Datasets:

math-ai

/

StackMathQA

like 77

Follow

math-ai 21

Computing an indefinite integral(2) $$\int \sqrt{1- \sin(2x)}dx =?$$ My attempt: $$\int \sqrt{1- \sin(2x)}dx = \int \sqrt{(\cos x - \sin x)^{2}} dx = \int|\cos x- \sin x| dx = ??$$

Notice, $$\cos x-\sin x=\sqrt 2\left(\frac{1}{\sqrt 2}\cos x-\frac{1}{\sqrt 2}\sin x\right)=\sqrt 2\sin\left(\frac{\pi}{4}-x\right)$$ but $\sin \theta\ge 0\iff 2k\pi\le \theta\le 2k\pi+\pi$ hence, solving the inequality * *$$2k\pi\le \left(\frac{\pi}{4}-x\right)\le 2k\pi+\pi$$ $$ 2k\pi\le \left(\frac{\pi}{4}-x\right)\implies x\leq -\left(2k\pi-\frac{\pi}{4}\right)$$ *$$\left(\frac{\pi}{4}-x\right)\le 2k\pi+\pi\implies x\ge -\left(2k\pi+\frac{3\pi}{4}\right)$$ hence, we get $$\color{red}{|\cos x-\sin x|}=\cases{\color{blue}{\cos -\sin x}\ \ \ \ \ \ \ \ \ \ \ \ \ \forall\ \ -\left(2k\pi+\frac{3\pi}{4}\right)\le x\le -\left(2k\pi-\frac{\pi}{4}\right)\\ \color{blue}{-(\cos -\sin x)}\ \ \ \ \ \ \ \forall\ \ -\left(2k\pi-\frac{\pi}{4}\right)\le x\le -\left(2k\pi-\frac{5\pi}{4}\right)}$$ where, $\color{red}{k}$ is any integer

Examples for $ \text{If } \; x-\lfloor x \rfloor + \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor = 1 \text{, then } x \text{ is irrational.}$ I saw the answers here about how to prove: $$ \text{If } \; x-\lfloor x \rfloor + \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor = 1 \text{, then } x \text{ is irrational.}$$ I understood the proof. But, can someone give me an example for an irrational $x$ that makes this equality true?

If $x>1$ then $0< \frac 1x < 1$ so $\lfloor \frac 1x \rfloor = 0$ and the equation reduces to $$x + \frac{1}{x} = 1 + \lfloor x \rfloor$$ Any number $x>1$ can be written on the form $x = n + r$ where $n$ is an integer and $0<r<1$. Take this form for $x$ in the equation above. Since $\lfloor n + r\rfloor = n$ the equation then reduces to $$ r^2 + (n-1)r - (n-1) = 0$$ Solve the quadratic equation for $r$ and pick the root that is in $(0,1)$. $x = n + r$ will then be an irrational number that satisfy your equation for any value of the integer $n > 1$.

How to evaluate $1234^{1234} \pmod{5379}$? Note: $5379 = 3 \times 11 \times 163$. I tried Chinese Remainder Theorem and Fermat's Little Theorem, got as far as: $$ 1234^{1234} = 1 \pmod{3} \\ 1234^{1234} = 5 \pmod{11} $$ With a bit more work: $$1234^{1234} = 93^{100} \pmod{163}$$ But $93^{100}$ doesn't really help? WolframAlpha tells me that $\phi(5379)=3240>1234$ So I can't use Euler's Theorem? N.B This appeared on a 1st year undergrad problem sheet. So presumably, not too much technology is required.

Well this is far from perfect,but it works if you have enough time or a calculator. $$93\equiv -70\pmod{163}$$ $$\begin{align} 93^{100}&\equiv(-70)^{100}\\ &= 490^{50}\cdot10^{50}\\ &\equiv 10^{50}\\ &= 2^{50}\cdot 5^{50}\\ &= 1024^5\cdot 3125^{10}\\ &\equiv 46^5\cdot 28^{10}\\ &=2^{25}\cdot 23^5\cdot 7^{10}\\ &= 2^{25}\cdot (23\cdot 7)^5\cdot 7^5\\ &= 2^{25}\cdot (161)^5\cdot 7^5\\ &\equiv 2^{10}\cdot2^{10}\cdot 2^5\cdot (-2)^5\cdot 7^5\\ &=1024\cdot 1024\cdot (-1024)\cdot 7^5\\ &\equiv -46^3\cdot 7^5\\ &= -(46\cdot 7)^3\cdot 7^2\\ &= -(322)^3\cdot 7^2\\ &\equiv -(-4)^3\cdot 49\\ &= 49\cdot 64\\ &\equiv 39 \end{align}$$

Solve the equation $1-\tan x + \tan^2 x - \tan^3 x + ... = \frac{\tan 2x}{1+\tan2x}$ How to solve this? Any advice? $$1-\tan x + \tan^2 x - \tan^3 x + ... = \frac{\tan 2x}{1+\tan2x}$$ Next step I do this $\sum\limits_{n=0}^\mathbb{\infty}(-1)^n \tan^nx = \frac{\tan 2x}{1+\tan2x} $ But I don't know next step. I am culeless, thanks for any advice.

Notice, for the sum of infinite series on LHS, $|\tan x|<1$ $$1-\tan x+\tan^2 x-\tan^3 x+\ldots =\frac{\tan 2x}{1+\tan 2x}$$ $$\frac{1}{1-(-\tan x)}=\frac{\frac{2\tan x}{1-\tan^ 2x}}{1+\frac{2\tan x}{1-\tan^ 2x}}$$ $$\frac{1}{1+\tan x}=\frac{2\tan x}{1-\tan^2 x+2\tan x}$$ $$1-\tan^2 x+2\tan x=2\tan x+2\tan^2 x$$ $$3\tan^2 x=1$$$$\tan^2 x=\frac{1}{3}$$ I hope you can solve further

Finding the derivative of $(\frac{a+x}{a-x})^{\frac{3}{2}}$ This is a very simple problem, but I am stuck on one step: Differentiate $(\frac{a+x}{a-x})^{\frac{3}{2}}$ Now, this is what I have done: $$ (\frac{a+x}{a-x})^{\frac{3}{2}} \\ \implies \frac{\delta}{\delta y}\frac{f}{g} \\ \implies gf' = (a-x)^{\frac{3}{2}} \times \frac{3}{2} (a+x)^{\frac{1}{2}} \times 2 \\ \implies fg' \implies (a+x)^{\frac{3}{2}} \times \frac{3}{2} (a-x)^{\frac{1}{2}} \times 0 = 0 \\ \implies \frac{(a-x)^{\frac{3}{2}} \times 3 (a+x)^{\frac{1}{2}}}{(a-x)^3}\\ \implies \frac{(a-x)^{\frac{3}{2}} - 3\sqrt{a+x}}{(a-x)^3} $$ But the answer is: $$ \frac{3\times a (a+x)^{\frac{1}{3}}}{(a-x)^{\frac{5}{2}}} $$ WolframAlpha shows: $$ \frac{3a \sqrt{\frac{a+x}{a-x}}}{(a-x)^2} $$ Another Answer (Somehow I got this): $$ \frac{3 \sqrt{\frac{a+x}{a-x}}}{2(a-x)} $$ ================== EDIT 1: What about: $$ y = (\frac{a+x}{a-x})^{\frac{3}{2}} \\ y = u^{\frac{3}{2}} \hspace{0.5cm} ; \hspace{0.5cm} u = \frac{a+x}{a-x}\\ \implies \frac{3}{2}u^{\frac{1}{2}} \hspace{0.5cm} ; \hspace{0.5cm} \frac{(0+1)\times (a-x) - [ -1 (a+x) ]}{(a-x)^2} \\ \implies \frac{2a}{(a-x)^2} \\ \implies \frac{3}{2}\sqrt{\frac{2a}{(a-x)^2}} = \frac{3}{2} \times \frac{\sqrt{2a}}{a-x} $$

i think the right answer is this here $$\frac{3}{2} \sqrt{\frac{a+x}{a-x}} \left(\frac{a+x}{(a-x)^2}+\frac{1}{a-x}\right)$$

Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve: $$ \left|\frac{x}{x+2}\right|\leq 2 $$ The answer is given to be $x\leq-4$ or $x \geq-1$ This is my attempt to solve the problem: By dividing, $\left|\frac{x}{x+2}\right|\leq 2$ is equivalent to $\left |1-\frac{2}{x+2}\right|\leq2$ which is also equivalent to $\left |\frac{2}{x+2}-1\right|\leq2$ So, $-2\leq\frac{2}{x+2}-1\leq2$ which is equivalent to $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$ Case 1: $x+2>0$. Solving $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$, I get $x\geq-4$ and $x\geq-\frac{4}{3}$ which is essentially $x\geq-\frac{4}{3}$. Case 2: $x+2<0$. Solving $-\frac{1}{2}\times(x+2)\geq{1}\geq\frac{3}{2}\times(x+2)$, I get $x\leq-4$ and $x\leq-\frac{4}{3}$ which is essentially $x\leq-4$. So, the solutions are: $x\leq-4$ or $x\geq-\frac{4}{3}$. I couldn't get $x \geq-1$ as a solution. Did I do anything wrong? The book I am using is Schaum's Outlines-Calculus. Another question I would like to ask is that am I using 'and' and 'or' correctly in the above attempt to solve the problem? I have had this problem many times.

the case $x+2>0$, you do not need to solve $-1/2\le 1/(x+2)\leftarrow x+2>0$. the case $x+2<0$, the inequality you solve it's not the original one. It should be $-1/2 \le 1/(x+2) \le 3/2$ and in case $x+2<0\rightarrow 1/(x+2)<3/2$, so it's just $-1/2 \le 1/(x+2)$. Think it simply.

How to compute this finite sum $\sum_{k=1}^n \frac{k}{2^k} + \frac{n}{2^n}$? I do not know how to find the value of this sum: $$\sum_{k=1}^n \frac{k}{2^k} + \frac{n}{2^n}$$ (Yes, the last term is added twice). Of course I've already plugged it to wolfram online, and the answer is $$2-\frac{1}{2^{n-1}}$$ But I do not know how to arrive at this answer. I am not interested in proving the formula inductively :)

Let $$S=\sum_{k=1}^{n}\frac{k}{2^k}$$ or, $$S=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+....+\frac{n}{2^n}$$ and $$\frac{S}{2}= \frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+....+\frac{n-1}{2^n}+\frac{n}{2^{n+1}}$$ Subtracting we get, $$\frac{S}{2}= \frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^n}-\frac{n}{2^{n+1}}$$ or, $$S= 1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{n-1}}-\frac{n}{2^n}=2\left(1-\frac{1}{2^n}\right)-\frac{n}{2^n}$$

prove polynomial division for any natural number Show that for any natural numbers $a$, $b$, $c~$ we have $~x^2 + x + 1|x^{3a+2} + x^{3b+1} + x^{3c}$. Any hints on what to use?

We have $$x^3-1=(x-1)\left(x^2+x+1\right)$$ This means that $$\begin{align}x^3-1&\equiv0\pmod{x^2+x+1}\\x^3&\equiv1\pmod{x^2+x+1}\end{align}$$ Now, substitute it $$\begin{align}x^{3a+2}+x^{3b+1}+x^{3c}&\equiv x^2\left(x^3\right)^a+x\left(x^3\right)^b+\left(x^3\right)^a\\&\equiv x^2\cdot1^a+x\cdot1^b+1^c\\&\equiv x^2+x+1\\&\equiv0\pmod{x^2+x+1}\end{align}$$

Show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ How can one show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ Assuming that : $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ So $(\sqrt{x-1}+\sqrt{y-1})^2\leq xy$ $\sqrt{(x-1)(y-1)} \leq xy-x-y+2$ $ (y-1)(x-1)+3 \leq \sqrt{(x-1)(y-1)}$ Here I'm stuck !

There are some errors in your calculation, e.g. a missing factor 2 in $$ (\sqrt{x-1}+\sqrt{y-1})^2 = x - 1 + y - 1 + 2\sqrt{x-1}\sqrt{y-1} $$ and in the last step the inequality sign is in the wrong direction and the number $3$ is wrong. For $x \ge 1$, $y \ge 1$ you can square the inequality (since both sides are non-negative): $$ \sqrt{\mathstrut x-1}+\sqrt{\mathstrut y-1}\leq \sqrt{\mathstrut xy} \\ \Longleftrightarrow (x-1) + (y-1) + 2 \sqrt{\mathstrut x-1}\sqrt{\mathstrut y-1} \le xy \\ \Longleftrightarrow 0 \le xy - x - y + 2 - 2 \sqrt{\mathstrut (x-1)(y-1)} \\ \Longleftrightarrow 0 \le (x-1)(y-1) - 2 \sqrt{\mathstrut (x-1)(y-1)} + 1 $$ With $t := \sqrt{(x-1)(y-1)}$ the right-hand side is $$ t^2 - 2 t + 1 = (t-1)^2 \ge 0 \, . $$ so that the inequality is true. It follows also that equality holds if and only if $t = 1$, i.e. if $(x-1)(y-1) = 1$.

how to prove that the sum of the power two of three reals is less than 5? $a, b$ and $c$ are three real number such that: $(\forall x \in [-1,1]): |ax^2+bx+c|\leq 1$ . Prove that: 1) $|c|\leq 1$ . 2) $-1\leq a+c\leq 1$ . 3)Deduce that : $a^2+b^2+c^2\leq 5$. The first and second questions are easy to prove ( just take $x=0, -1$ or $1$ ... ), but the third one ?!!! I'm sure that the responce must be easy since it's a deduction. I need your help to solve it, thank you.

$x=0$ gives us $|c| \le 1$. Similarly set $x = \pm1$ to get $|a \pm b+c| \le 1$. So we have the inequalities: $$-1 \le c \le 1 \iff -1 \le -c \le 1\tag{1}$$ $$-1 \le a+b+c \le 1 \iff -1\le -a-b-c \le 1 \tag{2}$$ $$-1 \le a-b+c \le 1 \iff -1 \le -a+b-c \le 1\tag{3}$$ Note the right side of the above three inequalities can be considered the "negatives" of the left sides. These should be enough for us to prove all sorts of things. Here, $$(2)-(1) \implies -2 \le a+b \le 2\implies (a+b)^2 \le 4$$ $$(3)-(1) \implies -2 \le a-b \le 2\implies (a-b)^2 \le 4$$ Now those two give $a^2+b^2=\frac12((a+b)^2+(a-b)^2) \le 4$. We just need to add to this $|c| \le 1 \implies c^2 \le 1$ to get what we want.

Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$ I would like to prove $$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$ * *I'm interested in more ways of proving it My thoughts: \begin{align} \sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\ \frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\ \frac{1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{1}{\sqrt{x +1}+\sqrt{x}}\\ \sqrt{x +1}+\sqrt{x} &\neq \sqrt{x+2}+\sqrt{x+1}\\ \sqrt{x} &\neq \sqrt{x+2}\\ \end{align} * *Is my proof correct? *I'm interested in more ways of proving it.

By the mean value theorem - MVT, for all $x >0$, it exists $\zeta_x \in (x, x+1)$ such that $$\frac{1}{2\sqrt{\zeta_x}}((x+1)-x)=\frac{1}{2\sqrt{\zeta_x}}=\sqrt{x+1}-\sqrt{x}$$ As $y \to \frac{1}{2\sqrt{y}}$ is stricly decreasing, you have $$\frac{1}{2\sqrt{\zeta_{x+1}}}=\sqrt{x+2} -\sqrt{x+1} < \sqrt {x+1}-\sqrt{x}=\frac{1}{2\sqrt{\zeta_x}}.$$

What is the method for finding $\int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}\mathrm{d}x$? $$ \int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}dx $$ A bit confused with how to integrate this question. I though it was partial fractions but was unsure about the what to do after that.

