Datasets:

math-ai

/

StackMathQA

like 76

Follow

math-ai 19

$\lim_{n \rightarrow \infty} n ((n^5 +5n^4)^{1/5} - (n^2 +2n)^{1/2})$ $$\lim_{n \rightarrow \infty} n ((n^5 +5n^4)^{1/5} - (n^2 +2n)^{1/2})$$ Please, help me to find the limit.

Use the general fact $$n(n+a-\sqrt[k]{n^k +akn^{k-1}})\rightarrow \frac{k+1}{2}a^2$$ as $n\to \infty$ to get a limit of $$\frac{2+1}{2}-\frac{5+1}{2}=-\frac{3}{2}$$ For proof of the above fact. If we let $A=\sqrt[k]{n^k +akn^{k-1}}$ then $$n((n+a)-A)=n\frac{(n+a)^k-A^k}{(n+a)^{k-1}+\cdots +A^{k-1}} =\frac{\binom{k}{2}a^2n^{k-1}+\cdots }{(n+a)^{k-1}+\cdots +A^{k-1}}\rightarrow \frac{k+1}{2}a^2$$

Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$. Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$. My attempt: $p\mid a^2+ab+b^2 \implies p\mid (a-b)(a^2+ab+b^2)\implies p\mid a^3-b^3$ So, we have, $a^{3k}\equiv b^{3k}\mod p$ and by Fermat's Theorem we have, $a^{3k+1}\equiv b^{3k+1}\mod p$ as $p$ is of the form $p=3k+2$. I do not know what to do next. Please help. Thank you.

Suppose that $p=3k+2$ is prime and $$ \left.p\ \middle|\ a^2+ab+b^2\right.\tag1 $$ then, because $a^3-b^2=(a-b)\left(a^2+ab+b^2\right)$, we have $$ \left.p\ \middle|\ a^3-b^3\right.\tag2 $$ Case $\boldsymbol{p\nmid a}$ Suppose that $p\nmid a$, then $(2)$ says $p\nmid b$. Furthermore, $$ \begin{align} a^3&\equiv b^3&\pmod{p}\tag3\\ a^{3k}&\equiv b^{3k}&\pmod{p}\tag4\\ a^{p-2}&\equiv b^{p-2}&\pmod{p}\tag5\\ a^{-1}&\equiv b^{-1}&\pmod{p}\tag6\\ a&\equiv b&\pmod{p}\tag7\\ \end{align} $$ Explanation $(3)$: $\left.p\ \middle|\ a^3-b^3\right.$ $(4)$: modular arithmetic $(5)$: $3k=p-2$ $(6)$: if $p\nmid x$, then $x^{p-2}\equiv x^{-1}\pmod{p}$ $(7)$: modular arithmetic Then, because of $(1)$ and $(7)$, $$ \begin{align} 0 &\equiv a^2+ab+b^2&\pmod{p}\\ &\equiv 3a^2&\pmod{p}\tag8 \end{align} $$ which, because $p\nmid 3$, implies that $p\mid a$, which contradicts $p\nmid a$ and leaves us with Case $\boldsymbol{p\mid a}$ If $p\mid a$, then $(2)$ says $p\mid b$ and we get $$ \left.p^2\ \middle|\ a^2+ab+b^2\right.\tag9 $$

Number theory proofs regarding perfect squares How do you prove that $3n^2-1$ is never a perfect square

Let $3n^2-1=b^2, \text{ for a } b \in \mathbb{Z}$ $$3n^2-1 \equiv -1 \pmod 3 \equiv 2 \pmod 3$$ $$b=3k \text{ or } b=3k+1 \text{ or } b=3k+2$$ Then: $$b^2=9k \equiv 0 \pmod 3 \text{ or } b^2=3n+1 \equiv 1 \pmod 3 \text{ or } b=3n+1 \equiv 1 \pmod 3$$ We see that it cannot be $b^2 \equiv 2 \pmod 3$,so the equality $3n^2-1=b^2$ cannot be true. Therefore, $3n^2-1$ is never a perfect square.

Double integral of a rational function Consider the region $D$ given by $1\leq x^2+y^2\leq2\land0\leq y\leq x$. Compute $$\iint_D\frac{xy(x-y)}{x^3+y^3}dxdy$$ Attempt: The region $D$ is part of a ring in the first quadrant below the line $y=x$ Any hints are wellcome.

Changing to polar coordinates, $x=\rho \cos\theta$, $y=\rho \sin\theta$, and the Jacobian of the transformation is $J=\rho$. Then: $$\int_1^\sqrt2 \rho d\rho\int_0^\frac{\pi}{4}\frac{\sin\theta\cos\theta(\cos\theta-\sin\theta)}{\cos^3\theta+\sin^3\theta}d\theta$$ The first integral is immediate and yields $\frac{1}{2}$, so we'll multiply the answer given by the trigonometric integral by one half. For the trigonometric integral, let's use the substitution $u=\cos^3\theta +\sin^3\theta$, $du=(-3\cos^2\theta\sin\theta+3\sin^2\theta\cos\theta)d \theta=-3(\cos^2\theta\sin\theta-\sin^2\theta\cos\theta)d\theta$. The integral becomes: $$-\frac{1}{3}\int_1^\frac{\sqrt2}{2}\frac{du}{u}=-\frac{1}{3}\log u\bigg|_{u=1}^{u=\frac{\sqrt2}{2}}=-\frac{1}{3} \log \frac{\sqrt 2}{2}$$ Multiplying by one half yields $I=-\frac{1}{6} \log \frac{\sqrt 2}{2}=\frac{\log 2}{12}$.

Surface area of sphere $x^2 + y^2 + z^2 = a^2$ cut by cylinder $x^2 + y^2 = ay$, $a>0$ The cylinder is given by the equation $x^2 + (y-\frac{a}{2})^2 = (\frac{a}{2})^2$. The region of the cylinder is given by the limits $0 \le \theta \le \pi$, $0 \le r \le a\sin \theta$ in polar coordinates. We need to only calculate the surface from a hemisphere and multiply it by two. By implicit functions we have: $$A=2\iint\frac{\sqrt{\left(\frac{\partial F}{\partial x}\right)^2 + \left(\frac{\partial F}{\partial y}\right)^2 + \left(\frac{\partial F}{\partial z}\right)^2}}{\left|\frac{\partial F}{\partial z} \right|} dA$$ where $F$ is the equation of the sphere. Plugging in the expressions and simplifying ($z \ge 0)$, we get: $$A=2a\iint\frac{1}{\sqrt{a^2 - x^2 - y^2}} dxdy$$ Converting to polar coordinates, we have: $$A = 2a \int_{0}^\pi \int_{0}^{a\sin(\theta)} \frac{r}{\sqrt{a^2 - r^2}} drd\theta$$ Calculating this I get $2\pi a^2$. The answer is $(2\pi - 4)a^2$. Where am I going wrong?

Given the equations $$ x^2+y^2+z^2=a^2, $$ and $$ x^2+y^2 = ay, $$ we obtain $$ ay + z^2 = a^2. $$ Using $$ \begin{eqnarray} x &=& a \sin(\theta) \cos(\phi),\\ y &=& a \sin(\theta) \sin(\phi),\\ z &=& a \cos(\theta),\\ \end{eqnarray} $$ we obtain $$ a^2 \sin(\theta) \sin(\phi) + a^2 \cos^2(\theta) = a^2 \Rightarrow \sin(\theta) = \sin(\phi) \Rightarrow \theta=\phi \vee \theta=\pi-\phi. $$ For the surface we have $$ \begin{eqnarray} \int d\phi \int d\theta \sin(\theta) &=& \int_0^{\pi/2}d\phi \int_0^\phi d\theta \sin(\theta) + \int_{\phi/2}^{\pi}d\phi \int_0^{\pi-\phi} d\theta \sin(\theta)\\ &=& 2 \int_0^{\pi/2}d\phi \int_0^\phi d\theta \sin(\theta). \end{eqnarray} $$ We can calculate the surface as $$ \begin{eqnarray} 4 a^2 \int_0^{\pi/2}d\phi \int_0^\phi d\theta \sin(\theta) &=& 4 a^2 \int_0^{\pi/2}d\phi \Big( 1 - \cos(\phi) \Big)\\ &=& 4 a^2 \Big( \pi/2 - 1 \Big)\\ &=& a^2 \Big( 2\pi - 4 \Big). \end{eqnarray} $$

Evaluate $\int \frac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x$. How to do this indefinite integral (anti-derivative)? $$I=\displaystyle\int \dfrac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x.$$ I tried doing some substitutions ($x^2+7=t^2$, $2x+1=t$, etc.) but it didn't work out.

Using Euler substitution by setting $t-x=\sqrt{x^2+7}$, we will obtain $x=\dfrac{t^2-7}{2t}$ and $dx=\dfrac{t^2+7}{2t^2}\ dt$, then the integral turns out to be \begin{align} \int \dfrac{1}{(2x+1)\sqrt {x^2+7}}\ dx&=\int\frac{1}{t^2+t-7}\ dt\\ &=\int\frac{1}{\left(t+\dfrac{\sqrt{29}+1}{2}\right)\left(t-\dfrac{\sqrt{29}-1}{2}\right)}\ dt\\ &=-\int\left[\frac{2}{\sqrt{29}(2t+\sqrt{29}+1)}+\frac{2}{\sqrt{29}(-2t+\sqrt{29}-1)}\right]\ dt. \end{align} The rest can be solved by using substitution $u=2t+\sqrt{29}+1$ and $v=-2t+\sqrt{29}-1$.

Minimum Value of $x_1+x_2+x_3$ For an Acute Triangle $\Delta ABC$ $$\begin{align}x_n=2^{n-3}\left(\cos^nA+\cos^nB+\cos^nC\right)+\cos A\,\cos B\,\cos C\end{align}$$ Then find the least value of $$x_1+x_2+x_3$$ My Approach: I have found $x_1$, $x_2$ and $x_3$ $$\begin{align}x_1=\frac{1}{4}\left(\cos A+\cos B+\cos C\right)+\cos A\,\cos B\,\cos C\\ =\frac{1}{4}\left(1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)+\cos A\,\cos B\,\cos C \tag{1}\end{align}$$ $$\begin{align}x_2=\frac{1}{4}\left(3+\cos 2A+\cos 2B+\cos 2C\right)+\cos A\,\cos B\,\cos C=\frac{1}{2} \tag{2}\end{align}$$ $$x_3=\frac{1}{4}\left(3\cos A+3\cos B+3\cos C+\cos 3A+\cos 3B+\cos 3C\right)+\cos A\,\cos B\,\cos C$$ $$\implies x_3=\frac{1}{2}+x_1+\frac{1}{4}\sum \cos 3A+2\prod \sin\frac{A}{2}\\ $$ $$\implies x_3=\frac{1}{2}+x_1-\prod \sin\frac{3A}{2}+2\prod \sin\frac{A}{2}\\ \tag{3}$$ $$\text{So}\;\;\;\;\;\;\;\begin{align}x_1+x_2+x_3=\frac{3}{2}+4\prod \sin\frac{A}{2}-\prod \sin\frac{3A}{2}+2\prod \cos A \end{align}$$ I cannot proceed any further.

Use AM-GM inequality,we have $$\cos^3{x}+\dfrac{\cos{x}}{4}\ge 2\sqrt{\cos^3{x}\cdot\dfrac{\cos{x}}{4}}=\cos^2{x}$$ then we have $$x_{1}+x_{3}\ge\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=2x_{2}$$ so $$x_{1}+x_{2}+x_{3}\ge 3x_{2}=\dfrac{3}{2}$$ because we have use this follow well know $$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$

Finding Cauchy principal value for: $ \int_1^\infty \frac{ a x^2 + c }{x^4 - b x^2 - c} \mathrm{d}x $ I need to solve the integral $ \displaystyle \mathcal{P} \int_1^\infty \frac{ a x^2 + c }{x^4 - b x^2 - c} \mathrm{d}x $, where $\mathcal{P}$ is the Cauchy principal value, $ - 1 \leq c \leq 1$ and $a, b$ are both real, but can be arbitrarily large, positive or negative. I'm not sure, whether this integral is solveable, but any hints or ideas are very welcome.

I actually ended up using a different solution, I found more direct and intuitively. It is completely equivalent with @Yves method, but I just state it for completeness. $$ \mathcal{P} \int_{1}^{\infty} \mathrm{d} x \frac{a x^{2} + c } { x^{4} - b x^{2} - c } = \mathcal{P} \int_{1}^{\infty} \mathrm{d} x \frac{a x^{2} + c }{ (x^{2} - d)^{2} - g^{2}} $$ $$ = \frac{1}{2g} \Bigg\{ \mathcal{P} \int_{1}^{\infty} a x^{2} \left[ \frac{1}{ x^{2} - d - g} - \frac{1}{ x^{2} - d - g} \right] \mathrm{d} x $$ $$ \qquad \quad + \mathcal{P} \int_{1}^{\infty} c \left[ \frac{1}{ x^{2} - d - g} - \frac{1}{ x^{2} - d - g} \right] \mathrm{d} x \Bigg\} $$ with $$ d = - \frac{b}{2}, \qquad g = \sqrt{c + d^2}. $$ Then dividing each fraction in two once again, we end with two integrals to solve $$ \mathcal{P} \int_{1}^{\infty} \mathrm{d} x \frac{x}{x \pm \sqrt{d \pm g}} = \lim_{\epsilon \rightarrow \infty} \left[ x \mp \sqrt{r} \ln( x \pm \sqrt{r} ) \right]_{x = 1}^{x = \epsilon}, $$ $$ \mathcal{P} \int_{1}^{\infty} \mathrm{d} x \frac{1}{x \pm \sqrt{d \pm g}} = \lim_{\epsilon \rightarrow \infty} \left[ \sqrt{r} \ln( x \pm \sqrt{r} ) \right]_{x = 1}^{x = \epsilon}. $$ Then, putting it all together, we end up with $$ \frac{1}{8 g} \left[ \frac{2 a (d + g) + c}{\sqrt{d + g}} \ln \left( \frac{1 + \sqrt{d+g}}{1 - \sqrt{d+g}} \right) - \frac{2 a (d - g) + c}{\sqrt{d - g}} \ln \left( \frac{1 + \sqrt{d-g}}{1 - \sqrt{d-g}} \right) \right], $$ which can then be rewritten in terms of $\tan^{-1}$ or $\tanh^{-1}$ depending on the properties of the quantitiy $\sqrt{d \pm g}$.

A transcendental number from the diophantine equation $x+2y+3z=n$ Let $\displaystyle n=1,2,3,\cdots.$ We denote by $D_n$ the number of non-negative integer solutions of the diophantine equation $$x+2y+3z=n$$ Prove that $$ \sum_{n=0}^{\infty} \frac{1}{D_{2n+1}} $$ is a transcendental number.

If I have not mistaken something, $$ [x^{2n+1}]\frac{1}{(1-x)(1-x^2)(1-x^3)} = \frac{(n+1)(n+3)}{3}, $$ when $n\equiv 0,2\pmod{3}$, and $$ [x^{2n+1}]\frac{1}{(1-x)(1-x^2)(1-x^3)} = \frac{(n+1)(n+3)+1}{3} $$ when $n\equiv 1\pmod{3}$, hence: $$\begin{eqnarray*}\sum_{n=0}^{+\infty}\frac{1}{D_{2n+1}}&=&\sum_{n=0}^{+\infty}\frac{3}{(n+1)(n+3)}-\!\!\!\!\sum_{n\equiv 1\!\!\pmod{\!\!3}}\frac{3}{(n+1)(n+3)(n+2)^2}\\&=&\frac{9}{4}-\frac{1}{3}\left(\frac{9}{2}-\frac{\sqrt{3}\,\pi}{2}-\frac{\pi^2}{6}\right)\\&=&\frac{1}{36}\left(27+2\pi\left(\pi+3\sqrt{3}\right)\right),\end{eqnarray*}$$ and we only need to show that $u=\pi(\pi+3\sqrt{3})$ is a trascendental number. But if $u$ were algebraic, then $$\pi = \frac{-3\sqrt{3}+\sqrt{27+4u}}{2}$$ would be algebraic too, contradiction.

if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let $$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$ Find the closed form $$a_{n}$$ since $$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$ so $$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)}$$ then I feel very ugly,can you someone have good partial fractions methods by hand? because I take an hour to solve this problem. ago I have solve $$x+2y+3z=n$$ the number of the positive integer solution $a_{n}$ I found $$a_{n}=\left[\dfrac{(n+3)^2}{12}\right]$$ Thank you

Hints : * *First, prove that $$\frac{1}{(1-x^2)(1-x^3)(1-x^4)} = \frac{7}{32(x+1)}-\frac{59}{288(x-1)}+\frac{1}{8(x-1)^2}+\frac{1}{16(x+1)^2}-\frac{1}{24(x-1)^3}+\frac{x+2}{9(x^2+x+1)}+\frac{1-x}{8(x^2+1)}.$$ *Then, use that $$ \frac{1-x}{x^2+1} = -\frac{1+i}{2(x-i)}+\frac{-1+i}{2(x+i)}.$$ and $$ \frac{x+2}{x^2+x+1} = \frac{j}{x+j}+\frac{\overline{j}}{x+\overline{j}}$$ where $j = \frac{1+i\sqrt{3}}{2}$. *Finally, use the classical series $$ \sum_{n=0}^{+\infty} x^k = \frac{1}{1-x},$$ $$ \sum_{n=0}^{+\infty} (k+1)x^k = \frac{1}{(1-x)^2},$$ and $$ \sum_{n=0}^{+\infty} \frac{1}{2}(k+1)(k+2)x^k = \frac{1}{(1-x)^3}.$$

Need help with this Geometric sequence problem First, sorry if Im not using the right syntax, im translating the problem and im not sure if im supposed to say "Sequence" or "series", and also thanks to who ever tries to help. The sum of a geometric sequence is 20, and the sum of its squared terms is 205. find how many terms are in the sequence if the first term is $\frac{1}{2}$. $a$1 + $a$2 + $a$3 +...+$a$$n$ = 20 $a$1 2 + $a$2 2 + $a$3 2 +...+$a$$n$ 2 = 205 $a$1 = $\frac{1}{2}$ find $n$. $20=\frac{a\left(q^n-1\right)}{q-1}$ ==> $205=\frac{10.25a\left(q^n-1\right)}{q-1}$ $205=\frac{a^2\left(q^{2n}-1\right)}{q^2-1}$ = $\frac{a^2\left(q^n-1\right)\left(q^n+1\right)}{\left(q-1\right)\left(q+1\right)}$ $\frac{10.25a\left(q^n-1\right)}{q-1}$ =$\frac{a^2\left(q^n-1\right)\left(q^n+1\right)}{\left(q-1\right)\left(q+1\right)}$ $10.25a\left(q^n-1\right)\left(q+1\right)$ = $a^2\left(q^n-1\right)\left(q^n+1\right)$ $20.5\left(q+1\right)\space =\space \left(q^n+1\right)$ This is basically where I tell myself i did something wrong.

You have found that $q^n+1=20.5(q+1)$. But since the sum of the terms is $20$, we have $q^n-1=40(q-1)$. (In essence you had written down this equation also.) Subtract, and find $q$, and then $n$. The numbers are disappointingly small, $3$ and $4$ respectively.

Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dx\, dy$ I need to evaluate the following integral: $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dx\, dy$$ I thought of evaluating the iterated integral $\displaystyle\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dy$, but because of the presence of $x^2$ and $y^2$ terms, I am not being able to do that. I tried substituting $x=r\cos \theta$ and $y=r\sin \theta$ but in that case I have some confusion regarding the limits of $r$ and $\theta$. Can I get some help?

Recall: $$x^2-xy+y^2=(x-\frac{1}{2}y)^2+\frac{3}{4}y^2$$ With that you get: $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}(x^2-xy+y^2)}dxdy=\\\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left((x-\frac{1}{2}y)^2+\frac{3}{4}y^2\right)}dxdy=\\ \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left(x'^2+\frac{3}{4}y^2\right)}dx'dy=\\ \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}x'^2}dx'e^{-\frac{3}{8}y^2}dy=\\ \int_{-\infty}^\infty\sqrt{2\pi}e^{-\frac{3}{8}y^2}dy=\\ \sqrt{2\pi}\sqrt{\frac{8}{3}\pi}=\frac{4}{\sqrt{3}}\pi $$

Is there a way to calculate the area of this intersection of four disks without using an integral? Is there anyway to calculate this area without using integral ?

