\frac{1}{4}e^2\end{align}$$ Since $\lim\frac{a_{n+1}}{a_n}=\frac{e^2}{4}$ it follows that $\lim a_n^{1/n}=\frac{e^2}{4}$.

How to calculate derivative of $f(x) = \frac{1}{1-2\cos^2x}$? $$f(x) = \frac{1}{1-2\cos^2x}$$ The result of $f'(x)$ should be equals $$f'(x) = \frac{-4\cos x\sin x}{(1-2\cos^2x)^2}$$ I'm trying to do it in this way but my result is wrong. $$f'(x) = \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} = \frac {1-2\cos^2x-(1-(2\cos^2x)')}{(1-2\cos^2x)^2} = $$ $$=\frac {-2\cos^2x + 2(2\cos x(\cos x)')}{(1-2\cos^2x)^2} = \frac {-2\cos^2x+2(-2\sin x\cos x)}{(1-2\cos^2x)^2} = $$ $$\frac {-2\cos^2x-4\sin x\cos x}{(1-2\cos^2x)^2}$$

The problem is in this step, from here $$f'(x) = \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} $$ to here $$\frac {1-2\cos^2x-(1-(2\cos^2x)')}{(1-2\cos^2x)^2}$$ because $$1'=0.$$

Limit of an unknown cubic expression Let $f(x)=ax^3+bx^2+cx+d$ and $g(x)=x^2+x-2$. If $$\lim_{x \to 1}\frac{f(x)}{g(x)}=1$$ and $$\lim_{x \to -2}\frac{f(x)}{g(x)}=4$$ then find the value of $$\frac{c^2+d^2}{a^2+b^2}$$ Since the denominator is tending to $0$ in both cases, the numerator should also tend to $0$, in order to get indeterminate form. But it led to more and more equations.

What about? $$f(x)=-x^3+x^2+4x-4$$ So: $$\frac{c^2+d^2}{a^2+b^2}=32/2=16$$ Let us make ansatz that $$f(x)=a(x+\alpha)(x-1)(x+2)$$ So: $$a(1+\alpha)=1,a(\alpha-2)=4\implies a=-1,\alpha=-2$$ so: $$f(x)=-(x-1)(x+2)(x-2)=-x^3+x^2+4x-4$$

Mathematical induction for inequalities: $\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1$ Prove by induction: $$\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1$$ adding $1/(3m+4)$ as the next $m+1$ value proves pretty fruitless. Can I make some simplifications in the inequality that because the $m$ step is true by the inductive hypothesis, the 1 is already less than all those values?

More generally (one of my favorite phrases), let $s_k(n) =\sum\limits_{i=n+1}^{kn+1} \frac1{i} $. I will show that $s_k(n+1)>s_k(n)$ for $k \ge 3$. In particular, for $n \ge 1$ $s_3(n) \ge s_3(1) =\frac1{2}+\frac1{3}+\frac1{4} =\frac{6+4+3}{12} =\frac{13}{12} > 1 $. $\begin{array}\\ s_k(n+1)-s_k(n) &=\sum\limits_{i=n+2}^{kn+k+1} \frac1{i}-\sum\limits_{i=n+1}^{kn+1} \frac1{i}\\ &=\sum\limits_{i=n+2}^{kn+1} \frac1{i}+\sum\limits_{i=kn+2}^{kn+k+1} \frac1{i} -\left(\frac1{n+1}+\sum\limits_{i=n+2}^{kn+1} \frac1{i}\right)\\ &=\sum\limits_{i=kn+2}^{kn+k+1} \frac1{i}-\frac1{n+1}\\ &=\sum\limits_{i=2}^{k+1} \frac1{kn+i}-\frac1{n+1}\\ &=\frac1{kn+2}+\frac1{kn+k+1}+\sum\limits_{i=3}^{k} \frac1{kn+i}-\frac1{n+1}\\ \end{array} $ $\sum\limits_{i=3}^{k} \frac1{kn+i} \ge \sum\limits_{i=3}^{k} \frac1{kn+k} = \frac{k-2}{kn+k} $. If we can show that $\frac1{kn+2}+\frac1{kn+k+1} \ge \frac{2}{kn+k} $, then $s_k(n+1)-s_k(n) \ge \frac{2}{kn+k}+\frac{k-2}{kn+k}-\frac1{n+1} = \frac{k}{kn+k}-\frac1{n+1} = \frac{1}{n+1}-\frac1{n+1} =0 $. But $\begin{array}\\ \frac1{kn+2}+\frac1{kn+k+1}-\frac{2}{kn+k} &=\frac{kn+k+1+(kn+2)}{(kn+2)(kn+k+1)}-\frac{2}{kn+k}\\ &=\frac{2kn+k+3}{(kn+2)(kn+k+1)}-\frac{2}{kn+k}\\ &=\frac{(2kn+k+3)(kn+k)-2(kn+2)(kn+k+1)}{(kn+2)(kn+k+1)(kn+k)}\\ \end{array} $ Looking at the numerator, $\begin{array}\\ (2kn+k+3)(kn+k)-2(kn+2)(kn+k+1) &=2k^2n^2+kn(k+3+2k)+k(k+3)\\ &-2(k^2n^2+kn(k+3)+2(k+1)\\ &=2k^2n^2+kn(3k+3)+k(k+3)\\ &-2k^2n^2-2kn(k+3)-4(k+1)\\ &=kn(3k+3)+k(k+3)\\ &-kn(2k+6)-4(k+1)\\ &=kn(k-3)+k(k+3)-4(k+1)\\ &=kn(k-3)+k^2-k-4\\ &> 0 \quad\text{for $k \ge 3$} \end{array} $ and we are done.

How can I demonstrate that $x-x^9$ is divisible by 30? How can I demonstrate that $x-x^9$ is divisible by $30$ whenever $x$ is an integer? I know that $$x-x^9=x(1-x^8)=x(1-x^4)(1+x^4)=x(1-x^2)(1+x^2)(1+x^4)$$ but I don't know how to demonstrate that this number is divisible by $30$.

Let's factor $x^9-x$ like you have done: $$ x^9-x=(x-1)x(x+1)(x^2+1)(x^4+1).\tag{$*$} $$ Let's look at the RHS. The product of the first 2 terms is divisible by $2$ because it consists of 2 consecutive integers. Similarly, the product of the first 3 terms is divisible by $3$. Now, if you had $$ (x-2)(x-1)x(x+1)(x+2) $$ then of course that would be divisible by $5$ as well. But note this $$ (x-1)x(x+1)(x^2+1)-(x-2)(x-1)x(x+1)(x+2)=5x(x^2-1)\equiv 0\pmod{5}. $$ So the product of the first 4 terms of the RHS of ($*$) is also divisible by $5$. Now you're done.

Proof of an Limit Using the formal definition of convergence, Prove that $\lim\limits_{n \to \infty} \frac{3n^2+5n}{4n^2 +2} = \frac{3}{4}$. Workings: If $n$ is large enough, $3n^2 + 5n$ behaves like $3n^2$ If $n$ is large enough $4n^2 + 2$ behaves like $4n^2$ More formally we can find $a,b$ such that $\frac{3n^2+5n}{4n^2 +2} \leq \frac{a}{b} \frac{3n^2}{4n^2}$ For $n\geq 2$ we have $3n^2 + 5n /leq 3n^2. For $n \geq 0$ we have $4n^2 + 2 \geq \frac{1}{2}4n^2$ So for $ n \geq \max\{0,2\} = 2$ we have: $\frac{3n^2+5n}{4n^2 +2} \leq \frac{2 \dot 3n^2}{\frac{1}{2}4n^2} = \frac{3}{4}$ To make $\frac{3}{4}$ less than $\epsilon$: $\frac{3}{4} < \epsilon$, $\frac{3}{\epsilon} < 4$ Take $N = \frac{3}{\epsilon}$ Proof: Suppose that $\epsilon > 0$ Let $N = \max\{2,\frac{3}{\epsilon}\}$ For any $n \geq N$, we have that $n > \frac{3}{\epsilon}$ and $n>2$, therefore $3n^2 + 5n^2 \leq 6n^2$ and $4n^2 + 2 \geq 2n^2$ Then for any $n \geq N$ we have $|s_n - L| = \left|\frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}\right|$ $ = \frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}$ $ = \frac{10n-3}{8n^2+4}$ Now I'm not sure on what to do. Any help will be appreciated.

Perhaps simpler: With the Squeeze Theorem: $$\frac34\xleftarrow[x\to\infty]{}\frac{3n^2}{4n^2}\le\frac{3n^2+5n}{4n^2+2}\le\frac{3n^2+5n}{4n^2}=\frac34+\frac54\frac1{n}\xrightarrow[n\to\infty]{}\frac34+0=\frac34$$ With arithmetic of limits: $$\frac{3n^2+5n}{4n^2+2}=\frac{3n^2+5n}{4n^2+2}\cdot\frac{\frac1{n^2}}{\frac1{n^2}}=\frac{3+\frac5n}{4+\frac2{n^2}}\xrightarrow[n\to\infty]{}\frac{3+0}{4+0}=\frac34$$

If $ f(x \cdot f(y) + f(x)) = y \cdot f(x) + x $, then $f(x)=x$ Let $ f : \mathbb{Q} \rightarrow \mathbb{Q} $ be a function which has the following property: $$ f(x \cdot f(y) + f(x)) = y \cdot f(x) + x \;,\; \forall \; x, y \in \mathbb{Q} $$ Prove that $ f(x) = x, \; \forall \; x, y \in \mathbb{Q} $. So far, I've found that $f(f(x)) = x$, $f(0) = 0$ and $f(-1) = -1$. (For $f(0)=0$, we substitute $x=0$ to arrive at $f(f(0))-yf(0)$ identically $0$ for all rational $y$; for $f(f(x))=x$, we substitute $y=0$ and use $f(0)=0$. For $f(-1) = -1$, substitute $x=y=-1$ to get $f(0)=-f(-1)-1$, and use $f(0)=0$.)

If you have $f(f(x))=x$ it means that the function is onto Furthermore, assume that $f(y_1) = f(y_2)$ for some $y_1 \neq y_2$ $$ f(-f(y_1)-1 ) =f(-f(y_1)-1 ) $$ $$\Rightarrow -y_1 -1 = -y_2 -1 $$ $$\Rightarrow y_1 = y_2 !! $$ Thus the function is $1$ $to$ $1$, its inverse $f^{-1}$ exists, in addition to $f(f(x))=x$, we have $f^{-1}=f$ From above, you get $f(f(1)+1) = f(1)+1 $ $f^{-1} = f \Rightarrow f(yf(x)+x) = f^{-1}(yf(x)+x) = xf(y) +f(x) $ For some $x_0=f(1)$, $f(x_0)=f(f(1))=1$ and putting $y=1$ $x_0 f(1) + f(x_0) = f(1 f(x_0) + x_0 )$ $\Rightarrow f(1)^2 +1 = f(f(1)+1) = 2f(1) \Rightarrow f(1)=1$ $\Rightarrow f(y+1)=f(y)+1$ by putting $x=1$ into $f(yf(x)+x) = xf(y) +f(x) $ From the original definition, putting $y=1$ gives $f(x+f(x)) = f(x) +x$ Since $f(x)$ is onto, $f(x)+x$ is also onto(to be proved), therefore we can find $x$ such that $y=x+f(x)$ for all $y$ then $f(y) =y$ I cannot prove that it is onto... However, I found another way by looking at the answer of Willard Zhan. He has proven $f(1/q) = 1/q$ for integers $q$ For all $\frac{p}{q}$, it can be always written as $\frac{m+1}{q}$, where $m$ is also an integer. putting $y=m$ and $x=\frac{1}{q}$ into $f(yf(x)+x)=xf(y)+f(x)$ $$f\left( \frac{m+1}{q}\right)=f\left( \frac{m}{q} +\frac{1}{q}\right) = \frac{f(m)+1}{q} = \frac{m+1}{q}$$ since we have proved $f(m)=m$ for integers. I think it completes the proof.