Partial Solution Well, assuming you did the partial fraction decomposition already (as you said you did), you should get the following integral $$\int\left(\frac{9-x}{2(x^2+1)}+\frac{1}{x-1}+\frac{1}{2 (x+1)}\right) dx$$ $$=\frac{9}{2}\int\frac {dx}{x^2+1} - \frac{1}{2}\int \frac {x}{x^2+1}dx + \int \frac{dx}{x-1}+ \frac{1}{2}\int \frac{dx}{x+1}$$ A.) $\quad \int\frac {dx}{x^2+1}$ = $\arctan x$ B.) $\quad\int \frac {x}{x^2+1}dx = \frac{1}{2}\int \frac {du}{u+1} = \frac{1}{2}\log (u+1) = \frac{1}{2}\log (x^2+1)$ C.)$\quad\int \frac{dx}{x-1} = \log(x-1)$ D.)$\quad\int \frac{dx}{x+1} = \log(x+1)$

Linear Transformations, Linear Algebra Let T:P3→P3 be the linear transformation such that $T(−2x^2)= −2x^2 − 2x$, $T(0.5x + 2)= 3x^2 + 4x−2$, and $T(2x^2 − 1)= 2x + 1$. Find $T(1), T(x), T(x^2)$, and $T(ax^2 + bx + c)$, where a, b, and c are arbitrary real numbers. I understand how to find $T(x^2)$ where you just divide the given $T(-2x^2)$ by $-2$ to get $T(x^2) = x^2 + x$. I'm not sure sure how to proceed in calculating the other transformation functions. Please list as many of the steps as possible in solving for the three other functions. Thank you!

We can write $1=-(-2x^2)-(2x^2-1), x=2(0.5x+2)-4$, so $$T(1)=T(-(-2x^2)-(2x^2-1))=-T(-2x^2)-T(2x^2-1)=2x^2-1$$ $$T(x)=T(2(0.5x+2)-4)=2T(0.5x+2)-4T(1)=-2x^2+8x$$ Then we get $$T(ax^2+bx+c)=a(x^2+x)+b(-2x^2+8x)+c(2x^2-1)$$ $$=(a-2b+2c)x^2+(a+8b)x-c$$

Find the minimum of $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ for $00$. Find the minimum of $\displaystyle (u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2$ for $0<u<\sqrt{2}$ and $v>0$. I think I have to use the Arithmetic and Geometric Means Inequalities. Or $\displaystyle \frac{1}{2}(u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2 \geq \frac{1}{4}\left(u-v+\sqrt{2-u^2}-\frac{9}{v}\right)^2$ by Chebychev inequality. Is anyone is able to give me a hint how to finish it? I think I have to use the Arithmetic and Geometric Means Inequalities.

You can simply think $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ is the square of distance between two points $a$ and $b$ in the plane, where $$a=(u,\sqrt{2-u^2}),b=(v,\frac{9}{v}).$$ Clearly, $a$ satisfies $x^2+y^2=2,(0<x<\sqrt{2})$ and $b$ satisfies $y=\frac{9}{x},(x>0)$. Then from the geometry graph, we easily know the minimum distance between the two curves is the distance between $(1,1)$ and $(3,3)$. So the minimum distance is $\sqrt{8}$, and hence the minimum solution is 8 at $u=1,v=3$.

What are the residues of $\frac{z^2 e^z}{1+e^{2z}}$? I hope to know the singularity and pole of $\frac{z^2 e^z}{1+e^{2z}}$. I try $\frac{z^2 e^z}{1+e^{2z}} = \frac{z^2}{e^{-z}+e^{z}}$ and observe that the denominator seems like cosine function. So I think the singularites are i(2n+1)$\pi$/2. But when I try to evaluate the poles, I fail and different values of $n$ has different residues. Could the residues be imaginary?

Since $\frac{z^2 e^z}{1+e^{2z}} = \frac{z^2}{2\cosh z}$, let's start with $$f(z)=\frac{z^2}{2\cos iz}$$ Now both $z^2$ and $\cos iz$ are entire. The function can be singular only where $z_{0_n} = iz = \frac{\pi +2\pi n}{2}$ and since the zeroes of the denominators are first order, we have $$\begin{align*} Res f(z_{0_n}) & =\frac{z^2}{(2\cos iz)'}\bigg|_{z_{0_n}} \\ \\ & =\frac{\left(\frac{\pi +2\pi n}{2}\right)^2}{-2i\sin \left(\frac{\pi +2\pi n}{2}\right)} \\ \\ & =\frac{i\left(\pi +2\pi n\right)^2}{8\sin \left(\frac{\pi}{2} +\pi n\right)} \\ \\ & =\frac{i\pi^2\left(1 +2 n\right)^2}{8\cos\left(\pi n\right)} \\ \\ \end{align*} $$ $$\boxed{Res f(z_{0_n}) =(-1)^n\frac{i\pi}{8}\left(1 +2 n\right)^2}$$ Compare with WolframAlpha...

What is the positive divisors of $n(n^2-1)(n^2+3)(n^2+5)$ I want to find the positive divisors of $n(n^2-1)(n^2+3)(n^2+5)$ from $n(n-1)(n+1)$ 2 and 3 should divide this expression for all positive n. how can I find the rest? which python says $(2, 3, 6, 7, 9, 42, 14, 18, 21, 63,126)$

If $$n\equiv 0 \mod 3,$$ then the factor $$n^2+3\equiv 0 \mod 3.$$ Otherwise $$n^2\equiv1 \mod 3$$ and thus $$n^2+5\equiv 0\mod3.$$ Therefore $$(n^2+3)(n^2+5)\equiv0\mod3.$$ Let us now look modulo $7$ and compute the number for all elements of $\mathbb Z_7$ (which means to check if the number is a multiple of $7$ for $n=0,\dots 6$. You can easily check that it is true. Therefore this number is a multiple of $2\times3\times3\times7=126$ and all its divisors. There are no others since for $n=2$, $n(n^2-1)(n^2+3)(n^2+5)=126\times3$ but for $n=3$, $n(n^2-1)(n^2+3)(n^2+5)=126\times 32$.

Homework: Sum of the cubed roots of polynomial Given $7X^4-14X^3-7X+2 = f\in R[X]$, find the sum of the cubed roots. Let $x_1, x_2, x_3, x_4\in R$ be the roots. Then the polynomial $X^4-2X^3-X+ 2/7$ would have the same roots. If we write the polynomial as $X^4 + a_1X^3 + a_2X^2 +a_3X + a_4$ then per Viete's theorem: $a_k = (-1)^k\sigma _k(x_1,x_2,x_3,x_4), k\in \{1,2,3,4\}$, where $\sigma _k$ is the $k$-th elementary symmetrical polynomial. Therefore: $x_1+x_2+x_3+x_4 = 2$ $x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 = 0\ (*)$ $x_1x_2x_3 +x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 = 1$ $x_1x_2x_3x_4 = 2/7$ Now how to determine the sum of the cubed roots? $2^3 = 8= (x_1+x_2+x_3+x_4)(x_1+x_2+x_3+x_4)^2 = (x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2 + 2(*))$ Here's where things go out of hand: $(x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2) = (x_1^3 + x_2^3 + x_3^3+x_4^3) + x_1^2(x_2+x_3+x_4)+x_2^2(x_1+x_3+x_4)+x_3^2(x_1+x_2+x_4)+x_4^2(x_1+x_2+x_3) = 8$ What should I do here?

Hint: as we can have only cubic terms in the symmetric polynomial sums, the only terms which can be used are of form $(\sum x)^3, \sum x \sum xy$ and $\sum xyz$. Then it is a matter of testing $3$ coefficients... $$\sum_{cyc} x_1^3 = \left(\sum_{cyc} x_1 \right)^3-3\left(\sum_{cyc} x_1 \right)\left(\sum_{cyc} x_1 x_2 \right)+3\sum_{cyc} x_1x_2x_3$$

How to show that $a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$ Let $$(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})=a_{0}+a_{1}x+\cdots+a_{200}x^{200}$$ show that $$a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$$ I have one methods to solve this problem: Let$$g(x)=(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})$$ Note $$g(x)=g(-x)$$ so $$a_{1}=a_{3}=\cdots=a_{199}=0$$ there exist other methods?

Let $A (x) = (1-x+x^2-x^3+\cdots-x^{99}+x^{100})$ and $B (x) = 1+x+x^2+\cdots+x^{100}$. Then we have $g(x) = A (x) B (x)$. We have $(x + 1) A (x) = x^{101} + 1$ and $(x - 1) B (x) = x ^ {101} - 1$. Then $$g (x) = \frac {x^ {202} - 1} {x^2 - 1} = x^{200} + x^{198} + \cdots + 1,$$ as desired.

Solve $x-\lfloor x\rfloor= \frac{2}{\frac{1}{x} + \frac{1}{\lfloor x\rfloor}}$ Could anyone advise me how to solve the following problem: Find all $x \in \mathbb{R}$ such that $x-\lfloor x\rfloor= \dfrac{2}{\dfrac{1}{x} + \dfrac{1}{\lfloor x\rfloor}},$ where $\lfloor *\rfloor$ denotes the greatest integer function. Here is my attempt: Clearly, $x \not \in \mathbb{Z}.$ $\dfrac{x}{\lfloor x \rfloor} - \dfrac{\lfloor x \rfloor}{x}= 2$ $ \implies x^2 -2x\lfloor x \rfloor - \lfloor x \rfloor ^2 = 0$ $\implies x = (1 \pm \sqrt2 )\lfloor x \rfloor $ $\implies \{x\} = \pm\sqrt2 \lfloor x \rfloor$ Thank you.

Completing the square gives $$ x^2-2x\lfloor x\rfloor+\lfloor x\rfloor^2=2\lfloor x\rfloor^2 $$ so $$ (x-\lfloor x\rfloor)^2=2\lfloor x\rfloor^2 $$ Since $0\le x-\lfloor x\rfloor<1$, we conclude $\lfloor x\rfloor=0$ that's disallowed by the starting equation.

Evaluate $\lim_\limits{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1}$ For all $n,k \in N a,b > 0$ $$\lim_{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1} = \lim_{x \to 0}\frac{x}{(1+ \frac{ax}{n})(1+ \frac{bx}{k})- 1}= \lim_{x \to 0}\frac{x}{x(\frac{a}{n} + \frac{b}{k}) + \frac{ab}{nk}x^2} = \lim_{x \to 0}\frac{1}{\frac{a}{n} + \frac{b}{k}} = \frac{nk}{ka+nb}$$ Am I right?

The limit is fine. Perhaps with a view to formalizing the procedure is appropriate to introduce an infinite series , and then Landau notation. $$O(g(x)) = \left\{\begin{matrix} f(x) : \forall x\ge x_0 >0 , 0\le |f(x)|\le c|g(x)| \end{matrix}\right\}$$ We will also use: $$f_1=O(g_1)\wedge f_2=O(g_2)\implies f_1f_2=O(g_1g_2)\,$$ Recalling, from MacLaurin series: $${(1+ax)^{1/n}}={\sum_{i=0}^{\infty} \binom{1/n}{i} (ax)^i }={ (1+\dfrac{ax}{n}+O(x^2))}\\~\\ {(1+bx)^{1/k}}={( 1+\dfrac{bx}{k}+O(x^2))} $$ Now, we have: $${(1+ax)^{1/n}}{(1+bx)^{1/k}}={ (1+\dfrac{ax}{n}+O(x^2))}{( 1+\dfrac{bx}{k}+O(x^2))}={(1+\dfrac{ax}{n} + \dfrac{bx}{k}+O(x^2))} $$ And finally $$ \lim_{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1} = \lim_{x \to 0}\frac{x}{{(1+\dfrac{ax}{n} + \dfrac{bx}{k}+O(x^2))}-1}\\~\\= \lim_{x \to 0}\frac{(x^{-1})x}{{(x^{-1})(\dfrac{ax}{n} + \dfrac{bx}{k}+O(x^2))}} = \lim_{x \to 0}\frac{1}{\frac{a}{n} + \frac{b}{k} +O(x)} = \frac{1}{\frac{a}{n} + \frac{b}{k}}= \frac{nk}{ak+nb} $$

What does it mean to represent elements of an ideal? Say I have the polynomial $x^9 + 1$ Then: $x^9 + 1 = (x+1)(x^2 + x + 1)(x^6 + x^3 + 1)$ is a complete factorization over $GF(2)$ of $x^9 + 1$ The dimension of each ideal is: length $n - deg(ideal)$ So for $n=9$, dimension of $(x+1)$ = $9-1=8$ of $(x^2 + x + 1) = 9 - 2 = 7$ of $(x^6 + x^3 + 1) = 9 - 6 = 3$ So let's use $(x^6 + x^3 + 1)$ as the example. The dimension is 3. So there should be $2^3 = 8$ elements. How do I find those elements?

Let’s call $f=x^6+x^3+1$. You want three linearly independent elements of the ideal $(f)$ of the ring $R=\Bbb F_2[x]/(x^9+1)$. Since $(f)$ is just the set of multiples of $f$, you certainly have $1\cdot f$, $xf$, and $x^2f$. Notice that $x^3f=x^9+x^6+x^3=1+x^6+x^3=1\cdot f$, already counted. I’ll leave it to you to show that those three polynomials are $\Bbb F_2$-linearly independent in $R$.

Evaluate $\lim_{x \to 0} \left(\frac{ \sin x }{x} \right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}}$$ The task should be solved by using Maclaurin series so I did some kind of asymptotic simplification $$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \to 0} \bigg(\frac{x - \frac{x^3}{6}}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \to 0} \bigg(1 - \frac{x^2}{6} \bigg)^{\frac{1}{x^2}}$$ How can we say answer that is $e$ in the power of $-\frac{1}{6}$. I want some proving of that fact. Thank you.

$$ \lim_{x \to 0} \left( 1 - \frac {x^2}6\right)^{\frac 1{x^2}} = \lim_{x \to 0} \left [ \left( 1 - \frac {x^2}6\right)^{\frac 6{x^2}} \right ]^{\frac 16} = \left [ \left (e^{-1} \right ) \right ]^{\frac 16} = e^{-\frac 16} $$ Here, I used somewhat modified limit regarding Euler's number $$ \lim_{t \to 0} \left( 1 - t\right )^{\frac 1t} = e^{-1} $$ More info, and proofs can be found here.

Positive equilibria for a system of eqautions I have the following system of equations \begin{align} \frac{dx}{d \tau} &= x \left(1-x-\frac{y}{x+b} \right) \\ \frac{dy}{d \tau} &= cy \left(-1+a\frac{x}{x+b} \right) \end{align} I am asked to show that if $a<1$, the only nonnegative equilibria are $(0,0), (1,0)$. So first it is obvious that in order to the equations become $0$ is $(x,y)=(0,0)$ Then I decided $y=0$ and $1-x-\displaystyle \frac{y}{x+b}=0 \Leftrightarrow x=1$ , hence $(x,y)=(1,0)$ In the same way I decided $x=0$ and $-1+a\displaystyle \frac{x}{x+b}=0$, but there is no $y$. Finally I decided $-1+a\displaystyle \frac{x}{x+b}=0$ and $1-x-\displaystyle \frac{y}{x+b}=0$, and this is difficult. I can't figure out what to do now. What about the fact that $a<1$? Can anyone help?

If $x \ne 0$, We have $$ 1 - a\frac{x}{x+b} = 0 $$ $$ a\frac{x}{x+b} = 1 $$ $$ ax = x + b$$ $$ (a-1)x = b$$ $$ x = \frac{b}{a-1} $$ If $a < 1$ and $b > 0$, $x < 0$ and therefore is not a non-negative solution

calculate $\lim\limits_{x \to 1}(1 - x)\tan \frac{\pi x}{2}$ I need to calculate $$\lim_{x \to 1}\left((1 - x)\tan \frac{\pi x}{2}\right)$$. I used MacLaurin for $\tan$ and got $\frac{\pi x} {2} + o(x)$. Then the full expression comes to $$\lim_{x \to 1}\left(\frac {\pi x} {2} - \frac {\pi x^2} {2} + o(x)\right) = 0$$But WolframAlpha says it should be $\frac 2 \pi$. What am I doing wrong?