Let $R$ be its radius and $D$ its diameter: $R = 5$, $D = 10$. $$\begin{align} \text{Area of big square} &= D^2 = 100 \\ \text{Area of circle} &= \frac{\pi D^2}{4} \approx 78.54 \\ \text{Area outside circle} &= 100 - 78.54 = 21.46 \\ \text{Area of 4 petals} &= 78.54 - 21.46 = 57.08 \\ \text{Area of single petal} &= \frac{57.08}{4} = 14.27 \\ \text{Area of small square} &= R^2 = 25 \end{align}$$ Let $x$ denote the area of the portion selected. $$\begin{align} \text{Area of 2 petals} &= 2 \cdot 14.27 = 28.54 \\ 0 &= 25 - 28.54 + x \\ \text{Area of 8 petals} &= 2 \cdot 57.08 = 114.16 \\ \text{OR}\\ 100 - 114.16 + 4 x &= 0 \end{align}$$ $$ x = 3.54 $$ Here we have a small area which needs to be added, that I found out by modeling to be $4.34$ $$\begin{align} \text{This gives us the desired area of}\\ \text{4.34+3.54} &\approx 7.88\\ \text{& a percentage of} &\approx 4 \times 7.88\\ \text{that is} &\approx 31.52\% \end{align}$$

If one number is thrice the other and their sum is $16$, find the numbers If one number is thrice the other and their sum is $16$, find the numbers. I tried, Let the first number be $x$ and the second number be $y$ Acc. to question $$ \begin{align} x&=3y &\iff x-3y=0 &&(1)\\ x&=16-3y&&&(2) \end{align} $$

Let the first number be $x$. Let the second number be $y$. According to question $$ \tag{1} x+y=16 $$ $$ \tag{2} x=3y $$ So, $x-3y=0 \tag{2}$ Multiply equation $(1)$ by $3$. Solve both equations: $$\tag{1} 3x+3y=48$$ $$\tag{2} x-3y=0$$ $$\tag{1) + (2}4x=48$$ $$\tag{3}x=12$$ Putting in equation $(1)$: $$\tag{1} x+y=16$$ $$\tag{1),(3} 12+y=16$$ $$\tag{4}y=16-12$$ $$\tag{5}y=4$$

Summation of Infinite Geometric Series Determine the sum of the following series: $$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} $$ My work: $$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} = \sum_{n=1}^{\infty } \frac{-1}{7} (\frac{3}{7})^{n-1}$$ $$\sum_{n=1}^{\infty } ar^{n-1} = \frac{a}{1-r} = \frac{\frac{-1}{7}}{1-\frac{3}{7}} = -\frac{1}{4}$$ Why does this not work? Sorry for the incorrect initial post!!! Edit: -3 changed to (-3)

$$\begin{align} \sum_{n=1}^{\infty } \frac{-3^{n-1}}{7^{n}} & = - \frac{1}{7} \sum_{n=1}^{\infty } (\frac{3}{7})^{n-1} \\ & = - \frac{1}{7} \frac{1}{1-\frac 3 7} \\ & = - \frac 1 4 \\[2ex] \sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} & = \frac{1}{7} \sum_{n=1}^{\infty } (-\frac{3}{7})^{n-1} \\ & = \frac{1}{7} \frac{1}{1+\frac 3 7} \\ & = \frac 1 {10} \end{align}$$

How find this sum $\sum\limits_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$ Find the sum close form $$f(x)=\sum_{i=0}^{2n}\dfrac{\binom{2n}{2i}\binom{2i}{i}x^{2i}}{2^{2i}}$$ if we let $$\dfrac{x}{2}=y$$ then $$f(y)=\sum_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$$ this PDF have this page 5 $$\sum_{k=j}^{n}\binom{n}{k}\binom{k}{j}=2^{n-j}\binom{n}{j}$$ the solution can see page 5 Maybe my problem can use this mathods?Thank you

Given any formal Laurent series $\;(???) = \sum \alpha_{k_1 k_2 \ldots k_n} t_1^{k_1} t_2^{k_2} \cdots t_n^{k_n}$, we will use the notation $[ t_1^{k_1} t_2^{k_2} \cdots t_n^{k_n} ](???)$ to denote the coefficient $\alpha_{k_1 k_2 \cdots k_n}$ in front of corresponding monomial. Instead of $f(y)$, let us denote the polynomial we wish to find a closed form as $p_{2n}(y)$. We have $$\begin{align} p_{2n}(y) &= \sum_{i=0}^{n}\binom{2n}{2i}\binom{2i}{i} y^{2i} = \sum_{i=0}^{n} \binom{2n}{2i} y^{2i}\bigg( [t^0](t + t^{-1})^{2i}\bigg)\\ &= \sum_{i=0}^{2n} \binom{2n}{i} \bigg( [t^0](y(t + t^{-1}))^i\bigg) = [\;t^0\;] \bigg( 1 + y(t+t^{-1})\bigg)^{2n} \end{align} $$ Substitute $t$ by $e^{i\theta}$ in above formal expression and notice for any $k \in \mathbb{Z}$, we have $$\frac{1}{2\pi}\int_0^{2\pi} e^{ik\theta} d\theta = \begin{cases}1,&k = 0\\0,&\text{ otherwise }\end{cases}$$ We obtain an integral representation for $p_{2n}(y)$, $$p_m(y) = \frac{1}{2\pi}\int_0^{2\pi} (1+2y\cos\theta)^{m} d\theta\quad\text{ for }\quad m = 2n$$ Treat this as a definition for $p_m(y)$ for general $m \in \mathbb{N}$ and consider following generating function: $$p(y,\rho) = \sum_{m=0}^\infty p_m(y)\rho^m$$ It is easy to see $$ p(y,\rho) = \frac{1}{2\pi}\int_0^{2\pi}\frac{d\theta}{1-\rho(1+2y\cos\theta)} = \frac{1}{4\pi y\rho}\int_0^{2\pi}\frac{d\theta}{\frac{1-\rho}{2y\rho}-\cos\theta} $$ Using the identity $$\frac{1}{2\pi}\int_0^{2\pi}\frac{d\theta}{a - \cos\theta} = \frac{1}{\sqrt{a^2-1}}\quad\text{ for } a > 1$$ We get $$\begin{align} p(y,\rho) &= \frac{1}{2y\rho}\frac{1}{\sqrt{\left(\frac{1-\rho}{2y\rho}\right)^2 - 1}} = \frac{1}{\sqrt{1-2\rho + (1- 4y^2)\rho^2}}\\ &= \frac{1}{\sqrt{1-2\frac{1}{\sqrt{1-4y^2}}(\rho\sqrt{1-4y^2}) + (\rho\sqrt{1-4y^2})^2}}\ \end{align} $$ Compare this with the generating function for Legendre polynomials, $$\frac{1}{\sqrt{1-2zt+t^2}} = \sum_{k=0}^\infty P_k(z) t^k$$ We find $$p(y,\rho) = \sum_{k=0}^\infty P_k\left(\frac{1}{\sqrt{1-4y^2}}\right) \left(\rho\sqrt{1-4y^2)}\right)^k$$ This leads to the expression we claimed in comment: $$p_{2n}(y) = (1-4y^2)^n P_{2n}\left(\frac{1}{\sqrt{1-4y^2}}\right)$$ For example, when $y = \frac12$, this leads to an interesting identity: $$\begin{align} \sum_{i=0}^{n} \frac{\binom{2n}{2i}\binom{2i}{i}}{2^{2i}} &= p_{2n}\left(\frac12\right) = \lim_{y\to\frac12^{-}} (1-4y^2)^n P_{2n}\left(\frac{1}{\sqrt{1-4y^2}}\right)\\ &= [ t^{2n} ] P_{2n}(t) = \frac{(4n-1)!!}{(2n)!} \end{align} $$

Algebraic proof of $\tan x>x$ I'm looking for a non-calculus proof of the statement that $\tan x>x$ on $(0,\pi/2)$, meaning "not using derivatives or integrals." (The calculus proof: if $f(x)=\tan x-x$ then $f'(x)=\sec^2 x-1>0$ so $f$ is increasing, and $f(0)=0$.) $\tan x$ is defined to be $\frac{\sin x}{\cos x}$ where these are defined by their infinite series. What I have so far: $$|z|\le1\implies\left|\sum_{n=4}^\infty\frac{z^n}{n!}\right|<\sum_{n=0}^\infty\frac{|z|^4}{4!\,5^n}=\frac{5|z|^4}{4\cdot 4!}$$ $$\left|\sin x-\Big(x-\frac{x^3}6\Big)\right|=\Im\left[\sum_{n=4}^\infty\frac{(ix)^n}{n!}\right]<\frac{5x^4}{4\cdot 4!}<\frac{x^3}6$$ $$\left|\cos x-\Big(1-\frac{x^2}2\Big)\right|=\Re\left[\sum_{n=4}^\infty\frac{(ix)^n}{n!}\right]<\frac{5x^4}{4\cdot 4!}<\frac{x^2}6$$ Thus $\sin x>x-\frac{x^3}3$ and $\cos x<1-\frac{x^2}3$, so $\tan x>x$. However, this only covers the region $x\le1$, and I still need to bound $\tan x$ on $(1,\pi/2)$. My best approximation to $\pi$ is the very crude $2<\pi<4$, derived by combining the above bounds with the double angle formulas (note that $\pi$ is defined as the smallest positive root of $\sin x$), so I can't quite finish the proof with a bound like $\sin x>1/\sqrt 2$, $\cos x\le\pi/2-x$ (assuming now $x\ge1\ge\pi/4$) because the bound is too tight. Any ideas?

Here is a sketch of what you might be looking for: Showing $\tan x > x$ is equivalent to showing $\sin x - x \cos x > 0$, since $\cos x > 0$ on $(0,\pi/2$). The series for $\sin x - x \cos x$ is $\displaystyle\sum_{j=1}^{\infty} \dfrac{(2j)x^{2j+1}}{(2j+1)!} = x^3/3 - x^5/30 + x^7/840 - x^9/45360 \ldots$ Group the terms in pairs: $(x^3/3 - x^5/30) + (x^7/840 - x^9/45360) + \ldots$. If $0 < x < \sqrt{10}$, the first difference is positive. The ratio of the terms in each difference is decreasing, so if the first difference is positive, all the rest are too, and the sum is positive. So $\sin x - x \cos x > 0$ on $0 < x < \sqrt{10}$, which gives you quite a bit of leeway since $\sqrt{10} > \pi/2$. (The first positive solution to $\sin x - x \cos x = 0$ happens at $x \approx 4.493$ according to WolframAlpha.)

Finding the perimeter of the room If the length and breadth of a room are increased by $1$ $m$, the area is increased by $21$ $m^2$. If the length is increased by $1$ $m$ and breadth is decreased by $1$ $m$ the area is decreased by $5$ $m^2$. Find the perimeter of the room. Let the length be $x$ and the breadth be $y$ Therefore, Area$=$$xy$ $m^2$ Accordingly, $(x+1) \cdot (y+1) \ = \ xy+21$ $m^2$ What should I do now? How should I find the second equation? Should the second equation look like: $(x+1) \ \cdot \ (y-1) \ = \ xy -5 \ $ $m^2$

$A=xy$ $(x+1)(y+1)=xy+21$ $(x+1)(y-1)=xy-5$ Foil out both equations to get: $xy+x+y+1=xy+21 \quad \to \quad x+y=20 \quad \to \quad y=20-x$ $xy-x+y-1=xy-5 \quad \to \quad -x+y=-4 \quad \to \quad y=-4+x$ Set them equal to each other: $20-x=-4+x$ $2x=24 \to x=12$ Since we know $x+y=20, y=8$. You can verify this solution by checking the conditions given. $A=12\cdot 8=96$ Adding $1$ to both the length and width: $A=13\cdot 9=117 \to 96+21=117$ Also, if you add $1$ to the length and subtract $1$ from the width: $A=13\cdot 7=91 \to 96-5=91$

Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$ If $a$, $b$ and $c$ are positive real numbers, prove that: $$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$ Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction. Things I have tried so far: Using Cauchy inequality I can write:$$\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)(a+b+c) \geq \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2$$ but I can't continue this.I tried expanded form:$$\sum \limits_{cyc} \frac{a^5c^2}{a^2b^2c^2} \geq \sum \limits_{cyc} \frac{a^3c}{abc}$$ Which proceeds me to this Cauchy:$$\sum \limits_{cyc} \frac{a^5c^2}{abc}\sum \limits_{cyc}a(abc)\geq \left(\sum \limits_{cyc}a^3c\right)^2$$ I can't continue this one too. The main Challenge is $3$ fraction on both sides which all of them have different denominator.and it seems like using Cauchy from first step won't leads to anything good.

Another way to do this would be the following (I'm doing Liu Gang's suggested generalization): We have to show $$\frac{a^{n+1}}{b^n} + \frac{b^{n+1}}{c^n} + \frac{c^{n+1}}{a^n} - \frac{a^n}{b^{n-1}} - \frac{b^n}{c^{n-1}} - \frac{c^n}{a^{n-1}} \ge 0.$$ The left hand side equals $$\frac{a^n(a - b)}{b^n} + \frac{b^n(b-c)}{c^n} + \frac{c^n(c-a)}{a^n},$$ and therefore it is enough to show that $$c^n a^{2n} (a-b) + a^n b^{2n} (b-c) + b^n c^{2n} (c-a) \ge 0.$$ Because the inequality is cyclic, we can assume that either $a \ge b \ge c$ or $a \ge c \ge b$. In the first case we have $c^n a^{2n} \ge b^n c^{2n}$ and $a^n b^{2n} \ge b^n c^{2n}$, so we get that the LHS is $\ge b^n c^{2n} (a - b + b - c + c - a) = 0$. In the second case we have $a^n b^{2n} \le c^n a^{2n}$ and $b^n c^{2n} \le c^n a^{2n}$, so we get that the LHS is $\ge c^n a^{2n} (a - b + b - c + c- a) = 0$. This proves the claim.

Recurrence of the form $2f(n) = f(n+1)+f(n-1)+3$ Can anyone suggest a shortcut to solving recurrences of the form, for example: $2f(n) = f(n+1)+f(n-1)+3$, with $f(1)=f(-1)=0$ Sure, the homogenous solution can be solved by looking at the characteristic polynomial $r^2-2x+1$, so that in general a solution for the homogenous equation is of the form $f^h(n) = c_1+c_2n$. But how does one deal with the constant 3 in this case?

Let $f_p = A + Bn + Cn^2$. $$ \begin{cases} 2f(n) = 2A +2Bn + 2Cn^2\\ -f(n+1) = -A - B(n+1) - C(n+1)^2\\ -f(n-1) = -A - B(n-1) - C(n-1)^2 \end{cases} \quad \Rightarrow $$ $$ 3 = Cn^2 - C(2n^2 + 2) = -2C \quad \Rightarrow \quad C = -\frac{3}{2} $$ Thus, $$ f(n) = C_1 + C_2n -\dfrac{3n^2}{2} $$ Now use the given initial conditions to find the constants $C_1$ and $C_2$.

Evaluating $ \int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx$ I am trying to evaluate the indefinite integral of $$\int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx.$$ The first thing I did was the substitution rule: $u=1+x^2$, so that $\displaystyle x \, dx=\frac{du}2$ and $1-x^2=2-u$. The integral then transforms to $$\int \sqrt{\frac{2-u}{u}} \, \frac{du}2$$ or $$\frac 12 \int \sqrt{\frac 2u - 1} \, du$$ I'm a bit stuck here. May I ask for help on how to proceed?

$\text {Let } x^{2}=2 \sin ^{2} \theta-1 \textrm{ for } \frac{\pi}{4} \leqslant \theta \leqslant \frac{\pi}{2}, \text {then } x d x=\sin \theta \cos \theta d \theta $. \begin{aligned}\int x \sqrt{\frac{1-x^{2}}{1+x^{2}}} d x&=\int \sqrt{\frac{2-2 \sin ^{2} \theta}{2 \sin ^{2} \theta} }\sin \theta \cos \theta d \theta \\ &=\int \cos ^{2} \theta d \theta\\&=\frac{1}{2} \int(1+\cos 2 \theta) d \theta \\ &=\frac{1}{2} \theta+\frac{\sin 2 \theta}{4}+C\\&=\frac{1}{2} \sin ^{-1} \sqrt{\frac{x^{2}+1}{2}}+\frac{1}{4} \sqrt{1-x^{4}}+C \end{aligned}

Solving the logarithimic inequality $\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$ I tried solving the logarithmic inequality: $$\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$$ several times but keeping getting wrong answers.

Here are the steps \[ \log_2 \frac{x}{2}+\frac{\log_2 x^{2}}{\log_2 \frac{2}{x}} \le 1 \] \[ \log_2 x -\log_2 2+\frac{2\log_2 x}{\log_2 2-\log_2 x} \le 1 \] \[ \log_2 x -1+\frac{2\log_2 x}{1-\log_2 x} \le 1 \] Let $\alpha= \log_2 x$, then \[ \alpha -1+\frac{2\alpha}{1-\alpha} \le 1 \] \[ \alpha -2+\frac{2\alpha}{1-\alpha} \le 0 \] \[ \frac{(1-\alpha)(\alpha -2)+2\alpha}{1-\alpha} \le 0 \] \[ \frac{5\alpha -\alpha^{2}-2}{1-\alpha} \le 0 \] After solving for $\alpha$, we have the solutions \[ 1<\alpha \le \frac{1}{2}(5+\sqrt{17}) \] \[ \alpha \le \frac{1}{2}(5-\sqrt{17}) \] Which is \[ 1<\log_2 x \le \frac{1}{2}(5+\sqrt{17}) \] \[ \log_2 x \le \frac{1}{2}(5-\sqrt{17}) \] Thus \[ 2< x \le (\sqrt{2})^{5+\sqrt{17}} \] \[ x \le (\sqrt{2})^{5-\sqrt{17}} \]

How many $3$ digit numbers with digits $a$,$b$ and $c$ have $a=b+c$ My question is simple to state but (seemingly) hard to answer. How many $3$ digit numbers exist such that $1$ digit is the sum of the other $2$. I have no idea how to calculate this number, but I hope there is a simple way to calculate it. Thank you in advance. EDIT: The first digit should not be $0$

Assuming a digit is an element of $\{0,1,2,3,4,5,6,7,8,9,10\}$ we have three cases for $a,b,c$ to see: * *$a=b=c=0$. All easy here, yields $1$ combination. *$b=c\ne 0$. $a=2b$, so $b<5$ giving us $4$ choices (digits $1$ to $4$). The position of $a$ uniquely determines the code, so multiply b $3$ to get $4\cdot 3 = 12$ combinations *$b\ne c$. We assume $a\ge b>c$ and chose $c$ first. Since $b+c < 10$ and $c<b$ $a \ge 2c$ so $c\le 4$. $$\begin{align*} c=4 & \Rightarrow b=5, a=9 & 1\\ c=3 & \Rightarrow b\in\{4,5,6\} & 3\\ c=2 & \Rightarrow b\in\{3,4,5,6,7\} & 5\\ c=1 & \Rightarrow b\in\{2,\ldots, 8\} & 7\\ c=0 & \Rightarrow b\in\{1,\ldots, 9\} & 9 (\text{only $2$ distinct digits here}) \end{align*}$$ totaling $16+9$ combinations, times $3! = 6$ for all but the $9$ we get $16\cdot 6 + 9 \cdot 3 = 123$ Summing up we have $1+12+123 = 136$ possibilities.

A nice trignometric identity How to prove that: $$\cos\dfrac{2\pi}{13}+\cos\dfrac{6\pi}{13}+\cos\dfrac{8\pi}{13}=\dfrac{\sqrt{13}-1}{4} $$ I have a solution but its quite lengthy, I would like to see some elegant solutions. Thanks!

I add another answer although it is not different in principle from some of the others. We have $$t=\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}$$ it is natural to then consider the other even divisions, $$s=\cos \frac{4\pi}{13}+\cos \frac{10\pi}{13}+\cos \frac{12\pi}{13}$$ Now $$t+s=\cos \frac{2\pi}{13}+\cos \frac{4\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}+\cos \frac{10\pi}{13}+\cos \frac{12\pi}{13}=\frac{\cos\frac{7\pi}{13}\sin\frac{6\pi}{13}}{\sin\frac{\pi}{13}}$$ $$\frac{\cos\frac{7\pi}{13}\sin\frac{6\pi}{13}}{\sin\frac{\pi}{13}}=\frac{1}{2}\frac{\sin\frac{13\pi}{13}-\sin \frac{\pi}{13}}{\sin\frac{\pi}{13}}=-\frac{1}{2}$$ Next we calculate $st$ using $\cos A \cos B=\frac{1}{2}(\cos (A+B)+\cos(A-B))$ we find $st=\frac{3}{2}(s+t)=-\frac{3}{4}$ So $s$ and $t$ are solutions to $4x^2+2x-3=0$ whose roots are $\frac{\sqrt{13}-1}{4}$ and $\frac{-\sqrt{13}-1}{4}$ Note that $$t=\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}-\cos \frac{5\pi}{13}$$ and since $\cos \frac{5\pi}{13}<\cos \frac{6\pi}{13}$ we have that $t>0$ and thus $$\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$$

Error in proving of the formula the sum of squares Given formula $$ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} $$ And I tried to prove it in that way: $$ \sum_{k=1}^n (k^2)'=2\sum_{k=1}^n k=2(\frac{n(n+1)}{2})=n^2+n $$ $$ \int (n^2+n)\ \text d n=\frac{n^3}{3}+\frac{n^2}{2}+C $$ But $$ \frac{n^3}{3}+\frac{n^2}{2}+C $$ is not equal to $$\frac{n(n+1)(2n+1)}{6}=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$$ Where I made a mistake?