Find two linearly independent solutions of the differential equation $(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$ I want to find two linearly independent solutions of the differential equation $$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$$ Previously I have seen that the following holds for the differential equation $y''+ \frac{1}{x}y'-\frac{1}{x^2}y=0, x>0$: * *We are looking for solutions of the differential equation of the form $x^r$. Then the function $x^r$ is a solution of the differential equation at $(0,+\infty)$ if: $$r(r-1)x^{r-2}+ \frac{1}{x} r x^{r-1}- \frac{1}{x^2}x^r=0 \forall x >0 \Rightarrow r=1 \text{ or } r=-1$$ * *So, the functions $y_1(x)=x, y_2= \frac{1}{x}$ are solutions of the differential equation and it also holds that they are linearly indepedent since $W(y_1, y_2) \neq 0$ For this differential equation $$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$$ I thought the following: $(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3} \Rightarrow y''+ \frac{3}{3x-1}y'-\frac{9}{(3x-1)^2}y=0$ * *We are looking for solutions of the differential equation of the form $\left( x- \frac{1}{3}\right)^r$. Then the function $\left( x- \frac{1}{3}\right)^r$ is a solution of the differential equation at $( \frac{1}{3},+\infty)$ if: $$r(r-1) \left( x- \frac{1}{3}\right)^{r-2}+ \frac{1}{ \frac{3x-1}{3}} r \left( \frac{3x-1}{3}\right)^{r-1}- \frac{9}{(3x-1)^2} \left( x- \frac{1}{3}\right)=0 \Rightarrow \dots \Rightarrow r= \pm 1$$ Therefore, the functions $z_1(x)=x- \frac{1}{3}, z_2(x)=\frac{1}{x- \frac{1}{3}}$ are solutions of the differential equation at $\left( \frac{1}{3}, +\infty\right)$. $$z_1(x) z_2'(x)-z_1'(x) z_2(x)=\frac{-2}{x- \frac{1}{3}} \neq 0$$ So, $z_1, z_2$ are linearly independent solutions of the differential equation. Thus, the general solution of $y''+ \frac{3}{3x-1}y'-\frac{9}{(3x-1)^2}y=0$ is of the form: $$c_1 \left( x- \frac{1}{3} \right)+ c_2 \left( \frac{1}{x- \frac{1}{3}}\right) | c_1, c_2 \in \mathbb{R}, x> \frac{1}{3}$$ EDIT: We set $t=x-\frac{1}{3}$ and we have: $$\frac{dy}{dt}=\frac{dy}{dx} \frac{dx}{dt}=\frac{dy}{dx}$$ $$\frac{d^2y}{dt^2}=\frac{d}{dt} \left( \frac{dy}{dt}\right)=\frac{d}{dt} \left( \frac{dy}{dx} \right)=\frac{dx}{dt} \frac{d}{dx} \left( \frac{dy}{dx} \right)=\frac{d^2y}{dx^2}$$ $$y''(x)+ \frac{1}{x-\frac{1}{3}}y'(x)-\frac{1}{\left( x-\frac{1}{3}\right)^2}y(x)=0 \\ \Rightarrow y''(t)+\frac{1}{t}y'(t)-\frac{1}{t^2}y(t)$$ Two linearly independent solutions are $y_1(t)=t$ and $y_2(t)=\frac{1}{t}, y \in (0,+\infty)$. Thus, two linearly independent solutions of $y''+\frac{1}{x-\frac{1}{3}}y'-\frac{1}{\left( x-\frac{1}{3} \right)^2}y=0, x> \frac{1}{3}$ are $y_1(x)=x-\frac{1}{3}, y_2(x)=\frac{1}{x-\frac{1}{3}}$ Is it right or have I done something wrong?

Hint I do not know how much this could help you; so, please forgive me if this is off-topic. If you define first $y=(3x-1)u$, $$(3x-1)^2 y''+(9x-3)y'-9y=0$$ rewrite $$(3 x-1)^2 \Big((3 x-1) u''+9 u'\Big)=0$$ for which order can be reduced and integration seems simple.

Proving that $7^n(3n+1)-1$ is divisible by 9 I'm trying to prove the above result for all $n\geq1$ but after substituting in the inductive hypothesis, I end up with a result that is not quite obviously divisible by 9. Usually with these divisibility induction problems, it falls apart nicely and we can easily factorise say a 9 if the question required us to prove that the expression is divisible by 9. However in this case, I do not end up with such a thing. My work so far below: Inductive Hypothesis: $7^k(3k+1)-1=9N$ where $N\in\mathbb{N}$ Inductive Step: $7^{k+1}(3k+4)-1 \\ =7\times 7^k(3k+1+3)-1 \\ =7\times \left [ 7^k(3k+1)+3\times 7^k \right ] -1 \\ = 7 \times \left [ 9N+1 + 3 \times 7^k \right ] -1 \\ = 63N+21\times 7^k+6 \\ = 3 \left [ 21N+7^{k+1}+2 \right ]$ So now I need to somehow prove that $21N+7^{k+1}+2$ is divisible by 3, but I'm not quite sure how to proceed from here...

First, show that this is true for $n=1$: $7^1\cdot(3\cdot1+1)-1=9\cdot3$ Second, assume that this is true for $n$: $7^n\cdot(3n+1)-1=9k$ Third, prove that this is true for $n+1$: $7^{n+1}\cdot(3(n+1)+1)-1=$ $7^{n+1}\cdot(3n+3+1)-1=$ $7^{n+1}\cdot(3n+1+3)-1=$ $7^{n+1}\cdot(3n+1)+7^{n+1}\cdot(3)-1=$ $7\cdot\color{red}{7^n\cdot(3n+1)}+3\cdot7^{n+1}-1=$ $7\cdot(\color{red}{9k+1})+3\cdot7^{n+1}-1=$ $63k+7+3\cdot7^{n+1}-1=$ $63k+3\cdot7^{n+1}+6=$ $\color{blue}{3(21k+7^{n+1}+2)}$ Now: * *$7\equiv1\pmod3\implies$ *$\forall{m}\in\mathbb{N}:7^m\equiv1\pmod3\implies$ *$7^{n+1}\equiv1\pmod3\implies$ *$7^{n+1}+2\equiv3\pmod3\implies$ *$7^{n+1}+2\equiv0\pmod3\implies$ *$3|7^{n+1}+2\implies$ *$\exists{p}\in\mathbb{N}:7^{n+1}+2=3p\implies$ *$3(21k+7^{n+1}+2)=3(21k+3p)=3(3(7k+p))\color{blue}{=9(7k+p)}$ Please note that the assumption is used only in the part marked red.

Prove that if $\lim\limits_{x\to 0}f\big(x\big(\frac{1}{x}-\big\lfloor\frac{1}{x}\big\rfloor\big)\big)$, then $\lim\limits_{x\to 0}f(x)=0$ Prove that if $\lim\limits_{x\to 0}f\bigg(x\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg)\bigg)$, then $\lim\limits_{x\to 0}f(x)=0$ My attempt: If I can show that $\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg) \to 1$ as $x\to 0$, then we are done. We know, $n\le \dfrac{1}{x}\le n+1$, so $\bigg\lfloor\dfrac{1}{x}\bigg\rfloor=n$. But, I cannot do anything more to it. Please help. Thank you.

Use the Squeeze Theorem. First note that $$\frac{1}{x}-1<\left\lfloor\frac{1}{x}\right\rfloor\le\frac{1}{x}$$ from which we obtain $$ \left|x\right|\ge \left|x\right|\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\ge 0$$ Therefore, as $x\to 0$, the Squeeze Theorem guarantees that $$\lim_{x\to 0} x\,\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)=0 $$ which implies $$\lim_{x\to 0}f\left(\,x\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right)=\lim_{x\to 0}f(x)$$ If $f$ is continuous, then $$\lim_{x \to 0} f\left(x \left(\frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right)=f\left(\lim_{x \to 0} \left[x \left(\frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right]\right)=f(0). $$

Integration: Area between curves Let $f(x)=x^2−c^2$ and $g(x)=c^2−x^2.$ Find $c>0$ such that the area of the region enclosed by the parabolas $f(x)$ and $g(x)$ is 9. The question above is what I am having trouble with. In order to solve this problem I use the formula given as: $\int_a^b f(x) - g(x) dx$ Here is what I have done so far: $9 = \int x^2−c^2 - (c^2−x^2) dx$ $9 = \int 2x^2−2c^2 dx$ $9 = 2\int x^2−c^2 dx$ $\frac 9 2 = \int x^2−c^2 dx$ To find a and b: $x^2 - c^2 = (x-c)(x+c)$ ; therefore $a = -c$ and $b = c$ Back to the original question, sub in a and b: $\frac 9 2 = \int_{-c}^c x^2−c^2 dx$ This is as far as I have gotten. I am not sure what to do next. Thanks in advance to anyone who can help.

Draw a picture. I would use symmetry and say that the area is $4$ times the integral from $0$ to $c$ of $c^2-x^2$. So we want $$\int_0^c (c^2-x^2)\,dx=\frac{9}{4}.$$ The integral is equal to $\frac{2}{3}c^3$. Now solve for $c$. Remark: Your approach will work, except that we want $\int_{-c}^c (c^2-x^2)\,dx=\frac{9}{2}$. Note that $\int_{-c}^c (x^2-c^2)\,dx$ is negative.

Symplectic lie algebra Can anyone explain me why, in the symplectic lie algebra, which is defined as $ sp(n)=\{X \in gl_{2n}:X^tJ+JX=0\}$ where $J=\begin{pmatrix} 0 & I \\ -I & 0 \\ \end{pmatrix} $ we can write its elements, in block form $X=\begin{pmatrix} A & B \\ C & -A^t \\ \end{pmatrix} $ where $ A,B,C \in M_{n\times n}$ and $B=B^t,C=C^t$ .How does it proved?

Hint: The block decomposition used for $J$ (as well as the answer) suggest writing a generic matrix $X \in \mathfrak{gl}_{2n}$ as $$X = \begin{pmatrix} A & B \\ C & D \end{pmatrix}.$$ To produce the block matrix description of $\mathfrak{sp}_{2n}$, simply substitute the block expressions for $X$ and $J$ in the definition of that algebra: \begin{align} X^t J + J X &= 0 \\ \begin{pmatrix} A & B \\ C & D \end{pmatrix}^t \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix} + \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix} \begin{pmatrix} A & B \\ C & D \end{pmatrix} &= 0 \textrm{.} \end{align} Now, simplify the l.h.s. of the equation to produce algebraic conditions that $A, B, C, D$ must satisfy.

If $p$ is prime and congruent to $1$, then show $((\frac{p-1}{2})!)^2 \equiv -1 \pmod p$ I got another one. Quadratic residues are completely new to me... Thanks!

Take the congruences, $p-1 \equiv -1 \pmod p$, $\text{ }\text{ } p-2 \equiv -2 \pmod p$ and so on upto, $\frac{p+1}2 \equiv -\frac{p-1}2 \pmod p$. Multiplying and rearranging, $$(p-1)!\equiv 1\cdot (-1)\cdot 2 \cdot (-2) ...\frac{p-1}2 \cdot (-\frac{p-1}2) \equiv (-1)^{\frac{p-1}2}[(\frac{p-1}{2})!]^2 \pmod p$$ Thus, by Wilson's theorem, $ -1 \equiv (-1)^{\frac{p-1}2}[(\frac{p-1}{2})!]^2 \pmod p$. If by "$p$ is congruent to $1$", you meant that $p \equiv 1 \pmod 4$, we are done as $(-1)^{\frac{p-1}2} =1$.

square root / factor problem $(A/B)^{13} - (B/A)^{13}$ Let $A=\sqrt{13+\sqrt{1}}+\sqrt{13+\sqrt{2}}+\sqrt{13+\sqrt{3}}+\cdots+\sqrt{13+\sqrt{168}}$ and $B=\sqrt{13-\sqrt{1}}+\sqrt{13-\sqrt{2}}+\sqrt{13-\sqrt{3}}+\cdots+\sqrt{13-\sqrt{168}}$. Evaluate $(\frac{A}{B})^{13}-(\frac{B}{A})^{13}$. By Calculator, I have $\frac{A}{B}=\sqrt{2}+1$ and $\frac{B}{A}=\sqrt{2}-1$. But, I don't know how. Has someone any idea about this.

Let $$A=\sum_{n=1}^{168}\sqrt{13+\sqrt{n}},B=\sum_{n=1}^{168}\sqrt{13-\sqrt{n}}$$ since $$\sqrt{2}A=\sum_{n=1}^{168}\sqrt{26+2\sqrt{n}}=\sum_{n=1}^{168}\left(\sqrt{13+\sqrt{169-n}}+\sqrt{13-\sqrt{169-n}}\right)=A+B$$ so we have $x=\dfrac{A}{B}=\sqrt{2}$,then we have $$x=\sqrt{2}+1,\dfrac{1}{x}=\sqrt{2}-1\Longrightarrow x+\dfrac{1}{x}=2\sqrt{2}$$ let $$a_{n}=x^n-x^{-n}$$ use this well know indentity $$a_{n+2}=(x+\dfrac{1}{x})a_{n+1}-a_{n}\Longrightarrow a_{n+2}=2\sqrt{2}a_{n+1}-a_{n}$$ $$a_{1}=2,a_{2}=4\sqrt{2}$$ so $$a_{3}=2\sqrt{2}a_{2}-a_{1}=16-2=14$$ $$a_{4}=2\sqrt{2}a_{3}-a_{2}=28\sqrt{2}-4\sqrt{2}=24\sqrt{2}$$ $$a_{5}=2\sqrt{2}a_{4}-a_{3}=96-14=82$$ $$\cdots$$

Need clarification on a Taylor polynomial question $$f(x) = 5 \ln(x)-x$$ second Taylor polynomial centered around $b=1$ is $-1 + 4(x-1) - (5/2)(x-1)^2$ let $a$ be a real number : $0 < a < 1$ let $J$ be closed interval $[1-a, 1+a]$ find upper bound for the error $|f(x)-T_2(x)|$ on interval $J$ answer in terms of a so i got to the point where i have $|f(x)-T_2(x)| <= (10/6)(x-1)^3$ however the answer is $(5/3)(a/(1-a))^3$ How did the $a/(1-a)$ get there? If i sub in for the max $x$ which would be $1 + a$, then i would get $(5/3)(a)^3$. Why is it divided by $(1-a)$?