$L=\lim\limits_{x \to 1}((1 - x)\tan \frac{\pi x}{2})$ $L=\lim\limits_{(x-1) \to 0}((1 - x)\cot(\frac{\pi }{2}- \frac{\pi x}{2}))$ $L=\frac{2 }{\pi}\lim\limits_{\frac{\pi }{2}(x-1) \to 0}(\frac{\pi }{2}(1 - x)(-)\frac{cos\frac{\pi }{2}(x- 1)}{sin\frac{\pi }{2}(x- 1)})$ $L=\frac{2 }{\pi}\lim\limits_{\frac{\pi }{2}(x-1) \to 0}(\frac{\pi }{2}(x - 1)\frac{cos\frac{\pi }{2}(x- 1)}{sin\frac{\pi }{2}(x- 1)})$ $L=\frac{2 }{\pi}\lim\limits_{\frac{\pi }{2}(x-1) \to 0}\frac{1}{1}$ $L=\frac{2 }{\pi}$

Limit of $\lim\limits_{n\to\infty}\frac{\sum_{m=0}^n (2m+1)^k}{n^{k+1}}$ I wanted to find the limit of: ($k \in N)$ $$\lim_{n \to \infty}{\frac{1^k+3^k+5^k+\cdots+(2n+1)^k}{n^{k+1}}}.$$ Stolz–Cesàro theorem could help but $\frac{a_n-a_{n-1}}{b_n-b_{n-1}}$ makes big mess here: $$\lim_{n \to \infty}{\frac{-0^k+1^k-2^k+3^k-4^k+5^k-6^k+\cdots-(2n)^k+(2n+1)^k}{n^{k+1}-(n-1)^{k+1}}}.$$ Is following statement true as well $$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_n-a_{n-2}}{b_n-b_{n-2}}$$?

Using Faulhaber's formula, $\lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+n^k}{n^{k+1}}} = \frac{1}{k+1}$. Then, $\lim_{n \to \infty}{\frac{1^k+3^k+5^k+\cdots+(2n+1)^k}{n^{k+1}}} = \lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+(2n+1)^k}{n^{k+1}}} - 2^k \lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+n^k}{n^{k+1}}}$ $ = \lim_{n \to \infty}\frac{(2n+1)^{k+1}}{(k+1)n^{k+1}}-2^k\frac{n^{k+1}}{(k+1)n^{k+1}}$. $ = \lim_{n \to \infty}\frac{2^{k+1} (n+1/2)^{k+1}-2^kn^{k+1}}{(k+1)n^{k+1}}$. $ = \lim_{n \to \infty}\frac{2^{k+1} n^{k+1}-2^kn^{k+1}}{(k+1)n^{k+1}}$. $ = \lim_{n \to \infty}\frac{2^{k+1}-2^k}{k+1}$. $ =\frac{2^k}{k+1}$.

How to manually calculate the sine? I started studying trigonometry and I'm confused. How can I manually calculate the sine? For example: $\sin(\frac{1}{8}\pi)$? I was told to start getting the sum of two values which will result the sine's value. For $\sin(\frac{7}{12}\pi)$, it would be $\sin(\frac{1}{4}\pi + \frac{1}{3}\pi)$. However, I find this way confusing. For example, I don't know which sum will result $\frac{1}{8}$ in the example above. Is there a better/easier way to do it? Please, can anyone explain step by step how to do it?

For $\frac{\pi}{8}$ you need to use double angle formula. Recall that $\sin[2\theta)=2\sin\theta\cos\theta=2\sin\theta\sqrt{1-\sin^2\theta}$ Then let $\theta=\frac{\pi}{8}$: $$\sin\left(\frac{\pi}{4}\right)=2\sin\frac{\pi}{8}\sqrt{1-\sin^2\frac{\pi}{8}}$$ $$\frac{1}{2}=4\sin^2\frac{\pi}{8}\left(1-\sin^2\frac{\pi}{8}\right)$$ $$\frac{1}{8}=4\sin^2\frac{\pi}{8}-\sin^4\frac{\pi}{8}$$ Solving as a quadratic in $\sin^2\frac{\pi}{8}$ gives: $$\sin^2\frac{\pi}{8}=\frac{1}{4}\left(2-\sqrt{2}\right)$$ (we can ignore the other solution to the quadratic as we know $\sin\frac{\pi}{8}<\sin\frac{\pi}{3}=\frac{1}{2}$) $$\sin\frac{\pi}{8}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$ (we can ignore the negative solution as we know $\sin\frac{\pi}{8}>0$) Note: this is just one way of doing it. You could have started from any of the multiple angle formula.

Converting a sum of trig functions into a product Given, $$\cos{\frac{x}{2}} +\sin{(3x)} + \sqrt{3}\left(\sin\frac{x}{2} + \cos{(3x)}\right)$$ How can we write this as a product? Some things I have tried: * *Grouping like arguments with each other. Wolfram Alpha gives $$\cos{\frac{x}{2}} + \sqrt{3}\sin{\frac{x}{2}} = 2\sin{\left(\frac{x}{2} + \frac{\pi}{6}\right)}$$but I don't know how to derive that myself or do a similar thing with the $3x$. *Write $3x$ as $6\frac{x}{2}$ and then using the triple and double angle formulas, but that is much too tedious and there has to be a more efficient way. *Rewriting $\sqrt{3}$ as $2\sin{\frac{\pi}{3}}$ and then expanding and trying to use the product-to-sum formulas, and then finally grouping like terms and then using the sum-to-product formulas, but that didn't work either. I feel like I'm overthinking this, so any help or insights would be useful.

First of all $$ A\cos\alpha+B\sin\alpha=\sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\cos\alpha+\frac{B}{\sqrt{A^2+B^2}}\sin\alpha\right)\\ =\sqrt{A^2+B^2}(\sin\beta\cos\alpha+\cos\beta\sin\alpha)=\sqrt{A^2+B^2}\sin(\beta+\alpha) $$ you only have to find $\beta$ such that $$ \left\{ \begin{align} \sin\beta&=\frac{A}{\sqrt{A^2+B^2}}\\ \cos\beta&=\frac{B}{\sqrt{A^2+B^2}} \end{align} \right. $$ Next, use Sum to product identities.

Why I am getting wrong answer to this definite integral? $$\int_0^{\sqrt3} \sin^{-1}\frac{2x}{1+x^2}dx $$ Obviously the substitution must be $x=tany$ $$2\int_0^{\frac{\pi}{3}}y\sec^2y \ dy $$ Taking $u=y $, $du=dy;dv=sec^2y \ dy, v=\tan y $ $$2\Big(y\tan y+\ln(\cos y)\Big)^{\frac{\pi}{3}}_{0} $$ Hence $$2\frac{\pi}{\sqrt 3}+2\ln\frac{1}{2} $$ But the answer given is $\frac{\pi}{\sqrt 3}$.

$$I=\int_{0}^{\sqrt{3}} \sin^{-1} \frac{2x}{1+x^2} dx$$ Notice that $$\frac{d}{dx} \sin^{1} \frac{2x}{1+x^2}= \frac{2}{1+x^2}~~ \frac{|1-x^2|}{(1-x^2)}= \frac{2}{1+x^2}, ~if~~ x^2<1,~~\frac{-2}{1+x^2}~if~ x^2>1.$$ So let us integrate by parts taking 1 as the second function. Then $$I=\left ( x \sin^{-1} \frac{2x}{1+x^2}\right) -\left(\int_{0}^{1} x\frac{d}{dx} \sin^{-1} \frac{2x}{1+x^2} dx+\int_{1}^{\sqrt{3}} x\frac{d}{dx} \sin^{-1} \frac{2x}{1+x^2} dx\right)$$ $$I=\left ( x \sin^{-1} \frac{2x}{1+x^2}\right) -\left(\int_{0}^{1} \frac{2x}{1+x^2} dx+\int_{1}^{\sqrt{3}} \frac{-2x}{1+x^2} dx\right).$$ $$I=\left ( x \sin^{-1} \frac{2x}{1+x^2}\right)_{(0,\sqrt{3})}-(\ln 2-\ln 4 +\ln 2)=\frac{\pi}{\sqrt{3}}.$$

What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$? Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$ I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.

HINT: Note that $$\color{Green}{x^2=-x+1}.$$ By multiplying the given expression by $x,$ we can obtained $$\color{Green}{x^3=2x-1}.$$ Again by multiplying $x$ we have $$\color{Green}{x^4=-3x+2}$$ and so on..

The sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$ Prove that the sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$. Can someone pls help and provide a solution for this and if possible explain the question

By mathematical induction, say $P(n):1+4+9+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ Now $P(1)$ is true as L.H.S. $= 1$ and R.H.S. $= 1$ Say $P(k)$ is true for some $k \in \mathbb{N}$ and $k>1$. Therefore $1+4+9+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}$ Now for $P(k+1)$, \begin{align} & 1+4+9+\cdots+k^2+(k+1)^2 \\[10pt] = {} & \frac{k(k+1)(2k+1)}{6}+(k+1)^2 \\[10pt] = {} & (k+1)\cdot \frac{2k^2+k+6(k+1)}{6} \\[10pt] = {} &\frac{(k+1)[k+1+1][2(k+1)+1]}{6} \end{align} So $P(k+1)$ is true. Hence by induction, $P(n):1+4+9+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ is true.

Find the characteristic polynomial of $(M^{-1})^3$ Given that M is a square matrix with characteristic polynomial $p_{m}(x) = -x^3 +6x^2+9x-14$ Find the characteristic polynomial of $(M^{-1})^3$ My attempt: x of $(M^{-1})^3$ is $1^3$, $(-2)^3$ , $7^3$ = $1$ , $-8$ , $343$ $p_{(m^-1)^3} = (x-1)(x+8)(x-343)$ or $-(x-1)(x+8)(x-343) $ = $x^3-336x^2-2409x+2744$ or $-x^3+336x^2+2409x-2744 $ Is my approach correct?

Note that $-x^3+6x^2+9x-14=-(x-1)(x+2)(x-7)$, so we may assume that $M$ is the diagonal matrix with entries $1,-2,7$ on the diagonal. Then it's easy to see that the characteristic polynomial of $M^{-3}$ is given by $(x-1)(x+(1/2)^3)(x-(1/7)^3)$.

How to eliminate $\theta$? While doing a sum I was stuck in a particular step: $$r_1= \frac{4a \cos \theta }{\sin^2 \theta}$$ and $$r_2=\frac{4a \sin \theta }{\cos^2 \theta}$$ How to eliminate $\theta$ ?

$$\begin{cases} \displaystyle r_1=\frac{4a\cos\theta}{\sin^2\theta}\\ \displaystyle r_2=\frac{4a\sin\theta}{\cos^2\theta}\\ \end{cases}$$ Try to find $\sin\theta$ and $\cos\theta$ as following: $$r_1\sin^2\theta=4a\cos\theta\tag{1}$$ $$r_2\cos^2\theta=4a\sin\theta\tag{2}$$ To equation $(1)$ plug computed $\sin\theta$ from equation $(2)$: $$\sin\theta=\frac{r_2\cos^2\theta}{4a}$$ $$r_1\left(\frac{r_2\cos^2\theta}{4a}\right)^2=4a\cos\theta$$ $$r_1\frac{r_2^2\cos^4\theta}{16a^2}=4a\cos\theta$$ $$r_1r_2^2\cos^3\theta=64a^3$$ $$\cos^3\theta=\frac{64a^3}{r_1r_2^2}$$ $$\cos\theta=\sqrt[3]{\frac{64a^3}{r_1r_2^2}}=\frac{4a}{\sqrt[3]{r_1r_2^2}}$$ Same thing to compute $\sin\theta$: $$\sin\theta=\frac{4a}{\sqrt[3]{r_2r_1^2}}$$ And use the Pythagorean identity: $$\sin^2\theta+\cos^2\theta=1$$ $$\left(\frac{4a}{\sqrt[3]{r_2r_1^2}}\right)^2+\left(\frac{4a}{\sqrt[3]{r_1r_2^2}}\right)^2=1$$ $$\frac{16a^2}{(r_2r_1^2)^{2/3}}+\frac{16a^2}{(r_1r_2^2)^{2/3}}=1$$

Show that the curve has two tangents I'm a little stuck on a math problem that reads as follows: Show that the curve $x = 5\cos(t), y = 3\sin(t)\, \cos(t)$ has two tangents at $(0, 0)$ and find their equations What I've Tried * *$ \frac{dx}{dt} = -5\sin(t) $ *$ \frac{dy}{dt} = 3\cos^2(t) - 3\sin^2(t) $ because of the product rule. We can simplify this to $ 3(\cos^2(t) - \sin^2(t)) \rightarrow 3\cos(2t) $ *In order to get the slope $ m $, we can write $$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $$ *Solving for $ \frac{dy}{dx} $ as follows: * *$ \frac{3\cos(2t)}{-5\sin(t)} $ can be rewritten as *$ \frac{-3}{5}(\cos(2t)\csc(t)) = m $ *Plugging $ (0,0) $ back into the equations of $ x $ and $ y $ we have as follows: * *$ 5\cos(t) = 0 \rightarrow t = \frac{\pi}{2} $ * *Note: I'm unsure what happens to the $ 5 $ *$ \frac{dx}{dt} $ evaluated at $ t = \frac{\pi}{2} $ gives us $ -5\sin(\frac{\pi}{2}) = -5 $ *$ \frac{dy}{dt} $ evaluated at $ t = \frac{\pi}{2} $ gives us $ 3\cos(\frac{2\pi}{2}) \rightarrow 3\cos(\pi) = -3 $ *$ \frac{dy}{dx} = \frac{3}{5} $ *Continuing on, if we add $ \pi $ to the value of $ t $ we get $ t = \frac{3\pi}{2} $. Plug the new value of $ t $ into the equations of $ x $ and $ y $ * *$ \frac{dx}{dt} $ evaluated at $ t = \frac{3\pi}{2} $ gives us $ -5sin(\frac{3\pi}{2}) = 5 $ *$ \frac{dy}{dt} $ evaluated at $ t = \frac{3\pi}{2} $ gives us $ 3cos(\frac{6\pi}{2}) \rightarrow 3cos(3\pi) = -3 $ *$ \frac{dy}{dx} = -\frac{3}{5} $ *We now have our two slopes of the tangent lines: * *$ y = -\frac{5}{3}x $ *$ y = \frac{5}{3}x $ The issue is that webassign is claiming that the slopes are wrong as can be seen here: Here is the solution in graph form that is correct: p.s. My apologies if this is a repost. I've seen this response Show that the curve x = 5 cos t, y = 4 sin t cos t has two tangents at (0, 0) and find their equations. and followed it already with no avail.

You computed the slopes as $\frac{dy}{dx}=\frac{3}{5}$ and $\frac{dy}{dx}=-\frac{3}{5}$, but for some reason when you wrote the equations of the tangent lines you took the reciprocal of these slopes and wrote $y=\frac{5}{3}x$ and $y=-\frac{5}{3}x$. They should be $y=\frac{3}{5}x$ and $y=-\frac{3}{5}x$. Other than that, your method looks correct.

Complex equation: $z^8 = (1+z^2)^4$ What's up with this complex equation? $ z^8 = (1+z^2)^4 $ To start with, there seems to be a problem when we try to apply root of four to both sides of the equation: $ z^8 = (1+z^2)^4 $ $ z^2 = 1 + z^2 $ which very clearly doesn't have any solutions, but we know there are solutions: the problem is from an exam, and, besides, wolphram alpha gladily gives them to us. We've tried to solve it using the trigonomectric form, but the sum inside of the parenthesis is killing all of our attempts. Any help? Ideas?

The solutions of $z^8 = (1+z^2)^4$, and the problem you see, are clearly seen from a factorization process: $$\begin{align} z^8 - (1+z^2)^4 &= 0 \,, \\ [z^4 + (1+z^2)^2] [z^4 - (1+z^2)^2] &= 0 \,, \\ [z^2 + i(1+z^2)] [z^2 - i (1+z^2)] [z^2 - (1+z^2)] [z^2 + (1+z^2)] &= 0 \,. \end{align}$$ From the last line, it is clear that you have only considered one case (the third term) over four other cases. This case, in particular, has no solutions, $$\begin{align} [z^2 + i(1+z^2)] [z^2 - i (1+z^2)] [z^2 - (1+z^2)] [z^2 + (1+z^2)] &= 0 \,, \\ [(1+i)z^2 + i] [(1-i)z^2 - i ] [ - 1 ] [2z^2 + 1] &= 0 \,, \\ \end{align}$$ I believe this clarify a bit the answers posted by Millikan and Levap.

Is This Matrix Diagonalizable? Consider the matrix A below: $$A = \begin{pmatrix} 3 & 2 & 2 \\ -1 & 0 & -2 \\ 1 & 2 & 4\end{pmatrix}$$ Is the matrix A diagonalizable? If so, then find a diagonal matrix D and an invertible matrix P such that $P^{-1}AP=D$. I know it is supposed to be diagonalizable but I have tried to solve it and haven't succeeded. The format I'm used to is $A=PDP^{-1}$. Thanks.