Here's an interesting approach using the summation of binomial coefficients. First, note that $$\sum_{i=1}^n {i+a\choose b} = {{n+a+1}\choose {b+1}}$$ and also that $$i^2=2\cdot \frac{(i+1)i}{1\cdot 2}-i=2{{i+1}\choose 2}-{i\choose 1}$$ Hence $$\begin{align} \sum_{i=1}^ni^2 &=\sum_{i=1}^n \left[2{{i+1}\choose 2}-{i\choose 1}\right]\\ &=2 \sum_{i=1}^n {{i+1}\choose 2}- \sum_{i=1}^n {i\choose 1}\\ &=2{{n+2}\choose 3}-{{n+1}\choose 2}\\ &=2\cdot \frac {(n+2)(n+1)n} {1\cdot 2\cdot 3} -\frac {(n+1)n} {1\cdot 2}\\ &=\frac {(n+1)n} 6 \cdot \left[ 2(n+2)-3 \right]\\ &={\frac 16}n(n+1)(2n+1) \end{align}$$

If $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then $|z|=$? If z is a complex number and $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then find the value of $|z|$ . I tried to put $ \frac {z^2 + z+ 1} {z^2 -z +1} =k $ then solve for $z$ and tried to find |z|, but it gets messy and I am stuck. The answer given is |z|=1

Let's come up with a more interesting question (which might be what you intended to ask). Find an $\alpha \in \mathbb{R}$ such that $\forall z \in \mathbb{C}: |z| = \alpha \implies \dfrac{z^2+z+1}{z^2-z+1} \in \mathbb{R} \cup \{\infty\}$. I'll show that $\alpha = 1$ works. If $|z| = 1$, then $z = e^{i\theta}$ for some $\theta \in [0,2\pi)$. Thus, $\dfrac{z^2+z+1}{z^2-z+1} = \dfrac{z+1+z^{-1}}{z-1+z^{-1}} = \dfrac{e^{i\theta}+e^{-i\theta}+1}{e^{i\theta}+e^{-i\theta}-1} = \dfrac{2\cos\theta+1}{2\cos\theta-1} \in \mathbb{R} \cup \{\infty\}$, as desired. Therefore, if $|z| = 1$, then $\dfrac{z^2+z+1}{z^2-z+1}$ is either purely real or undefined.

Proving or disproving inequality $ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \ge x + y + z $ Given that $ x, y, z \in \mathbb{R}^{+}$, prove or disprove the inequality $$ \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z $$ I have rearranged the above to: $$ x^2y(y - z) + y^2z(z - x) + z^2x(x - y) \ge 0 \\ \text{and, } \dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} \ge \dfrac{1}{xy} + \dfrac{1}{xz} + \dfrac{1}{yz} $$ What now? I thought of making use of the arithmetic and geometric mean properties: $$ \dfrac{x^2 + y^2 + z^2}{3} \ge \sqrt[3]{(xyz)^2} \\ \text{and, } \dfrac{x + y + z}{3} \ge \sqrt[3]{xyz} $$ but I am not sure how, or whether that'd help me at all.

Holder's inequality: $$u\cdot v \leq |u||v|$$ for any vectors $u,v$. Let $\mathbf u=(1/x,1/y,1/z)$ and $\mathbf v=(xz,xy,yz)$. Then show $|v| = xyz |u|$ and thus $$|\mathbf u||\mathbf v| = xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)= \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} $$ and: $$\mathbf u\cdot\mathbf v = x+y+z $$

Solving exponential equation $e^{x^2+4x-7}(6x^2+12x+3)=0$ How would you find $x$ in: $e^{x^2+4x-7}(6x^2+12x+3)=0$ I don't know where to begin. Can you do the following? $e^{x^2+4x-7}=1/(6x^2+12x+3)$ and then find $ln$ for both sides?

$e^{x^2+4x-7}(6x^2+12x+3)=0 \Rightarrow e^{x^2+4x-7}=0 \text{ or } \ 6x^2+12x+3=0$ $$\text{It is known that } e^{x^2+4x-7} \text{ is non-zero }$$ therefore,you have to solve : $$6x^2+12x+3=0$$ The solutions are: $$x=-1-\frac{1}{\sqrt{2}} \\ x=-1+\frac{1}{\sqrt{2}}$$

How to solve the recurrence relation $T(n) = T(\lceil n/2\rceil) + T(\lfloor n/2\rfloor) + 2$ I'm trying to solve a recurrence relation for the exact function (I need the exact number of comparisons for some algorithm). This is what i need to solve: $$\begin{aligned} T(1) &= 0 \\ T(2) &= 1 \\ T(n) & = T(\lceil n/2\rceil) + T(\lfloor n/2\rfloor) + 2 \qquad(\text{for each $n\ge2$}) \end{aligned}$$ Without the ceiling and floor I know that the solution is $T(n) = 3n/2 -2$ if $n$ is the power of 2, but how can I solve it with them? (and for every $n$, not just $n$ that is the power of 2). Thanks a lot.

As mentioned in comment, the first two conditions $T(1) = 0, T(2) = 1$ is incompatible with the last condition $$\require{cancel} T(n) = T(\lfloor\frac{n}{2}\rfloor) + T(\lceil\frac{n}{2}\rceil) = 2\quad\text{ for } \color{red}{\cancelto{\;\color{black}{n > 2}\;}{\color{grey}{n \ge 2}}} \tag{*1}$$ at $n = 2$. We will assume the condition $(*1)$ is only valid for $n > 2$ instead. Let $\displaystyle\;f(z) = \sum\limits_{n=2}^\infty T(n) z^n\;$ be the generating function for the sequence $T(n)$. Multiply the $n^{th}$ term of $(*1)$ by $z^n$ and start summing from $n = 3$, we obtain: $$\begin{array}{rrl} &f(z) - z^2 &= T(3) z^3 + T(4) z^4 + T(5) z^5 + \cdots\\ &&= (T(2) + 2)z^3 + (T(2) + T(2) + 2)z^4 + (T(2)+T(3)+2)z^5 + \cdots\\ &&= (1+z)^2 ( T(2)z^3 + T(3)z^5 + \cdots) + 2(z^3 + z^4 + z^5 + \cdots)\\ &&= \frac{(1+z)^2}{z}f(z^2) + \frac{2z^3}{1-z}\\ \implies & f(z) &= \frac{(1+z)^2}{z}f(z^2) + z^2\left(\frac{1+z}{1-z}\right)\\ \implies & \frac{(1-z)^2}{z} f(z) &= \frac{(1-z^2)^2}{z^2}f(z^2) + z(1-z^2) \end{array} $$ Substitute $z^{2^k}$ for $z$ in last expression and sum over $k$, we obtain $$f(z) = \frac{z}{(1-z)^2}\sum_{k=0}^\infty \left(z^{2^k} - z^{3\cdot2^k}\right) = \left( \sum_{m=1}^\infty m z^m \right)\sum_{k=0}^\infty \left(z^{2^k} - z^{3\cdot2^k}\right)$$ With this expression, we can read off $T(n)$ as the coefficient of $z^n$ in $f(z)$ and get $$T(n) = \sum_{k=0}^{\lfloor \log_2 n\rfloor} ( n - 2^k ) - \sum_{k=0}^{\lfloor \log_2(n/3)\rfloor} (n - 3\cdot 2^k)$$ For $n > 2$, we can simplify this as $$\bbox[4pt,border: 1px solid black;]{ T(n) = n \color{red}{\big(\lfloor \log_2 n\rfloor - \lfloor \log_2(n/3)\rfloor\big)} - \color{blue}{\big( 2^{\lfloor \log_2 n\rfloor + 1} - 1 \big)} + 3\color{blue}{\big( 2^{\lfloor \log_2(n/3)\rfloor +1} - 1\big)}}\tag{*2}$$ There are several observations we can make. * *When $n = 2^k, k > 1$, we have $$T(n) = n(k - (k-2)) - (2^{k+1} - 1) + 3(2^{k-1} - 1) = \frac32 n - 2$$ *When $n = 3\cdot 2^{k-1}, k > 0$, we have $$T(n) = n(k - (k-1)) - (2^{k+1} - 1) + 3(2^k - 1) = \frac53 n - 2$$ *For $2^k < n < 3\cdot 2^{k-1}, k > 1$, the coefficient for $n$ in $(*2)$ (i.e the factor in red color) is $2$, while the rest (i.e those in blue color) didn't change with $k$. So $T(n)$ is linear there with slope $2$. *For $3\cdot 2^{k-1} < n < 2^{k+1}, k > 1$, the coefficient for $n$ in $(*2)$ is now 1. Once gain $T(n)$ is linear there but with slope $1$ instead. Combine these, we find in general $$\frac32 n - 2 \le T(n) \le \frac53 n - 2 \quad\text{ for }\quad n > 2$$ and $T(n) = O(n)$ as expected. However, $\frac{T(n)}{n}$ doesn't converge to any number but oscillate "between" $\frac32$ and $\frac53$. Above is a picture ilustrating the behavior of $T(n)$. The blue pluses are the value of $T(n) - (\frac32 n - 2)$ computed for various $n$. The red line is $\frac{n}{6} = (\frac53 n - 2) - (\frac32 n - 2)$. As one can see, $T(n)$ doesn't converge to any straight line. Instead, it oscillate between lines $\frac32 n - 2$ and $\frac53 n - 2$ as discussed before.

Elementary algebra problem Consider the following problem (drawn from Stanford Math Competition 2014): "Find the minimum value of $\frac{1}{x-y}+\frac{1}{y-z}+ \frac{1}{x-z}$ for for reals $x > y > z$ given $(x − y)(y − z)(x − z) = 17.$" Method 1 (official solution): Combining the first two terms, we have $\frac{x−z}{(x-y)(y-z)} + \frac{1}{x-z}= \frac{(x-z)^2}{17}+ \frac{1}{x-z}.$ What remains is to find the minimum value of $f(a) = \frac{a^2}{17} + \frac{1}{a} = \frac{a^2}{17} + \frac{1}{2a}+ \frac{1}{2a}$ for positive values of $a.$ Using AM-GM, we get $f(a) \geq \frac{3}{68^{1/3}}$. Method 2: Let $x-y:=a, \; y-z:=b, \; x-z:=c.$ Then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= \frac{ab+bc+ac}{17}= \frac{3}{17}\frac{ab+bc+ac}{3} \geq \frac{3}{17} (a^2b^2c^2)^{1/3} =\frac{3}{17^{1/3}}.$ So Method 2 seems to give a sharper bound than the official solution. Have I done something wrong?

In the second solution, when $=$ in $\ge$ is satisfied, $a=b=c$. But then, $abc=17$ and $a+b=c$ makes the contradiction.

How prove $(\ln{\frac{1-\sin{xy}}{1+\sin{xy}}})^2 \geq \ln{\frac{1-\sin{x^2}}{1+\sin{x^2}}}\ln{\frac{1-\sin{y^2}}{1+\sin{y^2}}}$ How prove that if $x, y \in (0,\sqrt{\frac{\pi}{2}})$ and $x \neq y$, then $(\ln{\frac{1-\sin{xy}}{1+\sin{xy}}})^2 \geq \ln{\frac{1-\sin{x^2}}{1+\sin{x^2}}}\ln{\frac{1-\sin{y^2}}{1+\sin{y^2}}}$?

the right question is : $(\ln{\dfrac{1-\sin{xy}}{1+\sin{xy}}})^2 \le \ln{\dfrac{1-\sin{x^2}}{1+\sin{x^2}}}\ln{\dfrac{1-\sin{y^2}}{1+\sin{y^2}}}$ $f(x)=\ln{\dfrac{1-\sin{x}}{1+\sin{x}}}\le 0 , f(0)=0$ WLOG $y=ax,a\ge1 \implies $the inequality $\iff (f(ax^2))^2 \le f(x^2)f(a^2x^2) \iff (f(ax))^2 \le f(x)f(a^2x) \iff \dfrac{f(ax)}{f(x)} \le \dfrac{f(a^2x)}{f(ax)} $ $g(x)=\dfrac{f(ax)}{f(x)} \implies g(x) \le g(ax) \implies g'(x)> 0$ $g'(x)=\dfrac{2(\cos{(ax)}f(ax)-a\cos{x}f(x))}{\cos{x}\cos{(ax)}(fx)^2}$ $g'(x) >0\implies g_1(x)=\cos{(ax)}f(ax)-a\cos{x}f(x) \ge 0 $ $g'_1(x)=-a\sin{(ax)} f(ax)+a\cos{(ax)}f'(ax)+a\sin{x}f(x)-a\cos{x}f'(x)=a((\sin{x}f(x)-\cos{x}f'(x))-(\sin{(ax)} f(ax)-\cos{(ax)}f'(ax)))=a(g_2(x)-g_2(ax)) \\ g_2(x)=\sin{x}f(x)-\cos{x}f'(x)=\sin{x}f(x)+2$ $g'_2(x)=\cos{x}f(x) -2\tan{x}<0 \implies g_2(x) \ge g_2(ax) \implies g'_1(x) \ge 0$ $g_1(0)=0 \implies g_1(x)\ge 0$ the "=" will hold when $a=1$ or $x=0$

Determine variables that fit this criterion... There is a unique triplet of positive integers $(a, b, c)$ such that $a ≤ b ≤ c$. $$ \frac{25}{84} = \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc} $$ Just having trouble with this Canadian Math Olympiad question. My thought process going into this, is: Could we solve for $\frac{1}{a}$ in terms of the other variables? Then substitute that value in for each occurrence of $a$, to solve for $a$? That's all I can really think of right now. It's a question I'm not exactly used to... It's sort of the first of these kinds that I've faced. Thanks.

Factoring, we see $\displaystyle \frac{25}{84} = \frac{1}{a}(1+\frac{1}{b}(1+\frac{1}{c}))$ And we know the prime factoring of 84 gives $2\times2\times3\times7$ So we know $a,b,$ and $c$ are each going to be multiples of these primes. So we start with finding $a$: $\displaystyle \frac{25}{84}a = 1+\frac{1}{b}(1+\frac{1}{c})$ Now, $25a/84>0$, but $\frac{1}{b}(1+\frac{1}{c})>0$ too. Therefore $25a/84>1$. We want $a$ to be the smallest of the three factors, so we ask, what is the smallest it can be here? 2 won't work and neither will 3, but $2\times2=4$ will. So we provisionally say $a=4$. Then, $\displaystyle \frac{25}{21} = 1+\frac{1}{b}(1+\frac{1}{c})$ We go through the same process for $b$, remembering that $b>4$. Turns out that $b=3\times2=6$ is the smallest factor that will work. We provisionally say $b=6$. Finally, $\displaystyle \frac{8}{7} = 1+\frac{1}{c}$ $c=7$ follows immediately. So $a=4$, $b=6$, and $c=7$.

Inequality involving a finite sum this is my first post here so pardon me if I make any mistakes. I am required to prove the following, through mathematical induction or otherwise: $$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$$ I tried using mathematical induction through: $Let$ $P(n) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$ $Since$ $P(1) = \frac{1}{\sqrt1} < 2{\sqrt{1}}, and$ $P(k) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{k}} < 2{\sqrt{k}},$ $P(k+1) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{k}}+ \frac{1}{\sqrt{k+1}} < 2{\sqrt{k+1}}$ Unfortunately, as I am quite new to induction, I couldn't really proceed from there. Additionally, I'm not sure how to express ${\sqrt{k+1}}$ in terms of ${\sqrt{k}}$ which would have helped me solve this question much more easily. I am also aware that this can be solved with Riemann's Sum (or at least I have seen it being solved in that way) but I do not remember nor quite understand it.

If you want to take a look on the Riemann's sum method : $\forall n > 1$, we have $$\int_{n-1}^n \frac{dt}{\sqrt{t}} \ge (n-(n-1))\cdot \underset{x \in [n-1,n]}{\min} \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{n}} $$ Hence, $$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} \le 1+ \int_{1}^2 \frac{dt}{\sqrt{t}} + \int_{2}^3 \frac{dt}{\sqrt{t}} + ... + \int_{n-1}^n \frac{dt}{\sqrt{t}} = 1+\int_{1}^n\frac{dt}{\sqrt{t}} $$ $$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} \le 1+\left[2\sqrt{t}\right]_1^n = 1+ (2\sqrt{n} -2 )= 2\sqrt{n} -1 < 2\sqrt{n} $$ The difference can be seen by comparing the Right Riemann Sum of the function $t \rightarrow \frac{1}{\sqrt{t}} $ to its integral on $[1,n]$.

Proof by Induction: $(1+x)^n \le 1+(2^n-1)x$ I have to prove the following by induction: $$(1+x)^n \le 1+(2^n-1)x$$ for $n \ge 1$ and $0 \le x \le 1$. I start by showing that it's true for $n=1$ and assume it is true for one $n$. $$(1+x)^{n+1} = (1+x)^n(1+x)$$ by assumption: $$\le (1+(2^n-1)x)(1+x)$$ $$= 1+(2^n-1)x+x+(2^n-1)x^2$$ That is what I was able to do on my own. I do not understand the next step of the solution (they are now using $x$ instead of $x^2$ at the end): $$Because\,0 \le x \le 1$$ $$(1+x)^{n+1} \le 1 + (2^n-1)x+x+(2^n-1)x$$ $$=1+2(2^n-1)x+x$$ $$=1+(2^{n+1}-1)x$$ Because $0 \le x \le 1$ I assume that $(2^n-1)x^2 < (2^n-1)x$. So in my opinion, I just proved that $(1+x)^{n+1}$ is less than something greater than what I had to prove initially. Why is it enough to prove the original problem?

$$(1+x)^{n+1}=(1+x)(1+x)^{n}\le(1+x)(1+(2^{n}-1)x)=1+(2^{n}-1)x+x(1+(2^{n}-1)x)$$ $$\underbrace{\le}_{x\le1}1+(2^{n}-1)x+x(1+2^{n}-1)=1+(2^{n}-1)x+2^{n}x$$ $$=1+(2\cdot2^{n}-1)x=1+(2^{n+1}-1)x$$

Consecutive Prime Gap Sum (Amateur) List of the first fifty prime gaps: 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4. My conjecture is that the sum of consecutive prime gaps is always prime whenever a prime gap of 2 is added. $$ 1 + 2 = 3 $$ $$ 1 + 2 + 2 = 5 $$ $$ 1 + 2 + 2 + 4 + 2 = 11 $$ $$ 1 + 2 + 2 + 4 + 2 + 4 + 2 = 17 $$ $$ 1 + 2 + 2 + 4 + 2 + 4 + 2 + 4 + 6 + 2 = 29 $$ I don't know if this is meaningful or how to go about testing it completely (I've tested it up to 461) so I'll just leave this here and see what comes of it.

Set $g_n=p_{n+1}-p_n$, where $p_n$ is the series of prime numbers, with $p_1=2$. Then $$ p_1+\sum_{i=1}^n g_i=\sum_{i=1}^n g_i+2=p_{n+1}. $$ So the conjecture is obviously true, but not useful.

Evaluation of a sum of $(-1)^{k} {n \choose k} {2n-2k \choose n+1}$ I have some question about the paper of which name is Spanning trees: Let me count the ways. The question concerns about $\sum_{k=0}^{\lfloor\frac{n-1}{2} \rfloor} (-1)^{k} {n \choose k} {2n-2k \choose n+1}$. Could you recommend me how to prove $\displaystyle \sum_{k=0}^{\lfloor\frac{n-1}{2} \rfloor} (-1)^{k} {n \choose k} {2n-2k \choose n+1}=n 2^{n-1}$?