We are given: $$f(x) = 5 \ln(x) - x$$ The second Taylor polynomial centered around $b=1$ is given by: $$T_2(x) = -\frac{1}{2} 5 (x-1)^2+4 (x-1)-1$$ We are told to let $a$ be a real number such that $0 \lt a \lt 1$ and let $J$ be the closed interval $[1 −a, 1 +a]$. We are then asked to use the Quadratic Approximation Error Bound to find an upper bound for the error $|f(x) − T_2(x)|$ on the interval $J$. The error term is given by: $$R_{n+1} = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-b)^{n+1} \le \dfrac{M}{(n+1)!}(x-b)^{n+1}$$ We have $b=1, n= 2$, and have to find the max error for two items, thus: $$\dfrac{d^3}{dx^3} (5 \ln(x)-x) = \dfrac{10}{x^3}$$ So the maximum of $f^{(3)}(x)$ is given by: $$\displaystyle \max_{J} \left|f'''(x)\right| = \max_{ 1-a \le x \le 1+a} \left|\dfrac{10}{x^3}\right|$$ The max occurs at the left endpoint because $0 \lt a \lt 1$, so the maximum is given by: $$\dfrac{10}{(1-a)^3}$$ Next, we have to repeat this and find the maximum of $(x-1)^3$. In this case, the maximum occurs at the rightmost endpoint, hence the maximum is: $$((1+a) - 1)^3 = a^3$$ Putting this together yields: $$\left|f(x)-T_2(x)\right| = R_{3} = \dfrac{f^{(3)}(c)}{3!}(x-1)^{3} \le \dfrac{M}{3!}(x-1)^{3} = \dfrac{10}{3!(1-a)^3}a^3 = \dfrac{5}{3} \left(\dfrac {a}{1-a}\right)^3$$

For $x, y \in \Bbb R$ such that $(x+y+1)^2+5x+7y+10+y^2=0$. Show that $-5 \le x+y \le -2.$ I have a problem: For $x, y \in \Bbb R$ such that $(x+y+1)^2+5x+7y+10+y^2=0$. Show that $$-5 \le x+y \le -2.$$ I have tried: I write $(x+y+1)^2+5x+7y+10+y^2=(x+y)^2+7(x+y)+(y+1)^2+10=0.$ Now I'm stuck :( Any help will be appreciated! Thanks!

Since $(y+1)^2\ge 0$ we must have $(x+y)^2+7(x+y)+10=(x+y+5)(x+y+2)\leq 0$ then, solving this inequality for $x+y$, we get $$-5\le x+y \le -2$$

How to do multiplication in $GF(2^8)$? I am taking an Internet Security Class and we received some practice problems and answers, but I do not know how to do these problems , an explanation would be greatly appreciated Try to compute the following value:(the number is in hexadecimal and each represents a polynomial in GF(2^8) 1) {02} * {87} answer: {02} = {0000 0010} = x {87} = {1000 0111} = x -> {02} * {87} = (000 1110) XOR (0001 1011) = 0001 0101 = {15}

To carry out the operation, we need to know the irreducible polynomial that is being used in this representation. By reverse-engineering the answer, I can see that the irreducible polynomial must be $x^8+x^4+x^3+x+1$ (Rijndael's finite field). To carry out a product of any two polynomials then, what you want to do is multiply them and then use the relation $x^8+x^4+x^3+x+1\equiv 0$, or in other words $x^8\equiv x^4+x^3+x+1$, to eliminate any terms $x^k$ where $k\geq 8$, reducing modulo 2 as you go along. The binary {0000 0010} corresponds to the polynomial $x$ (i.e., $0x^7+0x^6+0x^5+0x^4+0x^3+0x^2+1x^1+0x^0$), while the binary {1000 0111} corresponds to $x^7+x^2+x+1$ (i.e., $1x^7+0x^6+0x^5+0x^4+0x^3+1x^2+1x^1+1x^0$). So to do the multiplication, we calculate \begin{align*} x*(x^7+x^2+x+1) &= x^8 +x^3+x^2+x \\ &\equiv (x^4+x^3+x+1) + x^3+x^2+x \\ &= x^4+2x^3+x^2+2x+1 \\ &\equiv x^4+x^2+1 \end{align*} which is represented in binary as {0001 0101}.

Solve the initial value problem $u_{xx}+2u_{xy}-3u_{yy}=0,\ u(x,0)=\sin{x},\ u_{y}(x,0)=x$ Solve the partial differential equation $$u_{xx}+2u_{xy}-3u_{yy}=0$$ subjet to the initial conditions $u(x,0)=\sin{x}$, $u_{y}(x,0)=x$. What I have done $$ 3\left(\frac{dx}{dy}\right)^2+2\frac{dx}{dy}-1=0 $$ implies $$\frac{dx}{dy}=-1,\frac{dx}{dy}=\frac{1}{3} $$ and so $$ x+y=c_{1},\ 3x-y=c_{2}. $$ Let $ \xi=x+y$, $\eta=3x-y$. Then \begin{align} u_{xx}&=u_{\xi\xi}+6u_{\xi\eta}+9u_{\eta\eta} \\ u_{yy}&=u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta} \\ u_{xy}&=u_{\xi\xi}-2u_{\xi\eta}-3u_{\eta\eta}. \end{align} Applying substitutions, $$ u_{\xi\eta}=0. $$ Thus, \begin{align} u(\xi,\eta)&=\varphi(\xi)+\psi(\eta) \\ u(x,y)&=\varphi(x+y)+\psi(3x-y). \end{align} Applying the initial value condition, \begin{align} u(x,0)&=\varphi(x)+\psi(3x)=\sin{x} \\ u_{y}(x,0)&=\varphi'(x)-\psi'(3x)=x \end{align} Therefore, \begin{align} \varphi(x)&= \frac{1}{2} \left(\sin{x}+\int_{x_{0}}^{x} \tau \, d\tau \right)+\frac k2 \\ ψ(3x)&=\frac{1}{2} \left(\sin{x}-\int_{x_{0}}^{x}\tau \, d\tau \right)-\frac k2. \end{align} I have no idea how to get $ψ(x)$. Does anyone could help me to continue doing this question? Thanks very much!

We use Laplace transform method and free CAS Maxima http://maxima.sourceforge.net/ Answer: $$u=\frac{\sin(x+y)}{4}+\frac{y^2}{3}+xy+\frac34\sin\left(x-\frac{y}{3}\right)$$ 2 method * *$D_x^2+2D_xD_y-3D_x^2=(D_x-D_y)(D_x+3D_y)$ *General solution of $u_x-u_y=0$ is $u_1=f(x+y)$ *General solution of $u_x+3u_y=0$ is $u_2=g\left(x-\frac{y}{3}\right)$ *$u=u_1+u_2=f(x+y)+g\left(x-\frac{y}{3}\right)$ *From initial conditions we get $$f(x)+g(x)=\sin(x),\\f'(x)-\frac13g'(x)=x$$ *After the integration of the second equation we get $$f(x)+g(x)=\sin(x),\\f(x)-\frac13g(x)=\frac{x^2}{2}+c$$ *Then $$f(x)=\frac{\sin{(x)}}{4}+\frac{3 {{x}^{2}}}{8}+\frac{3 c}{4},\\ g(x)=\frac{3 \sin{(x)}}{4}-\frac{3 {{x}^{2}}}{8}-\frac{3 c}{4} $$ *$$u=f(x+y)+g\left(x-\frac{y}{3}\right)\\=\frac{\sin{\left( x+y\right) }}{4}+\frac{3 {{\left( x+y\right) }^{2}}}{8}-\frac{3 {{\left( x-\frac{y}{3}\right) }^{2}}}{8}-\frac{3 \sin{\left( \frac{y}{3}-x\right) }}{4}\\= \frac{\sin(x+y)}{4}+\frac{y^2}{3}+xy+\frac34\sin\left(x-\frac{y}{3}\right) $$

Finding the order of an element in the dihedral group of order 4. How do I find the order of $$S_1=\left({\begin{array}{cc} \cos\frac{\pi}{3} & \sin\frac{\pi}{3}\\ \sin\frac{\pi}{3} & -\cos\frac{\pi}{3}\\ \end{array} }\right)$$ I know that $S_1$ is a dihedral group and is a reflection of the line that makes an angle of $\frac{\pi}{3}$ with the x-axis.. but finding its order is what I don't know how to do.. Please help.

$$S_1^2=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$$ In more detail $$\begin{align}S_1^2=&\begin{matrix} \cos^2{\frac{\pi}{3}}+\sin^2{\frac{\pi}{3}}&\cos{\frac{\pi}{3}}\sin{\frac{\pi}{3}}-\cos{\frac{\pi}{3}}\sin{\frac{\pi}{3}}\\\cos{\frac{\pi}{3}}\sin{\frac{\pi}{3}}-\cos{\frac{\pi}{3}}\sin{\frac{\pi}{3}}&\cos^2{\frac{\pi}{3}}+\sin^2{\frac{\pi}{3}}\end{matrix}\end{align}$$ And so the order is $2$. By the way reflection is always of order two

integration of definite integral involving sinx and cos x Evaluate $\int_0^{\pi}\frac{dx}{a^2\cos^2x +b^2 \sin^2x}$ I got numerator $\sec^2 x$ and denominator $b^2 ( a^2/b^2 + \tan^2x)$. I made substitution $u= \tan x$. That way $\sec^2 x$ got cancelled and the answer was of form $1/ab$ ($\tan^{-1} (bu/a)$) And then if I put limits answer is $0$ but answer is wrong. Where did I go wrong?

Are you talking about this?: $$\small\int\frac{dx}{a^2\cos^2x +b^2 \sin^2x}=\int\frac{\sec^2xdx}{a^2+b^2 \tan^2x}\stackrel{u=\tan x}=\frac1{b^2}\int\frac{du}{a^2/b^2+u^2}=\frac1{b^2}\frac1{a/b}\arctan\frac{\tan x}{a/b}$$ So: $$\int_0^{\pi}\frac{dx}{a^2\cos^2x +b^2 \sin^2x}=\frac1{ab}\arctan\frac{b\tan x}a\Bigg|_0^{\pi}=0$$ This is wrong because at $\pi/2$, $\cos x=0\iff \sec x\to\infty$!! Or you can say because $\tan x$(=u) misbehaves at $x=\pi/2$.It should rather be done like: $$\int_0^{\pi}\frac{dx}{a^2\cos^2x +b^2 \sin^2x}=\frac1{ab}\arctan\frac{b\tan x}a\Bigg|_0^{\displaystyle\pi/2^-}+\frac1{ab}\arctan\frac{b\tan x}a\Bigg|_{\displaystyle\pi/2^+}^{\displaystyle\pi}$$

Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ Let $a_1=2$ and we define $a_{n+1}=a_n+\sqrt {a_n},n\geq 1$. Is it possible to get a good aproximation of the $n$th term $a_n$? The first terms are $2,2+\sqrt{2}$, $2+\sqrt{2}+\sqrt{2+\sqrt{2}}$ ... Thanks in advance!

For the third term, we have the following. Let $c_n = \sqrt{a_n} -\frac{1}{2}n + \frac{1}{4}\ln n$, then $c_{n+1} -c_{n} = -\frac{1}{2} \cdot \frac{\sqrt{a_n}}{(\sqrt{a_n}+\sqrt{\sqrt{a_n}+a_n})^2} + \frac{1}{4} \ln(1+ \frac{1}{n})\quad (1)$. Note that $a_n = \frac{1}{4}n^2 - \frac{1}{4}n\ln n + o(n \ln n)$, the first term in R.H.S of (1) equals to $-\frac{1}{8} \cdot \frac{2n - \ln n + o(\ln n)}{n^2 -n \ln n + o(n \ln n)} = -\frac{1}{8n} \cdot(2+ \frac{ \ln n + o(\ln n)}{n - \ln n + o( \ln n)} )= -\frac{1}{4n} - \frac{\ln n}{8n^2}(1+o(1))$. Therefore (1) $= - \frac{\ln n}{8n^2}(1+o(1))$. Then we have $\lim_{n \rightarrow \infty} \frac{c_{n+1} -c_{n} }{- \frac{\ln n}{8n^2}} = 1$ and also have $\lim_{n \rightarrow \infty} \frac{c_{n+1} -c_{n} }{ \frac{\ln (n+1)+1}{8(n+1)} -\frac{\ln n+1}{8n} } = 1$. For $\frac{\ln n+1}{8n}$, we may integrate $-\frac{\ln x}{8x^2}$ with respect to $x$.

there exsit postive integer $x,y$ such $p\mid(x^2+y^2+n)$ For any give the postive integer $n$,and for any give prime number $p$. show that there exsit postive integer $x,y$ such $$p\mid(x^2+y^2+n)$$ My approach is the following: Assmue that $n=1,p=2$,we choose$(x,y)=(1,2)$ $$2\mid6=1^2+2^2+1$$ Assmue that $n=1,p=3$, we choose $(x,y)=(1,2)$ $$3\mid6=1^2+2^2+1$$ Assume that $n=1,p=5$,we choose $(x,y)=(2,5)$ $$5\mid30=2^2+5^2+1$$ Assume that $n=2,p=2$, we choose $(x,y)=(2,2)$ $$2\mid10=2^2+2^2+2$$ Assume that $n=2,p=3$ we choose $(x,y)=(2,3)$ $$3\mid15=2^2+3^2+2$$ Assume that $n=2,p=5$,we choose $(x,y)=(3,3)$ $$5\mid20=3^2+3^2+2$$ and so on Now I'm stuck and don't know how to proceed

The result is easy to prove if $p=2$, so we can assume from now on that $p$ is odd. Modulo $p$, there are $\frac{p+1}{2}$ squares, namely the $\frac{p-1}{2}$ quadratic residues of $p$, and $0$. So modulo $p$ there are $\frac{p+1}{2}$ distinct values of $x^2$. There are also (for fixed $n$) $\frac{p+1}{2}$ distinct values of $-y^2-n$, since there are $\frac{p+1}{2}$ distinct values of $y^2$, and hence of $-y^2$. Since $\frac{p+1}{2}+\frac{p+1}{2}=p+1\gt p$, by the Pigeonhole Principle there exist $x$ and $y$ such that $x^2\equiv -y^2-n\pmod{p}$. This implies that there are values of $x$ and $y$ such that $x^2-(-y^2-n)\equiv 0\pmod{p}$.