Assuming $A$ is an $n\times n$ matrix, $A$ diagonalizable if $P_A(\lambda)$ has $n$ distinct real roots or if, for each eigenvalue, the algebraic multiplicity equals the geometric multiplicity. So let's find out: $$P_A(\lambda) = det(A - \lambda I) = det \begin{pmatrix} 3-\lambda & 2 & 2 \\ -1 & -\lambda & -2 \\ 1 & 2 & 4-\lambda\end{pmatrix}$$ $$ = (3-\lambda)((-\lambda)(4-\lambda) + 4) - 2(\lambda - 4 + 2) + 2(-2+\lambda) = 0$$ $$\Rightarrow -\lambda^3+7 \lambda^2-16 \lambda+12 = 0$$ $$\Rightarrow -(\lambda-3) (\lambda-2)^2 = 0$$ $$\lambda = 3 \text{ (multiplicity $1$)}$$ $$\lambda = 2 \text{ (multiplicity $2$)}$$ We see that $\lambda=2$ has algebraic multiplicity 2, so we need to check for its geometric multiplicity. Geometric multiplicity is equal to $n - Rank(A - \lambda I)$ $$Rank(A - 2I) = dim\left(col\begin{pmatrix} 1 & 2 & 2 \\ -1 & -2 & -2 \\ 1 & 2 & 2\end{pmatrix}\right) $$ $$ = dim\left(col\begin{pmatrix} 1 & 2 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\right) = 1$$ So $$\text{geometric multiplicity} = 3 - Rank(A - 2I) = 3 - 1 = 2$$ $$ = \text{algebraic multiplicity when $\lambda=2$}$$ So the matrix is diaginalizable. Now we can diaginalize it: Let's find our eigenvectors. For $\lambda=2$: $$\vec{V_1} + \vec{V_2} = null(A - \lambda I) = null(A - 2I) = null\begin{pmatrix} 1 & 2 & 2 \\ -1 & -2 & -2 \\ 1 & 2 & 2\end{pmatrix}$$ $$ = null\begin{pmatrix} 1 & 2 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}$$ $$ = \begin{pmatrix} -2 \\ 1 \\ 0\end{pmatrix}x_1 + \begin{pmatrix} -2 \\ 0 \\ 1\end{pmatrix}x_2$$ So our eigenpairs so far are $\left(2, \begin{pmatrix} -2 \\ 1 \\ 0\end{pmatrix}\right)$ and $\left(2, \begin{pmatrix} -2 \\ 0 \\ 1\end{pmatrix}\right)$ Now, to find it for $\lambda=3$: $$\vec{V} = null(A-\lambda I) = null(A - 3I) = null\begin{pmatrix} 0 & 2 & 2 \\ -1 & -3 & -2 \\ 1 & 2 & 1\end{pmatrix}$$ $$ = null\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}x_3$$ So our eigenpair is $\left(3, \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\right)$ We now have everything we need to make our diaginalized matrix: $$PDP^{-1} = \begin{pmatrix}-2 & -2 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{pmatrix}\begin{pmatrix}-2 & -2 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 1 \end{pmatrix}^{-1}$$ $$ = \begin{pmatrix}-2 & -2 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 1\end{pmatrix}\begin{pmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{pmatrix}\begin{pmatrix}1 & 3 & 2 \\ -1 & -2 & -1 \\ 1 & 2 & 2\end{pmatrix}$$

How to prove $\sum_p p^{-2} < \frac{1}{2}$? I am trying to prove $\sum_p p^{-2} < \frac{1}{2}$, where $p$ ranges over all primes. I think this should be doable by elementary methods but a proof evades me. Questions already asked here (eg. What is the value of $\sum_{p\le x} 1/p^2$? and Rate of convergence of series of squared prime reciprocals) deal with the exact value of the above sum, and so require some non-elementary math.

Here's a solution that exploits a comment of Oscar Lanzi under my other answer (using an observation that I learned from a note of Noam Elkies [pdf]). In particular, it avoids both the identity $\sum_{n \in \Bbb N} \frac{1}{n^2} = \frac{\pi^2}{6}$ and using integration. Let $\Bbb P$ denote the set of prime numbers and $X$ the union of $\{2\}$ and the set of odd integers $> 1$; in particular $\Bbb P \subset X$, so where $E$ denotes the set of positive, even integers: $$\sum_{p \in \Bbb P} \frac{1}{p^2} \leq \sum_{n \in X} \frac{1}{n^2} = \color{#00af00}{\sum_{n \in \Bbb N \setminus E} \frac{1}{n^2}} - \frac{1}{1^2} + \frac{1}{2^2}.$$ Now, $$\sum_{n \in \Bbb N} \frac{1}{n^2} < 1 + \sum_{n \in \Bbb N \setminus \{1\}} \frac{1}{n^2 - \frac{1}{4}} = 1 + \sum_{n \in \Bbb N \setminus \{1\}} \left(\frac{1}{n - \frac{1}{2}} - \frac{1}{n + \frac{1}{2}} \right) = 1 + \frac{2}{3} = \frac{5}{3};$$ the second-to-last equality follows from the telescoping of the sum in the third expression. The sum over just the even terms satisfies $$\sum_{m \in E} \frac{1}{m^2} = \sum_{n \in \Bbb N} \frac{1}{(2 n)^2} = \frac{1}{4} \sum_{n \in \Bbb N} \frac{1}{n^2} ,$$ and thus $$\color{#00af00}{\sum_{n \in \Bbb N \setminus E} \frac{1}{n^2} = \left(1 - \frac{1}{4}\right) \sum_{n \in \Bbb N} \frac{1}{n^2} < \frac{3}{4} \cdot \frac{5}{3} = \frac{5}{4}}.$$ Substituting in the first display equation above yields $$\sum_{p \in \Bbb P} \frac{1}{p^2} \leq \sum_{n \in X} \frac {1}{n^2} < \color{#00af00}{\frac{5}{4}} - 1 + \frac{1}{4} = \frac{1}{2} .$$

Recurrence relation for the determinant of a tridiagonal matrix Let $$f_n := \begin{vmatrix} a_1 & b_1 \\ c_1 & a_2 & b_2 \\ & c_2 & \ddots & \ddots \\ & & \ddots & \ddots & b_{n-1} \\ & & & c_{n-1} & a_n \end{vmatrix}$$ Apparently, the determinant of the tridiagional matrix above is given by the recurrence relation $$f_n = a_n f_{n-1} - c_{n-1} b_{n-1}f_{n-2}$$ with initial values $f_0 = 1$ and $f_{-1} = 0$ (according to Wikipedia). Can anyone please explain to me how they came to this recurrence relation? I don't really understand how to derive it.

For $n \ge 2$ using Laplace expansion on the last row gives \begin{align} f_n &= \begin{vmatrix} a_1 & b_1 \\ c_1 & a_2 & b_2 \\ & c_2 & \ddots & \ddots \\ & & \ddots & \ddots & b_{n-3} \\ & & & c_{n-3} & a_{n-2} & b_{n-2} \\ & & & & c_{n-2} & a_{n-1} & b_{n-1} \\ & & & & & c_{n-1} & a_n \end{vmatrix} \\ &= (-1)^{2n-1} c_{n-1} \begin{vmatrix} a_1 & b_1 \\ c_1 & a_2 & b_2 \\ & c_2 & \ddots & \ddots \\ & & \ddots & \ddots & b_{n-3} \\ & & & c_{n-3} & a_{n-2} \\ & & & & c_{n-2} & b_{n-1} \end{vmatrix} + (-1)^{2n} a_n \begin{vmatrix} a_1 & b_1 \\ c_1 & a_2 & b_2 \\ & c_2 & \ddots & \ddots \\ & & \ddots & \ddots & b_{n-2} \\ & & & c_{n-2} & a_{n-1} \end{vmatrix} \\ &= - c_{n-1} (-1)^{2(n-1)} b_{n-1} \begin{vmatrix} a_1 & b_1 \\ c_1 & a_2 & b_2 \\ & c_2 & \ddots & \ddots \\ & & \ddots & \ddots & b_{n-3} \\ & & & c_{n-3} & a_{n-2} \end{vmatrix} + a_n f_{n-1} \\ &= a_n f_{n-1} - c_{n-1} b_{n-1} f_{n-2} \end{align} as recurrence relation. For the initial conditions: From comparing the above formula with matrices for $n=1$ and $n=2$ we get: $$ f_1 = a_1 f_0 - c_0 b_0 f_{-1} \overset{!}{=} a_1 \\ f_2 = a_2 f_1 - c_1 b_1 f_0 \overset{!}{=} a_1 a_2 - c_1 b_1 $$ The latter implies $f_0 = 1$ and the former $f_{-1} = 0$.

Evaluate $\int \frac{x+1}{(x^2-x+8)^3}\, dx$ Could you give me a hint on how to find $$\int \frac{x+1}{(x^2-x+8)^3}\, dx$$ It doesn't seem like partial fractions are the way to go with here and using the integration by parts method seems to be tedious. I have also tried substituting $(x^2-x+8)$ but it gets even more complicated then. Is there a way to solve this using only basic formulas?

We don't need to use any integration by parts. We start by completing the square $$\int \frac{x+1}{(x^2-x+8)^3} dx = \int \frac{x-\frac{1}{2}+\frac{3}{2}}{((x-\frac{1}{2})^2+\frac{31}{4})^3} dx =$$ $$\underbrace{ \int \frac{x-\frac{1}{2}}{((x-\frac{1}{2})^2+\frac{31}{4})^3} dx}_{=:I_1}+ \frac{3}{2}\underbrace{\int\frac{1}{((x-\frac{1}{2})^2+\frac{31}{4})^3}dx}_{=:I_2}$$ For $I_1$ substitution $(x-\frac{1}{2})^2 = t$ then $\frac{dt}{dx}= 2(x-\frac{1}{2})$ so $$I_1 = \frac{1}{2}\int \frac{1}{t^3}dt = -\frac{1}{4t^2}+C$$ For $I_2$ first substitution $x-\frac{1}{2}=t$ $$I_2 = \int \frac{1}{(t^2+\frac{31}{4})^3}dt$$ then rearrange $$ \int \frac{1}{(t^2+\frac{31}{4})^3}dt = \frac{4^3}{31^3}\int \frac{1}{{\left(1+\left(\frac{2t}{\sqrt{31}}\right)^2\right)}^3}dt.$$ Now another substitution $\frac{2t}{\sqrt{31}}=z$ gives us the following $$\frac{4^3}{31^3}\int \frac{1}{{\left(1+\left(\frac{2t}{\sqrt{31}}\right)^2\right)}^3}dt = \left(\frac{4}{31}\right)^\frac{5}{2}\int \frac{1}{(1+z^2)^3}dz.$$ The last substitution $z=\tan{y}$ and $dz= \frac{1}{\cos^2y}dy$, and we get something much simpler $$\left(\frac{4}{31}\right)^\frac{5}{2}\int \frac{1}{(1+z^2)^3}dz = \left(\frac{4}{31}\right)^\frac{5}{2}\int \frac{1}{\cos^2y}\cos^6(y)dy= \left(\frac{4}{31}\right)^\frac{5}{2}\int \cos^4(y)dy$$ Now we use that $\cos^2 y = {1 + \cos(2y) \over 2}$ to show that $\cos^4 y = \frac{\cos(4y) + 4\cos(2y)+3}{8}$ and thus $$\left(\frac{4}{31}\right)^\frac{5}{2}\int \cos^4(y)dy = \left(\frac{4}{31^\frac{5}{2}}\right)\int (\cos(4y) + 4\cos(2y)+3)dy$$ which is easy to find. Now you have to go backwards with all the substitutions to get the result with variable $x$ only.

Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)? Proposition 1: $\frac{0}{0} = 0$ Proof: Suppose that $\frac{0}{0}$ is not equal to $0$ $\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$ Therefore, $\frac{0}{0} = 0$. Q.E.D. Update (2015-12-01) after your answers: Proposition 2: $\frac{0}{0}$ is not a real number Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]: Suppose that $\frac{0}{0}= x$, where $x$ is a real number. Then, either $x = 0$ or $x$ is not equal to $0$. 1) Suppose $x = 0$, that is $\frac{0}{0} = 0$ Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $ Contradiction Therefore, it is not the case that $x = 0$. 2) Suppose that $x$ is not equal to $0$. $x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction Therefore, it is not the case that $x$ is a real number that is not equal to $0$. Therefore, $\frac{0}{0}$ is not a real number. Q.E.D. Update (2015-12-02) If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers. Proposition 3: $\frac{0}{0}$ is not a real number Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number. $\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$ $ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$ Q.E.D. Update (2015-12-07): How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)? Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$. Then, $\frac{0}{0} = 0$ Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$. $x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$ $\therefore \frac{0}{0}=0$ Q.E.D. Suggested definition of division of real numbers: If $b \ne 0$, then $\frac{a}{b}=c$ iff $a=bc$ If $a=0$ and $b=0$, then $\frac{a}{b}=0$ If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined. A somewhat more minimalistic version: Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$. Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$. $a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$ $\therefore \frac{0}{0}=0$ Q.E.D.

0/0 is indeterminate. Which means it has infinite number of values. (0)(1) = 0 => 0/0 = 1 (0)(2) = 0 => 0/0 = 2 (0)(4) = 0 => 0/0 = 3 ..... From above it can be shown that 0/0 has infinite number of values which make it meaningless. Hence the term.

What is the integral of $\frac{x-1}{(x+3)(x^2+1)}$? I've worked with partial fractions to get the integral in the form $$\int\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\,dx$$ Is there a quicker way?

Notice, for $(Bx+C)$ part, you should use separation as follows $$\int \frac{x-1}{(x+3)(x^2+1)}\ dx=\int \frac{-2}{5(x+3)}+\frac{2x-1}{5(x^2+1)}\ dx$$ $$=\frac{1}{5}\int \left(-\frac{2}{x+3}+\frac{2x}{x^2+1}-\frac{1}{x^2+1}\right)\ dx$$ $$=\frac{1}{5} \left(-2\int \frac{1}{x+3}\ dx+\int \frac{d(x^2)}{x^2+1}-\int\frac{1}{x^2+1}\ dx\right)$$

Linear Algebra - seemingly incorrect result when looking for a basis For the following matrix $$ A = \begin{pmatrix} 3 & 1 & 0 & 0 \\ -2 & 0 & 0 & 0 \\ -2 & -2 & 1 & 0 \\ -9 & -9 & 0 & -3 \end{pmatrix} $$ Find a basis for the eigenspace $E_{\lambda}(A)$ of each eigen value. First step is to find the characteristic polynomial to find the eigenvalues. $$\det(A- I\lambda) = 0 \implies \lambda ^4 - \lambda^3 - 7 \lambda^2 + 13 \lambda -6 = 0 $$ $$\lambda = 1 \lor \lambda = 2 $$ Now that we have the eigen values, to find a basis for $E_{\lambda}(A)$ for the eigen values $\lambda$, we find vectors $\mathbf{v}$ that satisfy $(A-\lambda I)\mathbf{v} = 0$: $$(A - 2I)\mathbf{v} = \begin{pmatrix} 1 & 1 & 0 & 0 \\ -2 & -2 & 0 & 0 \\ -2 & -2 & -1 & 0 \\ -9 & -9 & 0 & -5 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \implies \begin{cases} v_1 + v_2 = 0\\ -2v_1 -2v_2 = 0 \\ -2v_1 - 2v_2 -v_3 = 0 \\ -9v_1 -9v_2 - 5v_4 = 0 \end{cases}$$ $\implies v_2 = -v_1 \land v_3 = 0 \land v_4 = 0$. So $(v_1, v_2, v_3, v_4) = (v_1, -v_1, 0, 0) = v_1(1, -1, 0, 0)$ so $(1, -1, 0, 0)$ is the basis. However, this is apparently incorrect. What have I done wrong here? Also, for the eigenvalue $\lambda = 1$ I do get the correct answer, so only the part above gives me problems.