Since $\binom{m-2k}{n+1}$ is a degree $n+1$ polynomial in $k$ with lead term $\frac{(-2k)^{n+1}}{(n+1)!}$, we get that the multiple forward difference $\Delta_k^{n+1}\binom{m-2k}{n+1}=(-2)^{n+1}$. Multiply both sides by $(-1)^{n+1}$ to get $$ \begin{align} 2^{n+1} &=\sum_{k=0}^{n+1}(-1)^k\binom{n+1}{k}\binom{m-2k}{n+1}\\ &=\sum_{k=0}^{n+1}(-1)^k\left[\binom{n}{k}+\binom{n}{k-1}\right]\binom{m-2k}{n+1}\\ &=\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{m-2k}{n+1}-\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{m-2k-2}{n+1}\tag{1} \end{align} $$ Equation $(1)$ also says that for some $m_0$, $$ \sum_{k=0}^n(-1)^k\binom{n}{k}\binom{m-2k}{n+1}=2^n(m-m_0)\tag{2} $$ To determine $m_0$, consider the equation $$ \begin{align} \sum_{k=0}^n(-1)^k\binom{n}{k}\binom{m-2k}{n+1} &=\sum_{k=0}^n(-1)^{n+1-k}\binom{n}{k}\binom{n-m+2k}{n+1}\\ &=\sum_{k=0}^n(-1)^{k+1}\binom{n}{k}\binom{3n-m-2k}{n+1}\tag{3} \end{align} $$ If we set $m=\frac32n$, then the left and right sides of $(3)$ are negatives of each other yet equal, therefore, $0$. Thus, $m_0=\frac32n$. Therefore, $$ \sum_{k=0}^n(-1)^k\binom{n}{k}\binom{m-2k}{n+1}=2^n\left(m-\tfrac32n\right)\tag{4} $$ Plugging $m=2n$ into $(4)$ and noting that $2n-2k\ge n+1\implies k\le\frac{n-1}{2}$, we get $$ \sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^n\binom{n}{k}\binom{2n-2k}{n+1}=n2^{n-1}\tag{5} $$

Finding the asymptotes of a general hyperbola I'm looking to find the asymptotes of a general hyperbola in $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ form, assuming I know the center of the hyperbola $(h, k)$. I came up with a solution, but it's too long for me to be confident that I didn't make a mistake somewhere, so I was wondering if I could run it by someone and see if it works. It's mostly algebraic, and I'm prone to making tiny errors in algebra that throw off the entire problem. So to start, since we know the center $(h, k)$, we can first translate the hyperbola by $(-h, -k)$ using the transform $x_0 = x - \Delta{x}, y_0 = y - \Delta{y}$ with $\Delta{x} = -h$ and $\Delta{y} = -k$. Assuming $F'$ is the translated $F$, we can divide the entire equation by $-F'$ to put it in the following form: $$ ax^2 + bxy + cy^2 + dx + ey = 1 $$ With $a = -A'/F'$ and $A'$ the translated $A$, $b = -B'/F'$ and $B'$ the translated $B$, and so on. Next we convert to polar coordinates to get the following: $$ r^2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta}) + r(d\cos{\theta} + e\sin{\theta}) - 1 = 0 $$ Solving for $r$ will give us $$ r = \frac{-d\cos{\theta} - e\sin{\theta} \pm \sqrt{(d\cos{\theta} + e\sin{\theta})^2 + 4(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})}}{2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})} $$ Now assume $\theta_0 = 2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})$, this means that as $\theta_0 \rightarrow 0^{\pm}$, $r \rightarrow \pm\infty$. The angles at which $r \rightarrow \pm\infty$ are the asymptotes of the hyperbola, so now it's just a matter of solving for where $\theta_0 = 0$. This is where the majority of the algebra takes place and this is where I'm worried I made some miniscule mistake. $$ \begin{align} & 2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta}) = 0\\ & \Longleftrightarrow a(1 - \sin^2{\theta}) + b\cos{\theta}\sin{\theta} + c\sin^2{\theta} = 0 \\ & \Longleftrightarrow \sin^2{\theta}(c - a) + b\cos{\theta}\sin{\theta} = -a \\ & \Longleftrightarrow \frac{1 - \cos{2\theta}}{2}(c - a) + \frac{b}{2}\sin{2\theta} = -a \\ & \Longleftrightarrow (c - a)(1 - \cos{2\theta}) + b\sin{2\theta} = -2a \\ & \Longleftrightarrow (c - a)(1 - \cos{2\theta}) + 2a = -b\sqrt{1 - \cos^2{2\theta}} \\ & \Longleftrightarrow \frac{a - c}{b}(1 - \cos{2\theta}) - \frac{2a}{b} = \sqrt{1 - \cos^2{2\theta}} \\ & \Longleftrightarrow (\frac{a - c}{b}(1 - \cos{2\theta}))^2 - 4a\frac{a - c}{b^2}(1 - \cos{2\theta}) + (\frac{2a}{b})^2 = 1 - \cos^2{2\theta} \\ & \Longleftrightarrow (\frac{a - c}{b})^2(1 - 2\cos{2\theta} + \cos^2{2\theta}) - 4a\frac{a - c}{b^2}(1 - \cos{2\theta}) + (\frac{2a}{b})^2 = 1 - \cos^2{2\theta} \\ & \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + [2(\frac{a - c}{b})(\frac{a}{b} - \frac{a - c}{b})]\cos{2\theta} + [(\frac{2a}{b})^2 + (\frac{a - c}{b})^2 - 4a\frac{a - b}{b^2}) - 1] = 0 \\ & \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + 2(\frac{a - c}{b})(\frac{c}{b})\cos{2\theta} + [(\frac{2a}{b})^2 + (\frac{a - c}{b})(\frac{-3a - c}{b}) - 1] = 0 \\ & \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + 2(\frac{a - c}{b})(\frac{c}{b})\cos{2\theta} + [\frac{a^2 + 2ac + c^2}{b^2} - 1] = 0 \\ \end{align} $$ Now let $$ U = [(\frac{a - c}{b})^2 + 1] \\ V = 2(\frac{a - c}{b})(\frac{c}{b}) \\ W = \frac{a^2 + 2ac + c^2}{b^2} - 1 $$ So that the above equation becomes $$ U\cos^2{2\theta} + V\cos{2\theta} + W = 0. $$ Solving for $\cos{2\theta}$ gives us $$ \cos{2\theta} = \frac{-V \pm \sqrt{V^2 - 4UW}}{2U}. $$ Finally we may solve for theta like so, $$ \theta = \frac{1}{2}\arccos{\frac{-V \pm \sqrt{V^2 - 4UW}}{2U}}. $$ This gives us two numbers, $\theta_1$ and $\theta_2$, each corresponding to the slopes $m_1$ and $m_2$ of the asymptotes. The relationship between the slope of a line $m$ and the angle $\theta$ between the line and the positive x-axis is $m = \tan{\theta}$. You can use the identity $\tan{(\frac{1}{2}\arccos{x})} = \sqrt{\frac{1 - x}{x + 1}}$ to solve for $m_1$ and $m_2$ in terms of non-trig functions, but I think this answer is sufficient enough. Now that we have the slopes of the asymptotes, we can find the y-intercepts $b_1$ and $b_2$ of each line by simply plugging in the original center $(h, k)$ into each equation for the line and solving. $$ b_1 = k - m_1h \\ b_2 = k - m_2h $$ Thus, the asymptotes of the hyperbola with general equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ have equations $y = m_1x + b_1$ and $y = m_2x + b2$. Does this look correct? Also, did I overcomplicate things? Was there an easy solution all along that I was missing?

For a conic, $$ax^2+2hxy+by^2+2gx+2fy+\color{blue}{c}=0$$ which is a hyperbola when $ab-h^2<0$. Its asymptotes can be found by replacing $\color{blue}{c}$ by $\color{red}{c'}$ where $$\det \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & \color{red}{c'} \end{pmatrix} =0$$ That is $$\color{red}{c'}=\frac{af^2+bg^2-2fgh}{ab-h^2}$$ The asymptotes are $$\fbox{$ax^2+2hxy+by^2+2gx+2fy+\frac{af^2+bg^2-2fgh}{ab-h^2}=0 \,$}$$ Alternatively, using the centre of the conics $$ \left( \frac{bg-fh}{h^2-ab}, \frac{af-gh}{h^2-ab} \right)$$ and the slope $m$ of an asymptote is given by $$a+2hm+bm^2=0$$ On solving, $$m=\frac{-h \pm \sqrt{h^2-ab}}{b}=\frac{a}{-h \mp \sqrt{h^2-ab}}$$ Therefore $$\fbox{$ y-\frac{af-gh}{h^2-ab}= \frac{-h \pm \sqrt{h^2-ab}}{b} \left( x-\frac{bg-fh}{h^2-ab} \right) $}$$ See another answer here for your interest.

Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem: $$y'=\text{Ax}^2+\text{Bx}+c,$$ where $y(1)=1$, I get: $$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$ But the answer to this is: $$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$ Could someone show me what has been done and explain why?

$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$ where $d$ is a cosntant to fix is equivalent to $$\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+d$$ where $d$ is a constant to fix. And the second one is surely more handy to apply the initial condition.

find formula to count of string I have to find pattern of count of series. Lenght of series is $2n$. It is neccessary to use exaclty double times every number in range $[1...n]$ and all of neighboring numbers are different. Look at example. $a_0 = 0$ $a_1 = 0$ $a_2 = 2$ (because of $1212$ and $2121$) And my idea is: $$ a_{n+1}= {2n+1\choose 2}a_n $$ Is it properly ?

For $a_3$, you'll add two $3$'s somewhere to each one of those so they're not next to each other. There are $5$ possible places: the two ends, and three in between two numbers, so there are ${5 \choose 2}$ ways of putting the two $3$'s in so they're not next to each other. For case $a_n$: There are $2n+1$ possible places to place $n+1$ in each of the ${a_n}$ series of length $2n$, so there are ${2n+1 \choose 2}$ ways to place the two $n$'s in each series. Or, $$a_{n+1} = {2n+1 \choose 2} a_n.$$ We can see the closed form by inspection: $$a_{n} = {2n-1 \choose 2} a_{n-1} \\ = {2n-1 \choose 2}{2n-3 \choose 2}{2n-5 \choose 2} \cdots {7 \choose 2}{5 \choose 2}a_2 \\ = \frac{(2n-1)(2n-2)}{2}\frac{(2n-3)(2n-4)}{2}\frac{(2n-5)(2n-6)}{2} \cdots \frac{7 \cdot 6}{2}\frac{5 \cdot 4}{2}\cdot 2,$$ so $$a_n =\frac{(2n-1)!}{3 \cdot 2^{n-2}}, n \geq 3.$$

Simple use of log I am struggling to see how we can go from the first expression to the second: $$\begin{align} 2\log_3 12 - 4\log_3 6 &= \log_3 \left ( \frac{4^2 \cdot 3^2}{2^4 \cdot 3^4} \right )\\ &= \log_3 (3^{-2}) = -2 \end{align}$$

First note that \[ \log_{b}(x^{n})=n\log_{b}(x) \] \[ \log_{b}(x)-\log_{b}(y)=\log_{b}\left(\frac{x}{y}\right) \] Here are the steps \[ 2\log_{3} 12 - 4\log_{3} 6 = 2\log_{3} (4\cdot 3) - 4\log_{3} (2\cdot 3) \] \[ =\log_{3} (4\cdot 3)^{2} - \log_{3} (2\cdot 3)^{4} = \log_{3} (4^{2}\cdot 3^{2}) - \log_{3} (2^{4}\cdot 3^{4}) \] \[ = \log_{3} \frac{4^{2}\cdot 3^{2}}{2^{4}\cdot 3^{4}} = \log_{3} \frac{16}{16\cdot 3^{2}}=\log_{3} \frac{1}{3^{2}}=\log_{3} 3^{-2}=-2\log_{3} 3 = -2 \]

How to find $\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$ $$I=\int x.\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$$ Try 1: Put $z= \ln(x+\sqrt{1+x^2})$, $\mathrm dz=1/\sqrt{1+x^2}\mathrm dx$ $$I=\int \underbrace{x}_{\mathbb u}\underbrace{z}_{\mathbb v}\mathrm dz=x\int zdz-\int (z^2/2)\mathrm dz\tag{Wrong}$$ Try 2: Put $z= x+\sqrt{1+x^2}$ $$\implies x-z =\sqrt{1+x^2}\implies x^2+z^2-2xz =1+x^2\implies x =\frac{z^2-1}{2z}$$ $$\mathrm dz =\left(1+\frac{x}{\sqrt{1+x^2}}\right)\mathrm dx =\frac{z\mathrm dx}{x-z}=\frac{-2z^2\mathrm dx}{1+z^2}$$ $$I =\int\frac{(z^2-1)\ln z}{2z}.\frac{(1+z^2)\mathrm dz}{-2z^2}$$ $$=\int\frac{(z^4-1)\mathrm dz}{4z^3} =\frac14\int\left(z-\frac1{z^3}\right)\mathrm dz =z^2/2+2/z^2+C\tag{Wrong}$$ Try 3: Put $z =\sqrt{1+x^2},\mathrm dx =x/\sqrt{1+x^2}\mathrm dx$ $$I =\int \ln(x+z)\mathrm dz =\int \ln(z+\sqrt{z^2-1})\mathrm dz$$ Don't know how to solve this integral. [Note that if I take $u=z+\sqrt{z^2-1}$, it is $=\sqrt{1+x^2}+\sqrt{1+x^2-1}=x+\sqrt{1+x^2}$; same as first try.] What's wrong in try 1 & 2, how to further solve try 3 and the best method to solve this question? Update: Sorry, I don't know hyperbolic/inverse hyperbolic trigonometry.

Hint: $$ \int x\frac{\ln({x+\sqrt{1+x^2})}}{\sqrt{1+x^2}}dx=\int\ln({x+\sqrt{1+x^2})}d\sqrt{1+x^2} $$ and $$ (\ln({x+\sqrt{1+x^2})})'=\frac{1}{\sqrt{1+x^2}}. $$

Indefinite integral of $\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}$ How do I find $$\int\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}\mathrm dx$$ I used partial fractions by breaking up $x^2 + 2x - 2$ into $(x+1)^2 - 3$ and split it into $(a+b)(a-b)$ but as u can see it's extreme tedious. I was wondering if there is a faster technique to resolve this.

HINT: Write $$\frac{2x^3 + 5x^2 +2x +2)}{(x^2 +2x + 2)(x^2 + 2x - 2)}=\frac{Ax+B}{x^2+2x+2}+\frac{Cx+D}{x^2+2x-2}$$ For the ease of calculation, we can write $$\frac{Ax+B}{x^2+2x+2}=\frac A2\cdot\frac{\left(\dfrac{d(x^2+2x+2)}{dx}\right)}{x^2+2x+2}+\frac C{(x+1)^2+1^2}.$$

How can I show why this equation has no complex roots? I've been asked to show why an equation has no complex roots but i'm at a complete loss. The equation is $F_{n+2}=F_n$ Where $F_n=(x-1)(x-2)...(x-n)$ and n is a positive integer. I'd really appreciate if someone could explain how I could go about showing this because I'd really like to understand. Thanks in advance.

$F_{n}=(x-1)(x-2) \cdots (x-n)$ $F_{n+2}=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2))$ $F_{n} = F_{n+2}$ $(x-1)(x-2) \cdots (x-n)=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2))$ $(x-1)(x-2) \cdots (x-n)[(x-(n+1))(x-(n+2)) - 1] = 0$ $(x-1)(x-2) \cdots (x-n)[(x^2-[(n+2)+(n+1)]x+(n+1)(n+2) - 1] = 0$ $(x-1)(x-2) \cdots (x-n)[x^2-(2n+3)x+n^2+3n+ 1] = 0$ Therefore $x = 1,2,3,...,(n-1),n$ and $x^2-(2n+3)x+n^2+3n+ 1 = 0$ $$x=\frac{(2n+3)\pm \sqrt{(2n+3)^2-4(1)(n^2+3n+1)}}{2(1)}$$ $$x=\frac{(2n+3)\pm \sqrt{5}}{2}$$ $$x=\frac{2n+3+ \sqrt{5}}{2} x=\frac{2n+3- \sqrt{5}}{2}$$ Therefore the roots are not complex

Sum the series $\sum_{n = 1}^{\infty}\{\coth (n\pi x) + x^{2}\coth(n\pi/x)\}/n^{3}$ This sum is from Ramanujan's letters to G. H. Hardy and Ramanujan gives the summation formula as \begin{align} &\frac{1}{1^{3}}\left(\coth \pi x + x^{2}\coth\frac{\pi}{x}\right) + \frac{1}{2^{3}}\left(\coth 2\pi x + x^{2}\coth\frac{2\pi}{x}\right) \notag\\ &\, \, \, \, \, \, \, \, + \frac{1}{3^{3}}\left(\coth 3\pi x + x^{2}\coth\frac{3\pi}{x}\right) + \cdots\notag\\ &\, \, \, \, \, \, \, \, = \frac{\pi^{3}}{90x}(x^{4} + 5x^{2} + 1)\notag \end{align} Since $$\coth x = \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} = \frac{1 + e^{-2x}}{1 - e^{-2x}} = 1 + 2\frac{e^{-2x}}{1 - e^{-2x}}$$the above sum is transformed into $$(1 + x^{2})\sum_{n = 1}^{\infty}\frac{1}{n^{3}} + 2\sum_{n = 1}^{\infty}\frac{e^{-2n\pi x}}{n^{3}(1 - e^{-2n\pi x})} + 2x^{2}\sum_{n = 1}^{\infty}\frac{e^{-2n\pi/x}}{n^{3}(1 - e^{-2n\pi/x})}$$ If we put $q = e^{-\pi x}$ we get sums like $\sum q^{2n}/\{n^{3}(1 - q^{2n})\}$ which I don't know how to sum. It seems I am going on a wrong track. Please provide some alternative approach. Update: All the answers given below so far use complex analyis (transforms and residues) to evaluate the sum. I am almost certain that Ramanujan did not evaluate the sum using complex analysis. Perhaps the method by Ramanujan is more like the one explained in this question. Do we have any approach based on real-analysis only?

Recall the well known Mittag-Leffler expansion of hyperbolic cotangent function (denote $\mathbb{W}=\mathbb{Z}/\{0\}$) : $$\sum_{m\in\mathbb{W}}\frac{1}{m^2+z^2}=\frac{\pi\coth\pi z}{z}-\frac{1}{z^2}\tag{ML}$$ Hence, your sum is by its symmetry : $$\begin{align} S&=\frac{1}{2}\sum_{n \in \mathbb{W}}\{\coth (n\pi x) + x^{2}\coth(n\pi/x)\}/n^{3} \\ \\ &=\frac{1}{2\pi x}\sum_{n \in \mathbb{W}}\left(\frac{1}{n^4}+\sum_{m\in\mathbb{W}}\frac{x^2/n^2}{m^2+n^2x^2}\right) +\left(\frac{x^4}{n^4}+\sum_{m\in\mathbb{W}}\frac{x^2/n^2}{m^2+n^2/x^2}\right)\tag{1}\\ \\ &=\frac{1}{2\pi x}\left(\zeta(4)+x^4\zeta(4)+\sum_{n,m \in \mathbb{W}^2}\frac{x^2}{n^2}\frac{1}{m^2+n^2x^2}+\frac{x^2}{n^2}\frac{1}{m^2+n^2/x^2}\right)\tag{2}\\ \\ &=\frac{1}{2\pi x}\left(2\zeta(4)+2x^4\zeta(4)+x^2\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2m^2}\frac{m^2+n^2x^2}{m^2+n^2x^2}\right)\tag{3}\\ \\ &=\frac{1}{2\pi x}\left(2\zeta(4)+2x^4\zeta(4)+4x^2\zeta^2(2)\right)\tag{4}\\ \\ &=\frac{1}{2\pi x}\left(2\frac{\pi^4}{90}+2x^4\frac{\pi^4}{90}+4x^2\frac{\pi^4}{36}\right)\tag{5}\\ \\ &=\frac{\pi^3}{90x}\left(1+x^4+5x^2\right) \end{align}$$ Explanations $(1)$ Use the Mittag-Leffler formula (ML) with $z=nx$ and $z=n/x$ $(2,4)$ Recall $\zeta(s)=\sum_{n=1}^{\infty}1/n^s$ $(3)$ In the second sum rename $n \longleftrightarrow m$ $(5)$ Zetas for $s=2$ and $4$ are well known, i.e. $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$

How to express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k=\log_2 (\sqrt{9} + \sqrt{5})$? If $$k=\log_2 (\sqrt{9} + \sqrt{5})$$ express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k$.

Adding the logarithms $\log_2{(\sqrt{9}-\sqrt{5})}$ and $\log_2{(\sqrt{9}+\sqrt{5})}$ we get the following: $$\log_2{(\sqrt{9}-\sqrt{5})}+\log_2{(\sqrt{9}+\sqrt{5})}=\log_2{(\sqrt{9}-\sqrt{5})(\sqrt{9}+\sqrt{5})}=\log_2{(9-5)}=\log_2{4}=\log_2{2^2}=2 \cdot \log_2{2}=2$$ Knowing that $k=\log_2{(\sqrt{9}+\sqrt{5})}$ we have: $$\log_2{(\sqrt{9}-\sqrt{5})}+\log_2{(\sqrt{9}+\sqrt{5})}=\log_2{(\sqrt{9}-\sqrt{5})}+k$$ Therefore, $$2=\log_2{(\sqrt{9}-\sqrt{5})}+k \Rightarrow \log_2{(\sqrt{9}-\sqrt{5})}=2-k$$

How to deduce a closed formula given an equivalent recursive one? I know how to prove that a closed formula is equivalent to a recursive one with induction, but what about ways of deducing the closed form initially? For example: $$ f(n) = 2 f(n-1) + 1 $$ I know how to use induction to prove that $\forall n \ge1$: $$ f(n) = 2^n f(0) + 2^n - 1 $$ And I was able to come up with that formula in the first place just by examining it informally. It seems like there should have been a straightforward, formal way to deduce the closed form from the recursive definition in the first place, but I'm blanking. I know it's not always easy (e.g. Fibonacci) but it seems like it ought to be here. For example, how might I go about procedurally deducing the closed form for: $$ f(n) = 2*f(n-1) + k $$ Thanks.

$$f(n) = 2f(n-1) + k \\ f(n-1) = 2f(n-2) + k \\ f(n-2) = 2f(n-3) + k \\ \dots \\ f(1) = 2f(0) + k$$ $$$$ $$\Rightarrow f(n) = 2f(n-1) + k= \\ 2(2f(n-2) + k)+k= \\ 2^2f(n-2)+(2+1)k=2^2(2f(n-3) + k)+(2+1)k= \\ 2^3f(n-3)+(2^2+2+1)k= \\ \dots \overset{*}{=} \\ 2^nf(0)+(2^{n-1}+2^{n-2}+\dots +2+1)k$$ Therefore, $$f(n)=2^nf(0)+k\sum_{i=0}^{n-1}2^i= \\ 2^nf(0)+k\frac{2^n-1}{2-1}= 2^nf(0)+(2^n-1)k$$ EDIT: $(*):$ We see that $2^3f(n-3)+(2^2+2+1)k$ is of the form $2^jf(n-j)+(2^{j-1}+2^{j-2}+\dots+2^2+2+1)k$ After some steps we have the following: $2^nf(n-n)+(2^{n-1}+2^{n-2}+\dots+2^2+2+1)k= \\ 2^nf(0)+(2^{n-1}+2^{n-2}+\dots+2^2+2+1)k$

Check if two vector equations of parametric surfaces are equivalent Give the vector equation of the plane through these lines: $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}\,\,\,$ and $\,\,\,\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\0\\3\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}$. My answer is: $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\-1\\2\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\-3\\1\end{pmatrix}$. The solutions manual suggests the following equation, which is the equation of a straight line (probably a typo): $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}$ Is my solution the right solution? Could someone provide a general way to check?