Finding for what $x$ values the error of $\sin x\approx x-\frac {x^3} 6$ is smaller than $10^{-5}$ Find for what $x$ values the error of $\sin x\approx x-\frac {x^3} 6$ is smaller than $10^{-5}$ I thought of two ways but got kinda stuck: * *Since we know that $R(x)=f(x)-P(x)$ then we could solve: $\sin x-x+\frac {x^3} 6<10^{-2}$ but I have no idea how to do it, deriving the expression to a more simple expression makes it loose the information on $10^{-5}$. *Find for what $x,c$ we have $R_5(x)=\frac {\cos (c)x^5}{5!}<10^{-5}$ but can I simply choose $c=0$? then the answer would be: $x<\frac {(5!)^{1/5}}{10}$ which looks like it's right from the graph of $R(x)$ in 1. But why can I choose $c$? what if it was a less convenient function? Note: no integrals.

Consider $$ \sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots. $$ This series is alternating, and the terms are strictly decreasing in magnitude if $|x| < 1$. So we get \begin{align*} \text{estimate} - \text{ reality} &= x - \frac{x^3}{6} - \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots\right) \\ &=\frac{x^5}{120} - \ldots \\ &\leq \frac{x^5}{120} \end{align*} The last estimate works only because the terms are strictly decreasing in magnitude. So it suffices to take $x$ satisfying $$ -10^{-5} < \frac{x^5}{120} < 10^{-5} $$ But this is not a necessary condition.

Evaluate the following indefinite integral Evaluate the integral : $$\int\frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}\,dx$$ I tried through putting $x=\tan \theta$ as well as $x=\tan^2\theta$ .but I am unable to remove the square root. I also tride by putting $x+x^2+x^3=z^2$. But I could not proceed anyway...Please help... Update : putting $x=\frac{1-t}{1+t}$ , I get , $$\sqrt{x+x^2+x^3}=\sqrt{\frac{2-3t+t^2-t^3}{(1+t)^3}}$$ How you got $\sqrt{x+x^2+x^3}=\sqrt{(t^3+3)(1-t^2)}/(t+1)^2$ ?

Hint 1: $t \mapsto (1-x)/(1+x)$ $$\begin{equation}\displaystyle\int\frac{x-1}{(x+1)\sqrt{x+x^2+x^3}}\,\mathrm{d}x = 2\arccos\left(\frac{\sqrt{x}} {x+1}\right) + \mathcal{C}\end{equation}$$ Hint 2: One can show that $t = (1-x)/(1+x)$ is it's own inverse. In other words $x = (1-t)/(1+t)$. Hence the derivative becomes. $\mathrm{d}x = -2t \,\mathrm{d}t/(t+1)^2$. Similarly we have $\sqrt{x^3+x^2+x} = \sqrt{(t^3+3)(1-t^2)}/(t+1)^2$ so we get a nice cancelation. Explicitly we have $$ \begin{align*} \sqrt{x^3+x^2+x} & = \sqrt{ \left(\frac{1-t}{1+t}\right)^3 + \left(\frac{1-t}{1+t}\right)^2 + \left(\frac{1-t}{1+t}\right) } \\ & = \sqrt{ \frac{-t^3+t^2-3t+3}{(1+t)^3}} = \sqrt{ \frac{1+t}{1+t}\frac{(1-t)(t^2+3)}{(1+t)^3}} = \frac{\sqrt{(1-t^2)(t^3+3)}}{(t+1)^2} \end{align*} $$ Hint 3: Use the substitution $\cos u \mapsto t$, what happens?

$f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x) $ $f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x)$ Determine the greatest and least values of $\frac{39}{f(x)+14}$ and state a value of x at which greatest values occurs. Do I just use a graphing calculator for this? Is there a way I could do this without a graphing calculator?

HINT: $$f(x)=17\cos^2x+7\sin^2x-24\sin x\cos x=\dfrac{17(1+\cos2x)+7(1-\cos2x)-24\sin2x}2$$ $$=12+5\cos2x-12\sin2x=12+\sqrt{12^2+5^2}\cos\left(2x+\arctan\dfrac{12}5\right)$$ Now for real $x,-1\le\cos\left(2x+\arctan\dfrac{12}5\right)\le1$

Divergence of $\prod_{n=2}^\infty(1+(-1)^n/\sqrt n)$. Looking looking for a verification of my proof that the above product diverges. $$\begin{align} \prod_{n=2}^\infty\left(1+\frac{(-1)^n}{\sqrt n}\right) & =\prod_{n=1}^\infty\left(1+\frac1{\sqrt {2n}}\right)\left(1-\frac1{\sqrt{2n+1}}\right)\\ & =\prod_{n=1}^\infty\left(1-\frac1{\sqrt{n_1}}+\frac1{\sqrt {2n}}-\frac1{\sqrt{2n(2n+1)}}\right)\\ & =\prod_{n=1}{\sqrt{2n(2n+1)}-\sqrt{2n}+\sqrt{2n+1}-1\over\sqrt{2n(2n+1)}}\\ & \ge\prod_{n=1}^\infty{\sqrt{2n(2n+1)+2n+1}-\sqrt{2n}-1\over\sqrt{2n(2n+1)}},\quad\sqrt x+\sqrt y\ge\sqrt{x+y}\\ & = \prod_{n=1}^\infty{2n-\sqrt{2n}\over\sqrt{2n(2n+1)}}\\ & = \prod_{n=1}^\infty{2n-1\over\sqrt{2n+1}} \end{align}$$ This last product diverges since $$\lim_{n\to\infty}{2n-1\over\sqrt{2n+1}}=\infty.$$ I'm suspicious because $$\lim_{n\to\infty}\left(1+{(-1)^n\over\sqrt n}\right)=1.$$

I would like for some approval to my answer, please. Show that $$\prod_2^\infty(1+\frac{(-1)^k}{\sqrt k})$$ diverges even though$$\sum_2^\infty \frac{(-1)^k}{\sqrt k}$$ converges. $\prod_2^\infty(1+\frac{(-1)^k}{\sqrt k})$ diverges if and only if $\sum _2^\infty\log(1+\frac{(-1)^k}{\sqrt k})$ diverges. Indeed, by Leibnitz' converges test $$\sum_2^\infty \frac{(-1)^k}{\sqrt k}$$ converges. Thus, $\sum _2^\infty\log(1+\frac{(-1)^k}{\sqrt k})$ diverges if and only if $\sum _2^\infty(\log(1+\frac{(-1)^k}{\sqrt k})+\frac{(-1)^k}{\sqrt k})$ diverges. We observe that for all $k\geq 2$, $$ |\log(1+\frac{(-1)^k)}{\sqrt k})-\frac{(-1)^k)}{\sqrt k}|\\ =|\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3} + \frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5}+-...| $$ But since we know that the last series is convergen (to $|\log(1+\frac{(-1)^k)}{\sqrt k})-\frac{(-1)^k)}{\sqrt k}|$), then it is equal to $$ =|(\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}) + (\frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5})+-...|=:A $$ but every addend in this series is positive because for all $l\in\mathbb{N}$: $$ (-1)^{2l}(2l+1)\sqrt k^{2l+1}>(-1)^{2l+1}\cdot 2l\cdot\sqrt k^{2l} $$ ($\sqrt k>1$). Thus, the previos series $A$ obtains: $$ A=(\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}) + (\frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5})+-... \\ >\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}\geq \frac{1}{2k}-\frac{1}{3k^{1.5}} $$ But because $\sum \frac{1}{2k}$ diverges and $\sum\frac{1}{3k^{1.5}}$ converges then $\sum(\frac{1}{2k}-\frac{1}{3k^{1.5}})$ diverges then $A$ diverges.

If $A,B>0$ and $A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$. If $A,B>0$ and $\displaystyle A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$. $\bf{My\; Try::}$ Given $$\displaystyle A+ B = \frac{\pi}{3}$$ and $A,B>0$. So we can say $$\displaystyle 0< A,B<\frac{\pi}{3}$$. Now taking $\tan $ on both side, we get $$\displaystyle \tan(A+B) = \tan \left(\frac{\pi}{3}\right).$$ So $\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B} = \sqrt{3}$. Now Let $\displaystyle \tan A\cdot \tan B=y\;,$ Then $\displaystyle \tan B = \frac{y}{\tan A}.$ So $$\displaystyle \frac{\tan A+\frac{y}{\tan A}}{1-y}=\sqrt{3}\Rightarrow \tan^2 A+y=\sqrt{3}\tan A-y\sqrt{3}\tan A$$ So equation $$\tan^2 A+\sqrt{3}\left(y-1\right)\tan A+y=0$$ Now for real values of $y\;,$ Given equation has real roots. So its $\bf{Discrimnant>0}$ So $$\displaystyle 3\left(y-1\right)^2-4y\geq 0\Rightarrow 3y^2+3-6y-4y\geq 0$$ So we get $$3y^2-10y+3\geq 0\Rightarrow \displaystyle 3y^2-9y-y+3\geq 0$$ So we get $$\displaystyle y\leq \frac{1}{3}\cup y\geq 3$$, But above we get $\displaystyle 0<A,B<\frac{\pi}{3}$ So We Get $$\bf{\displaystyle y_{Max.} = \left(\tan A \cdot \tan B\right)_{Max} = \frac{1}{3}}.$$ My Question is can we solve above question using $\bf{A.M\geq G.M}$ Inequality or Power Mean equality. Thanks

$$ \tan A\tan B=\tan A\tan\big(\frac{\pi}{3}-A\big)=\tan A.\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}=1-1+\frac{\sqrt{3}\tan A-\tan^2A}{1+\sqrt{3}\tan A}\\ =1+\frac{-1-\sqrt{3}\tan A+\sqrt{3}\tan A-\tan^2A}{1+\sqrt{3}\tan A}=1-\frac{1+\tan^2A}{1+\sqrt{3}\tan A}=1-\frac{\sec^2A}{1+\sqrt{3}\tan A}\\ =1-\frac{2}{2\cos^2A+\sqrt{3}.2\sin A\cos A}=1-\frac{2}{1+2\big[\frac{1}{2}\cos 2A+\frac{\sqrt{3}}{2}\sin 2A\big]}\\ =1-\frac{2}{1+2\sin\big(\frac{\pi}{6}+2A\big)} $$ $\tan A\tan B$ is maximum $\implies \frac{2}{1+2\sin\big(\frac{\pi}{6}+2A\big)}$ is minimum $\implies \sin\big(\frac{\pi}{6}+2A\big)$ is maximum$\implies \sin\big(\frac{\pi}{6}+2A\big)=1$ $$ (\tan A\tan B)_\text{max}=1-\frac{2}{1+2.1}=1-\frac{2}{3}=\frac{1}{3} $$

Inequality and Induction: $\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$ I needed to prove that $\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$, $\forall n \geq 1$ . I've atempted by induction. I proved the case for $n=1$ and assumed it holds for some $n$. The left-side of the n+1 case is $\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2}$. Using the inductive hypothesis, i could reach the result that $\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2} <\frac{2n+1}{\sqrt{2n+1}(2n+2)}=\frac{\sqrt{2n+1}}{(2n+2)}$. Now, i'm wondering how should i connect it to my goal : $\frac{1}{\sqrt{2n+2+1}}=\frac{1}{\sqrt{2n+3}} $ I know one way to prove that $\frac{\sqrt{2n+1}}{(2n+2)}<\frac{1}{\sqrt{2n+3}} $ We just square things, then eventually reach $ (n+1).(n+3)<(n+2)^2$ Which is easily provable because $3<4$ .... But I was wondering if there was another way to show that ... perharps a more direct way to show that last bit ... a way that was not so direct and brute as to involve squaring both sides. A way of gradually manipulating the left-side until reaching the inequality with the right side. Thanks in advance.

$$\begin{align}\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2} &<\frac{2n+1}{\sqrt{2n+1}(2n+2)}\\~\\&=\frac{\sqrt{2n+1}}{(2n+2)}\\~\\&=\frac{\sqrt{(2n+1)(2n+3)}}{(2n+2)\sqrt{2n+3}}\\~\\&=\frac{\sqrt{4n^2+8n+3}}{(2n+2)\sqrt{2n+3}}\\~\\&\lt \frac{\sqrt{4n^2+8n+3+\color{blue}{1}}}{(2n+2)\sqrt{2n+3}}\\~\\&= \frac{\sqrt{(2n+2)^2}}{(2n+2)\sqrt{2n+3}}\\~\\&=\frac{1}{\sqrt{2n+3}}\end{align}$$ Alternatively you may also use AM-GM inequality and conclude directly that $\sqrt{(2n+1)(2n+3)}\lt 2n+2$

Taylor Expansion of Inverse of Difference of Vectors I am trying to derive the multipole moment of a gravitational potential, but I'm getting stuck on some math I believe. So basically the problem is finding the Taylor Expansion for $$\frac{1}{|\mathbf{x}-\mathbf{x'}|}=\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}.$$ The expansion is supposed to be around $\mathbf{x'}=0$. I have two questions: 1) Do I take the partial derivatives in terms of x or x' (I'm thinking it should be just x, but I'm not sure)? 2) When I'm supposed to multiply by a factor that is the equivalent of $(x-a)$, what should that be? I was thinking I should just dot with x', but that doesn't give me the correct answer.