You're missing an eigenvalue! I think you've made a typo in your characteristic polynomial, writing $-\lambda^4$ instead of $\lambda^4$. The characteristic polynomial factorizes to $(x-2)(x+3)(x-1)^2$, hence the eigenvalues are $-3$, $2$, and $1$. The vector you have found is indeed a basis for the 2-eigenspace, and you can easily to compute that the basis vectors of the eigenspaces of $-3$ and $1$ are $\left( \begin{array}{ccc} 0 \\ 0 \\ 0 \\ 1 \end{array} \right)$ and $\left( \begin{array}{ccc} 0 \\ 0 \\ 1 \\ 0 \end{array} \right)$ respectively.

Adding and multiplying piecewise functions How do I add and multiply two piecewise functions? $$ f(x)= \begin{cases} x+3 &\text{if }x<2\\ \dfrac{x+13}{3} &\text{if }x>2 \end{cases} $$ $$ g(x)= \begin{cases} x-3 &\text{if }x<3\\ x-5 &\text{if }x>3 \end{cases} $$

It helps to think about the fact that you can only add and multiply stuff that actually exists (i.e. is defined): We can only add and multiply these functions in places where they both exist at the same time, namely: $x<2\\ 2<x<3\\ x>3$ The final step is to check what the functions equal at each of these segments, then put it all together: $$ f(x)\times g(x)= \begin{cases} (x+3)\times(x-3) &\mbox{if } \quad x<2\\ (\frac{x+13}{3})\times(x-3) &\mbox{if } \quad 2<x<3\\ (\frac{x+13}{3})\times(x-5) &\mbox{if } \quad x>3 \end{cases} $$ $$ f(x)+g(x)= \begin{cases} (x+3)+(x-3) &\mbox{if } \quad x<2\\ (\frac{x+13}{3})+(x-3) &\mbox{if } \quad 2<x<3\\ (\frac{x+13}{3})+(x-5) &\mbox{if } \quad x>3 \end{cases} $$ BONUS ROUND: What about when functions have different arguments, like $f(x)$ and g(y)? Since there is no $x$ in $g()$ and there is no $y$ in $f()$, we don't have to worry about one function being defined when the other one isn't. They live in different worlds (i.e. different domains) so we just have to add or multiply each case of one with each case of the other. For example: $$ f(x)= \begin{cases} x+3 &\mbox{if } \quad x<2\\ \frac{x+13}{3} &\mbox{if } \quad x>2 \end{cases} $$ $$ g(y)= \begin{cases} y-3 &\mbox{if } \quad y<3\\ y-5 &\mbox{if } \quad y>3 \end{cases} $$ $$ f(x)\times g(y)= \begin{cases} (x+3)\times(y-3) &\mbox{if } \quad x<2,y<3\\ (x+3)\times(y-5) &\mbox{if } \quad x<2,y>3\\ (\frac{x+13}{3})\times(y-3) &\mbox{if } \quad x>2,y<3\\ (\frac{x+13}{3})\times(y-5) &\mbox{if } \quad x>2,y>3 \end{cases} $$

Prove $a^2+b^2\geq \frac{c^2}{2}$ and friends if $a+b\geq c\geq0$ Sorry for my inequality spam, but I got to prepare for my exams today :( Here's another: Problem: Prove$$a^2+b^2\geq \frac{c^2}{2}$$$$a^4+b^4\geq \frac{c^4}{8}$$$$a^8+b^8\geq \frac{c^8}{128}$$ if $a+b\geq c\geq0$ Attempt: Working backwards: $$a^2+b^2\geq \frac{c^2}{2}$$ $$\implies 2a^2+2b^2\geq c^2$$ I am stuck on the first one, let alone the others. I know by AM-GM, $\frac{a^2+b^2}{2}\geq ab$ but how can I use it here?

Using the information you supplied, you could also just square both sides of the original inequality to obtain $a^2+2ab+b^2 \geq c^2$. Divide the inequality by 2 and subtract $ab$ to the other side. Given the AM-GM, we know that $\frac{a^2+b^2}{2} \geq ab $. Adding this inequality to the previous one we obtain $2\frac {a^2+b^2}{2} \geq \frac {c^2}{2} +ab-ab $. Simplifying gives us our desired result of $a^2+b^2\geq \frac {c^2}{2}$. As macavity said, you obtain the rest of the inequalities by substituting $a^2$ for $a $, $b^2$ for $b $, and so on.

Area bounded between the curve $y=x^2 - 4x$ and $y= 2x/(x-3)$ I've determined the intersects to be $x = 0, 2, 5$ and that $\frac{2x}{x-3}$, denoted as $f(x)$, is above $x(x-4)$, denoted as $g(x)$, so to find the area, I'll need to find the integral from $0$ to $2$ of $f(x) - g(x)$. But I've been stuck for a while playing around with this question.

Notice, the area bounded by the curves from $x=0$ to $x=2$, is given as $$\int_{0}^{2}\left(\frac{2x}{x-3}-(x^2-4x)\right)\ dx$$ $$=\int_{0}^{2}\left(\frac{2(x-3)+6}{x-3}-x^2+4x\right)\ dx$$ $$=\int_{0}^{2}\left(2+\frac{6}{x-3}-x^2+4x\right)\ dx$$ $$=\left(2x+6\ln|x-3|-\frac{x^3}{3}+2x^2\right)_{0}^{2}$$ $$=4-\frac{8}{3}+8-6\ln (3)$$$$=\color{red}{\frac{28}{3}-6\ln (3)}$$

Series' convergence - making my ideas formal Find the collection of all $x \in \mathbb{R}$ for which the series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges. My first step was the use the ratio test: $$ \lim_{n \to \infty} \dfrac{(3^{n+1}+n+1) \cdot |x|^{n+1}}{(3n+n) \cdot |x|^n} = \lim_{n \to \infty} \dfrac{3^{n+1}|x|+n|x|+|x| }{3^n+n} = \lim_{n \to \infty} ( \dfrac{3^{n+1}|x|}{3^n+n} + \dfrac{n|x|}{3^n+n} + \dfrac{|x|}{3^n+n}) $$ $$ = \lim_{n \to \infty} (\dfrac{3|x|}{1+\dfrac{n}{3^n}} + \dfrac{|x|}{\dfrac{3^n}{n} + 1} + \dfrac{|x|}{3^n + n}) = 3|x| + 0 + 0 = 3|x| $$ So we want $3|x| < 1$, i.e. $|x| < \dfrac{1}{3}$. But we have to also check the points $x=\dfrac{1}{3}, -\dfrac{1}{3}$, where the limit equals $1$. For $x=\dfrac{1}{3}$ we have $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot (\dfrac{1}{3})^n = \displaystyle \sum_{n=1}^\infty (1 + n \cdot \dfrac{1}{3}^n) $ which obviously diverges. For $x=-\dfrac{1}{3}$ we have $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot (-\dfrac{1}{3})^n = \displaystyle \sum_{n=1}^\infty ((-1)^{n} + n \cdot (-\dfrac{1}{3})^n)$. Now I know this diverges because $(-1)^n$ and $(-\dfrac{1}{3})^n$ have alternating coefficients. But is there a theorem that I can use here? So we conclude that $|x| < \dfrac{1}{3}$, but is the above work enough to show it conclusively?

The series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges if $|x|<\frac{1}{3}$ Indeed $$(3^n + n)\cdot x^n\sim_{\infty} 3^n \cdot x^n$$ but this is a geometric series that converges only if $|x|<\frac {1}{3}$

Sum of all values of $b$ if the difference between the largest and smallest values of the function $f(x)=x^2-2bx+1$ in the segment $[0,1]$ is $4$ Find sum of all possible values of the parameter $b$ if the difference between the largest and smallest values of the function $f(x)=x^2-2bx+1$ in the segment $[0,1]$ is $4$. I found that the smallest value of $f(x)=x^2-2bx+1$ is $1-b^2$ But i do not know what will be the largest value of the quadratic expression,whether it is at $x=0$ or $x=1$. Please help me.Thanks.

Since $f(x) = (x-b)^2 + (1 - b^2)$, the vertex will be at $(b, 1-b^2)$. Also $f(1) = 2-2b$ and $f(0) = 1$. When the vertex is at or to the left of $x=0$, then $f(x)$ is increasing on $[0, 1]$. So $$ \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) = f(1) - f(0) = 1-2b$$ and $1-2b=4$ when $b = -\frac 32$ When the vertex is at or to the right of $x=1$, then $f(x)$ is decreasing on $[0, 1]$. So $$ \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) = f(0) - f(1) = 2b-1$$ and $2b-1=4$ when $b = \frac 52$ When the x-coordinate of the vertex is in the interval $[0,\frac 12]$ then $\begin{align} \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) &= \max\{f(0),f(1)\} - (1-b^2)\\ &= f(1) - (1-b^2)\\ &= (b-1)^2 \end{align}$ This does not equal $4$ when the x-coordinate of the vertex, b, is in the interval $[0,\frac 12]$ Finally, when the x-coordinate of the vertex is in the interval $(\frac 12, 1]$ then $\begin{align} \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) &= \max\{f(0),f(1)\} - (1-b^2)\\ &= f(0) - (1-b^2)\\ &= b^2 \end{align}$ This does not equal $4$ when the x-coordinate of the vertex, b, is in the interval $(\frac 12, 0]$. So the requested sum is -$\frac 32 + \frac 52 = 1$

If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$ Prove that there exists integers $S,T$ such that $a_{n+1}=Sa_n+Ta_{n-1}$. Attempt I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd have the system of diophantine equations: $$ 7=3S+2T\\ 18=7S+3T\\ 47=18S+7T $$ Now: it seems that all of the $a_i$ are pairwise coprime, so these equations should always have solutions, but how could I check that the intersection of all the solutions is not $\emptyset$?

The result is $a_{n + 1} = 3 a_n - a_{n - 1}$. HINT: Let $a_{k + 1} = a_k + b_k$. Then, by the original equation, we get $$a_{n - 1} + b_{n - 1} + b_n = \frac {(a_{n - 1} + b_{n - 1})^2 + 5} {a_{n - 1}} = a_{n - 1} + 2 b_{n - 1} + \frac {b_{n - 1}^2 + 5} {a_{n - 1}}.$$ Hence, $$b_n = b_{n - 1} + \frac {b_{n - 1}^2 + 5} {a_{n - 1}}.$$

Trig : With point $(X,Y)$ and a end point $(x_2,y_2)$, work out the 2 angles So imagine your arm, you have a shoulder point be $X,Y (0,0)$ you have a angles off this, then you have a elbow point $(x_1,y_1)$ and another angles which ends at your hand. If you know the arm length (4 and 4), you know your starting point $(0,0)$ and you want to get the hand to say $(6,2)$ is there some maths to work out what the angles at the elbow and shoulder should be?

Draw a line from the origin to the hand, called the distance line, then you have (based on your image, I assumed $y = -2$): $$ \begin{eqnarray} l &=& 4 & \textrm{length of an arm segment} \\ d &=& \sqrt{x^2 + y^2} = 2\sqrt{10} & \textrm{distance from the origin to the hand} \\ \tan \theta_1 &=& \frac{y}{x} = -\frac{1}{3} & \textrm{angle between x-axis and distance line} \\ \cos \theta_2 &=& \frac{d/2}{l} = \frac{\sqrt{10}}{4} & \textrm{angle between distance line and arm} \\ \end{eqnarray} $$ Express $\cos \theta_2$ in terms of $\tan$ and then apply the tangent sum identity to get the angle at the shoulder: $$ \begin{eqnarray} \tan \theta_2 &=& \frac{-\sqrt{1 - \cos^2 \theta_2}}{\cos \theta_2} &=& -\frac{\sqrt{15}}{5} & \textrm{negative in fourth quadrant} \\ \tan (\theta_1 + \theta_2) &=& \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} &=& -\frac{12 + 5\sqrt{15}}{21} & \textrm{tangent sum identity} \\ \theta_1 + \theta_2 &=& \arctan -\frac{12 + 5\sqrt{15}}{21} &\approx& -0.98 \, \textrm{rad} \end{eqnarray} $$ Apply the double angle formula to get the angles with the distance line at the shoulder and hand, then remains the angle at the elbow: $$ \begin{eqnarray} \cos 2\theta_2 &=& 2 \cos^2 \theta_2 - 1 &=& \frac{1}{4} & \textrm{double angle formula} \\ \theta_3 &=& \pi - \arccos \frac{1}{4} &\approx& 1.82 \, \textrm{rad} \\ \end{eqnarray} $$

Write $\frac {1}{1+z^2}$ as a power series centered at $z_0=1$ I'm trying to solve a question where I need to write $\frac {1}{1+z^2} $ as a power series centered at $z_0=1$ I'm not allowed to use taylor expansion. So my first thought was to rewrite the function in a form where I can apply the basic identity: $$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n $$ So let's rewrite the original function $f(z) = \frac {1}{1+z^2} $ and use above identity: $$\frac {1}{1+z^2} = \frac {1}{1-(-z^2)} $$ But observe that I need to center it at $z_0=1$ so I rewrote it to: $$\frac {1}{1-(-z^2)}=\frac {1}{2z-(-(z-1)^2)} $$ I can factor out $\frac {1}{2z}$ and use the basic identity to get : $$\frac {1}{2z}\frac {1}{1-\frac{(z-1)^2}{2z}} = \frac {1}{2z}\sum_{n=0}^\infty \left(\frac{-(z-1)^2}{2z}\right)^n =\frac {1}{2z}\sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^n} = \sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^{n+1}}$$ The problem is if I ask wolfram alpha: power series$ \frac{1}{1+z^2} $centered at z = 1 it will give me a different answer that doesn't match the sum I found. Can anyone help me figure this one out? What am I doing wrong?

$$ \begin{align} \frac1{1+z^2} &=\frac1{2+2w+w^2}\tag1\\[9pt] &=\frac1{2i}\frac1{1-i+w}-\frac1{2i}\frac1{1+i+w}\tag2\\[6pt] &=\frac{1-i}4\frac1{1+\frac{1+i}2w}+\frac{1+i}4\frac1{1+\frac{1-i}2w}\tag3\\ &=\sum_{k=0}^\infty\frac{(-1)^k}4\left[\left(\frac{1+i}2\right)^{k-1}+\left(\frac{1-i}2\right)^{k-1}\right](z-1)^k\tag4\\ &=\frac1{\sqrt2}\sum_{k=0}^\infty\left(-\frac1{\sqrt2}\right)^k\cos\left(\frac{\pi(k-1)}4\right)(z-1)^k\tag5 \end{align} $$ Explanation: $(1)$: substitute $z=w+1$ $(2)$: partial fractions $(3)$: use the forms $\frac1{1+aw}$ for easier series $(4)$: write $\frac1{1+aw}$ as geometric series and undo $(1)$ $(5)$: convert to cosines

Arc length of the squircle The squircle is given by the equation $x^4+y^4=r^4$. Apparently, its circumference or arc length $c$ is given by $$c=-\frac{\sqrt[4]{3} r G_{5,5}^{5,5}\left(1\left| \begin{array}{c} \frac{1}{3},\frac{2}{3},\frac{5}{6},1,\frac{4}{3} \\ \frac{1}{12},\frac{5}{12},\frac{7}{12},\frac{3}{4},\frac{13}{12} \\ \end{array} \right.\right)}{16 \sqrt{2} \pi ^{7/2} \Gamma \left(\frac{5}{4}\right)}$$ Where $G$ is the Meijer $G$ function. Where can I find the derivation of this result? Searching for any combination of squircle and arc length or circumference has led to nowhere.