But the really simple way is to pick three arbitrary points (not on the same line) from the first plane and check if they all lie in the second one.

Equation $3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$ Solve the equation $$3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$$ Having a complex root of modulus $1$. To get the solution, I tried to take a complex root $\sqrt{\frac{1}{2}} + i \sqrt{\frac{1}{2}}$ but couldn't get the solution right. Please help me.

Let the root be $\cos y+i\sin y,$ Using Complex conjugate root theorem, $\cos y-i\sin y$ must be another root So, if the four roots are $\cos y\pm i\sin y,u, v$ using Vieta's formula, $(\cos y+i\sin y)(\cos y-i\sin y)u\cdot v=\dfrac63\implies v=\dfrac2u$ So we have $$3[x-(\cos y+i\sin y)][x-(\cos y-i\sin y)](x-u)\left(x-\dfrac2u\right)=3x^4 + 2x^3 + 9x^2 + 4x + 6$$ $$\iff3[(x^2-2x\cos y+1)]\left[x^2-\left(u+\frac2u\right)+2\right]=3x^4 + 2x^3 + 9x^2 + 4x + 6$$ $$\iff3\left[x^4-x^3\left(2\cos y+u+\frac2u\right)+x^2\left[1+2+2\cos y\left(u+\frac2u\right)\right]+\cdots\right]=3x^4 + 2x^3 + 9x^2 + 4x + 6$$ Equating the coefficients of $x^3,x^2$ $$2\cos y+u+\frac2u=-\frac23\ \ \ \ \ (1)$$ and $$1+2+2\cos y\left(u+\frac2u\right)=\frac93\iff2\cos y\left(u+\frac2u\right)=0$$ If $\cos y=0\implies\sin y=\pm1\implies \cos y\pm\sin y=\pm i$ which does not satisfy the given equation So, $u+\dfrac2u$ must be $0$ and from $(1),2\cos y=-\dfrac23\iff\cos y=-\dfrac13\implies\sin y=\pm\dfrac{2\sqrt2}3$ Observation: $x^2+2$ is a factor of $$3x^4 + 2x^3 + 9x^2 + 4x + 6$$

general of partial sum of sequence I am trying to find the limit of an infinite series given as $$\sum\frac{1}{n^2-1}.$$ I came across the following general term of the sequence of partial sums $$3/4-\left(\frac{1}{2n}-\frac{1}{2(n+1)}\right).$$ I would appreciate assistance to understand how this expression is arrived at. I have tried breaking down the original expression into partial fractions, but cannot get to the given result.

You have $$ \frac{1}{n^2-1} = \frac 1 2 \left(\ \underbrace{\frac{1}{n-1}-\frac{1}{n+1}\ }_{\text{Call this $\{$A$\}$}} \right) $$ Consequently \begin{align} & \frac12\left( \left(\frac{1}{2-1} - \frac{1}{2+1}\right) + \left(\frac{1}{3-1} - \frac{1}{3+1}\right) + \left( \frac{1}{4-1} - \frac{1}{4+1} \right) + \cdots + \{\text{A}\} \right) \\[10pt] = {} & \frac 1 2 \left( \left(\frac 1 1 - \frac 1 3\right) + \left( \frac 1 2 - \frac 1 4 \right) + \left( \frac 1 3 - \frac 1 5 \right) + \left( \frac 1 4 - \frac 1 6 \right) + \left( \frac 1 5 - \frac 1 7 \right) + \cdots +\{\text{A}\} \right). \end{align} So $+1/3$ cancels $-1/3$; $+1/4$ cancels $-1/4$; $+1/5$ cancels $-1/5$, and so on. Only a few terms at the beginning and the end do not cancel, and you're left with an expression for the partial sum whose complexity does not grow with the number of terms.

Simplifying the sum of powers of the golden ratio I seem to have forgotten some fundamental algebra. I know that: $(\frac{1+\sqrt{5}}{2})^{k-2} + (\frac{1+\sqrt{5}}{2})^{k-1} = (\frac{1+\sqrt{5}}{2})^{k}$ But I don't remember how to show it algebraicly factoring out the biggest term on the LHS gives $(\frac{1+\sqrt{5}}{2})^{k-2}(1+(\frac{1+\sqrt{5}}{2}))$ which doesn't really help

$$ \left (\frac{1+\sqrt{5}}{2} \right )^{k-2} + \left (\frac{1+\sqrt{5}}{2} \right )^{k-1} = \left ( \frac{1+ \sqrt{5}}{2}\right )^{k-2} \left ( 1+ \frac{1+ \sqrt{5}}{2}\right)$$ It is known that the Greek letter phi (φ) represents the golden ratio,which value is: $$\phi=\frac{1+ \sqrt{5}}{2}$$ One of its identities is: $$\phi^2=\phi+1$$ Therefore: $$ \left ( 1+ \frac{1+ \sqrt{5}}{2}\right)= \left ( 1+ \frac{\sqrt{5}}{2}\right)^2$$ So: $$ \left ( \frac{1+ \sqrt{5}}{2}\right )^{k-2} \left ( 1+ \frac{1+ \sqrt{5}}{2}\right)= \left ( 1+ \frac{\sqrt{5}}{2}\right)^k$$

Evaluating Summation of $5^{-n}$ from $n=4$ to infinity The answer is $\frac1{500}$ but I don't understand why that is so. I am given the fact that the summation of $x^{n}$ from $n=0$ to infinity is $\frac1{1-x}$. So if that's the case then I have that $x=\frac15$ and plugging in the values I have $\frac1{1-(\frac15)}= \frac54$.

The problem you have is that you do not know why the formula works to begin with. If you did the situation would be clear. Here's the thing: $$\sum_{n=0}^{\infty}r^n=\frac{1}{1-r} \; \;\;\;\;\; |r|<1.$$ Let $r=\frac{1}{5}$, then you really have the following situation: $$\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....$$ Let's call this infinite sum $S$ and proceed as follows, $$S=\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....$$ then $$ \left(\frac{1}{5}\right)S=\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+\left(\frac{1}{5}\right)^6....\;\;\;\;$$ Subtract the second from the first, $$S-\left(\frac{1}{5}\right)S=1$$ $$S(1-\left(\frac{1}{5}\right))=1$$ $$S=\frac{1}{1-\left(\frac{1}{5}\right)}$$ $$S=\frac{5}{4}.$$ Now recall what $S$ was and realize, $$S=\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}$$ But, you want the powers to start at $n=4$, so subtract the first 4 powers(0,1,2,3) to get, $$S-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)$$ which means $$\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)=\frac{1}{500}.$$

Evaluating $\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}$ How to evaluate the following limit? $$\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}.$$

You may write $$ \begin{align} \frac{(1+x)^{1/x} - e}{x} &= \frac{\large e^{\large\frac{\log (1+x)}{x}} - e}{x}\\\\ &= \frac{e^{\large \frac{x-\frac{x^2}{2}+{\mathcal{O}}(x^3)}{x}} - e}{x}\\\\ &= \frac{ e^{1-\frac{x}{2}+{\mathcal{O}}(x^2)} - e}{x}\\\\ &= \frac{ e \:e^{-\frac{x}{2}+{\mathcal{O}}(x^2)} - e}{x}\\\\ &= \frac{ e \:\left(1-\frac{x}{2}+{\mathcal{O}}(x^2)\right)- e}{x}\\\\ &=-\frac{e}{2}+{\mathcal{O}}(x) \end{align} $$ giving $$\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}=-\frac e2$$ where we have used $$ \begin{align} &\log(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3}+ \cdots, \quad |x|<1, \\ &e^x =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+ \cdots . \end{align} $$

Are these proofs logically equivalent? Here are two proofs, firstly: x = 0.999... 10x = 9.999... = 9 + 0.999... = 9 + x 9x = 9 x = 1 And secondly: x = 1 - 1 + 1 - 1 + 1 - 1 ... = 1 - (1 - 1 + 1 - 1 + 1 ... = 1 - x 2x = 1 x = 1/2 The fourth line in the first and the third line in the second use the same trick. Does the legitimacy of the first proof guarantee that of the second?

No. The first series is absolutely convergent, so its terms can be rearranged without changing the sum, while the second series is divergent. The number $0.\overline{9}$ can be expressed in the form \begin{align*} 0.\overline{9} & = \sum_{k = 1}^{\infty} \frac{9}{10^k}\\ & = \frac{9}{10} \sum_{k = 0}^{\infty} \frac{1}{10^k} \end{align*} which is a convergent geometric series, so \begin{align*} 0.\overline{9} & = \frac{9}{10} \cdot \frac{1}{1 - \frac{1}{10}}\\ & = \frac{9}{10} \cdot \frac{1}{\dfrac{9}{10}}\\ & = \frac{9}{10} \cdot \frac{10}{9}\\ & = 1 \end{align*} In fact, the series $$0.\overline{9} = \sum_{k = 1}^{\infty} \frac{9}{10^k}$$ is absolutely convergent since the series $$\sum_{k = 1}^{\infty} \bigg| \frac{9}{10^K} \bigg| = \sum_{k = 1}^{\infty} \frac{9}{10^k}$$ converges. Any rearrangement of an absolutely convergent series has the same sum. The series $$1 - 1 + 1 - 1 + 1 - 1 + \cdots = \sum_{k = 0}^{\infty} (-1)^k$$ diverges since the sequence of partial sums $\{1, 0, 1, 0, 1, 0, \ldots\}$ alternates between $1$ and $0$. Since the series never converges to a limit, it is meaningless to say that $$x = 1 - 1 + 1 - 1 + 1 - 1 + \cdots$$ A series must be absolutely convergent to guarantee that rearranging the terms of the series does not change the sum.

Show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$ If $a^3+b^3+c^3=3abc$ and $a+b+c=0$ show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$

$a+b+c=0\iff b+c=-a\implies (b+c)^2=a^2$ $\implies\dfrac{(b+c)^2}{3bc}=\dfrac{a^3}{3abc}$ Actually, $b+c=-a\implies(b+c)^3=(-a)^3$ $\implies -a^3=b^3+c^3+3bc(b+c)=b^3+c^3+3bc(-a)\iff\sum a^3=3abc$

How to show that $a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$ How do you show that $$a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$$ I could write $\sqrt{a^2+b^2-ab}=\sqrt{(a+b)^2-3ab}$, but this seems to lead nowhere.

$$a+b> \sqrt{a^2+b^2-ab} \iff (a+b)^2>\left (\sqrt{a^2+b^2-ab}\right )^2$$ $$ \iff\quad a^2+2ab+b^2>a^2+b^2-ab$$ $$\iff\quad 3ab>0 \quad \iff\quad ab>0$$ which is correct because of the hypothesis $a,b >0$.

How to solve the integral $\int \frac {(x^2 +1)}{x^4- x^2 +1} dx$ I have started this problem but I'm not completely sure I'm going down the right path with it. So far I have completed the square in the denominator. $x^4-x^2+1= (x^2-1/2)^2+\frac{3}{4}$ Then, let $u=x^2-\frac{1}{2}$ so $x=\sqrt(u+\frac{1}{2})$ $\int\frac{x^2+1}{x^4-x^2+1}dx = \int\frac{u+\frac{1}{2}+1}{u^2+\frac{3}{4}}du =\int\frac{u}{u^2+\frac{3}{4}} +\frac{\frac{3}{2}}{u^2+\frac{3}{4}}du$

$$\frac{x^2+1}{x^4-x^2+1}=\frac{1+\dfrac1{x^2}}{x^2-1+\dfrac1{x^2}}$$ Now $\displaystyle\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x$ and $\displaystyle x^2-1+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2-1$ Hope you can take it from here

$\nabla \cdot f + w \cdot f = 0$ Let $w(x,y,z)$ be a fixed vector field on $\mathbb{R}^3$. What are the solutions of the equation $$ \nabla \cdot f + w \cdot f = 0 \, ? $$ Note that if $w = \nabla \phi $, then the above equation is equivalent to $$ \nabla \cdot (e^\phi f) = 0, $$ for which the solutions are of the form $f = e^{-\phi} \nabla \times g$ for some arbitrary $g$.

You can try to proceed along the following lines: \begin{equation*} (\partial _{\mathbf{x}}+\mathbf{w})\cdot \mathbf{f}=0 \end{equation*} Special case $\mathbf{w}$ is constant. Then \begin{equation*} \exp [-\mathbf{w\cdot x}]\partial _{\mathbf{x}}\exp [+\mathbf{w\cdot x} ]=\partial _{\mathbf{x}}+\mathbf{w} \end{equation*} so \begin{eqnarray*} \exp [-\mathbf{w\cdot x}]\partial _{\mathbf{x}}\exp [+\mathbf{w\cdot x}% ]\cdot \mathbf{f} &=&0 \\ \partial _{\mathbf{x}}\cdot \{\exp [+\mathbf{w\cdot x}]\mathbf{f}\} &=&0 \\ \exp [+\mathbf{w\cdot x}]\mathbf{f} &\mathbf{=a}&+\partial _{\mathbf{x}% }\times \mathbf{b(x)} \\ \mathbf{f(x)} &=&\exp [-\mathbf{w\cdot x}]\{\mathbf{a}+\partial _{\mathbf{x}% }\times \mathbf{b(x)}\} \end{eqnarray*} In general you have to find $\mathbf{U}(\mathbf{x})$ ($3\times 3$ matrix) such that \begin{equation*} \mathbf{U}^{-1}(\mathbf{x})\cdot \partial _{\mathbf{x}}\cdot \mathbf{U}(% \mathbf{x})=\partial _{\mathbf{x}}+\mathbf{w(x}) \end{equation*} Let \begin{eqnarray*} \mathbf{U}(\mathbf{x}) &=&\exp [\mathbf{A(x)}] \\ \partial _{\mathbf{x}}\cdot \mathbf{U}(\mathbf{x}) &=&\mathbf{U}(\mathbf{x}% )[\partial _{\mathbf{x}}+\{\partial _{\mathbf{x}}\cdot \mathbf{A(x)\}]} \\ \partial _{\mathbf{x}}\cdot \mathbf{A(x)} &=&\mathbf{w(x}) \end{eqnarray*} Then \begin{eqnarray*} (\partial _{\mathbf{x}}+\mathbf{w(x)})\cdot \mathbf{f} &=&\mathbf{U}^{-1}(% \mathbf{x})\cdot \partial _{\mathbf{x}}\cdot \{\mathbf{U}(\mathbf{x})\cdot \mathbf{f}\}=0 \\ \partial _{\mathbf{x}}\cdot \{\mathbf{U}(\mathbf{x})\cdot \mathbf{f}\} &=&0 \\ \mathbf{U}(\mathbf{x})\cdot \mathbf{f(x)} &=&\{\mathbf{a}+\partial _{\mathbf{ x}}\times \mathbf{b(x)}\} \\ \mathbf{f(x)} &=&\mathbf{U}^{-1}(\mathbf{x})\cdot \{\mathbf{a}+\partial _{ \mathbf{x}}\times \mathbf{b(x)}\} \end{eqnarray*}

If $a+b+c=1$ and $abc>0$, then $ab+bc+ac<\frac{\sqrt{abc}}{2}+\frac{1}{4}.$ Question: For any $a,b,c\in \mathbb{R}$ such that $a+b+c=1$ and $abc>0$, show that $$ab+bc+ac<\dfrac{\sqrt{abc}}{2}+\dfrac{1}{4}.$$ My idea: let $$a+b+c=p=1, \quad ab+bc+ac=q,\quad abc=r$$ so that $$\Longleftrightarrow q<\dfrac{\sqrt{r}}{2}+\dfrac{1}{4}$$ Note this $a,b,c\in \mathbb{R}$, so we can't use schur inequality such $$p^3-4pq+9r\ge 0, \quad pq\ge 9r$$ and so on maybe can use AM-GM inequality to solve it.

Edit: Incomplete approach. Only works if $a,b,c\geq 0$. By replacing $\frac{1}{4}$ on the RHS with $\frac{(a+b+c)^2}{4}$, the inequality you seek is equivalent to $$ a^2+b^2+c^2+2\sqrt{abc}>2(ab+bc+ca).\tag{I} $$ To prove (I), we use the following result $$ a^2+b^2+c^2+3(abc)^{2/3}\geq 2(ab+bc+ca)\tag{II} $$ the proof of which can be found here. Because of (II), it is enough to verify now that $$ 2\sqrt{abc}>3(abc)^{2/3}\iff abc<\left(\frac{2}{3}\right)^6 $$ but this last inequality follows from the AM-GM inequality $$ \sqrt[3]{abc}\leq\frac{a+b+c}{3}=\frac{1}{3}\implies abc\leq\left(\frac{1}{3}\right)^3<\left(\frac{2}{3}\right)^6. $$

The area not covered by six pointed star In a circle with radius $r$, two equi triangles overlapping each other in the form of a six pointed star touching the circumference is inscribed! What is the area that is not covered by the star? Progress By subtracting area of the star from area of circle , the area of the surface can be found! But how to calculate the area of the star?

The length of a side of an equilateral triangle is $$\sqrt{r^2+r^2-2\cdot r\cdot r\cdot \cos (120^\circ)}=\sqrt 3r.$$ The distance between the center of the circle and each side of an equilateral triangle is $$\sqrt 3r\cdot \frac{\sqrt 3}{2}\cdot \frac{1}{3}=\frac 12r.$$ Hence, the length of a side of the smaller equilateral triangle, which is the 'corner' of the star, is $$\frac 12r\cdot \frac{2}{\sqrt 3}=\frac{1}{\sqrt 3}r.$$ Hence, the area of the star is $$\frac{\sqrt 3}{4}\cdot (\sqrt 3r)^2+3\times \frac{\sqrt 3}{4}\left(\frac{1}{\sqrt 3}r\right)^2=\sqrt 3r^2.$$ So, the answer is $$\pi r^2-\sqrt 3r^2=(\pi-\sqrt 3)r^2.$$

Prove that $\frac{1}{1^4}+\frac{1}{2^4}+\cdots+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ I have to show: $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ for natural $n$ I tried to show it by induction (but I think it could be possible to show it using some ineqaulity of means) so for $n=1$ we have $1=1$ so inequality holds then I assume it's true for $n$ and for $n+1$ my thesis is $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{(n+1)^4} \le 2-\frac{1}{\sqrt{n+1}}$ I know that: $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{(n+1)^4} \le 2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}$ but later I'm not sure if I have to show $\displaystyle2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}\le2-\frac{1}{\sqrt{n+1}}$ or should be $\ge$

You should show its: $\displaystyle2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}\le2-\frac{1}{\sqrt{n+1}}$ The other ways does not help your proof (think about it for a while)

Can someone help me to find a counter example that shows that $a \equiv b \mod m$ does not imply $(a+b)^m \equiv a^m +b^m \mod m$ Can someone help me to find a counter example that shows that $a \equiv b \mod m$ does not imply $(a+b)^m \equiv a^m +b^m \mod m$. I have tried many different values but I can't seem to find one. I tried to prove it but I also cannot do it. Thanks.

Take $a=1$, $b=2$ and $m=6$. $(a+b)^6 = a^6 + 6 a^5 b + 15 a^4 b^2 + 20 a^3 b^3 + 15 a^2 b^4 + 6 a b^5 + b^6$ $(a+b)^6 \equiv a^6 - a^3 b^3 + b^6 \bmod 3$ If $(a+b)^6 \equiv a^6 + b^6 \bmod 6$, then $3$ must divide $a$ or $b$. It is enough to take $a=1$ and $b=2$ to have $(a+b)^6 \not\equiv a^6 + b^6 \bmod 6$. In general, $(a+b)^6 \equiv a^6 + b^6 \bmod 6$ iff $2$ divides $ab(a+b)$ and $3$ divides $ab$.