The multivariable Taylor expansion is probably most easily written as $$ f(\mathbf{x+h}) = \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{i_1,i_2,\dotsc,i_k=1}^n (\partial_{x_{i_1}} \partial_{x_{i_2}} \dotsm \partial_{x_{i_k}} f(\mathbf{x}) ) h_{i_1} h_{i_2} \dotsm h_{i_k} $$ (which keeps all terms of the same order together), so the first few terms are $$ f(\mathbf{x}) + \mathbf{h} \cdot \nabla f(\mathbf{x}) + \mathbf{h} \cdot (Hf(\mathbf{x}))\mathbf{h} + \dotsb $$ However, probably the simpler way in this case is to write $$ \lvert \mathbf{x-x'} \rvert^{-1} = (|\mathbf{x}|^2+2\mathbf{x\cdot x'}+|\mathbf{x'}|^2)^{-1/2}. $$ Setting $|\mathbf{x}|=R$, $|\mathbf{x}|=r$, and $\mathbf{x\cdot x'} = 2rR\cos{\theta}$, where $\theta$ is the angle between the vectors, we have $$ \lvert \mathbf{x-x'} \rvert^{-1} = R^{-1}(1+2r/R \cos{\theta}+r^2/R^2)^{-1/2}. $$ Setting $r/R=s$, we can now expand this as a function of $s$ using the binomial theorem: $$\begin{align*} \frac{1}{\lvert \mathbf{x-x'} \rvert} &= \frac{1}{R}(1+2s \cos{\theta}+s^2)^{-1/2} \\ &= \frac{1}{R} \left( 1 - \frac{1}{2}(2s \cos{\theta}+s^2) + \frac{-1}{2}\frac{-3}{2}\frac{1}{2!}(2s \cos{\theta}+s^2)^2 + O(s^3) \right) \\ &= \frac{1}{R} \left( 1 - s\cos{\theta} - \frac{1}{2}s^2 + \frac{3}{2}s^2 \cos^2{\theta} + O(s^3) \right) \\ &= \frac{1}{R} - \frac{s\cos{\theta}}{R} - \frac{1}{2R}(3 \cos^2{\theta}-1)s^2 + O(s^3) \end{align*}$$ Resubstituting, we get the multipole expansion $$ \frac{1}{\lvert \mathbf{x-x'} \rvert} = \frac{1}{\lvert \mathbf{x} \rvert} - \frac{\mathbf{x \cdot x'}}{\lvert \mathbf{x} \rvert^2} + \frac{1}{2} \frac{3(\mathbf{x \cdot x'})^2-\lvert \mathbf{x'} \rvert^2}{\lvert \mathbf{x} \rvert^5} + O(\lvert \mathbf{x'} \rvert^3/\lvert \mathbf{x} \rvert^4) $$ To actually answer your questions, * *Since $\lvert \mathbf{x -x'} \rvert=\lvert \mathbf{x'-x} \rvert$, here it doesn't actually matter. You're taking derivatives of the function $\lvert \mathbf{x} \rvert^{-1}$ and evaluating them at $\mathbf{x}$. *You are correct in dotting with $\mathbf{x'}$: notice that the third term in the expansion, for example, is $$ \sum_{i,j=1}^3 \frac{3x_i x_j- \delta_{ij}}{2\lvert \mathbf{x} \rvert} x'_i x'_j. $$

Another optimization problem I am having trouble figuring out a next step in an optimization problem the question is to find the max and min values of $f(x,y)=\frac{x+y}{2+x^2+y^2}$ I calculated $f_x$ and $f_y$ and set both them equal to zero, and the only possibility you get is x=y. I dont know how else to find it after this. But the back of the book says the answer is a max at $f(1,1)$ and a min at $f(-1,-1)$ but I dont know how? $$f_x=\frac{-x^2-2xy+y^2+2}{(2+x^2+y^2)^2}$$ $$f_y= \frac{-y^2-2xy+x^2+2}{(2+x^2+y^2)^2}$$ Can anyone see why please? Thankyou

Setting $f_x=0$ and $f_y=0$ gives $-x^2-2xy+y^2+2=0$ and $-y^2-2xy+x^2+2=0$, so subtracting these equations gives $2y^2-2x^2=0, \;\;y^2=x^2,\; $ and so $y=\pm x$. 1) If $y=x$, substituting into the first equation gives $x^2=1$ so $x=\pm 1$. 2) If $y=-x$, substituting into the first equation gives $x^2=-1$, so there is no real solution. Therefore $(1,1)$ and $(-1,-1)$ are the only critical points. Since $\;\displaystyle f_{xx}=(2+x^2+y^2)^{-2}(-2x-2y)-4x(-x^2-2xy+y^2+2)(2+x^2+y^2)^{-3}$, $\;\displaystyle \hspace{.36 in}f_{xy}=(2+x^2+y^2)^{-2}(-2x+2y)-4y(-x^2-2xy+y^2+2)(2+x^2+y^2)^{-3}$, $\;\displaystyle \hspace{.36 in}f_{yy}=(2+x^2+y^2)^{-2}(-2y-2x)-4y(-y^2-2xy+x^2+2)(2+x^2+y^2)^{-3}$, A) $\;D=f_{xx}f_{yy}-(f_{xy})^2=(-\frac{1}{4})(-\frac{1}{4})-0^2=\frac{1}{16}>0\;$ and $\;f_{xx}=-\frac{1}{4}<0$ at $(1,1)$, $\hspace{1.4 in}$so $f$ has a relative maximum at $(1,1)$. B) $\;D=f_{xx}f_{yy}-(f_{xy})^2=(\frac{1}{4})(\frac{1}{4})-0^2=\frac{1}{16}>0\;$ and $\;f_{xx}=\frac{1}{4}>0$ at $(-1,-1)$, $\hspace{1.4 in}$so $f$ has a relative minimum at $(-1,-1)$.

Limit as $x$ tend to zero of: $x/[\ln (x^2+2x+4) - \ln(x+4)]$ Without making use of LHôpital's Rule solve: $$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)}$$ $ x^2+2x+4=0$ has no real roots which seems to be the gist of the issue. I have attempted several variable changes but none seemed to work.

An approach without L'Hopital's rule. $$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)}=\lim_{x\to 0} {1\over {1\over x}\ln {x^2+2x+4\over x+4}}=\lim_{x\to 0} {1\over \ln \big ({x^2+2x+4\over x+4}\big)^{1\over x}}$$ but $$({x^2+2x+4\over x+4}\big)^{1\over x}=({x^2+x+x+4\over x+4}\big)^{1\over x}=\big(1+{x^2+x\over x+4}\big)^{1\over x}=\big(1+{ x(x+1)\over x+4}\big)^{1\over x}=\big(1+\color\red x({x+1\over x+4})\big)^{1\over \color\red x}$$ which tends to $e^{1\over 4}$ as $x$ goes to $0$. So the original limit is ${1\over \ln{e^{1\over 4}}}=\color\red 4$

Problem when $x=\cos (a) +i\sin(a),\ y=\cos (b) +i\sin(b),\ z=\cos (c) +i\sin(c),\ x+y+z=0$ Problem: If $$x=\cos (a) +i\sin(a),\ y=\cos (b) +i\sin(b),\ z=\cos (c) +i\sin(c),\ x+y+z=0$$ then which of the following can be true: 1) $\cos 3a + \cos 3b + \cos 3c = 3 \cos (a+b+c)$ 2) $1+\cos (a-b) + \cos (b-c) =0$ 3) $\cos 2a + \cos 2b +\cos 2c =\sin 2a +\sin 2b +\sin 2c=0$ 4) $\cos (a+b)+\cos(b+c)+\cos(c+a)=0$ Try: I tried taking $x=e^{ia},y=e^{ib},z=e^{ic}$ and then i tried expressing each option in euler form FOR EXAMPLE: 1) $-3/2 e^{-i a-i b-i c}-3/2 e^{i a+i b+i c}+1/2 e^{-3 i a}+1/2 e^{3 i a}+1/2 e^{-3 i b}+1/2 e^{3 i b}+1/2 e^{-3 i c}+1/2 e^{3 i c}$ 2) $1/2 e^{i a-i b}+1/2 e^{i b-i a}+1/2 e^{i b-i c}+1/2 e^{i c-i b}+1$ 3) $1/2 e^{-2 i a}+1/2 e^{2 i a}+1/2 e^{-2 i b}+1/2 e^{2 i b}+1/2 e^{-2 i c}+1/2 e^{2 i c}$ 4) $1/2 e^{-i a-i b}+1/2 e^{i a+i b}+1/2 e^{-i a-i c}+1/2 e^{i a+i c}+1/2 e^{-i b-i c}+1/2 e^{i b+i c}$ Now after all this i'm stuck!!Please help!! How should i proceed?

HINT: For $(1),$ Using If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$., we have $x^3+y^3+z^3=3xyz$ Now $x^3=e^{i(3a)}=\cos3a+i\sin3a$ and $xy=e^{i(a+b)}=\cos(a+b)+i\sin(a+b),xyz=\cdots$ For $(3),(4)$ $x+y+z=0\implies\cos a+\cos b+\cos c=\sin a+\sin b+\sin c=0$ So, $x^{-1}+y^{-1}+z^{-1}=?$ Now $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=(x+y+z)^2-2xyz(x^{-1}+y^{-1}+z^{-1})=?$

Prove that equation $x^6+x^5-x^4-x^3+x^2+x-1=0$ has two real roots Prove that equation $$x^6+x^5-x^4-x^3+x^2+x-1=0$$ has two real roots and $$x^6-x^5+x^4+x^3-x^2-x+1=0$$ has two real roots I think that: $$x^{4k+2}+x^{4k+1}-x^{4k}-x^{4k-1}+x^{4k-2}+x^{4k-3}-..+x^2+x-1=0$$ and $$x^{4k+2}-x^{4k+1}+x^{4k}+x^{4k-1}-x^{4k-2}-x^{4k-3}-..+x^2+x-1=0$$ has two real roots but i don't have solution

HINT: you can write this equation: $x^6+x^5-x^4-x^3=-x^2-x+1$ $y=x^6+x^5-x^4-x^3$ $y=-x^2-x+1$ and draw two functions noting that there are two points where two fnctions intersect.

Quadratic Integers in $\mathbb Q[\sqrt{-5}]$ Can someone tell me if $\frac{3}{5}$, $2+3\sqrt{-5}$, $\frac{3+8\sqrt{-5}}{2}$, $\frac{3+8\sqrt{-5}}{5}$, $i\sqrt{-5}$ are all quadratic integers in $\mathbb Q[\sqrt{-5}]$. And if so why are they in $\mathbb Q[\sqrt{-5}]$.

Only one of them is. There are a number of different ways to tell, and of these the easiest are probably the minimal polynomial and the algebraic norm. In a quadratic extension of $\mathbb{Q}$, the minimal polynomial of an algebraic number $z$ looks something like $a_2 x^2 + a_1 x + a_0$, and if $a_2 = 1$, we have an algebraic integer. If $z = m + n\sqrt{d}$, then the norm is $N(z) = m^2 - dn^2$. In $\mathbb{Q}(\sqrt{-5})$ for $z = m + n\sqrt{-5}$, this works out to $N(z) = m^2 + 5n^2$. If $z$ is an algebraic integer, then $N(z) \in \mathbb{Z}$. Let's take the five numbers one by one: * *$\frac{3}{5}$ is obviously in $\mathbb{Q}$ so it's also in the extension $\mathbb{Q}(\sqrt{-5})$. But its minimal polynomial is $5x - 3$, so $a_1 = 5$ but $a_2 = 0$. Also, the norm is $\frac{9}{25}$. Clearly $\frac{3}{5}$ is not an algebraic integer. *$2 + 3\sqrt{-5}$ has a minimal polynomial of $x^2 - 4x + 49$, so our $a_2$ here is indeed 1. Also, its norm is 49. Here we have an algebraic integer in $\mathbb{Q}(\sqrt{-5})$. *$\frac{3}{2} + 4\sqrt{-5}$ has a minimal polynomial of $4x^2 - 12x + 329$, so $a_2 = 4$. And the norm is $\frac{329}{4}$, which is not an integer. So $\frac{3}{2} + 4\sqrt{-5}$ is not an algebraic integer in $\mathbb{Q}(\sqrt{-5})$. (Side note: $\frac{3}{2} + 4\sqrt{5}$ is not an algebraic integer in $\mathbb{Q}(\sqrt{5})$ either, but $\frac{3}{2} + \frac{7\sqrt{5}}{2}$ is). *$\frac{3}{5} + \frac{8\sqrt{-5}}{5}$... you can do this one on your own. *$i \sqrt{-5}$ works out to $-\sqrt{5}$, which is an algebraic integer, but it comes from a different domain.