By your definition, $\mathcal{C} = \{(x,y) \in \mathbb{R}^{2}: x^4 + y^4 = r^4\}$. Which can be parametrized as \begin{align} \mathcal{C} = \begin{cases} \left(+\sqrt{\cos (\theta )},+\sqrt{\sin (\theta )} \right)r\\ \left(+\sqrt{\cos (\theta )},-\sqrt{\sin (\theta )} \right)r\\ \left(-\sqrt{\cos (\theta )},+\sqrt{\sin (\theta )} \right)r\\ \left(-\sqrt{\cos (\theta )},-\sqrt{\sin (\theta )} \right)r \end{cases} , \qquad 0 \leq \theta \leq \frac{\pi}{2}, \, 0<r \end{align} Now, look at this curve in $\mathbb{R}^{2}_{+}$ as $y = \sqrt[4]{r^4-x^4}$, then observe that symmetry with both axis. It yields the arc length is just: $$c = 4 \int_{0}^{r} \sqrt{1+\left(\dfrac{d}{dx}\sqrt[4]{r^4-x^4}\right)^2} \,dx = 4 \int_{0}^{r} \sqrt{1+\frac{x^6}{\left(r^4-x^4\right)^{3/2}}} \,dx$$

Is there a special value for $\frac{\zeta'(2)}{\zeta(2)} $? The answer to an integral involved $\frac{\zeta'(2)}{\zeta(2)}$, but I am stuck trying to find this number - either to a couple decimal places or exact value. In general the logarithmic deriative of the zeta function is the dirichlet series of the van Mangolt function: $$\frac{\zeta'(s)}{\zeta(s)} = \sum_{n \geq 0} \Lambda(n) n^{-s} $$ Let's cheat: Wolfram Alpha evaluates this formula as: $$ \frac{\zeta'(2)}{\zeta(2)} = - 12 \log A + \gamma + \log 2 + \log \pi \tag{$\ast$}$$ This formula features some interesting constants: * *$A$ is the Glaisher–Kinkelin constant 1.2824271291006226368753425688697... *$\gamma$ is the Euler–Mascheroni constant 0.577215664901532860606512090082... *$\pi$ is of course 3.14... Wikipedia even says that $A$ and $\pi$ are defined in similar ways... which is an interesting philosophical point. Do we have a chance of deriving $(\ast)$?

By differentiating both sides of the functional equation$$ \zeta(s) = \frac{1}{\pi}(2 \pi)^{s} \sin \left( \frac{\pi s}{s} \right) \Gamma(1-s) \zeta(1-s),$$ we can evaluate $\zeta'(2)$ in terms of $\zeta'(-1)$ and then use the fact that a common way to define the Glaisher-Kinkelin constant is $\log A = \frac{1}{12} - \zeta'(-1)$. Differentiating both sides of the functional equation, we get $$\begin{align} \zeta'(s) &= \frac{1}{\pi} \log(2 \pi)(2 \pi)^{s} \sin \left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s) + \frac{1}{2} (2 \pi)^{s} \cos \left(\frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s)\\ &- \frac{1}{\pi}(2 \pi)^{s} \sin \left(\frac{\pi s}{2} \right)\Gamma^{'}(1-s) \zeta(1-s) - \frac{1}{\pi}(2 \pi)^{s} \sin \left(\frac{\pi s}{2} \right)\Gamma(1-s) \zeta'(1-s). \end{align}$$ Then letting $s =-1$, we get $$\zeta'(-1) = -\frac{1}{2\pi^{2}}\log(2 \pi)\zeta(2) + 0 + \frac{1}{2 \pi^{2}}(1- \gamma)\ \zeta(2) + \frac{1}{2 \pi^{2}}\zeta'(2)$$ since $\Gamma'(2) = \Gamma(2) \psi(2) = \psi(2) = \psi(1) + 1 = -\gamma +1. \tag{1}$ Solving for $\zeta'(2)$, $$ \begin{align} \zeta'(2) &= 2 \pi^{2} \zeta'(-1) + \zeta(2)\left(\log(2 \pi)+ \gamma -1\right) \\ &= 2 \pi^{2} \left(\frac{1}{12} - \log (A) \right) + \zeta(2)\left(\log(2 \pi)+ \gamma -1\right) \\ &= \zeta(2) - 12 \zeta(2) \log(A)+ \zeta(2) \left(\log(2 \pi)+ \gamma -1\right) \tag{2} \\ &= \zeta(2) \left(-12 \log(A) + \gamma + \log(2 \pi) \right). \end{align}$$ $(1)$ https://en.wikipedia.org/wiki/Digamma_function $(2)$ Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ EDIT: If you want to show that indeed $$\zeta'(-1)= \frac{1}{12}- \lim_{m \to \infty} \left( \sum_{k=1}^{m} k \log k - \left(\frac{m^{2}}{2}+\frac{m}{2} + \frac{1}{12} \right) \log m + \frac{m^{2}}{4} \right) = \frac{1}{12}- \log(A),$$ you could differentiate the representation $$\zeta(s) = \lim_{m \to \infty} \left( \sum_{k=1}^{m} k^{-s} - \frac{m^{1-s}}{1-s} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} \right) \ , \ \text{Re}(s) >-3. $$ This representation can be derived by applying the Euler-Maclaurin formula to $\sum_{k=n}^{\infty} {k^{-s}}$.

Differentiating $x^2=\frac{x+y}{x-y}$ Differentiate: $$x^2=\frac{x+y}{x-y}$$ Preferring to avoid the quotient rule, I take away the fraction: $$x^2=(x+y)(x-y)^{-1}$$ Then: $$2x=(1+y')(x-y)^{-1}-(1-y')(x+y)(x-y)^{-2}$$ If I were to multiply the entire equation by $(x-y)^2$ then continue, I get the solution. However, if I continue the following, I don't. Likely some place I erred, but I cannot figure out where: Expansion: $$2x=(x-y)^{-1}+y'(x-y)^{-1}-(x+y)(x-y)^{-2}+y'(x+y)(x-y)^{-2}$$ Preparing to isolate for $y'$: $$2x-(x-y)^{-1}+(x+y)(x-y)^{-2}=y'(x-y)^{-1}+y'(x+y)(x-y)^{-2}$$ $$2x-(x-y)^{-1}+(x+y)(x-y)^{-2}=y'[(x-y)^{-1}+(x+y)(x-y)^{-2}]$$ Isolating $y'$: $$y'=\frac{2x-(x-y)^{-1}+(x+y)(x-y)^{-2}}{(x-y)^{-1}+(x+y)(x-y)^{-2}}$$ Multiple top and bottom by $(x-y)$: $$y'=\frac{2x(x-y)-1+(x+y)(x-y)^{-1}}{1+(x+y)(x-y)^{-1}}$$ Then, inserting $x^2$ into $(x+y)(x-y)^{-1}$, I get: $$y'=\frac{2x(x-y)-1+x^2}{1+x^2}$$ While the answer states: $$y'=\frac{x(x-y)^2+y}{x}$$ Which I do get if I multiplied the entire equation by $(x-y)^2$ before. It does not seem to be another form of the answer, as putting $x=2$, the denominator cannot match each other. Where have I gone wrong?

It's worth noting that you haven't actually avoided the quotient rule, at all. Rather, you've simply written out the quotient rule result in a different form. However, we can avoid the quotient rule as follows. First, we clear the denominator to give us $$x^2(x-y)=x+y,$$ or equivalently, $$x^3-x^2y=x+y.$$ Gathering the $y$ terms on one side gives us $$x^3-x=x^2y+y,$$ or equivalently, $$x^3-x=(x^2+1)y.\tag{$\heartsuit$}$$ Noting that $x^2+1$ cannot be $0$ (assuming that $x$ is supposed to be real), we have $$\frac{x^3-x}{x^2+1}=y.\tag{$\star$}$$ Now, differentiating $(\heartsuit)$ with respect to $x$ gives us $$3x^2-1=2xy+(x^2+1)y',$$ or equivalently $$3x^2-1-2xy=(x^2+1)y'.$$ Using $(\star)$ then gives us $$3x^2-1-2x\cdot\frac{x^3-x}{x^2+1}=(x^2+1)y',$$ which we can readily solve for $y'.$ As for what you did wrong, the answer is: concluding that different denominators meant different values! Indeed, if $x=2,$ then solving $x^2=\frac{x+y}{x-y}$ for $y$ means that $y=\frac65.$ Substituting $x=2$ and $y=\frac65$ into $y'=\frac{2x(x-y)-1+x^2}{1+x^2}$ yields $$y'=\cfrac{\frac{31}5}5=\frac{31}{25},$$ while substituting $x=2$ and $y=\frac65$ into $y'=\frac{x(x-y)^2+y}{x}$ yields $$y'=\cfrac{\frac{62}{25}}2=\frac{31}{25}.$$ Hence, the answer is the same in the $x=2$ case! Now, more generally, using $(\star)$ in the equation $y'=\frac{2x(x-y)-1+x^2}{1+x^2}$ yields $$y'=\frac{x^4+4x^2-1}{(x^2+1)^2}.$$ The same result is achieved by using $(\star)$ in the equation $y'=\frac{x(x-y)^2+y}{x}.$ Hence, your answer is the same in both cases, though it doesn't look like it!

The value of double integral $\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dx\,dy$? Given double integral is : $$\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dx\,dy$$ My attempt : We can't solve since variable $x$ can't remove by limits, but if we change order of integration, then $$\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dy\,dx$$ $$\implies\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dy\,dx = \frac{1}{2}$$ Can you explain in formal way, please? Edit : This question was from competitive exam GATE. The link is given below on comments by Alex M. and Martin Sleziak(Thanks).

$$\begin{align}\int_{0}^{1}\int_{0}^{\frac{1}{x}}\frac{x}{1+y^2}\space\text{d}x\text{d}y&= \int_{0}^{1}\left(\int_{0}^{\frac{1}{x}}\frac{x}{1+y^2}\space\text{d}x\right)\text{d}y\\&=\int_{0}^{1}\left(\frac{1}{1+y^2}\int_{0}^{\frac{1}{x}}x\space\text{d}x\right)\text{d}y\\&=\int_{0}^{1}\left(\frac{\left[x^2\right]_{0}^{\frac{1}{x}}}{2\left(1+y^2\right)}\right)\text{d}y\\&=\int_{0}^{1}\left(\frac{\left(\frac{1}{x}\right)^2-0^2}{2\left(1+y^2\right)}\right)\text{d}y\\&=\int_{0}^{1}\left(\frac{\frac{1}{x^2}}{2\left(1+y^2\right)}\right)\text{d}y\\&=\int_{0}^{1}\frac{1}{2x^2(1+y^2)}\text{d}y\\&=\frac{1}{2x^2}\int_{0}^{1}\frac{1}{1+y^2}\text{d}y\\&=\frac{\left[\arctan(y)\right]_{0}^{1}}{2x^2}\\&=\frac{\arctan(1)-\arctan(0)}{2x^2}\\&=\frac{\frac{\pi}{4}-0}{2x^2}\\&=\frac{\frac{\pi}{4}}{2x^2}\\&=\frac{\pi}{8x^2}\end{align}$$

Division in finite fields Let's take $GF(2^3)$ as and the irreducible polynomial $p(x) = x^3+x+1$ as an example. This is the multiplication table of the finite field I can easily do some multiplication such as $$(x^2+x)\cdot(x+1) = x^3 + x^2 + x +1 = x+1+x^2+x+2 = x^2$$ I am wondering how to divide some random fields such as $x^2 / (x+1)$. The result is $x^2+x$ (compare above). But how do I actually calculate this. Polynomial long division does not help be: * *Why don't I get $x+1$ as result? *How can I calculate $x / (x^2+x+1)$? The result should be $x+1$

If we have a field $K = F[X]/p(X)$, then we can compute the inverse of $\overline{q(X)}$ in $K$ as follows. Since $p$ is irreducible, either $q$ is zero in $K$ or $(p,q)=1$. By the Euclidean algorithm, we can find $a,b$ such that $a(X)p(X) + b(X)q(X) = 1$. Then $\overline{b(X)} \cdot \overline{q(X)} = \overline{1}$, so $\overline{b(X)}$ is the inverse of $\overline{q(X)}$. In your example, the Euclidean algorithm gives us $(x+1)(x^3 + x + 1) + (x^2)(x^2 + x+1) = 1$ (modulo 2), so $(x^2+x+1)^{-1} = x^2$ and $\frac{x}{x^2+x+1} = x(x^2) = x^3 = x+1$.

What is the probability that, at the end of the game, one card of each color was turned over in each of the three rounds? Three players each have a red card, blue card and green card. The players will play a game that consists of three rounds. In each of the three rounds each player randomly turns over one of his/her cards without replacement. What is the probability that, at the end of the game, one card of each color was turned over in each of the three rounds? Express your answer as a common fraction. I thought the answer would just be $1*(\dfrac{1}{3})^2*1*(\dfrac{1}{2})^2$ since the first person can have any card shown and the next $2$ must match, then the next round the first person can have any card shown and the others must match. The last time everyone matches so the probability is $1$.

The probability that the three players choose different colors on the first turn is $\frac{3}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}=\frac{2}{9}$. Given that they do, consider the generating function for the colors played on the second turn: $(r+b)(b+g)(r+g)=\color{red}1b^2 2g+\color{red}1b^2 r+\color{red}1b g^2+\color{green}2 b g r+\color{red}1b r^2+\color{red}1g^2 r+\color{red}1g r^2$. The probability that the players choose different colors in the second round is therefore $\frac{\color{green}2}{\color{green}2+\color{red}6}=\frac{1}{4}$. Given that they do, which happens with probability $\frac{2}{9}\cdot\frac{1}{4}=\frac{1}{18}$, the probability they choose different colors on the last turn is $1$, so the final probability is $\frac{1}{18}$.

Writing a summation as the ratio of polynomial with integer coefficients Write the sum $\sum _{ k=0 }^{ n }{ \frac { { (-1) }^{ k }\left( \begin{matrix} n \\ k \end{matrix} \right) }{ { k }^{ 3 }+9{ k }^{ 2 }+26k+24 } } $ in the form $\frac { p(n) }{ q(n) }$, where $p(n)$ and $q(n)$ are polynomials with integral coefficients. I am not able to progress in this problem.Please help. Thank you.

$\sum _{ k=0 }^{ n }{ \frac { { (-1) }^{ k }\left( \begin{matrix} n \\ k \end{matrix} \right) }{ { k }^{ 3 }+9{ k }^{ 2 }+26k+24 } } = \sum _{k=0}^{n} {\frac { (-1)^k \dbinom{n}{k}}{2(k+2)} }-\sum _{k=0}^{n} {\frac { (-1)^k \dbinom{n}{k}}{(k+3)} }+\sum _{k=0}^{n} {\frac { (-1)^k \dbinom{n}{k}}{2(k+4)} }$ Now consider the binomial theorem, $x(1-x)^n = \sum_{k=0}^{n} (-1)^k \dbinom{n}{k}x^{k+1}$ Now integrating both sides from 0 to 1 $\int_{0}^{1} x(1-x)^n = \sum _{k=0}^{n} {\frac { (-1)^k \dbinom{n}{k}}{2(k+2)} } $ similarly for the other parts. so the sum is equal to $ \frac{1}{2} \left( \int_{0}^{1}( x(1-x)^n -2x^2(1-x)^n +x^3(1-x)^n) \right)=-\frac{1}{2}\int_{0}^{1} (1-x)^{n+3}+\frac{1}{2}\int_{0}^{1} (1-x)^{n+2} =\frac{1}{2(n+4)(n+3)} $

Behaviour a function when input is small If I have the function: $P(N)=\frac{P_0N^2}{A^2+N^2}$, with $P_0, A$ positive constants For small $N$, am I right in thinking that because $A$ dominates $N$ we have that $P(N) \approx \frac{P_0N^2}{A^2}$

More precisely, we can write for $A^2>N^2$ $$\begin{align} P(N)&=P_0\left(\frac{N^2}{A^2+N^2}\right)\\\\ &=P_0\left(\frac{(N/A)^2}{1+(N/A)^2}\right)\\\\ &=P_0\frac{N^2}{A^2}\sum_{k=0}^\infty (-1)^k\left(\frac{N^2}{A^2}\right)^{k}\\\\ &=\frac{P_0N^2}{A^2}-P_0\frac{N^4}{A^4}+P_0\frac{N^6}{A^6}+O\left(\frac{N^8}{A^8}\right) \end{align}$$ Therefore, if we retain only the first term in the expansion we can formally write $$P(N)\approx \frac{P_0N^2}{A^2}$$ where the approximation error is of order $(N/A)^4$.