Evaluating an indefinite integral $\int\sqrt {x^2 + a^2} dx$ indefinite integral $$\int\sqrt {x^2 + a^2} dx$$ After some transformations and different substitution, I got stuck at this $$a^2\ln|x+(x^2+a^2)| + \int\sec\theta\tan^2\theta d\theta$$ I am not sure I am getting the first step correct. Tried substituting $ x=a\tan \theta$ but that doesn't help either.

Here we have another way to see this: $$ \int \sqrt{x^2+a^2} dx $$ using the substitution $$ t=x+\sqrt{x^2+a^2}\\ \sqrt{x^2+a^2}=t-x $$ and squaring we have $$ a^2 =t^2-2tx\\ x=\frac{t^2-a^2}{2t}. $$ Finally we can use: $$ dx=\frac{2t(t)-(t^2-a^2)(1)}{2t^2}dt = \frac{t^2+a^2}{2t^2}dt\\ \sqrt{x^2+a^2}=t-\frac{t^2-a^2}{2t}=\frac{t^2+a^2}{2t}. $$ Thus: $$ \int \sqrt{x^2+a^2} dx = \int \frac{t^2+a^2}{2t} \frac{t^2+a^2}{2t^2}dt=\int \frac{(t^2+a^2)^2}{4t^3}dt $$ which is elementary, if we expand the square of the binomial: $$ \int\frac{t}{4}dt+\int\frac{a^2}{2t}dt+\int\frac{a^4}{4t^3}dt=\frac{t^2}{8}+a^2\ln\sqrt{t}-\frac{a^4}{16t^4}, $$ where as stated $t=x+\sqrt{x^2+a^2}.$

Question of trigonometry If $\cos^2 A=\dfrac{a^2-1}{3}$ and $\tan^2\left(\dfrac{A}{2}\right)=\tan^{2/3} B$. Then find $\cos^{2/3}B+\sin^{2/3}B $. I tried componendo and dividendo to write the second statement as cos A but i couldnt simplify it

$$\cos^{2/3}B+\sin^{2/3}B=\cos^{2/3}B\left(1+\tan^{2/3}B\right)=\cos^{2/3}B\left(1+\tan^2\frac{A}{2}\right)=\left(\cos^2B\right)^{1/3}\left(1+\tan^2\frac{A}{2}\right)=\left(\frac{1}{1+\tan^6\left(\frac{A}{2}\right)}\right)^{1/3}\left(1+\tan^2\frac{A}{2}\right)=\left(1+\frac{3\tan^2 \left(\frac{A}{2}\right)\left(1+\tan^2\left(\frac{A}{2}\right)\right)}{1+\tan^6\left(\frac{A}{2}\right)}\right)^{1/3}$$ I hope you can continue from here (you just need to find $\tan^2\frac{A}{2}$ as a function of $a$).

How do I solve $\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$? I'm having trouble finding this limit: $$\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$$ I tried multiplying by the conjugate: $$\lim_{x\to -\infty}(\frac{\sqrt{x^2 + x + 1} + x}{1} \times \frac{\sqrt{x^2 + x + 1} - x}{\sqrt{x^2 + x + 1} - x}) = \lim_{x\to -\infty}(\frac{x + 1}{\sqrt{x^2 + x + 1} - x})$$ And multiplying by $\frac{\frac{1}{x}}{\frac{1}{x}}$ $$\lim_{x\to -\infty}(\frac{x + 1}{\sqrt{x^2 + x + 1} - x} \times \frac{\frac{1}{x}}{\frac{1}{x}}) = \lim_{x\to -\infty}(\frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - 1})$$ That gives me $\frac{1}{0}$. WolframAlpha, my textbook, and my estimate suggest that it should be $-\frac{1}{2}$. What am I doing wrong? (Problem from the 2nd chapter of Early Transcendentals by James Stewart)

Little mistake: $$ \lim_{x\to -\infty}\frac{x + 1}{\sqrt{x^2 + x + 1} - x} \times \frac{\frac{1}{x}}{\frac{1}{x}} = \lim_{x\to -\infty}\frac{1 + \frac{1}{x}}{{\color{red}-}\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - 1}=-\frac{1}{2} $$

How find this integral $\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$ let $$D=\{(x,y)|y\ge x^3,y\le 1,x\ge -1\}$$ Find the integral $$I=\dfrac{1}{2}\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$$ My idea: $$I=\int_{0}^{1}dx\int_{x^3}^{1}(x^2y+2y^2)dy+\int_{-1}^{0}dx\int_{0}^{-x^3}(xy^2+2x+2y^2)dy$$ so $$I=\int_{0}^{1}[\dfrac{1}{2}x^2y^2+\dfrac{2}{3}y^3]|_{x^3}^{1}dx+\int_{-1}^{0}[\dfrac{1}{3}xy^3+2xy+\dfrac{2}{3}y^3]|_{0}^{-x^3}dx$$ $$I=\int_{0}^{1}[\dfrac{1}{2}x^2+\dfrac{2}{3}-\dfrac{1}{2}x^8-\dfrac{2}{3}x^9]dx+\int_{-1}^{0}[-\dfrac{1}{3}x^{10}-2x^4-\dfrac{2}{3}x^9]dx$$ so $$I=\dfrac{5}{6}-\dfrac{1}{18}-\dfrac{2}{30}+\dfrac{1}{33}+\dfrac{1}{10}-\dfrac{2}{30}=\dfrac{67}{90}$$ My question: my reslut is true? can you someone can use computer find it value? because I use Tom methods to find this reslut is $$\dfrac{79}{270}$$ which is true? so someone can use maple help me?Thank you

$\frac{67}{90}$ doesn't look correct. Here is what wolfram computes

Closed Form of Recursion $a_n = \frac{6}{a_{n-1}-1}$ Given that $a_0=2$ and $a_n = \frac{6}{a_{n-1}-1}$, find a closed form for $a_n$. I tried listing out the first few values of $a_n: 2, 6, 6/5, 30, 6/29$, but no pattern came out.

Start with $a_n=\frac6{a_{n-1}-1}$, and replace $a_{n-1}$ with $\frac6{a_{n-2}-1}$. We obtain $$a_n=\frac6{\frac6{a_{n-2}-1}-1}=\frac{6(1-a_{n-2})}{a_{n-2}-7}$$ Doing this again with $a_{n-2}=\frac6{a_{n-3}-1}$ and so forth, we get $$a_n=\frac{6(7-a_{n-3})}{7a_{n-3}-13}=\frac{6(13-7a_{n-4})}{13a_{n-4}-55}=\frac{6(55-13a_{n-5})}{55a_{n-5}-133}$$ It would seem then, that $$a_n=\frac{6(F_{k-1}-F_{k-2}a_{n-k})}{F_{k-1}a_{n-k}-F_k}$$ For some coefficients $F_k$. But what is $F_k$? To find out replace $a_{n-k}$ with $\frac6{a_{n-k-1}-1}$, then $$a_n=\frac{6(F_{k-1}-F_{k-2}a_{n-k})}{F_{k-1}a_{n-k}-F_k}=\frac{6(F_{k-1}-F_{k-2}\frac6{a_{n-k-1}-1})}{F_{k-1}\frac6{a_{n-k-1}-1}-F_k}=\frac{6((6F_{k-2}+F_{k-1})-F_{k-1}a_{n-k-1})}{F_{k}a_{n-k-1}-(6F_{k-2}+F_{k-1})}=\frac{6(F_{k}-F_{k-1}a_{n-k-1})}{F_{k}a_{n-k-1}-F_{k+1}}$$ So it would seem that $F_k=F_{k-1}+6F_{k-2}$. By noting that $F_1=1,F_2=7$, we have the solution (I presume you can solve it yourself) $$F_k=\frac25(-2)^k+\frac353^k$$ By plugging in $k=n$, we now have $$a_n=\frac{6(F_{n-1}-F_{n-2}a_{0})}{F_{n-1}a_{0}-F_n}=\frac{6(F_{n-1}-2F_{n-2})}{2F_{n-1}-F_n}$$ which eventually simplifies to $$a_n=3-\frac{5}{4\left(-\frac23\right)^n+1}$$ which fits both $a_0=2$ and $a_n=\frac6{a_{n-1}-1}$, so the expression is indeed correct.

How to simplify $(\sin\theta-\cos\theta)^2+(\sin\theta+\cos\theta)^2$? Simplify: $(\sin \theta − \cos \theta)^2 + (\sin \theta + \cos \theta)^2$ Answer choices: * *1 *2 *$ \sin^2 \theta$ *$ \cos^2 \theta$ I am lost on how to do this. Help would be much appreciated.

$$\begin{align} &\phantom{=}\left(\sin x-\cos x\right)^2+\left(\sin x+\cos x\right)^2\\ &=\sin^2x-2\sin x\cos x+\cos^2x+\sin^2x+2\sin x\cos x+\cos^2x\\ &=2\sin^2x+2\cos^2x\\ &=2\left(\sin^2x+\cos^2x\right)\\ &=2 \end{align}$$

Infinitely Many Circles in an Equilateral Triangle In the figure there are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles. I need to find the total area of the circles. I know this is going to have something to do with summation as a value approaches infinity, but I'm not entirely sure how to approach the problem. Here's what I have so far: I know that the radius of the central inscribed circle is $ \frac{\sqrt{3}}{6} $. As such, the area of the first circle is $$ A = \pi\left(\frac{\sqrt{3}}{6}\right)^2. $$ Because there are three "branches" of infinite circles, I'm assuming that the answer will look something like: $$ A = \pi\left(\frac{\sqrt{3}}{6}\right)^2 + 3 \sum_{1}^{\infty}\text{something}.$$

Look at the following figure carefully, As the triangle is equilateral ($AC$ is the angle bisector). So, $\angle ACD = 30^{\circ}$ $$\tan 30^{\circ} = \frac{AD}{DC} = 2AD\ (\because DC = 1/2) $$ $$\therefore AD = \frac{1}{2\sqrt{3}}$$ This is the radius of the bigger circle, let its area be $A_1$ $$\therefore A_1 = \frac{\pi}{12}$$ To calculate the radius of the next smaller circle (let it be $x$), please note that $$AC = \frac{1}{\sqrt{3}}$$ $$AB =\frac{1}{2\sqrt{3}} +x$$ $$\therefore BC = AC - AB =\frac{1}{2\sqrt{3}} -x $$ Note that triangles $ADC$ and $BCE$ are similar. $$\therefore \frac{AD}{AC} = \frac{BE}{BC}$$ $$\frac{1}{2\sqrt{3}} \times \sqrt{3} = x \times \left( \frac{2\sqrt{3}}{1-2\sqrt{3}x} \right)$$ $$\therefore x = \frac{1}{6\sqrt{3}}$$ Similarly we can find the radii of the next circles. They would be $\frac{1}{18\sqrt{3}}$, $\frac{1}{54\sqrt{3}}, ...$ Now, the main answer, The sequence $\frac{1}{6\sqrt{3}},\frac{1}{18\sqrt{3}}, \frac{1}{54\sqrt{3}}, ... $ can be generally written as $\frac{1}{6\sqrt{3}(3)^{n-1}}$ Total area of these circles, $$T = \frac{\pi}{12} + 3\sum_{n=1}^{\infty} \pi {\left(\frac{1}{6\sqrt{3}(3)^{n-1}} \right)}^2 $$ Notice that, $$\sum_{n=1}^{\infty} \pi {\left(\frac{1}{6\sqrt{3}(3)^{n-1}}\right)}^2 = \sum_{n=1}^{\infty} \pi {\left( \frac{1}{108}\right)}{\left(3^{-(n-1)}\right)}^2$$ =$$\sum_{n=1}^{\infty} \pi {\left( \frac{1}{108}\right)}{\left(3^{-2(n-1)}\right)}$$ =$$\sum_{n=1}^{\infty} \pi {\left( \frac{1}{108}\right)}{\left(\frac{1}{9}\right)}^{n-1}$$ This is a GP with $a = \frac{\pi}{108}$ and $r = \frac{1}{9}$ For infinite terms, the sum of this GP = $\frac{a}{1-r} = \frac{\pi}{96}$ Now, finally, $$T = \frac{\pi}{12} + 3 \times \frac{\pi}{96} = \frac{11\pi}{96}$$

If the product $(x+2)(x+3)(x+4)\cdots(x+9)(x+10)$ expands to $a_9x^9+a_8x^8+\dots+a_1x+a_0$, then what is the value of $a_1+a_3+a_5+a_7+a_9$? When expanded, the product $(x+2)(x+3)(x+4)\cdots(x+9)(x+10)$ can be written as $a_9x^9+a_8x^8+\dots+a_1x+a_0$. What is the value of $a_1+a_3+a_5+a_7+a_9$?

Let $P(x) = (x+2)(x+3)..(x+10)$. Then what is $\dfrac{P(1)-P(-1)}2=?$

finding a function from given function here is a function for: $f(x-\frac{\pi}{2})=\sin(x)-2f(\frac{\pi}{3})$ what is the $f(x)$? I calculate $f(x)$ as follows: $$\begin{align} x-\frac{\pi}{2} &= \frac{\pi}{3} \Rightarrow x= \frac{5\pi}{6} \\ f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6}-2f(\frac{\pi}{3}) \\ 3f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6} \\ f(\frac{\pi}{3}) &=(1/3)\sin\frac{5\pi}{6} \end{align}$$ $f(x)=(1/3)\sin\frac{5x}{2}$

Assuming $f$ is defined for all $x\in\mathbb{R}$. First, note that for any $x$, $$ f(x) = \sin\!\left(x+\frac{\pi}{2}\right)-2f\!\left(\frac{\pi}{3}\right) = \cos x -2f\!\left(\frac{\pi}{3}\right) $$ so it only remains to compute $f\!\left(\frac{\pi}{3}\right)$. From the expression above $$ f\!\left(\frac{\pi}{3}\right) = \cos \frac{\pi}{3} -2f\!\left(\frac{\pi}{3}\right) = \frac{1}{2} -2f\!\left(\frac{\pi}{3}\right) $$ and therefore rearranging the terms gives $f\!\left(\frac{\pi}{3}\right) = \frac{1}{6}$. Putting it all together, $$ \forall x\in \mathbb{R}, \quad f(x)=\cos x - \frac{1}{3}\;. $$ (It then only remains to check this expression satisfies the original functional equation, to be certain. It does; but even a quick sanity check for $x=0$, $x=\frac{\pi}{2}$ and $x=\pi$ will be enough to build confidence.)

Closed form of $\sum_{n=1}^{\infty}(-1)^{n+1} n (\log(n^2+1)-\log(n^2))$ How would you start computing this series? $$\sum_{n=1}^{\infty}(-1)^{n+1} n (\log(n^2+1)-\log(n^2))$$ One of the ways to think of would be Frullani integral with the exponential function , but it's troublesome due to the power of $n$ under logarithm. What else might I try?

I'm getting the same answer as you. $$ \sum_{n=1}^{\infty} (-1)^{n+1} n \log \left( \frac{n^{2}+1}{n^{2}}\right) = \sum_{n=1}^{\infty} (-1)^{n+1} n \int_{0}^{1} \frac{1}{n^{2}+x} \ dx$$ Then since $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}n}{n^{2}+x}$ converges uniformly on $[0,1]$, $$ \begin{align} &\sum_{n=1}^{\infty} (-1)^{n+1} n \int_{0}^{1} \frac{1}{n^{2}+x} \ dx \\ &= \int_{0}^{1} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}n}{n^{2}+x} \ dx \\ &= \frac{1}{4} \int_{0}^{1} \left[ \psi \left(\frac{i \sqrt{x}}{2} \right)- \psi \left(\frac{1}{2} + \frac{i \sqrt{x}}{2} \right) + \psi \left(-\frac{ i \sqrt{x}}{2} \right)- \psi \left(\frac{1}{2} - \frac{i \sqrt{x}}{2} \right) \right] \ dx .\end{align}$$ The above series can be derived by working backwards and using the series representation of the digamma function (14) . Now let $t = \sqrt{x}$. Then $$ \begin{align} &\sum_{n=1}^{\infty} (-1)^{n+1} n \log \left(\frac{n^{2}+1}{n^{2}} \right) \\&= \frac{1}{2} \int_{0}^{1} \left[t \psi \left(\frac{it}{2} \right) - t \psi \left(\frac{1}{2} + \frac{i t}{2} \right) + t \psi \left(-\frac{it}{2} \right) - t \psi \left(\frac{1}{2} - \frac{it}{2} \right) \right] \ dt . \end{align}$$ And integrating by parts, $$ \begin{align} &\sum_{n=1}^{\infty} (-1)^{n+1} n \log \left(\frac{n^2+1}{n^{2}} \right) \\ &= \frac{1}{2} \Bigg[ 4 \psi^{(-2)} \left(\frac{it}{2} \right) - 2i t \log \Gamma \left( \frac{it}{2}\right) - 4 \psi^{(-2)} \left(\frac{1}{2}+ \frac{it}{2} \right) + 2i t \log \Gamma \left( \frac{1}{2} + \frac{it}{2}\right) \\ &+ 4 \psi^{(-2)} \left(-\frac{it}{2} \right) + 2i t \log \Gamma \left(- \frac{it}{2}\right) - 4 \psi^{(-2)} \left(\frac{1}{2} -\frac{it}{2} \right) - 2i t \log \Gamma \left(\frac{1}{2} - \frac{it}{2}\right)\Bigg] \Bigg|^{1}_{0} \\ &= 2 \psi^{(-2)} \left(\frac{i}{2} \right) - i \log \Gamma \left(\frac{i}{2} \right) - 2 \psi^{(-2)} \left(\frac{1}{2} + \frac{i}{2} \right) + i \log \Gamma \left(\frac{1}{2} + \frac{i}{2} \right) + 2 \psi^{(-2)} \left(-\frac{i}{2} \right) \\ &+ i \log \Gamma \left(-\frac{i}{2} \right) - 2 \psi^{(-2)} \left(\frac{1}{2} - \frac{i}{2} \right) - i \log \Gamma \left(\frac{1}{2} - \frac{i}{2} \right) + 4 \psi^{(-2)} \left(\frac{1}{2} \right). \end{align}$$ In terms of simplification, $\psi^{(-2)} \left( \frac{1}{2}\right)$ can be expressed in terms of the Glaisher-Kinkelin constant. And further simplification is possible using the Schwarz reflection principle. $$\begin{align} \sum_{n=1}^{\infty} (-1)^{n+1} n \log \left(\frac{n^{2}+1}{n^{2}} \right) &= 4 \ \text{Re} \ \psi^{(-2)} \left(\frac{i}{2} \right) -4 \ \text{Re} \ \psi^{(-2)} \left(\frac{1}{2} + \frac{i}{2} \right) + 2 \ \text{Im} \ \log \Gamma \left( \frac{i}{2}\right) \\ &-2 \ \text{Im} \ \log \Gamma \left(\frac{1}{2} + \frac{i}{2} \right) + 6\log A + \frac{5}{6} \log 2 + \log \pi \end{align}$$ which according to Wolfram Alpha is approximately $0.4277662568$

Prove $\frac{n^2+2}{(2 \cdot n^2)-1} \to \frac{1}{2}$ Prove $\frac{n^2+2}{(2 \cdot n^2)-1} \to \frac{1}{2}$ for $n \to \infty$. I've been looking at this for hours! Also, sorry I don't have the proper notation. This is where I'm at: $$ \left| \frac{n^2 + 2}{2 \cdot n^2 - 1} - \frac{1}{2}\right| = \left| \frac{5}{4 \cdot n^2 - 2} \right| $$ I thought I was supposed to get to the point where I can say $1/n <n$, but I can only get to $1/n-1$ so that can't be the right approach or I'm missing something. A friend says to make $n > \frac{5}{\varepsilon^2}$ but I''m not sure what to do with that tip. Any help would be greatly appreciated!

$\left|\frac{n^2+2}{2n^2-1} - \frac {1}{2}\right| = \frac{5}{4n^2-2}$ Let $\epsilon>0$ be given. Let $n_0$ be the smallest integer such that $n\geq n_0> \sqrt{\frac{5}{4\epsilon} + \frac 12}$. Equivalently, $\epsilon>\frac{5}{4n^2-2}$. Thus, $\left|\frac{n^2+2}{2n^2-1} - \frac {1}{2}\right|< \epsilon$, for all $n\geq n_0$.

3 Variable Diophantine Equation Find all integer solutions to $$x^4 + y^4 + z^3 = 5$$ I don't know how to proceed, since it has a p-adic and real solution for all $p$. I think that the only one is (2, 2, -3) and the trivial ones that come from this, but I can't confirm it.