Infinite Product - Seems to telescope Evaluate $$\left(1 + \frac{2}{3+1}\right)\left(1 + \frac{2}{3^2 + 1}\right)\left(1 + \frac{2}{3^3 + 1}\right)\cdots$$ It looks like this product telescopes: the denominators cancel out (except the last one) and the numerators all become 3. What would my answer be?

we have the following identity (which affirms that the product telescopes): $$\left (1+\frac{2}{3^n+1}\right)=3\cdot\frac{3^{n-1}+1}{3^n+1}=\frac{1+3^{-(n-1)}}{1+3^{-n}}$$ (as denoted in the comment by Thomas Andrews)and as a result: $$\prod_{k=1}^n \left (1+\frac{2}{3^k+1}\right)=\frac{2\cdot 3^n}{3^n+1}=\frac{2}{1+3^{-n}}$$

Calculation of real root values of $x$ in $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}.$ Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $ $\bf{My\; Solution::}$ Here domain of equation is $\displaystyle x\geq 1$. So squaring both sides we get $\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$. $\displaystyle (1-2x)^2=4(x^2-1)\Rightarrow 1+4x^2-4x=4x^2-4\Rightarrow x=\frac{5}{4}.$ But when we put $\displaystyle x = \frac{5}{4}\;,$ We get $\displaystyle \frac{3}{2}-\frac{1}{2}=2\Rightarrow 1=2.$(False.) So we get no solution. My Question is : Can we solve above question by using comparision of expressions? Something like $\sqrt{x+1}<\sqrt{x-1}+\sqrt{4x-1}\; \forall x\geq 1?$ If that way possible, please help me solve it. Thanks.

For $x\ge1$ we have $$\sqrt{4x-1}\ge \sqrt {3x} $$ and $$\sqrt{x+1}\le \sqrt {2x}$$ hence $$\sqrt{x+1}-\sqrt{x-1}\le \sqrt{2x}<\sqrt{3x}\le\sqrt{4x-1} $$

differentiation of a matrix function In statistics, the residual sum of squares is given by the formula $$ \operatorname{RSS}(\beta) = (\mathbf{y} - \mathbf{X}\beta)^T(\mathbf{y} - \mathbf{X}\beta)$$ I know differentiation of scalar functions, but how to I perform derivatives on this wrt $\beta$? By the way, I am trying to take the minimum of RSS wrt to $\beta$, so I am setting the derivative equal to 0. I know somehow product rule has to hold. So here I have the first step $$-\mathbf{X}^T(\mathbf{y}-\mathbf{X}\beta) + (\mathbf{y}-\mathbf{X}\beta)^T(-\mathbf{X})= 0$$

First you can remove the transposition sign from the first bracket: $RSS=(\mathbf{y}^T - \beta ^T \mathbf{X} ^T)(\mathbf{y} - \mathbf{X}\beta)$ Multiplying out: $RSS=y^Ty-\beta ^T \mathbf{X} ^Ty-y^TX\beta+\beta^TX^T X\beta $ $\beta ^T \mathbf{X} ^Ty$ and $y^TX\beta$ are equal. Thus $RSS=y^Ty-2\beta ^T \mathbf{X} ^Ty+\beta^TX^T X\beta$ Now you can differentiate with respect to $\beta$: $\frac{\partial RSS}{\partial \beta}=-2X^Ty+2X^T X\beta=0$ Dividing by 2 and bringing the first summand to the RHS: $X^T X\beta=X^Ty$ Multiplying both sides by $(X^T X)^{-1}$ $(X^T X)^{-1}X^T X\beta=(X^T X)^{-1}X^Ty$ $(X^T X)^{-1}X^T X= I$ (Identity matrix). Finally you get $\beta=(X^T X)^{-1}X^Ty$ Equality of $\beta ^T \mathbf{X} ^Ty$ and $y^TX\beta$ I make an example: $\left( \begin{array}{c c} b_1 & b_2 \end{array} \right) \cdot \left( \begin{array}{c c c} x_{11} & x_{21} \\ x_{12} & x_{22}\end{array} \right) \cdot \left( \begin{array}{c c} y_1 \\ y_2 \end{array} \right) $ $=\left( \begin{array}{c c} b_1x_{11}+b_2x_{12} & b_1x_{21}+b_2x_{22} \end{array} \right) \cdot \left( \begin{array}{c c} y_1 \\ y_2 \end{array} \right)$ $=b_1 x_{11}y_1+b_2 x_{12}y_1+b_1x_{21}y_2+b_2x_{22}y_2\quad (\color{blue}{I})$ $\left( \begin{array}{c c} y_1 & y_2 \end{array} \right) \cdot \left( \begin{array}{c c c} x_{11} & x_{12} \\ x_{21} & x_{22}\end{array} \right) \cdot \left( \begin{array}{c c} b_1 \\ b_2 \end{array} \right) $ $=\left( \begin{array}{c c} y_1x_{11}+y_2x_{21} & y_1x_{12}+y_2x_{22} \end{array} \right) \cdot \left( \begin{array}{c c} b_1 \\ b_2 \end{array} \right)$ $=y_1 x_{11}b_1+y_2 x_{21}b_1+y_1x_{12}b_2+y_2x_{22}b_2 \quad (\color{blue}{II})$ $\color{blue}{I}$ and $\color{blue}{II}$ are equal. Derivative Rules $\frac{\partial \beta ^T X ^T y }{\partial \beta }=X^Ty$ $\frac{\partial \beta^T X^T X \beta }{\partial \beta }=2X^TX\beta$

All real values $a$ for a $2$-dimensional vector? Find all real numbers $a$ for which there exists a $2D$, nonzero vector $v$ such that: $\begin{pmatrix} 2 & 12 \\ 2 & -3 \end{pmatrix} {v} = a {v}$. I substituted $v$ with $\begin{pmatrix} c \\ d \end{pmatrix}$ and multiplied to obtain the system of equations: $2x+12y = ax$ $2x-3y= ay$ Since the value $a$ is only of importance, I added the two equations to obtain $4x+9y = ax + ay$. Would that mean that $a = 4, 9$ is correct?

you can go from $$2x+12y = kx,\, 2x-3y = ky $$ to $$\frac{y}{x} = \frac{k-2}{12} = \frac2{k+3}.$$ therefore $k$ satisfies the characteristic equation $$0=(k+3)(k-2) - 24 = k^2+k-30 = (k+6)(k-5).$$ therefore $$k = 5, -6 $$

Expressing ${}_2F_1(a, b; c; z)^2$ as a single series Is there a way to express $${}_2F_1\bigg(\frac{1}{12}, \frac{5}{12}; \frac{1}{2}; z\bigg)^2$$ as a single series a la Clausen? Note that Clausen's identity is not applicable here.

Using Maple, I get $$ \sum_{n=0}^\infty \frac{\Gamma \left( {\frac {7}{12}} \right) \Gamma \left( {\frac {11}{12}} \right) {\mbox{$_4$F$_3$}(\frac{1}{12},{\frac {5}{12}},-n,\frac{1}{2}-n;\,1/2,-n+{\frac {7}{12}},-n+{\frac {11}{12}};\,1)\;(4z)^n }} {16\,\Gamma \left( -n+{\frac {11}{12}} \right) \Gamma \left( -n+{ \frac {7}{12}} \right) \Gamma \left( 2\,n+1 \right) \sin^2 \left( {\frac {5}{12}}\,\pi \right) \sin^2 \left( \frac{1}{12}\,\pi \right)} \\ = 1+{\frac {5}{36}}z+{\frac {295}{3888}}{z}^{2}+{\frac {5525}{104976}}{ z}^{3}+{\frac {4281875}{105815808}}{z}^{4}+{\frac {564921305}{ 17142160896}}{z}^{5}+O \left( {z}^{6} \right) $$ I don't know how much use that is.

How to calculate $\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$ I came across this strange limit whilst showing convergence of a series: $$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$$ How can I calculate this limit?

Squeeze theorem gives you the proof that the limit is $\frac{3}{4}$. Since you mentioned you were looking for another way to verify that the limit is correct, here is one way (although not rigorous like the squeeze theorem) $$\begin{align}\frac{2^n+3^n}{3^n+4^n} = \frac{2^n}{3^n+4^n}+\frac{3^n}{3^n+4^n} \\ = \left(\frac{\frac{1}{2^n}}{\frac{1}{2^n}}\right)\frac{2^n}{3^n+4^n}+\left(\frac{\frac{1}{3^n}}{\frac{1}{3^n}}\right)\frac{3^n}{3^n+4^n} \\ = \frac{1}{\left(\frac{3}{2}\right)^n+2^n}+\frac{1}{1+\left(\frac{4}{3}\right)^n}\end{align}$$ You should be able to see that $\lim_{n \to \infty} \left(\frac{3}{2}\right)^n+2^n = \infty$ so $\lim_{n \to \infty} \frac{1}{\left(\frac{3}{2}\right)^n+2^n} = 0$. Then notice that for large $n$ the quantity $$\frac{1}{1+\left(\frac{4}{3}\right)^n} \approx \frac{1}{\left(\frac{4}{3}\right)^n} = \left(\frac{3}{4}\right)^n$$ so for large values of $n$, $$\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}} \approx \sqrt[n]{\left(\frac{3}{4}\right)^n} = \frac{3}{4}$$

How to integrate $\int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx$ How would I do the following integral? $$\int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx$$ Where $x > 0$ and $k$ is a constant greater than $0$

Consider the integral \begin{align} I = \int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx \end{align} Let $t = x^{k/2}$ for which $x = t^{2/k}$ and $dx = (2/k) t^{(2/k) - 1} \, dt$ for which the integral becomes \begin{align} I = \frac{2}{k} \int \frac{dt}{1 + t^{2}}. \end{align} This is the integral for $\tan^{-1}(t)$ leading to \begin{align} I = \frac{2}{k} \, \tan^{-1}(t) + c_{0} \end{align} and upon backward substitution \begin{align} \int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx = \frac{2}{k} \, \tan^{-1}\left(x^{\frac{k}{2}}\right) + c_{0} \end{align}

Stoppage time for sequence of uniform random numbers with a recursively shrinking domain Define $x_n = U(x_{n-1})$ where $U(x)\in\lbrace 0,1,\ldots,x\rbrace$ is a uniformly distributed random integer. Given $x_0$ as some large positive integer, what is the expected value of $n$ for which $x_n=0$? The answer I came up with uses iterated conditioning, writing $y_n=\mathbb{E}[x_n]$, $y_n = \mathbb{E}[\mathbb{E}[U ( x_{n-1} ) | x_{n-1}]] = \mathbb{E}[\mathbb{E}[x_{n-1}/2 | x_{n-1}]] = \frac{1}{2}y_{n-1}$ with $y_0=x_0$ so $y_n=x_0/2^n$. To me, the question can then be recast as finding the value of $n$ for which $y_n < 1/2$, i.e. $$n = \left\lceil \frac{\ln x_0}{\ln 2}\right\rceil+1$$ But I tried simulating this in R for some values of $x_0$ and this answer seems to consistently overestimate the simulated result, have I made a blunder in my reasoning? For reference, here is my R code: simStopTime <- function(x1) { i <- 1 x <- x1 while (x[i] > 0 ) { i <- i+1 x[i] <- sample(0:x[i-1], 1) } return(i - 1) # Subtract 1 because indexing in R starts at 1 } samples <- replicate(5000, simStopTime(1000)) mean(samples) # Fluctuates around 8.47 on repeated runs log(1000)/log(2) + 1 #Gives 10.96578

The expected waiting time $T_k$ to get down from $k$ to $0$ is $T_0 = 0$ for the base case, and otherwise it is $T_k = 1 + H_k$, where $H_k$ is the $k$th harmonic number $1 + 1/2 + 1/3 + \cdots + 1/k$. For large $k$ this is approximately $1 + \gamma + \ln k$, where $\gamma \doteq 0.57722$ is the Euler-Mascheroni constant. By inspection $T_0 = 0$. For $k > 0$, we observer that it takes one step to get to a number that is uniformly distributed in the interval $[0, k]$, and the expected time can therefore be written recursively: $$ T_k = 1 + \frac{1}{k+1} (T_0 + T_1 + T_2 + \cdots + T_k) $$ Recognize that $T_0 = 0$ and multiply both sides by $k+1$: $$ (k+1)T_k = k+1 + T_1 + T_2 + \cdots + T_k $$ Subtract $T_k$ from both sides: $$ kT_k = k+1 + T_1 + T_2 + \cdots + T_{k-1} $$ In particular $$ T_1 = 1+1 = 2 $$ We now proceed by induction. Suppose that we know already that $T_i = 1+H_i$ for $1 \leq i \leq k-1$. We can then write $$ \begin{align} kT_k & = k+1 + (1+H_1) + (1+H_2) + \cdots + (1+H_{k-1}) \\ & = k+1 + (k-1) + H_1 + H_2 + \cdots + H_{k-1} \\ & = 2k + (1) + (1 + 1/2) + \cdots + (1 + 1/2 + 1/3 + \cdots + 1/(k-1)) \end{align} $$ After the $2k+1$ in the last line, we have $k-1$ terms of $1$, $k-2$ terms of $1/2$, $k-3$ terms of $1/3$, and so on, until $1$ term of $1/(k-1)$. We can therefore write $$ \begin{align} kT_k & = 2k + \frac{k-1}{1} + \frac{k-2}{2} + \frac{k-3}{3} + \cdots + \frac{1}{k-1} \\ & = 2k + \Bigl(\frac{k}{1}-1\Bigr) + \Bigl(\frac{k}{2}-1\Bigr) + \cdots + \Bigl(\frac{k}{k-1}-1\Bigr) \\ & = 2k - (k-1) + \frac{k}{1} + \frac{k}{2} + \frac{k}{3} + \cdots + \frac{k}{k-1} \\ & = k+1 + kH_{k-1} \end{align} $$ Divide both sides by $k$ to get: $$ T_k = 1 + 1/k + H_{k-1} = 1+H_k $$ For $k = 1000$, we have $T_{1000} = 1+H_{1000} \doteq 1 + 0.5772 + 6.9078 = 8.4850$. (A symbolic math package gives us more directly the value $T_{1000} = 1+H_{1000} \doteq 8.48547$.)