$I=3\sqrt2\int_{0}^{x}\frac{\sqrt{1+\cos t}}{17-8\cos t}dt$.If $0Let $I=3\sqrt2\int_{0}^{x}\frac{\sqrt{1+\cos t}}{17-8\cos t}dt$.If $0<x<\pi$ and $\tan I=\frac{2}{\sqrt3}$.Find $x$. $\int\frac{\sqrt{1+\cos t}}{17-8\cos t}dt=\int\frac{\sqrt2\cos\frac{t}{2}}{17-8\times\frac{1-\tan^2\frac{t}{2}}{1+\tan^2\frac{t}{2}}}dt=\int\frac{\sqrt2\cos\frac{t}{2}(1+\tan^2\frac{t}{2})}{17+17\tan^2\frac{t}{2}-8+8\tan^2\frac{t}{2}}dt=\int\frac{\sqrt2\cos\frac{t}{2}(\sec^2\frac{t}{2})}{9+25\tan^2\frac{t}{2}}dt$ $=\int\frac{\sqrt2(\sec^2\frac{t}{2})}{(\sec\frac{t}{2})(9+25\tan^2\frac{t}{2})}dt$ $=\int\frac{\sqrt2(\sec^2\frac{t}{2})}{\sqrt{1+\tan^2\frac{t}{2}}(9+25\tan^2\frac{t}{2})}dt$ Put $\tan\frac{x}{2}=t$ $=\int\frac{\sqrt2(\sec^2\frac{t}{2})}{\sqrt{1+\tan^2\frac{t}{2}}(9+25\tan^2\frac{t}{2})}dt$ $=\int\frac{\sqrt2 }{2\sqrt{1+p^2}(9+25p^2)}dp$ I am stuck here.Please help me.Thanks.

HINT: $$\frac{\sqrt{1+\cos t}}{17-8\cos t}=\dfrac{\sqrt2\left|\cos\dfrac t2\right|}{17-8\left(1-2\sin^2\dfrac t2\right)}$$ Set $\sin\dfrac t2=u$

Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual". I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series and seems to me of the form: $-3-7-11-15\ldots $ I feel like its of the closed form: $\sum(-4i+1)$ So how do I prove that the equality is right?

It can be easily shown that the series on the left reduces to the sum of integers, multiplied by $-1$ if $k$ is even. For even $k$: $$\begin{align} &1^2-2^2+3^3-4^2+\cdots+(k-1)^2-k^2\\ &=(1-2)(1+2)+(3-4)(3+4)+\cdots +(\overline{k-1}-k)(\overline{k-1}+k)\\ &=-(1+2+3+4+\cdots+\overline{k-1}+k)\\ &=-\frac {k(k+1)}2\end{align}$$ Similar, for odd $k$, $$\begin{align} &1^2-2^2+3^3-4^2+5^2+\cdots-(k-1)^2+k^2\\ &=1+(-2+3)(2+3)+(-4+5)(4+5)+\cdots +(-\overline{k-1}+k)(\overline{k-1}+k)\\ &=1+2+3+4+\cdots+\overline{k-1}+k\\ &=\frac {k(k+1)}2\end{align}$$ Hence, $$1^2-2^2+3^3-4^2+5^2+\cdots+(-1)^{k-1}k^2=(-1)^{k-1}\frac {k(k+1)}2\quad\blacksquare$$

Remainder of the numerator of a harmonic sum modulo 13 Let $a$ be the integer determined by $$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{23}=\frac{a}{23!}.$$ Determine the remainder of $a$ when divided by 13. Can anyone help me with this, or just give me any hint?

By Wilson's theorem, $12!\equiv -1\pmod{13}$. By Wolstenholme's theorem, $H_{12}\equiv 0\pmod{13}$. Since: $$ a = \sum_{k=1}^{12}\frac{23!}{k} + (12!)(14\cdot 15\cdot\ldots\cdot 23)+\sum_{k=14}^{23}\frac{23!}{k} $$ we have: $$ a\equiv 0-10!+0 \equiv \frac{-12!}{11\cdot 12}\equiv \frac{1}{(-2)\cdot(-1)}\equiv\frac{1}{2}\equiv\color{red}{7}\pmod{13}$$ since $11\equiv -2\pmod{13}$ and $12\equiv -1\pmod{13}$.

$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$ The other day I came across this problem: Let $x$, $y$, $z$ be real numbers. Prove that $$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$$ The first thought was power mean inequality, more exactly : $AM \leq SM$ ( we noted $AM$ and $SM$ as arithmetic and square mean), but I haven't found anything helpful. (To be more specific, my attempts looked like this : $\frac{x+1}{2} \leq \sqrt{\frac{x^2+1}{2}}$) I also take into consideration Cauchy-Buniakowsky-Scwartz or Bergström inequality, but none seems to help. Some hints would be apreciated. Thanks!

For real $x$,we have(or Use Cauchy-Schwarz inequality) $$x^2+1\ge\dfrac{1}{2}(x+1)^2$$ the same we have $$y^2+1\ge\dfrac{1}{2}(y+1)^2$$ $$z^2+1\ge\dfrac{1}{2}(z+1)^2$$ so $$(x^2+1)(y^2+1)(z^2+1)\ge\dfrac{1}{8}[(x+)(y+1)(z+1)]^2$$ Use AM-GM inequality $$\dfrac{1}{8}[(x+1)(y+1)(z+1)]^2+8\ge 2\sqrt{\dfrac{1}{8}[(x+)(y+1)(z+1)]^2\cdot 8}=2(x+1)(y+1)(z+1)$$

Geometric Inequality $(a+b)(b+c)(c+a)(s-a)(s-b)(s-c)\leq (abc)^{2}$ everyone. $a$,$b$,$c$ are three sides of a triangle. Prove or disprove the following. $(a+b)(b+c)(c+a)(a+b-c)(b+c-a)(c+a-b)\leq 8(abc)^{2}$ I know two inequalities. $8(s-a)(s-b)(s-c)\leq abc~$ , $~(a+b)(b+c)(c+a)\geq 8abc$ But for the above combination of them, I have no idea. Thanks in advance.

Use Heron's formula $$(a+b-c)(b+c-a)(c+a-b)(a+b+c)=\dfrac{a^2b^2c^2}{R^2}$$ where $R$ be the center of the circumcircle of $\Delta ABC$. your inequality can write as $$8R^2\ge\dfrac{(a+b)(b+c)(a+c)}{a+b+c}$$ since $$9R^2\ge a^2+b^2+c^2$$ it suffices to prove $$8(a+b+c)(a^2+b^2+c^2)\ge 9(a+b)(b+c)(a+c)\tag{1}$$ use AM-GM inequality $$27(a+b)(b+c)(a+c)\le 8(a+b+c)^3$$ then it easy to prove (1)

Solving $\lim_{x\to+\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})$ Do you have any tips on how to solve the limit in the title? Whatever I think of doesn't lead to the solution. I tried using: $\sin{x}-\sin{y}=2\cos{\frac{x+y}{2}}\sin{\frac{x-y}{2}}$ and I got: $$\lim_{x\to+\infty}\bigg(2\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\bigg)=$$ $$\lim_{x\to+\infty}\bigg(2\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}}\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\bigg)=$$ $$\lim_{x\to+\infty}\bigg(2\cos{\frac{1}{2(\sqrt{x+1}-\sqrt{x})}}\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\bigg)$$ but, as you can see, this leads me to $\infty-\infty$ in $\cos$ term. How can I get rid of that?

Notice, $$\lim_{x\to +\infty}\left(2\cos\left(\frac{1}{2(\sqrt{x+1}-\sqrt x)}\right)\sin\left(\frac{ \sqrt{x+1}-\sqrt x}{2}\right) \right)$$ $$=2\lim_{x\to +\infty}\cos\left(\frac{1}{2(\sqrt{x+1}-\sqrt x)}\right)\sin\left(\frac{1}{2(\sqrt{x+1}+\sqrt x)}\right)$$ $$=2\lim_{x\to +\infty}\cos\left(\frac{1}{2(\sqrt{x+1}-\sqrt x)}\right)\cdot \lim_{x\to +\infty}\sin\left(\frac{1}{2(\sqrt{x+1}+\sqrt x)}\right)$$ since, $-1\le \cos y\le 1\ \ \ \forall \ \ \ y\in R$, $$=2(k)\cdot \sin\left(0\right)=\color{red}{0}$$ where, $-1\le k\le 1$

Prove for every odd integer $a$ that $(a^2 + 3)(a^2 + 7) = 32b$ for some integer $b$. I've gotten this far: $a$ is odd, so $a = 2k + 1$ for some integer $k$. Then $(a^2 + 3).(a^2 + 7) = [(2k + 1)^2 + 3] [(2k + 1)^2 + 7]$ $= (4k^2 + 4k + 4) (4k^2 + 4k + 8) $ $=16k^4 + 16k^3 + 32k^2 + 16k^3 + 16k^2 + 32k + 16k^2 + 16k + 32$ $=16k^4 + 32k^3 + 64k^2 + 48k + 32$ But this isn't a multiple of 32, at most I could say $(a^2 + 3)(a^2 + 7) = 16b$ for some integer $b$

Notice $$16k^4 +48k = 16k(k^3+3).$$ If $k$ is even, we are done. If $k$ is odd, then $k^3 +3$ is even, and we are done.

Need help with $\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$ Please help me to evaluate this integral: $$\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$$ Using substitution $x=2\arctan t$ it can be transformed to: $$\int_0^\infty\frac{2}{1+t^2}\arctan^2\left(\frac{2t}{3+t^2}\right)dt$$ Then I tried integration by parts, but without any success...

A Fourier analytic approach. If $x\in(0,\pi)$, $$\begin{eqnarray*}\arctan\left(\frac{\sin x}{2+\cos x}\right) &=& \text{Im}\log(2+e^{ix})\\&=&\text{Im}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 2^n}\,e^{inx}\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 2^n}\,\sin(nx),\end{eqnarray*}$$ hence by Parseval's theorem: $$ \int_{0}^{\pi}\arctan^2\left(\frac{\sin x}{2+\cos x}\right)\,dx=\frac{\pi}{2}\sum_{n\geq 1}\frac{1}{n^2 4^n}=\color{red}{\frac{\pi}{2}\cdot\text{Li}_2\left(\frac{1}{4}\right)}.$$ As a side note, we may notice that $\text{Li}_2\left(\frac{1}{4}\right)$ is quite close to $\frac{1}{4}$. By applying summation by parts twice we get: $$ \sum_{n\geq 1}\frac{1}{n^2 4^n} = \color{red}{\frac{1}{3}-\frac{1}{12}}+\sum_{n\geq 1}\frac{1}{9\cdot 4^n}\left(\frac{1}{n^2}-\frac{2}{(n+1)^2}+\frac{1}{(n+2)^2}\right)$$ and the last sum is positive but less than $\frac{11}{486}$, since $f:n\mapsto \frac{1}{n^2}-\frac{2}{(n+1)^2}+\frac{1}{(n+2)^2}$ is a positive decreasing function on $\mathbb{Z}^+$.

Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$. Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$. I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work here. The $ab+bc+ca$ reminds of a cyclic expression, so that may help by factoring the inequality and getting a true statement.

It is actually very simple. Use nothing but AM-GM. $$\frac{a^3}{b} + ab \geq 2a^2$$ $$\frac{b^3}{c} + bc \geq 2b^2$$ $$\frac{c^3}{a} + ac \geq 2c^2$$ $$LHS + (ab+bc+ac) \geq 2(a^2+b^2+c^2) \geq 2(ab + bc +ac)$$ We are done.

Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral: $$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$ I know I must solve it by substitution, but I don't know how exactly.

Using long division, $$ \frac{x^2+4}{x^2+6x+10} = 1 -6\frac{x+1}{x^2+6x+10}= 1 -\frac{6x}{x^2+6x+10}-\frac{6}{x^2+6x+10}$$ Integrating $$1 -\frac{6x}{x^2+6x+10}-\frac{6}{x^2+6x+10} $$ should be straight forward Note that $x^2+6x+10= (x+3)^2 +1$

If $2$ is subtracted from each root,the results are reciprocals of the original roots.Find the value of $b^2+c^2+bc.$ The equation $x^2+bx+c=0$ has distinct roots .If $2$ is subtracted from each root,the results are reciprocals of the original roots.Find the value of $b^2+c^2+bc.$ Let $\alpha$ and $\beta$ are the roots of the equation $x^2+bx+c=0$. According to the question, $\alpha-2=\frac{1}{\alpha}$ and $\beta-2=\frac{1}{\beta}$ $\alpha^2-2\alpha-1=0$ and $\beta^2-2\beta-1=0$ Adding the two equations, $\alpha^2+\beta^2-2\alpha-2\beta-2=0$ $(\alpha+\beta)^2-2\alpha\beta-2\alpha-2\beta-2=0$ $(-b)^2-2c-2(-b)-2=0$ $b^2+2b-2c-2=0$ But i am not able to find $b^2+c^2+bc=0.$What should i do now?I am stuck here.

So from equation we get quadratic which is $x^2-2x-1$ so roots if this are $1+\sqrt{2},1-\sqrt{2}$ so sum of roots is $-b/1=2,$ and product is $c/1=-1$ so $b^2+c^2+bc=4+1-2=3$ hope its clear.

Evaluate the definite integral $\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}dx$ Problem : Determine the value of $$\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx$$ My approach: using $\int^a_0f(x)\ \text dx = \int^a_0 f(a-x)\ \text dx$, $$ \begin{align} \frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx &= \frac{105}{19}\int^{\pi/2}_0 \frac{\sin (4\pi -8x)}{\cos x}\ \text dx\\ &= \frac{105}{19}\int^{\pi/2}_0 -\frac{\sin 8x}{\cos x}\ \text dx \end{align} $$ But it seems it won't work please help thanks

Method $1$ $1.$ Use the identity $$2\sin x \sum_{k=1}^{n}\cos(2k-1)x = \sin2nx$$ which can easily be verified by using $$2\sin x \cos(2k-1)x = \sin 2k x - \sin 2(k-1) x$$ and the telescoping property of the sums. $2.$ Use the above formula to get $$\frac{\sin 2nx}{\sin x} = 2 \sum_{k=1}^{n}\cos(2k-1)x$$ $3.$ Integrate to obtain $$\begin{align} F(x) &=\int \frac{\sin 2nx}{\sin x} \, dx \\ &= 2 \sum_{k=1}^{n} \int \cos(2k-1)x \, dx \\ &= 2 \sum_{k=1}^{n} \frac{1}{2k-1} \sin(2k-1)x + C \end{align}$$ $4.$ Evaluating the definite integral results in harmonic partial sums $$\begin{align} I &= \int_{0}^{\frac{\pi}{2}} \frac{\sin 2nx}{\sin x} \\ &= F(\frac{\pi}{2})-F(0)\\ &= 2 \sum_{k=1}^{n} \frac{\sin (2k-1) \frac{\pi}{2}}{2k-1} - 0\\ &= \boxed{2 \sum_{k=1}^{n} \frac{(-1)^{k+1}}{2k-1}} \end{align}$$ In your case, you may choose $n=4$.

If $f(x)$ is a continous function such that $f(x) +f(\frac{1}{2}+x) =1; \forall x\in [0, \frac{1}{2}]$ then $4\int^1_0 f(x) dx =$ . Problem : If $f(x)$ is a continous function such that $f(x) +f(\frac{1}{2}+x) =1; \forall x\in [0, \frac{1}{2}]$ then $4\int^1_0 f(x) dx = ?$ My approach : $$f(x) +f(\frac{1}{2}+x) =1 \tag{1}\label{1}$$ Let us put $x = \frac{1}{2}+x$ we get $$f(\frac{1}{2}+x) +f(1+x) =1 \tag{2}\label{2}$$ $\eqref{1}-\eqref{2}\Rightarrow f(x) -f(1+x) = 0$ $\Rightarrow f(x) = f(1+x)$ Is it the correct method to find $f(x)$? Please guide thanks.

\begin{align*} &4\int_0^1 f \\ =& 4\int_{0}^{0.5} f(t) dt + 4 \int_{0.5}^1 f(t) dt\\ =& 4\int_{0}^{0.5} f(t) dt + 4 \int_0^{0.5} f(x+\frac12) dx \quad\text{($x = t - \frac12$)}\\ =& 4\int_{0}^{0.5} (f(t) + f(t+\frac12)) dt\\ =& 4\int_{0}^{0.5} 1 dt\\ =& 4\cdot 0.5\\ =& 2 \end{align*}

Prove that $\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$ Prove that $$\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$$ I don't have an idea about how to start.