After a careful investigation I present some results which might be helpful in the final resolution of this problem. Lets start with the original equation i.e. $$x^4+y^4+z^3=5$$ It is easy to see that there is no solution of this equation where $x,y$ and $z$ are all positive. Second if $(x,y,z)$ is a solution then so is $(-x,-y,z)$. From this observations we get that $z<0$. Let rewrite the equation in terms of positive values i.e. $$x^4+y^4-z^3=5$$ where $x,y$ and $z$ are all positive. This is equivalent to $$x^4+y^4=5+z^3$$ From this, one gets $$x^4+y^4\equiv z^3\mod(5)$$ First we show that $x\equiv 0\mod(5)$ and $y\equiv 0\mod(5)$ iff $z\equiv 0\mod(5)$. If $x\equiv 0\mod(5)$ and $y\equiv 0\mod(5)$ then $$x^4+y^4\equiv0\mod(5)\Rightarrow 5+z^3\equiv z^3\equiv0\mod(5)\Rightarrow z\equiv0\mod(5)$$ Now let $z\equiv0\mod(5)$ then $$5+z^3\equiv0\mod(5)\Rightarrow x^4+y^4\equiv0\mod(5)$$ However for any $x\in\mathbb{Z}$ one has $x^4\equiv0\mod(5)$ or $x^4\equiv1\mod(5)$. Therefore $$x^4+y^4\equiv0\mod(5)\Leftrightarrow x\equiv y\equiv0\mod(5)$$ However an inspection of our modified equation yields that we can not have simultaneously $x$ and $y$ divisible by $5$ for otherwise $z\equiv0\mod(5)$ and $$x^4+y^4\equiv0\mod(5^3)\Rightarrow 5+z^3\equiv0\mod(5^3)$$ which is impossible as $5+z^3\equiv 5\mod(5^3)$. Knowing that $x$ and $y$ can not be simultaneously divisible by $5$ then $z$ is not divisible by $5$ either. Applying Fermat's little theorem we can rewrite the modified equation as $$zx^4+zy^4\equiv1\mod(5)$$ Let say without loss of generality $y\equiv0\mod(5)$ and $x^4\equiv1\mod(5)$ then $$z\cdot 1+z\cdot 0\equiv1\mod(5)\Rightarrow z\equiv1\mod(5)$$ The other exhaustive case would be $x^4\equiv1\mod(5)$ and $y^4\equiv1\mod(5)$ in which case $$z\cdot 1+z\cdot 1\equiv1\mod(5)\Rightarrow 2z\equiv1\mod(5)\Rightarrow z\equiv3\mod(5)$$ In this case a direct inspection for $z=3$ would yield $x=\pm 2$ and $y=\pm 2$.

2nd order homogeneous ODE I am trying to solve a system of differential equations (for the full system see below) and I am stuck the following 2nd order ODE (with $a$ and $b$ being constants and $\dot x = \frac{dx}{dt}$): $$\ddot x - \frac34 a\dot x^2 -b \dot x + 2 a b = 0$$ I tried to substitute $v := \dot x$, which leads me to a different ODE which I was not able to solve. And now I have no idea what else I could try here and would love to get some pointers. And here the original system / IVP I want to solve: $$ \dot x = 2 a + 2 e^{ax}y \\ \dot y = \frac32 e^{-ax}a^3+(a^2+b)y-\frac12 a e^{ax}y^2 \\ x(0)=0 \\ y(T)=0$$ In which I solved the first equation for $y$ and substituted $y$ and $\dot y$ in the 2nd which gave me the equation above.

As I said in a comment, start defining $z=x'$; so the differential equation becomes $$\frac{dz}{dt} - \frac34 a z^2 -b z + 2 a b = 0$$ that is to say $$\frac{dz}{dt} = \frac34 a z^2 +b z - 2 a b $$ then, as Semsem suggested, it is separable; so $$\frac{dt}{dz} = \frac{1}{\frac34 a z^2 +b z - 2 a b}$$ so $$t+C=-\frac{2 \tanh ^{-1}\left(\frac{3 a z+2 b}{2 \sqrt{b} \sqrt{6 a^2+b}}\right)}{\sqrt{b} \sqrt{6 a^2+b}}$$ from which $$z=-\frac{2 \left(\sqrt{b} \sqrt{6 a^2+b} \tanh \left(\frac{1}{2} \sqrt{b} \sqrt{6 a^2+b} (C+t)\right)+b\right)}{3 a}$$ Now, one more unpleasant integration leads to $$x=-\frac{2 \left(2 \log \left(\cosh \left(\frac{1}{2} \sqrt{b} \sqrt{6 a^2+b} (C+t)\right)\right)+b t\right)}{3 a}+D$$

How do I solve this geometric series I have this geometric series $2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$to solve. So I extract the number two and get $2(\frac{1}{2}^0+ \frac{1}{2}^1+...+ \frac{1}{2}^7)$ I use the following formula $S_n= \frac{x^{n+1}-1}{x-1}$ so I plug in the values in this formula and get $S_n= 2\frac{\frac{1}{2}^{7+1}-1}{\frac{1}{2}-1}$ but the result is not correct. What did I do wrong? Thanks!!

Hint: Its $$2 + (1+\frac12+ \frac14+ \cdots + \frac{1}{128})$$ not multiplied with $2$. You can also think of it as follows: The first term is $a_1=2$ and the common ratio is $r=1/2$ and then you sum it using the formula where you last term is $a_9=1/128$. Edit: If you do want to factor out a $2$, then you get $$2(1+\frac12 + \frac14 + \cdots + \frac{1}{256})= 2(\frac{1}{2^0} + \frac{1}{2^1} + \frac{1}{2^2} + \cdots + \frac{1}{2^8})$$

Finding cartesian equation for trigonometric parametric forms I'm trying to find the cartesian equation for these parameteric forms: $$ x = sin\theta + 2 cos \theta \\ y = 2 sin\theta + cos\theta $$ I tried: $$\begin{align} x^2 & = sin^2\theta + 4cos^2\theta \\ & = 1 - cos^2\theta + 4cos^2\theta \\ & = 1 + 3cos^2\theta \\ \\ y^2 & = 4sin^2\theta+cos^2\theta\\ & = 4(1 - cos^2\theta) + cos^2\theta \\ & = 4 - 3cos^2\theta \\ \\ \therefore & \space4 - y^2 = x^2 - 1\\ \space & x^2 + y^2 = 5 \end{align}$$ Which differs from the given answer of $5x^2 + 5y^2 - 8xy = 9$. Where am I going wrong?

In general, $(a+b)^2\ne a^2+b^2$ unless $ab=0$ Solve for $\sin\theta,\cos\theta$ in terms of $x,y$ Then use $\sin^2\theta+\cos^2\theta=1$ to eliminate $\theta$

Finding Solutions to Trigonometric Equation Find all $x$ in the interval (0, $\frac{\pi}{2}$) such that $$\frac{\sqrt{3}-1}{\sin x} + \frac{\sqrt{3}+1}{\cos x} = 4\sqrt{2}.$$

Rewrite it in the form $$2\sqrt2\left(\frac{\sqrt3-1}{2\sqrt2}\cos x+\frac{\sqrt3+1}{2\sqrt2}\sin x\right)=2\sqrt2\sin 2x.$$ For $\phi=\arcsin\frac{\sqrt3-1}{2\sqrt2}$ it implies $$\sin(x+\phi)=\sin 2x,$$ i.e. $x+\phi=2x+2\pi n$ or $x+\phi=\pi-2x+2\pi n$, $n\in\Bbb Z$. Therefore, the only solutions in $(0,\pi/2)$ are $\phi$ and $\frac{\pi-\phi}{3}$.

$\int\frac{2x+1}{x^2+2x+5}dx$ by partial fractions $$\int\frac{2x+1}{x^2+2x+5}dx$$ I know I'm supposed to make the bottom a perfect square by making it $(x+1)^2 +4$ but I don't know what to do after that. I've tried to make $x+1= \tan x$ because that's what we did in a class example but I keep getting stuck.

$$ \int\frac{2x+1}{x^2+2x+5}dx $$ I would first write $w=x^2+2x+5$, $dw=(2x+2)\,dx$, and then break the integral into $$ \int\frac{2x+2}{x^2+2x+5}dx + \int\frac{-1}{x^2+2x+5}dx. $$ For the first integral I would use that substitution. Then $$ \overbrace{\int\frac{-dx}{x^2+2x+5} = \int\frac{-dx}{(x+1)^2 + 2^2}}^{\text{completing the square}} = \frac{-1}2\int \frac{dx/2}{\left(\frac{x+1}{2}\right)^2+1} = \frac{-1}2 \int\frac{du}{u^2+1}. $$ Then we get an arctangent.

Convergence of $\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$ Does the series $$\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$$ converges? My attempt: Since the ratio test is inconclusive, my idea is to use the Stirling Approximation for n! $$\frac{(2n)!}{n!n!4^n} \sim (\frac{1}{4^n} \frac{\sqrt{4\pi n}(\frac{2n}{e})^{2n}}{\sqrt{2 n \pi} \sqrt{2n \pi} (\frac{n}{e})^{2n}} =\frac{(2)^{2n}}{4^n \sqrt{n \pi}}$$ The series of the secomd term diverges. It is correct to conclude thatthe series diverges? Another ideas are welcome! Thanks

A much simpler way: $$ \frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{4(n+1)(n+1)}=\frac{2n+1}{2n+2}\ge \sqrt{\frac{n}{n+1}}, $$ since $$ \left(\frac{2n+1}{2n+2}\right)^2=\left(1-\frac{1}{2n+2}\right)^2\ge 1-\frac{2}{2n+2} =\frac{n}{n+1}, $$ and hence $$ a_n=a_1\prod_{k=2}^n\frac{a_{k}}{a_{k-1}}\ge a_1\prod_{k=2}^n\sqrt{\frac{k-1}{k}} =\frac{a_1}{\sqrt{n}}, $$ and hence $$ \sum_{n=1}^\infty a_n=\infty. $$ Note that $$ \frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{4(n+1)(n+1)}=\frac{2n+1}{2n+2}=1-\frac{1}{2n+2}=1-\frac{1}{2n}+{\mathcal O}\left(\frac{1}{n^2}\right). $$ This series diverges due to Gauss Test otherwise known as Raabe's Test.

Given $a+b+c=4$ find $\max(ab+ac+bc)$ $a+b+c = 4$. What is the maximum value of $ab+ac+bc$? Could this be solved by a simple application of Jensen's inequality? If so, I am unsure what to choose for $f(x)$. If $ab+ac+bc$ is treated as a function of $a$ there seems no easy way to express $bc$ in terms of $a$. EDIT: The context of the question is maximising the surface area of a rectangular prism. Also I might have misinterpreted the question, because it says "the sum of the length of the edges (side lengths are a,b,c) is 4", and gives the options $\frac1{3}, \frac{2}{3}, 1, \frac{4}{3}$. Otherwise, how would this be done?

Jensen's Inequality gives $$ \left[\frac13(a+b+c)\right]^2\le\frac13(a^2+b^2+c^2) $$ and we know that $$ \begin{align} ab+bc+ca &=\frac12\left[(a+b+c)^2-(a^2+b^2+c^2)\right]\\ &\le\frac12\left[(a+b+c)^2-\frac13(a+b+c)^2\right]\\ &=\frac13(a+b+c)^2 \end{align} $$ and equality can be achieved when $a=b=c$. Therefore, if $a+b+c=4$, the maximum of $ab+bc+ca$ is $\frac{16}{3}$ which can be achieved if $a=b=c=\frac43$.

Integration without substitution How to i integrate this with out substitutions or Partial fraction decomposition ? ($3x^2$+$2$)/[$x^6$($x^2$+1)] I've got to : 2/x^6(x^2+1),but after this i haven't been able to eliminate the 2.

You can use $\displaystyle\frac{3x^2+2}{x^6(x^2+1)}=\frac{2(x^2+1)}{x^6(x^2+1)}+\frac{x^2}{x^6(x^2+1)}=\frac{2}{x^6}+\frac{1}{x^4(x^2+1)}$ $\displaystyle=\frac{2}{x^6}+\left(\frac{x^2+1}{x^4(x^2+1)}-\frac{x^2}{x^4(x^2+1)}\right)=\frac{2}{x^6}+\frac{1}{x^4}-\frac{1}{x^2(x^2+1)}$ $\displaystyle=\frac{2}{x^6}+\frac{1}{x^4}-\left(\frac{x^2+1}{x^2(x^2+1)}-\frac{x^2}{x^2(x^2+1)}\right)=\frac{2}{x^6}+\frac{1}{x^4}-\frac{1}{x^2}+\frac{1}{x^2+1}.$ [Notice that this gives the partial fraction decomposition of this particular function.]

Find the value of $\sqrt{10\sqrt{10\sqrt{10...}}}$ I found a question that asked to find the limiting value of $$10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$$If you make the substitution $x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$ it simplifies to $x=10\sqrt{x}$ which has solutions $x=0,100$. I don't understand how $x=0$ is a possible solution, I know that squaring equations can introduce new, invalid solutions to equations and so you should check the solutions in the original (unsquared) equation, but doing that here doesn't lead to any non-real solutions or contradictions. I was wondering if anyone knows how $x=0$ turns out as a valid solution, is there an algebaric or geometric interpretation? Or is it just a "special case" equation? A similar question says to find the limiting value of $\sqrt{6+5\sqrt{6+5\sqrt{6+5\sqrt{...}}}}$, and making a similar substituion for $x$ leads to $$x=\sqrt{6+5x}$$ $$x^2=6+5x$$ which has solutions $x=-1,6$. In this case though, you could substitute $x=-1$ into the first equation, leading to the contradiction $-1=1$ so you could satisfactorily disclude it. Is there any similar reasoning for the first question? I know this might be a stupid question but I'm genuinely curious :)

Denote the given problem as $x$, then \begin{align} x&=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}}}\\ &=10\cdot10^{\large\frac{1}{2}}\cdot10^{\large\frac{1}{4}}\cdot10^{\large\frac{1}{8}}\cdot10^{\large\frac{1}{16}}\cdots\\ &=10^{\large1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots}\\ &=10^{\large y} \end{align} where $y$ is an infinite geometric series in which its value is \begin{align} y &=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots\\ &=\frac{1}{1-\frac{1}{2}}\\ &=2 \end{align} Therefore \begin{equation} x=10^{\large 2}=100 \end{equation}

Prove that $ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x $ In a related question the following integral was evaluated $$ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{\mathrm{d}x/2}{1 + \cos x \sin x} =\int_0^{2\pi} \frac{\mathrm{d}x/2}{2 + \sin x} \,\mathrm{d}x =\int_{-\infty}^\infty \frac{\mathrm{d}x/2}{1+x+x^2} $$ I noticed something interesting, namely that $$ \begin{align*} \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x & = \int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\ & = \int_0^{\pi} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x = \int_0^{\pi} \frac{(\sin x)^2}{1 - \cos x \sin x} \,\mathrm{d}x \end{align*} $$ The same trivially holds if the upper limits are changed to $\pi/2$ as well ($x \mapsto \pi/2 -u$). But I had problems proving the first equality. Does anyone have some quick hints?

$$ \frac{\cos(\pi/2-x)^2}{1+\cos(\pi/2-x)\sin(\pi/2-x)} = \frac{\sin(x)^2}{1+\sin(x)\cos(x)} $$

Any shorter way to solve trigonometric problem? If $10 \sin^4\theta + 15 \cos^4 \theta=6$, then find value of $27 \csc^2 \theta + 8\sec^2 \theta$ I know the normal method o solve this problem in which we need to multiply L.H.S. of $10 \sin^4\theta + 15 \cos^4 \theta=6$ by $(\sin^2\theta + \cos \theta^2)^2$ and then simplifying it. It is kind of simplification in which we can commit silly mistakes. So, is there any shorter way to solve this trigonometric problem?

How can $20\sin^2\theta + 15\cos^2\theta = 6$? $20\sin^2\theta + 15 \cos^2\theta = 15 + 5\sin^2\theta$, implying that $5 \sin^2\theta = -9$, but $\sin^2\theta$ should be nonnegative.

Sum the series (real analysis) $$\sum_{n=1}^\infty {1 \over n(n+1)(n+2)(n+3)(n+4)}$$ I tried to sum the above term as they way I can solve the term $\sum_{n=1}^\infty {1 \over (n+3)}$ by transforming into ${3\over n(n+3)} ={1\over n}-{1\over(n+3)}$ but I got stuck while trying to transform $12\over n(n+1)(n+2)(n+3)(n+4)$ into something solvable.

By working it out, by various methods, it is seen that \begin{align} \frac{1}{n(n+1)(n+2)(n+3)(n+4)} = \frac{1}{4!} \, \sum_{r=0}^{4} (-1)^{r} \binom{4}{r} \, \frac{1}{n+r} \end{align} Now, \begin{align} S &= \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)(n+3)(n+4)} \\ &= \frac{1}{4!} \, \sum_{r=0}^{4} (-r)^{r} \binom{4}{r} \, \sum_{n=1}^{\infty} \frac{1}{n+r} \\ &= \frac{1}{4!} \left[ \zeta(1) - 4 \left( \zeta(1) - 1\right) + 6 \left( \zeta(1) - 1 - \frac{1}{2} \right) - 4 \left( \zeta(1) - 1 - \frac{1}{2} - \frac{1}{3} \right) + \left(\zeta(1) - 1 - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} \right) \right] \\ &= \frac{1}{4!} \left[ 4 - 6 \left( 1 + \frac{1}{2} \right) + 4 \left( 1 + \frac{1}{2} + \frac{1}{3} \right) - \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) \right] \\ S &= \frac{1}{4 \cdot 4!} \end{align}

Sums with squares of binomial coefficients multiplied by a polynomial It has long been known that \begin{align} \sum_{n=0}^{m} \binom{m}{n}^{2} = \binom{2m}{m}. \end{align} What is being asked here are the closed forms for the binomial series \begin{align} S_{1} &= \sum_{n=0}^{m} \left( n^{2} - \frac{m \, n}{2} - \frac{m}{8} \right) \binom{m}{n}^{2} \\ S_{2} &= \sum_{n=0}^{m} n(n+1) \binom{m}{n}^{2} \\ S_{3} &= \sum_{n=0}^{m} (n+2)^{2} \binom{m}{n}^{2}. \end{align}

HINT: As $\displaystyle n\binom mn=m\binom{m-1}{n-1},$ $\displaystyle\sum_{n=0}^mn\binom mn^2=m\sum_{n=0}^m\binom mn\binom{m-1}{n-1}=m\binom{2m-1}m$ comparing the coefficient of $x^{2m-1}$ in $(1+x)^m(1+x)^{m-1}=(1+x)^{2m-1}$

Prove by contradiction that $(x-y)^3+(y-z)^3+(z-x)^3 = 30$ has no integer solutions By factorizing it I found that $(x-y)(y-z)(z-x) = 10$

Just as Karvens comments: Let $a=x-y$ and $b=y-z$. Then $-(a+b)=z-x$. Clearly, $a,b,c \in \Bbb Z$. So $$30=a^3+b^3-(a+b)^3=(a+b)(a^2-ab+b^2)-(a+b)^3=-3ab(a+b)$$ And hence $$10=-ab(a+b).$$ Therefore $a=\pm 1$ or $a=\pm 2$ or $a=\pm 5$ or $a=\pm 10$. Claerly they cannot be the solution.

Distance between powers of 2 and 3 As we know $3^1-2^1 = 1$ and of course $3^2-2^3 = 1$. The question is that whether set $$ \{\ (m,n)\in \mathbb{N}\quad |\quad |3^m-2^n| = 1 \} $$ is finite or infinite.

Note first that if $3^{2n}-1=2^r$ then $(3^n+1)(3^n-1)=2^r$. The two factors in brackets differ by $2$ so one must be an odd multiple of $2$, and this is only possible if $n=1$ (the only odd number we can allow in the factorisation is $1$) Now suppose that $3^n-1=2^r$ and $n$ is odd. Now $3^n\equiv -1$ mod $4$ so $3^n-1$ is not divisible by $4$. Now suppose $3^n=2^{2r}-1=(2^r+1)(2^r-1)$. The two factors differ by $2$ and cannot therefore both be divisible by $3$. Only $r=1$ is possible. The final case is $3^n=2^{k}-1$ where $k$ is odd. Now the right hand side is $\equiv -2$ mod $3$, so only $k=1$ is possible, and $n=0$ (if permitted). __ The previous version of this answer was overcomplicated - trying to do things in a hurry.

Calculation of $\int\frac1{\tan \frac{x}{2}+1}dx$ Calculation of $\displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$ $\bf{My\; Try}::$ Let $\displaystyle I = \displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$, Now let $\displaystyle \tan \frac{x}{2}=t\;,$ Then $\displaystyle dx=\frac{2}{1+t^2}dt$ So $\displaystyle I = 2\int\frac{1}{(1+t)\cdot (1+t^2)}dt$ Now Using Partial fraction, $\displaystyle \frac{1}{(1+t)\cdot (1+t^2)} = \frac{A}{1+t}+\frac{Bt+C}{1+t^2}\Rightarrow 1=A(1+t^2)+(Bt+C)(1+t)$ Now put $(1+t)=0\Rightarrow t=-1\;,$ We get $\displaystyle 1=2A\Rightarrow A = \frac{1}{2}.$ Now Put $(1+t^2)=0\Rightarrow t^2=-1\;,$ We Get $1=Bt^2+(B+C)t+C$ So $\displaystyle 1=\left(-B+C\right)+(B+C)$. So Solving equation...$B+C=0$ and $-B+C=1$ So $\displaystyle B=-\frac{1}{2}$ and $\displaystyle C=\frac{1}{2}$ So $\displaystyle I = 2\int\frac{1}{(1+t)\cdot (1+t^2)}dt = \int\frac{1}{1+t}dt+\int\frac{-t+1}{1+t^2}dt$ So $\displaystyle I = \frac{1}{1+t}dt-\frac{1}{2}\int\frac{2t}{1+t^2}dt+\int \frac{1}{1+t^2}dt$ So $\displaystyle I = \ln \left|1+t\right|-\frac{1}{2}\ln \left|1+t^2\right|+\tan^{-1}(t)+\mathcal{C}$ So $\displaystyle I = \ln \left|1+\tan \frac{x}{2}\right|-\frac{1}{2}\ln \left|1+\tan^2 \frac{x}{2}\right|+\frac{x}{2}+\mathcal{C}$ Can we solve it without using partial fraction? If yes then please explain to me. Thanks

Hint: Multiply the denominator and nominator of the fraction by $\cos x/2(\sin x/2-\cos x/2)$. At the bottom you get $-\cos x$, on the top you get $-\cos^2 x/2+1/2\sin x$. What can you do with $\cos^2 x/2$?