Tangent line of a lemniscate at (0,0) I need to find the tangent line of the function $y=g(x)$ implicitly defined by $(x^2+y^2)^2-2a^2(x^2-y^2)=0$ at $(0,0)$, but I don't know how. I can't use implicit differentiation and evaluate at $(0,0)$, because when $y=0$ I can't use the Implicit Function Theorem to calculate the derivative and, therefore, the slope of the tangent line. I'd appreciate your help. Thanks.

$$(x^2+y^2)^2-2a^2(x^2-y^2)=0$$ Solving for $y$ we do substitution $t=y^2$ $$x^4+x^2t+t^2-2a^2x^2+2a^2t=0$$ $$t^2+t(2x^2+2a^2)+x^4-2a^2x^2=0$$ $$t=\pm a\sqrt{4x^2+a^2}-x^2-a^2$$ As $t=-a\sqrt{4x^2+a^2}-x^2-a^2$ is not positive we get solutions $$y=\pm \sqrt{a\sqrt{4x^2+a^2}-x^2-a^2}$$ Let $f(x)=\sqrt{a\sqrt{4x^2+a^2}-x^2-a^2}$. We have $y=\pm f(x)$, $y'=\pm f'(x)$. The derivative of $f$ is the following $$f'(x)=\frac{1}{2\sqrt{a\sqrt{4x^2+a^2}-x^2-a^2}}\cdot \left( a\frac{1}{2\sqrt{4x^2+a^2}}\cdot(8x)-2x \right)$$ $$=\frac{x\left(a\frac{2}{\sqrt{4x^2+a^2}}-1\right)}{\sqrt{a\sqrt{4x^2+a^2}-x^2-a^2}}$$ Function $f$ is continuous and differentiable at any small neighborhood of $0$ excluding $0$ itself. The limits of $f'(x)$ as $x\to0^\pm$ exist and they are given by $$\lim\limits_{x\to0^+}f'(x)=1$$ $$\lim\limits_{x\to0^-}f'(x)=-1$$ Thus at $(0,0)$ there are two tangent lines with equations $$t_1(x)=x$$ $$t_2(x)=-x$$

If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of m. Problem : If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of $m$. My Approach: $|x^2+4x+3|-mx+2m=0$ Case I : $x^2+4x+3-mx+2m=0$ $\Rightarrow x^2+ x (4-m) + 3+2m=0 $ Discriminant of above qudratic is $D = (4-m)^2 -4(3+2m) \geq 0$ $D = 16+m^2-8m-12-8m$ Solving for $m$ we get the values $-8 \pm 2\sqrt{15}$ Case II : Similarly solving for the given equation taking negative sign of modulus we get the solution for $m =$$8 \pm 2\sqrt{15}$ Can we take all the values of m to satisfy the given condition of the problem , please suggest which value of m should be neglected in this. Thanks.

$$m(x-2)=|(x+3)(x+1)|\ge0$$ If $m=0,$ there are two real solutions Else $m(x-2)=|(x+3)(x+1)|=0$ has no solution So, $$m(x-2)=|(x+3)(x+1)|>0$$ Now $|(x+3)(x+1)|=-(x+3)(x+1)$ if $-3\le x\le-1$ $=+(x+3)(x+1)$ otherwise If $m>0,x-2>0\iff x>2\implies m(x-2)=x^2+4x+3$ which has exactly two solutions If $m<0,x-2<0\iff x<2$ If $-1<x<2$ or if $x<-3;$ $m(x-2)=(x+3)(x+1)\ \ \ \ (1)$ If $-3\le x\le-1,m(x-2)=-(x+3)(x+1)\ \ \ \ (2)$ We need the discriminant of $(1)$ or $(2)$ to be zero honoring the range.

How to integrate $\frac{1}{(1+a\cos x)}$ from $-\pi$ to $\pi$ How to solve the following integration?$$\int_{-\pi}^\pi\frac{1}{1+a \cos x}$$

You can look at my earlier answer here (on my previous avatar). Below is another method. We have $$I = \int_{-\pi}^{\pi} \dfrac{du}{1+a\cos(u)} = 2 \int_0^{\pi} \dfrac{du}{1+a \cos(u)} = 2\int_0^{\pi/2} \dfrac{du}{1+a\cos(u)} + 2\int_{\pi/2}^{\pi} \dfrac{du}{1+a\cos(u)}$$ Hence, $$\dfrac{I}2 = \int_0^{\pi/2} \dfrac{du}{1+a\cos(u)} + \int_0^{\pi/2}\dfrac{du}{1-a\cos(u)} = \int_0^{\pi/2} \dfrac{2du}{1-a^2\cos^2(u)} = \int_0^{\pi/2}\dfrac{2\sec^2(u)du}{\sec^2(u)-a^2}$$ This gives us $$\dfrac{I}4 = \int_0^{\pi/2}\dfrac{\sec^2(u)du}{1+\tan^2(u)-a^2}$$ Setting $t=\tan(u)$, we obtain $$\dfrac{I}4 = \int_0^{\infty} \dfrac{dt}{t^2+(1-a^2)}=\dfrac1{\sqrt{1-a^2}} \left.\arctan\left(\dfrac{t}{\sqrt{1-a^2}}\right) \right \vert_0^{\infty} \implies I = \dfrac{2\pi}{\sqrt{1-a^2}}$$

Express $\ln(3+x)$ and $\frac{1+x}{1-x}$ as Maclaurin series its probably a lot to ask.. but how can I obtain the Maclaurin series for the two functions $f(x)=\ln(3+x)$ and $g(x)=\frac{1+x}{1-x}$ ? as far as I know I cant use any commonly known series to help me with this one ? so finding the derivatives at 0: $f(0) = \ln(3)$ $f'(x)= 1/(x+3) \implies f'(0) = 1/3$ $f''(x)= -1/(x+3)^2 \implies f''(0)= -1/ 3^2$ $f'''(x)= 2/(x+3)^3 \implies f'''(0)= 2/3^3$ $f''''(x)= -6/(x+3)^4 \implies f''''(0)= -6/3^4$ and now I guess Ill have to use the maclaurin formula: $$f(x)=\sum \frac{f^{(n)}(0) x^n}{n!}$$ but now how do I continue from here? I've got: $$(\ln 3) + \frac{x}{3} - \frac{x^2}{3^2 \cdot 2!} + \frac{2x^3}{3^3 \cdot 3!} - \frac{6x^4}{3^4 \cdot 4!} $$ same question goes for $g(x)$ honestly, on that one upon simplifying the numerator from $g(0)$ with the factorial when I plug it into the maclaurien series formula I just get $g(x)= 1+2x+2x^2 + 2x^3 + 2x^4 + \cdots$

for $(x+1)/(1-x)$ write it as $$ x \frac{1}{1-x} + \frac{1}{1-x} $$ substitute the series for $1/(1-x)$ and proceed. To get a single series, add the terms one by one. Also $$ \log(3 -x) = \log(3) + \log(1 - x/3) $$ so substitute $x/3$ in the series for $\log(1-y)$ which can be found be integrating the series for $1/(1-y).$ switch the sign of $x$ to get the series for $\log(3+x).$ Ie multiply the coefficient of $x^n$ by $(-1)^n.$

Giving integer images (bis) Prove the statement below (the restriction $x < y < z$ is to avoid apparent uncertainties but the property is valid for all $x, y, z$ really). $$F(x,y,z) = \frac{(y+z)x^n}{(z-x)(y-x)} +\frac{(z+x)y^n}{(z-y)(x-y)}+\frac{(x+y)z^n}{(x-z)(y-z)} \in \mathbb Z.$$

If $n = 1$, then the whole expression is equal to $-1$. If $n=2$, it is equal to $0$. Let us therefore assume that $n > 2$. We have: $$F(x,y,z) = \frac{y^nz^2 - y^2z^n + x^n(y-z)(y+z) + x^2(z^n-y^n)}{(x-y)(x-z)(y-z)} = \frac{f(x,y,z)}{(x-y)(x-z)(y-z)}.$$ What we want to know is if the numerator is divisible by $x-y$, $x-z$ and $y-z$. Let's check (for example) $y-z$. It is clear that $y-z$ divides the second term. Finally, observe that $$y^nz^2 - y^2z^n + x^2(z^n-y^n) = y^2z^2(y^{n-2}-z^{n-2}) - x^2(y^n-z^n)$$ and $$a^k-b^k = (a-b)(a^{k-1} + a^{k-2}b + \dots + b^{k-1}).$$ We apply this with $a = y$ and $b = z$ after noticing that if $a,b \in \mathbb Z$, then $a+b$ and $ab$ is an integer (so the expression in brackets is indeed integer).

How many non prime factors are in the number $N=2^5 \cdot 3^7 \cdot 9^2 \cdot 11^4 \cdot 13^3$. to find non prime factors in the number $N=2^5 \cdot 3^7 \cdot 9^2 \cdot 11^4 \cdot 13^3$. First I tried finding all the factors by adding 1 to each of the exponents and then multiplying them and then finding the prime factors of the given number and then subtracting the prime factors from the total factors but I'm not getting the answer. Answer is $1436$.

Write it as $2^5 * 3^{11} * 11^4 * 13^3$. Thus, as you said, adding one to each exponent and multiplying, we get the total number of divisors, which is $6* 12 * 5* 4 = 1440$. Subtracting the four prime factors ($2,3,11,13$) leaves us with 1436.

Real Numbers are Roots $r, s$. Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$, and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$. Find the sum of all possible values of $|b|$. Using Vieta's Formulas, $r+s+x_1$ $=0$ $\Rightarrow x_1$ $=-r-s$, where $x_1$ is the third root. Similarly, $x_2=-r-s-1$ $=x_1-1$, where $x_2$ is the third root of $q(x)$. I have the list here: $a_n = a_n$ $a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$ $a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$ $\vdots$ $a_0 = (-1)^n a_n r_1r_2\cdots r_n$ Obviously, $b = a_0$ so for $p(x)$: $$b = (-1)^{3}(1)(r)(s)(-1)(r + s) = (r)(s)(r+s) = r^2s + s^2r$$ For $q(x)$ then, $$a_0 = (b + 240) = (-1)^{3}(1)(r+4)(s-3)(-1)(r + s + 1) = (r+4)(s-3)(r + s + 1) $$ $$= r^2s - 3r^2 + rs^2 + 2rs - 15r + 4s^2 -8s - 12$$ $$b = r^2s - 3r^2 + rs^2 + 2rs - 15r + 4s^2 -8s - 252$$ But that leaves an awfully weird system of equations. PLEASE DO NOT GIVE ME A FULL ANSWER, Just help.

Since $r,s$ are two roots of $x^3+ax+b=0$, we have $$ r^3+ar+b=0,s^3+as+b=0 \tag{1}$$ and hence $$ (r^3-s^3)+a(r-s)=0. $$ Assuming $r-s\neq0$, we have $$ r^2+rs+s^2+a=0.\tag{2}$$ Similarly since $r+4,s-3$ are two roots of $x^3+ax+b+240=0$, we have $$ (r+4)^3+a(r+4)+b+240=0,(s-3)^3+a(s-3)+b+240=0\tag{3}$$ and hence $$ [(r+4)^3-(s-3)^3]+a(r-s+7)=0.$$ Assuming $r-s+7\neq 0$, we have $$ (r+4)^2+(r+4)(s-3)+(s-3)^2+a=0.\tag{4}$$ From (4)-(3) gives $$ 13+5r-2s=0 $$ from which we find that $$s=\frac{1}{2}(5r+13).\tag{5}$$ Putting this in (2), we obtain $a$ in terms of $r$. Putting them in the first equation of (1), we have $b$ in terms of $r$. Putting $s,a,b$ in the first equation of (3), we will have an equation of $r$ which you can get the values $r$. I think you can do the rest.