How to derive the closed form of this recurrence? For the recurrence, $T(n) = 3T(n-1)-2$, where $T(0)= 5$, I found the closed form to be $4\cdot 3^n +1$(with help of Wolfram Alpha). Now I am trying to figure it out for myself. So far, I have worked out: $T(n-1) = 3T(n-2)-2, T(n-2) = 3T(n-3)-2, T(n-3) = 3T(n-4)-2$ leading me to: $T(n)=81\cdot T(n-4)-54-18-6-2 $ etc. I have noticed the constants follow a Geometric Series whose sum is given by $1-3^n$. I am having trouble putting this together to end up with the final closed form.

By making use of the generating function method the difference equation $T_{n} = 3 T_{n-1} - 2$, $T_{0} = 5$, can be seen as follows \begin{align} T(x) &= \sum_{n=0}^{\infty} T_{n} x^{n} = 3 \sum_{n=0}^{\infty} T_{n-1} x^{n} - 2 \sum_{n=0}^{\infty} x^{n} \\ T(x) &= 3 \left( T_{-1} + \sum_{n=1}^{\infty} T_{n-1} x^{n} \right) - \frac{2}{1-x} \\ &= 3 \left( T_{-1} + x T(x) \right) - \frac{2}{1-x} \end{align} which yields \begin{align} (1-3x) T(x) = 3 T_{-1} - \frac{2}{1-x}. \end{align} Since $T_{-1} = \frac{7}{3}$ then \begin{align} T(x) &= \frac{5 - 7x}{(1-x) (1-3x)} = \frac{1}{1-x} + \frac{4}{1-3x} \\ &= \sum_{n=0}^{\infty} \left( 4 \cdot 3^{n} + 1 \right) x^{n} \end{align} and provides $T_{n} = 4 \cdot 3^{n} + 1$.

Integral of $\int \frac{\cos \left(x\right)}{\sin ^2\left(x\right)+\sin \left(x\right)}dx$ What is the integral of $\int \frac{\cos \left(x\right)}{\sin ^2\left(x\right)+\sin \left(x\right)}dx$ ? I understand one can substitute $u=\tan \left(\frac{x}{2}\right)$ and one can get (1) $\int \frac{\frac{1-u^2}{1+u^2}}{\left(\frac{2u}{1+u^2}\right)^2+\frac{2u}{1+u^2}}\frac{2}{1+u^2}du$ but somehow this simplifies to (2) $\int \frac{1}{u}-\frac{2}{u+1}du$ to get: (3) $\ln \left(\tan \left(\frac{x}{2}\right)\right)-2\ln \left(\tan \left(\frac{x}{2}\right)+1\right)+C$ as the final answer. But how does one simplify (1) to (2)?

Consider the integral \begin{align} I = \int \frac{\cos(x) \, dx}{\sin^{2}(x) + \sin(x) } . \end{align} Method 1 Make the substitution $u = \tan\left(\frac{x}{2}\right)$ for which \begin{align} \cos(x) &= \frac{1-u^{2}}{1+u^{2}} \\ \sin(x) &= \frac{2u}{1+u^{2}} \\ dx &= \frac{2 \, du}{1+u^{2}} \end{align} and the integral becomes \begin{align} I &= \int \frac{ \frac{1-u^{2}}{1+u^{2}} \cdot \frac{2 \, du}{1+u^{2}} }{ \left( \frac{2u}{1+u^{2}} \right)^{2} + \frac{2u}{1+u^{2}} } \\ &= \int \frac{2(1-u^{2})}{(1+u^{2})^{2}} \frac{ (1+u^{2})^{2} }{2u} \, \frac{du}{2u+ 1 + u^{2} } \\ &= \int \frac{(1-u^{2})^{2} \, du}{ u (1+u)^{2} } = \int \frac{(1-u) \, du}{u(1+u)} \\ &= \int \left( \frac{1}{u} - \frac{2}{1+u} \right) \, du\\ &= \ln(u) - 2 \ln(1+u) + c_{1}. \end{align} This leads to \begin{align} \int \frac{\cos(x) \, dx}{\sin^{2}(x) + \sin(x) } = \ln \left( \frac{\tan\left(\frac{x}{2}\right)}{ \left( 1 + \tan\left(\frac{x}{2}\right) \right)^{2} } \right) + c_{1} \end{align} Method 2 Make the substitution $u = \sin(x)$ to obtain \begin{align} I &= \int \frac{du}{u^{2} + u} = \int \left( \frac{1}{u} - \frac{1}{1+u} \right) \, du \\ &= \ln(u) - \ln(1+u) + c_{2} \end{align} for which \begin{align} \int \frac{\cos(x) \, dx}{\sin^{2}(x) + \sin(x) } = \ln\left( \frac{\sin(x)}{1+\sin(x) } \right) + c_{2} \end{align}

Can't get this implicit differentiation I've been working at this implicit differentiation problem for a little over an hour now, and I, nor my friends can figure it out. The question reads "Find the equation of the tangent line to the curve (a lemniscate) $2(x^2+y^2)^2=25(x^2−y^2)$ at the point (3,1). Write the equation of the tangent line in the form $y=mx+b.$" Every time that we do it we get a ridiculous number for the slope (${150}/{362}$)

We have $2(x^2+y^2)^2=25(x^2−y^2)$. Since $(x^2)' = 2 x \ dx$, I would differentiate this as $2(2(x^2+y^2)(x^2+y^2)') =25(2x\ dx - 2y\ dy) $ or $4(x^2+y^2)(2x\ dx-2 y\ dy) =25(2x\ dx - 2y\ dy) $ or $8(x(x^2+y^2)dx-y(x^2+y^2)dy) =50x\ dx-50y\ dy $ or $(8x(x^2+y^2)-50x)dx =(8y(x^2+y^2)-50y)dy $ or $\dfrac{dy}{dx} =\dfrac{8x(x^2+y^2)-50x}{8y(x^2+y^2)-50y} $. Putting $x=3$ and $y=1$, $\begin{array}\\ \dfrac{dy}{dx} &=\dfrac{8x(x^2+y^2)-50x}{8y(x^2+y^2)-50y}\\ &=\dfrac{24(9+1)-150}{8(8+1)-50}\\ &=\dfrac{90}{22}\\ &=\dfrac{45}{11}\\ \end{array} $. If the line is $y = mx+b$, $m = 45/11$ and $b = y-mx =1-3m =1-135/11 =-124/11 $. All errors are (obviously) my fault, and I will accept all appropriate punishments.

Proving $\frac{\sin x + \sin 2x + \sin3x}{\cos x + \cos 2x + \cos 3x} = \tan2x$ I need to prove: $$ \frac{\sin x + \sin 2x + \sin3x}{\cos x + \cos 2x + \cos 3x} = \tan2x $$ The sum and product formulae are relevant: $$ \sin(A + B) + \sin (A-B) = 2 \sin A \cos B \\ \sin(A + B) - \sin (A-B) = 2 \cos A \sin B \\ \cos(A + B) + \cos (A-B) = 2 \cos A \cos B $$ Taking the numerator first: $$ \sin x + \sin 2x + \sin3x = 2\sin\left(\frac{3}{2}x\right) \cos\left(\frac{1}{2}x\right) + \sin 3x \\ \sin 3x = \sin\left(\frac{3}{2}x + \frac{3}{2}x\right) - \sin\left(\frac{3}{2}x - \frac{3}{2}x\right) = 2\cos\left(\frac{3}{2}x\right)\sin\left(\frac{3}{2}x\right) \\ \therefore \sin x + \sin 2x + \sin3x=2\sin\left(\frac{3}{2}x\right)\left[\cos\left(\frac{1}{2}x\right) + \cos\left(\frac{3}{2}x\right)\right] $$ and now the the denominator: $$ \cos x + \cos 2x + \cos 3x = 2\cos\left(\frac{3}{2}x\right) \cos\left(\frac{1}{2}x\right) + \cos 3x $$ But I don't know how to express $\cos3x$ in terms of a product, so I can't factorize and cancel. Can someone help me make the next few steps?

\begin{align} \frac{\sin x + \sin 2x + \sin3x}{\cos x + \cos 2x + \cos 3x}&=\frac{\sin x + \sin 3x + \sin2x}{\cos x + \cos 3x + \cos 2x}\\ &=\frac{2\sin 2x\cos x+ \sin 2x}{2\cos 2x\cos x + \cos 2x}\\ &=\frac{\sin2x(2\cos x+1)}{\cos 2x(2\cos x + 1)}\\ & = \tan2x \end{align}

implicit equation for elliptical torus I just wondering what the implicit equation would be if an ellipse with major axis a and minor axis b, rotating about the Z axis with a distance of $R_0$. The $R_0$>a and $R_0$>b which means the rotation will result in a non-degenerate torus. My aim is to determine if some points are inside the toroidal surface. The surface is shown in the image found online.

You can obtain this as follows. If you start with a slice where $y = 0$, you begin with the equation $$ \frac{z^2}{a^2} + \frac{(x - R_0)^2}{b^2} - 1 = 0 $$ However, this doesn't give you the rotated version; to rotate it about the $z$-axis, simply replace the $x$ by $\sqrt{x^2 + y^2}$, yielding $$ \frac{z^2}{a^2} + \frac{\Big(\sqrt{x^2 + y^2} - R_0\Big)^2}{b^2} - 1 = 0 $$ This is a little unsatisfying though, since polynomials are much nicer than radicals. However, a little bit of manipulation yields $$ \frac{z^2}{a^2} + \frac{x^2 + y^2 + R_0^2}{b^2} - \frac{2R_0}{b^2}\sqrt{x^2 + y^2} - 1 = 0 $$ which, if you isolate the radical and square both sides yields $$ \bigg(\frac{z^2}{a^2} + \frac{x^2 + y^2 + R_0^2}{b^2} - 1\bigg)^2 - \frac{4R_0^2}{b^4}(x^2 + y^2) = 0 $$ Voila, a polynomial!

How to find all solutions of $\tan(x) = 2 + \tan(3x)$ without a calculator? Find all solutions of the equation $\tan(x) = 2 + \tan(3x)$ where $0<x<2\pi$. By replacing $\tan(3x)$ with $\dfrac{\tan(2x) + \tan(x)}{1-\tan(2x)\tan(x)}$ I've gotten to $\tan^3 (x) - 3 \tan^2 (x) + \tan(x) + 1 =0$. I am not sure how to proceed from there without the use of a calculator.

Let $\tan x=t$. Then, we have $$t=2+\frac{3t-t^3}{1-3t^2}\Rightarrow (t-2)(1-3t^2)=3t-t^3$$$$\Rightarrow t^3-3t^2+t+1\Rightarrow (t-1)(t^2-2t-1)=0.$$ Hence, we have $$\tan x=t=1,1\pm\sqrt 2\Rightarrow x=\frac{\pi}{4}+n\pi,\frac{3}{8}\pi+n\pi,-\frac{\pi}{8}+n\pi$$ where $n\in\mathbb Z$.

Given the following derivatives, find the integrals Find the derivatives of $\ln(x+\sqrt{x^2+1})$ and $\arcsin(x)$, and use the result to find the integrals of the following functions: * *$$ \dfrac{1}{ \sqrt{ \pm x^2 \pm a^2 }} $$ *$$ \sqrt{\pm x^2 \pm a^2} $$ Except for the cases where both are minuses. $a$ is a positive constant. So for the two derivatives, I just found the following derivatives $$[\ln (x+\sqrt{x^2 \pm a^2}) ]' = \dfrac{1}{\sqrt{x^2 \pm a^2}}$$ And also: $$ [ b\arcsin(\dfrac{x}{a} + c)]' = \dfrac{b}{\sqrt{a^2-(x+ac)^2}}$$ These formules make the first part easy. We get $\int \dfrac{1}{ \sqrt{ x^2 + a^2 }} = \ln (x+\sqrt{x^2 + a^2})$, $\int \dfrac{1}{ \sqrt{ x^2 - a^2 }} = \ln (x+\sqrt{x^2 - a^2})$ and $\int \dfrac{1}{ \sqrt{ a^2 - x^2 }} = \arcsin(\dfrac{x}{a})$ However, I am not able to figure out what the easiest way is to get the second part of the question, using the knowledge we have. Can someone help out.

Hint. You may use an integration by parts for the second family: $$ \begin{align} \int \sqrt{\pm x^2 \pm a^2} \:{\rm{d}}x&=x\sqrt{\pm x^2 \pm a^2}-\int \frac{x \times\pm x }{\sqrt{\pm x^2 \pm a^2}}\:{\rm{d}}x\\ &=x\sqrt{\pm x^2 \pm a^2}-\int \frac{\pm x^2}{\sqrt{\pm x^2 \pm a^2}}\:{\rm{d}}x\\ &=x\sqrt{\pm x^2 \pm a^2}-\int \frac{\pm x^2 \pm a^2 -\pm a^2}{\sqrt{\pm x^2 \pm a^2}}\:{\rm{d}}x\\ &=x\sqrt{\pm x^2 \pm a^2}-\int \sqrt{\pm x^2 \pm a^2} \:{\rm{d}}x+\pm a^2\int \frac{1}{\sqrt{\pm x^2 \pm a^2}}\:{\rm{d}}x \end{align} $$ giving $$ \int \sqrt{\pm x^2 \pm a^2} \:{\rm{d}}x=\frac x2\sqrt{\pm x^2 \pm a^2}+\frac{\pm a^2}{2}\int \frac{1}{\sqrt{\pm x^2 \pm a^2}}\:{\rm{d}}x $$ then you conclude with the first family.

Sum of roots of an equation $\sqrt{x-1}+\sqrt{2x-1}=x$ Find the sum of the roots of the equation $\sqrt{x-1}+\sqrt{2x-1}=x$ My attempt: Squaring the equation: $(x-1)+(2x-1) +2\sqrt{(x-1)(2x-1)}=x^2$ $\implies x^2-3x+2=2\sqrt{(x-1)(2x-1)} $ $\implies (x-1)(x-2)=2\sqrt{(x-1)(2x-1)} $ $\implies (x-2)=2\sqrt{\displaystyle \frac{(2x-1)}{(x-1)}} $ Squaring, $(x^2-4x+4)(x-1)=8x-4$ $\implies x^2(x-5)=0$. So, the sum of roots should be five. The given answer is 6. Could anyone look at my attempt to find where I went wrong. Thanks.

The problem is that you divided by $x-1$, so, you lost a root.Strating from $$ (x-1)(x-2)=2\sqrt{(x-1)(2x-1)}$$ as you properly wrote and squaring $$(x-1)^2(x-2)^2=4{(x-1)(2x-1)}$$ Expanding and grouping leads to $$x^4-6 x^3+5 x^2=0$$ so the sum of the roots is $6$ (you can check that the roots are $0,0,1,5$.

Integrate $\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$ I would like some guidance regarding the following integral: $$\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$$ EDIT: The upper problem was derived from the following integral $$\int\frac{\sqrt{x^2+2}}{x^2+1}dx$$ Where I rationalized the numerator which followed into: $$\int\frac{dx}{\sqrt{x^2+2}}+\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$$

There is a general formula for it. $$\int \frac{dx}{(x^2+1)\sqrt{x^2+a}}=\frac{1}{\sqrt{a-1}}\tan^{-1}\left(\frac{\sqrt{a-1}x}{\sqrt{x^2+a}}\right)+C\tag{1}$$ $a=2$ gives $$\int \frac{dx}{(x^2+1)\sqrt{x^2+2}}=\tan^{-1}\left(\frac{x}{\sqrt{x^2+2}}\right)+C$$ Formula $(1)$ can be proven by substitution : $t=1/x$, $s=\sqrt{at^2+1}$. \begin{align}\int \frac{dx}{(x^2+1)\sqrt{x^2+a}}&=-\int\frac{t}{(t^2+1)\sqrt{at^2+1}}dt\\&=-\int\frac{1}{s^2+a-1}ds\\&=\frac1{\sqrt{a-1}}\tan^{-1}\left(\frac{\sqrt{a-1}}{s}\right)+C\\&=\frac{1}{\sqrt{a-1}}\tan^{-1}\left(\frac{\sqrt{a-1}x}{\sqrt{x^2+a}}\right)+C\end{align} For a more general method of computing these types of integrals, one can use the Euler substitution to transform into the integration of a rational function.

Estimating the behavior for large $n$ I want to find how these coefficients increase/decrease as $n$ increases: $$ C_n = \frac{1}{n!} \left[(n+\alpha)^{n-\alpha-\frac{1}{2}}\right]$$ with $\alpha=\frac{1}{br-1}$ and $0\leq b,r \leq 1$. I used the Stirling's Approximation factorial $n!\sim \sqrt{2\pi n} n^n e^{-n}$ and got: $$ C_n = \frac{1}{\sqrt{2\pi n} n^n e^{-n}} \left[(n+\alpha)^{n-\alpha-\frac{1}{2}}\right]$$ I can't proceed any further. I would greatly appreciate any comment!

Taking the reciprocal of Stirling's Asymptotic expansion as derived in this answer: $$ n!=\frac{n^n}{e^n}\sqrt{2\pi n}\left(1+\frac{1}{12n}+\frac{1}{288n^2}-\frac{139}{51840n^3}+O\left(\frac{1}{n^4}\right)\right) $$ we get $$ \frac1{n!}=\frac{e^n}{n^n}\frac1{\sqrt{2\pi n}}\left(1-\frac{1}{12n}+\frac{1}{288n^2}+\frac{139}{51840n^3}+O\left(\frac{1}{n^4}\right)\right) $$ Applying this to $\dfrac{(n+\alpha)^{n-\alpha-1/2}}{n!}$ and using the the log and exponential series for $\left(1+\frac\alpha{n}\right)^{n-\alpha-1/2}$ yields $$ \begin{align} &\frac{(n+\alpha)^{n-\alpha-1/2}}{n!}\\ &=\frac{e^nn^{-\alpha-1}}{\sqrt{2\pi}}\left(1+\frac\alpha{n}\right)^{n-\alpha-1/2}\left(1-\frac{1}{12n}+\frac{1}{288n^2}+\frac{139}{51840n^3}+O\left(\frac{1}{n^4}\right)\right)\\[4pt] &=\small\frac{e^{n+\alpha}n^{-\alpha-1}}{\sqrt{2\pi}}\left(1-\frac{1+6\alpha+18\alpha^2}{12n}+\frac{1+12\alpha+144\alpha^2+456\alpha^3+324\alpha^4}{288n^2}+O\left(\frac{1}{n^3}\right)\right) \end{align} $$ Approximating $\boldsymbol{\left(1+\frac\alpha{n}\right)^{n-\alpha-1/2}}$ $$ \begin{align} &\left(n-\alpha-\frac12\right)\log\left(1+\frac\alpha{n}\right)\\ &=\left(n-\alpha-\frac12\right)\left(\frac\alpha{n}-\frac{\alpha^2}{2n^2}+\frac{\alpha^3}{3n^3}+O\left(\frac1{n^4}\right)\right)\\ &=\alpha-\frac{\alpha+3\alpha^2}{2n}+\frac{3\alpha^2+10\alpha^3}{12n^2}+O\left(\frac1{n^3}\right) \end{align} $$ Exponentiating, we get $$ \begin{align} \left(1+\frac\alpha{n}\right)^{n-\alpha-1/2} &=e^\alpha\left(1-\frac{\alpha+3\alpha^2}{2n}+\frac{9\alpha^2+38\alpha^3+27\alpha^4}{24n^2}+O\left(\frac1{n^3}\right)\right) \end{align} $$

If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$ If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$ Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$ I've attempted the question but I don't think I've done it correctly: $$ \begin{align*} b^2 &= 4 - a^2\\ b &= \sqrt{4-a^2} \end{align*} $$ Therefore, $$ \begin{align*} (a + ib)^3 &= 8\\ a + \sqrt{4-a^2} &= 2\\ \sqrt{4-a^2} &= 2 - a\\ 2 - a &= 2 - a \end{align*} $$ Therefore if $(a + ib)^3 = 8$, then $a^2 + b^2 = 4$.

Let $z=a+ib,$ then it is given that $z^3=8.$ Therefore taking the modulus of both sides $|z|^3=8.$ Hence $|z|=2$ and $|z|^2=a^2+b^2=4.$