How many words with letters from the word ABRACADABRA if they must end in a consonant and $d$ must be after $r$. How many words with letters from the word ABRACADABRA if they must end in a consonant and $d$ must be after $r$. What I did: I have $A:5$ $B:2$ $R:2$ $C:1$ $D:1$ If the words must end in a consonant and d must be after r I have only two cases: 1)$D$ at the end. 2)$C$ at the end. 3)$B$ at the end. Case 1: $$ \_\ \_\ \_\ \_\ \_\ \_\ \_\ \_\ \_\ \_\ \text{D} $$ So I have to choose $10$ letters for the remaining slots with $A$ repeated $5$ times, $2$ $B$s and 2 $R$s: $$ \frac{10!}{5!2!2!} $$ Case 2: I set $D=R$ and same thought process as before, giving me: $$ \frac{10!}{5!3!2!} $$ Case 3: Same as case 2. $$Total= \frac{10!}{5!2!2!}+\frac{10!}{5!3!} $$ Is this correct?

Since D must appear after both R's, the last letter can be a B, C, or D. Case 1: The last letter is D. We have ten places to fill with five A's, two B's, two R's, and one C. We can fill five of the ten places with A's in $\binom{10}{5}$ ways. We can fill two of the remaining five places with B's in $\binom{5}{2}$ ways. We can fill two of the remaining three places with R's in $\binom{3}{2}$ ways. Finally, we can fill the last place with a C in $\binom{1}{1}$ way, so there are $$\binom{10}{5}\binom{5}{2}\binom{3}{2}\binom{1}{1} = \frac{10!}{5!5!} \cdot \frac{5!}{3!2!} \cdot \frac{3!}{2!1!} \cdot \frac{1!}{1!0!} = \frac{10!}{5!2!2!1!}$$ permutations that end in a D, as you found. Case 2: The last letter is C. Then we have ten places to fill with five A's, two B's, two R's, and one D. If, at first, we ignore the requirement that D must appear after the two R's, using the same procedure as above yields $$\binom{10}{5}\binom{5}{2}\binom{3}{2}\binom{1}{1}$$ However, in only one third of these permutations does D appear after both R's. Thus, the number of permutations in which the last letter is a C and D appears after both R's is $$\frac{1}{3}\binom{10}{5}\binom{5}{2}\binom{3}{2}\binom{1}{1} = \frac{1}{3} \cdot \frac{10!}{5!2!2!1!}$$ Case 3: The last letter is a B. Then we have ten places to fill with five A's, two R's, one B, one C, and one D. If, at first, we ignore the requirement that D must appear after both R's, then using the same procedure as above yields $$\binom{10}{5}\binom{5}{2}\binom{3}{1}\binom{2}{1}\binom{1}{1}$$ However, in only one third of these permutations does the letter D appear after both R's. Thus, the number of permutations in which the last letter is a B and D appears after both R's is $$\frac{1}{3}\binom{10}{5}\binom{5}{2}\binom{3}{1}\binom{2}{1}\binom{1}{1} = \frac{1}{3} \cdot \frac{10!}{5!2!1!1!1!}$$ To find the number of words that can be formed from the word ABRACADABRA in which the last letter is a consonant and D appears after both R's, add the totals for the three disjoint cases.

Basis for the vector space P2 I am trying to wrap my head around vector spaces of polynomials in P2. If I represent the polynomial $ ax^2 + bx + c $ with the matrix $ A = \begin{bmatrix} 1,0,0 \\ 0,1,0 \\ 0,0,1 \\ \end{bmatrix} $ and the vector $ \begin{bmatrix} 1 \\ x \\ x^2 \\ \end{bmatrix} $ what corresponds to $a$, $b$, and $c$ in the matrix $A$?

I think you need to be clear about what you mean by "representing" the polynomial. You can if you like make the assignments $$ x^2 \;\; \to \;\; \left [ \begin{array}{c} 0\\ 0\\ 1\\ \end{array} \right ] \hspace{2pc} x \;\; \to \;\; \left [ \begin{array}{c} 0\\ 1\\ 0\\ \end{array} \right ] \hspace{2pc} 1 \;\; \to \;\; \left [ \begin{array}{c} 1\\ 0\\ 0\\ \end{array} \right ] $$ Then your polynomial can be represented by the vector $$ ax^2 + bx + c \;\; \to\;\; \left [ \begin{array}{c} c\\ b\\ a\\ \end{array} \right ]. $$ To describe a linear transformation in terms of matrices it might be worth it to start with a mapping $T:P_2 \to P_2$ first and then find the matrix representation. Edit: To answer the question you posted, I would take each basis vector listed above and apply the matrix to it: \begin{eqnarray*} \left [ \begin{array}{ccc} 3 & 2 & 7 \\ 0 & 1 & 0 \\ 4 & 0 & 1 \\ \end{array} \right ] \left [ \begin{array}{c} 1 \\ 0\\ 0\\ \end{array} \right ] & = & \left [ \begin{array}{c} 3 \\ 0 \\ 4 \\ \end{array} \right ] \;\; \to \;\; 4x^2 + 3 \\ \left [ \begin{array}{ccc} 3 & 2 & 7 \\ 0 & 1 & 0 \\ 4 & 0 & 1 \\ \end{array} \right ] \left [ \begin{array}{c} 0 \\ 1\\ 0\\ \end{array} \right ] & = & \left [ \begin{array}{c} 2 \\ 1\\ 0\\ \end{array} \right ] \;\; \to \;\; x+ 2 \\ \left [ \begin{array}{ccc} 3 & 2 & 7 \\ 0 & 1 & 0 \\ 4 & 0 & 1 \\ \end{array} \right ] \left [ \begin{array}{c} 0 \\ 0\\ 1\\ \end{array} \right ] & = & \left [ \begin{array}{c} 7 \\ 0\\ 1\\ \end{array} \right ] \;\; \to \;\; x^2 + 7. \end{eqnarray*}

Prove that $x$ has order $5$. let $ x \in G$ such that $(a^{-1})*(x^2)*(a) = x^3$ for some self inverse $a.$ Prove that $x$ has order $5.$ I don't know how to start this proof. Seems really difficult.

$a^{-1}x^2a=x^3 \implies a^{-1}x^4a=x^6 \implies a^{-1}x^6a=x^9$ but $x^6=x^3x^3=(a^{-1}x^2a)(a^{-1}x^2a)=a^{-1}x^4a \implies a^{-1}(a^{-1}x^4a)a=x^9 \implies a^{-1}a^{-1}x^4aa=x^9$ now using $a^2=a^{-2}=e$ we have $x^4=x^9$ so $x^5=e$ and because $5$ is prime $x=e$ or $x$ has order $5$.

Show that: $\sinh^{-1}(x) = \ln(x + \sqrt{x^2 +1 } )$ could someone Please give me some hint of how to do this question thanks

Hint: Assuming we already know these hyperbolic functions are invertible: $$\sinh(\log(x+\sqrt{x^2+1})):=\frac12\left(e^{\log(x+\sqrt{x^2+1})}-e^{-\log(x+\sqrt{x^2+1})}\right)=$$ $$=\frac12\left(x+\sqrt{x^2+1}-\frac1{x+\sqrt{x^2+1}}\right)=\frac12\left(\frac{x^2+x^2+1+2x\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}}\right)=$$ $$=\frac12\left(\frac{2x(x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}}\right)=\frac{2x}2=x$$

How to find a function that is the upper bound of this sum? The Problem Consider the recurrence $ T(n) = \begin{cases} c & \text{if $n$ is 1} \\ T(\lfloor(n/2)\rfloor) + T(\lfloor(n/4)\rfloor) + 4n, & \text{if $n$ is > 1} \end{cases}$ A. Express the cost of all levels of the recursion tree as a sum over the cost of each level of the recursion tree B. Give a function $g(n)$ and show that it is an upper bound on the sum My Work I was able to do part A. I drew the first six levels of the recursion tree and expressed the expressed the cost of all levels as $\sum_{i=0}^{log_2n} \frac{4n}{2^i}f(i+2) $ where $f(n)$ is the $n$th term in the Fibonacci sequence(0, 1, 1, 2, 3, 5, 8) How would I come up with a function that would be an upper bound of this sum?

Suppose we start by solving the following recurrence: $$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + 4n$$ where $T(1) = c$ and $T(0) = 0.$ Now let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$ be the binary representation of $n.$ We unroll the recursion to obtain an exact formula for $n\ge 2$ $$T(n) = c [z^{\lfloor \log_2 n \rfloor}] \frac{1}{1-z-z^2} + 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} [z^j] \frac{1}{1-z-z^2} \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}.$$ We recognize the generating function of the Fibonacci numbers, so the formula becomes $$T(n) = c F_{\lfloor \log_2 n \rfloor +1} + 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}.$$ We now compute lower and upper bounds which are actually attained and cannot be improved upon. For the lower bound consider a one digit followed by a string of zeroes, to give $$T(n) \ge c F_{\lfloor \log_2 n \rfloor +1} + 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} 2^{\lfloor \log_2 n \rfloor-j} \\ = c F_{\lfloor \log_2 n \rfloor +1} + 8 \times 2^{\lfloor \log_2 n \rfloor} \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} 2^{-j-1}.$$ Now since $$|\varphi|=\left|\frac{1+\sqrt{5}}{2}\right|<2$$ the sum term converges to a number, we have $$\frac{1}{2} \le \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} 2^{-j-1} \lt \sum_{j=0}^{\infty} F_{j+1} 2^{-j-1} = 2.$$ For an upper bound consider a string of one digits to get $$T(n) \le c F_{\lfloor \log_2 n \rfloor +1} + 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} \sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^{k-j} \\ = c F_{\lfloor \log_2 n \rfloor +1} + 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} (2^{\lfloor \log_2 n \rfloor+1-j} - 1) \\ = c F_{\lfloor \log_2 n \rfloor +1} - 4 (F_{\lfloor \log_2 n \rfloor +2} -1) + 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} 2^{\lfloor \log_2 n \rfloor+1-j} \\ = c F_{\lfloor \log_2 n \rfloor +1} - 4 (F_{\lfloor \log_2 n \rfloor +2} -1) + 4 \times 2^{\lfloor \log_2 n \rfloor+1} \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} 2^{-j} \\ = c F_{\lfloor \log_2 n \rfloor +1} - 4 (F_{\lfloor \log_2 n \rfloor +2} -1) + 16 \times 2^{\lfloor \log_2 n \rfloor} \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} 2^{-j-1}.$$ The same constant appears as in the lower bound. Now since the term $F_{\lfloor \log_2 n \rfloor}$ is asymptotically dominated by $2^{\lfloor \log_2 n \rfloor}$ (we have $F_{\lfloor \log_2 n \rfloor}\in o(2^{\lfloor \log_2 n \rfloor})$ because $F_{\lfloor \log_2 n \rfloor} \in\Theta(\varphi^{\lfloor \log_2 n \rfloor}))$ joining the upper and the lower bound we get for the asymptotics of this recurrence that it is $$T(n)\in\Theta\left(2^{\lfloor \log_2 n \rfloor}\right) = \Theta\left(2^{\ \log_2 n}\right) = \Theta(n),$$ which, let it be said, could also have been obtained by inspection. Remark. The evaluation of the constant is done by noting that the generating function of $$F_{j+1} 2^{-j-1}\quad \text{is}\quad\frac{1/2}{1-z/2-z^2/4}$$ which at $z=1$ evaluates to $\frac{1/2}{1-1/2-1/4} = 2.$ We have a certain flexibility as to what power of two to use in the constant but this does not affect the asymptotics. This MSE link has a similar calculation.

Epsilon and Delta proof of $\lim_{x\to0} \frac{2-\sqrt{4-x}}{ x}$ I need to prove $\lim_{x\to0} \frac{2-\sqrt{4-x}}{ x}$ I first found the limit to be $\frac{1}{4}$ by using l'hopital's rule. By definition i need to find a $\delta > 0$ for every $\epsilon >0$ Then i will have $|x-0|<\delta$ and $$|\frac{2-\sqrt{4-x}}{ x}-\frac{1}{4}|<\epsilon$$ I have tried multiple ways to simplify, but I can't seem to get it in the form of just $x$. And I am a bit confused on how to pick my delta in this case. Any help would be much appreciated.

It gets much simple when you write $$ \frac{2−\sqrt{4−x}}x = \frac{(2-\sqrt{4−x})(2+\sqrt{4−x})}{x(2+\sqrt{4−x})} =\frac1{2+\sqrt{4−x}} $$ then: \begin{align} \left| \frac1{2+\sqrt{4−x}} - \frac 14 \right| &= \frac{|2 - \sqrt{4-x}|}{4(2+\sqrt{4−x})} \le \frac{|2 - \sqrt{4-x}|}8 \\ &= \frac{|2 - \sqrt{4-x}|(2 + \sqrt{4-x})}{8(2 + \sqrt{4-x})} \\ &= \frac{|x|}{8(2 + \sqrt{4-x})} \le \frac{|x|}{16} \end{align} so just take $\delta = 16\epsilon$.