What we have is $$\frac1{1^3} - \frac1{7^3} + \frac1{9^3} - \frac1{15^3} + \cdots $$ Imagine if we used negative summation indices (e.g., $n=0, -1, -2, \cdots$). Then we would have $$\frac1{(-7)^3} - \frac1{(-1)^3} + \frac1{(-15)^3} - \frac1{(-9)^3} + \cdots$$ You should see then that the sum is $$\frac12 \sum_{n=-\infty}^{\infty} \left [\frac1{(8 n-7)^3} - \frac1{(8 n-1)^3} \right ] $$ The significance of this is that we may use a very simple result from residue theory to evaluate the sum: $$\sum_{n=-\infty}^{\infty} f(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k} [f(z) \cot{\pi z} ]$$ where the $z_k$ are the non-integer poles of $f$. Here $$f(z) = \frac12 \left [\frac1{(8 z-7)^3} - \frac1{(8 z-1)^3} \right ] = \frac1{2 \cdot 8^3} \left [\frac1{(z-7/8)^3} - \frac1{(z-1/8)^3} \right ]$$ $f$ has poles at $z_1=1/8$ and $z_2=7/8$. At $z_1$, the residue is $$\frac1{2 \cdot 8^3} \frac1{2!} \left [ \frac{d^2}{dz^2} \cot{\pi z} \right ]_{z=1/8} = \frac1{2 \cdot 8^3} \frac1{2!} (2 \pi ^2) \cot \left(\frac{\pi }{8}\right) \csc ^2\left(\frac{\pi }{8}\right)$$ The calculation for the other pole is similar. Then our sum is $$\frac1{2 \cdot 8^3} \frac1{2!} \pi (2 \pi ^2) \left [\cot \left(\frac{\pi }{8}\right) \csc ^2\left(\frac{\pi }{8}\right)-\cot \left(\frac{7 \pi }{8}\right) \csc ^2\left(\frac{7 \pi }{8}\right) \right ] = \frac{\pi^3}{\cdot 8^3} \cot \left(\frac{\pi }{8}\right) \csc ^2\left(\frac{\pi }{8} \right )$$ Simplifying... $$\cot \left(\frac{\pi }{8}\right) = 1+\sqrt{2}$$ $$\csc ^2\left(\frac{\pi }{8} \right ) = \frac{2}{1-\frac{\sqrt{2}}{2}} = \frac{4}{2-\sqrt{2}} = 2 \left (2+\sqrt{2} \right)$$ Thus: $$\sum_{n=1}^{\infty} \left [\frac1{(8 n-1)^3} - \frac1{(8 n-7)^3} \right ] = \frac{\pi^3}{256} \left (4+3 \sqrt{2} \right )$$

Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$ What is the value of the following sum? $$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$ The possible answers are: A. $e$ B. $\frac{e}{2}$ C. $\frac{3e}{2}$ D. $1 + \frac{e}{2}$ I tried to expand the options using the series representation of $e$ and putting in $x=1$, but I couldn't get back the original series. Any ideas?

Clearly the $r^{th}$ numerator is $1+2+3+...+r= \frac{r(r+1)}{2}$ . And the $r^{th}$ denominator is $r!$. Thus $$\displaystyle U_r=\frac{\frac{r(r+1)}{2}}{r!}=\frac{r(r+1)}{2r!}$$ Since the degree of the numerator is $2$ , use partial fractions to find $A,B,C$ such that (If you use partial fractions up to $(r-3)!$ , its' coefficient will be zero when comparing coefficients.) $\displaystyle U_r=\frac{r(r+1)}{2r!}=\frac{A}{(r-2)!}+\frac{B}{(r-1)!}+\frac{C}{r!}$ $\displaystyle (2r!)\times U_r=(2r!)\times \frac{r(r+1)}{2r!}=(2r!)\times \frac{A}{(r-2)!}+(2r!)\times \frac{B}{(r-1)!}+(2r!)\times \frac{C}{r!}$ So $\displaystyle r(r+1)=r!\times \frac{2A}{(r-2)!}+r!\times \frac{2B}{(r-1)!}+r!\times \frac{2C}{r!}$ ............................................................................. Now observe that $r!=1\times 2\times 3\times .... \times (r-2)\times(r-1)\times r $ $\Rightarrow r!=(r−2)! ×(r−1)r $ and $ \Rightarrow r!=(r−1)!×r $ ............................................................................... So $\displaystyle r(r+1)=(r−2)! ×(r−1)r \times \frac{2A}{(r-2)!}+(r−1)!×r\times \frac{2B}{(r-1)!}+r!\times \frac{2C}{r!}$ So $\displaystyle r^2+r = 2A(r-1)r+2Br+ 2C $ Clearly $C=0$ , $B=1$ and $A=\frac{1}{2}$ So $\displaystyle U_r=\frac{r(r+1)}{2r!}=\frac{1}{2(r-2)!}+\frac{1}{(r-1)!}$ $\displaystyle \sum_{r=2}^{\infty}U_r= \frac{1}{2} \left( \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+.....\right)+\left( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.....\right)$ $\displaystyle \sum_{r=2}^{\infty}U_r= \frac{1}{2} \left( e\right)+\left( e-1\right)$ $\displaystyle \sum_{r=1}^{\infty}U_r= U_1+\frac{1}{2} \left( e\right)+\left( e-1\right)=1+\frac{e}{2}+e-1 =\frac{3e}{2}$

If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer I am trying to solve: If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer. If I write $n=2k+1$ and $m=2l+1$ I get stuck at $$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$

$$((2k+1)(2l+1))^2-1=16k^2l^2+4k^2+16kl^2+16kl+4k+4l^2+16lk^2+4l.$$ Dropping all the terms with coefficient $16$, $$4(k^2+k+l^2+l)=4(k(k+1)+l(l+1))$$ must be a multiple of $8$. With a slightly simpler evaluation: $$((2k+1)(2l+1))^2-1=(4kl+2k+2l)(4kl+2k+2l+2)=4(2kl+k+l)(2kl+k+l+1).$$ This is a multiple of $8$.

Can we find $x_{1}, x_{2}, ..., x_{n}$? Consider this. $$x_{1}+x_{2}+x_{3}+....+x_{n}=a_{1}$$ $$x_{1}^2+x_{2}^2+x_{3}^2+....+x_{n}^2=a_{2}$$ $$x_{1}^4+x_{2}^4+x_{3}^4+....+x_{n}^4=a_{3}$$ $$x_{1}^8+x_{2}^8+x_{3}^8+....+x_{n}^8=a_{4}$$ $$.............................$$ $$x_{1}^{2^{n-1}}+x_{2}^{2^{n-1}}+x_{3}^{2^{n-1}}+....+x_{n}^{2^{n-1}}=a_{n}$$ Can we find $x_{1}, x_{2}, ..., x_{n}$ by knowing $a_{1}, a_{2}, ..., a_{n}$?

If you mean the diophantine-equation tag and the solutions are supposed to be integers, searching will be easy because high powers are spaced far apart. Because of the symmetry you can insist that $x_1 \ge x_2 \ge \dots \ge x_n$. You can focus just on the last equation $x_{1}^{2^{n-1}}+x_{2}^{2^{n-1}}+x_{3}^{2^{n-1}}+....+x_{n}^{2^{n-1}}=a_{n}$. We have $x_{1}^{2^{n-1}} \le a_n \le nx_{1}^{2^{n-1}}$ or $(\frac{a_n}n)^{2^{1 \le -n}} x_1 \le a_n^{2^{1-n}}$ Since $n^{2^{1-n}}$ is just a bit greater than $1$, this is a tight bound. For example, let $n=5$ and the final equation be $x_{1}^{16}+x_{2}^{16}+x_{3}^{16}+x_{4}^{16}+x_{5}^{16}=46512832447930819$ We get $9.955 \le a_1 \lt 11.008$, so $a_1$ must be $10$ or $11$. If we take $a_1=10$, then $9.94 \lt a_2$, so $a_2=10$ as well. It turns out we get driven to $10,10,10,10,10$, which is too large, or $10,10,10,10,9$, which is too small. If we take $a_1=11$ things work better (because I made them). We get $7.66 \lt a_2 \lt 8.35$, so $a_2=8$, then $7.47 \lt a_3 \lt 8.002,$ so $a_3=8$ and we find $a_4=5, a_5=3$ works.

Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2 \ge 1$ Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2 \ge 1$. My solution: since $a+b+c=1$ we have to show that $a^2+3b^2+5c^2\ge1=a+b+c$ Since $a,b,c \ge 0 $ the inequality is true given that every term on the left hand side of the inequality is greater or equal to the corresponding term on the right. However I am not sure if I am reasoning correctly, as the hint from my book seems to depict the problem in a harder way than I am ,as it suggests to square the expression $a+b+c=1$ and so on... So my question is wheter I am overlooking some detail in the problem which makes my solution inadequate.

By means of rearrangement inequality, it can be shown that, is $a_1\le a_2\le\cdots\le a_n$ and $b_1\le b_2\le\cdots\le b_n$, then $\dfrac{a_1b_1+a_2b_2+\cdots+a_nb_n}{n}\ge \dfrac{a_1+a_2+\cdots+a_n}{n}\dfrac{b_1+b_2+\cdots+b_n}{n}$ Then, $\dfrac{a^2+3b^2+5c^2}{3}\ge\dfrac{a^2+b^2+c^2}{3}\dfrac{1+3+5}{3}$. Thus $a^2+3b^2+5c^2\ge 3(a^2+b^2+c^2)\ge 3\dfrac{(a+b+c)^2}{3}=1$ (last follows from Jensen's inequality).

Finding the common integer solutions to $a + b = c \cdot d$ and $a \cdot b = c + d$ I find nice that $$ 1+5=2 \cdot 3 \qquad 1 \cdot 5=2 + 3 .$$ Do you know if there are other integer solutions to $$ a+b=c \cdot d \quad \text{ and } \quad a \cdot b=c+d$$ besides the trivial solutions $a=b=c=d=0$ and $a=b=c=d=2$?

First, note that if $(a, b, c, d)$ is a solution, so are $(a, b, d, c)$, $(c, d, a, b)$ and the five other reorderings these permutations generate. We can quickly dispense with the case that all of $a, b, c, d$ are positive using an argument of @dREaM: If none of the numbers is $1$, we have $ab \geq a + b = cd \geq c + d = ab$, so $ab = a + b$ and $cd = c + d$, and we may as well assume $a \geq b$ and $c \geq d$. In particular, since $a, b, c, d > 1$, we have $a b \geq 2a \geq a + b = ab$, so $a = b = 2$ and likewise $c = d = 2$, giving the solution $$(2, 2, 2, 2).$$ On the other hand, if at least one number is $1$, say, $a$, we have $b = c + d$ and $1 + b = cd$, so $1 + c + d = cd$, and we may as well assume $c \leq d$. Rearranging gives $(c - 1)(d - 1) = 2$, so the only solution is $c = 2, d = 3$, giving the solution $$(1, 5, 2, 3).$$ Now suppose that at least one of $a, b, c, d$, say, $a$ is $0$. Then, we have $0 = c + d$ and $b = cd$, so $c = -d$ and $b = -d^2$. This gives the solutions $$A_s := (0, -s^2, -s, s), \qquad s \in \Bbb Z .$$ We are left with the case for which at least one of $a, b, c, d$, say, $a$, is negative, and none is $0$. Suppose first that none of the variables is $-1$. If $b < 0$, we must have $cd = a + b < 0$, and so we may assume $c > 0 > d$. On the other hand, $c + d = ab > 0$, and so (using a variation of the argument for the positive case) we have $$ab = (-a)(-b) \geq (-a) + (-b) = -(a + b) = -cd \geq c > c + d = ab,$$ which is absurd. If $b > 0$, we have $c + d = ab < 0$, so at least one of $c, d$, say, $c$ is negative. Moreover, we have $cd = a + b$, so $d$ and $a + b$ have opposite signs. If $d < 0$, then since $c, d < 0$, we are, by exploiting the appropriate permutation, in the above case in which $a, b < 0$, so we may assume that $d > 0$, and hence that $a + b < 0$. Now, $$ab \leq a + b = cd < c + d = ab,$$ which again is absurd, so there no solutions in this case. This leaves only the case in which at least one of $a, b, c, d$ is $-1$, say, $a$. Then, we have $-b = c + d$ and $-1 + b = cd$, so $-1 + (- c - d) = cd$. Rearranging gives $(c + 1)(d + 1) = 0$, so we may assume $c = -1$ giving (up to permtuation) the $1$-parameter family of solutions $$B_t := (-1, t, -1, 1 - t), \qquad t \in \Bbb Z,$$ I mentioned in my comment (this includes two solutions, $B_0$ and $B_1$, which are equivalent by a permutation, that include a zero entry). This exhausts all of the possibilities; in summary: Any integer solution to the system $$\left\{\begin{array}{rcl}a + b \!\!\!\!& = & \!\!\!\! cd \\ ab \!\!\!\! & = & \!\!\!\! c + d \end{array}\right.$$ is equal (up to the admissible permutations mentioned at the beginning of this answer) to exactly one of * *$(1, 5, 2, 3)$ *$(2, 2, 2, 2)$ *$A_s := (0, -s^2, -s, s)$, $s \geq 0$, and *$B_t := (-1, t, -1, 1 - t)$, $t \geq 2$. The restrictions on the parameters $s, t$ are consequences of the redundancy in the solutions we found: $A_{-s}$ is an admissible permutation of $A_s$, $B_{1 - t}$ an admissible permutation of $B_t$, and $B_1$ one of $A_1$.

Finding the sum of the infinite series whose general term is not easy to visualize: $\frac16+\frac5{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\cdots$ I am to find out the sum of infinite series:- $$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...............$$ I can not figure out the general term of this series. It is looking like a power series as follows:- $$\frac{1}{6}+\frac{5}{6^2\cdot2!}+\frac{5\cdot8}{6^3\cdot3!}+\frac{5\cdot8\cdot11}{6^4\cdot4!}+.....$$ So how to solve it and is there any easy way to find out the general term of such type of series?

Let us consider $$\Sigma=\frac{1}{6}+\frac{5}{6\times 12}+\frac{5\times8}{6\times12\times18}+\frac{5\times8\times11}{6\times12\times18\times24}+\cdots$$ and let us rewrite it as $$\Sigma=\frac{1}{6}+\frac 16\left(\frac{5}{ 12}+\frac{5\times8}{12\times18}+\frac{5\times8\times11}{12\times18\times24}+\cdots\right)=\frac{1}{6}+\frac 16 \sum_{n=0}^\infty S_n$$ using $$S_n=\frac{\prod_{i=0}^n(5+3i)}{\prod_{i=0}^n(12+6i)}$$ Using the properties of the gamma function, we have $$\prod_{i=0}^n(5+3i)=\frac{5\ 3^n \Gamma \left(n+\frac{8}{3}\right)}{\Gamma \left(\frac{8}{3}\right)}$$ $$\prod_{i=0}^n(12+6i)=6^{n+1} \Gamma (n+3)$$ which make $$S_n=\frac{5\ 2^{-n-1} \Gamma \left(n+\frac{8}{3}\right)}{3 \Gamma \left(\frac{8}{3}\right) \Gamma (n+3)}$$ $$\sum_{n=0}^\infty S_n=\frac{10 \left(3\ 2^{2/3}-4\right) \Gamma \left(\frac{2}{3}\right)}{9 \Gamma \left(\frac{8}{3}\right)}=3\ 2^{2/3}-4$$ $$\Sigma=\frac{1}{\sqrt[3]{2}}-\frac{1}{2}$$

If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following: If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$

Hint At least for $(a, b) \neq (-1, 0)$, which is not realized by any value $t$, draw the line through $(-1, 0)$ and $(a, b)$ in the $xy$-plane. Writing the point $(a, b)$ of intersection of the line and the unit circle $x^2 + y^2 = 1$ in terms of the slope $t$ of the line gives exactly $$(a, b) = \left(\frac{1 - t^2}{1 + t^2}, \frac{2 t}{1 + t^2}\right),$$ so the claim is true for all $(a, b) \neq (-1, 0)$. On the other hand, substituting shows that there is no value $t$ for which $(a, b) = (-1, 0)$.

Geometric progression in an inequality Problem: Show that if $a>0$ and $n>3$ is an integer then $$\frac{1+a+a^2 \cdots +a^n}{a^2+a^3+ \cdots a^{n-2}} \geq \frac{n+1}{n-3}$$ I am unable to prove the above the inequality. I used the geometric progression summation formula to reduce it to proving $\frac{a^{n+1}-1}{a^2(a^{n-3}-1)} \geq \frac{n+1}{n-3}$. Also writing it as $$\frac{1+a+a^2 \cdots +a^n}{n+1} \geq \frac{a^2+a^3+ \cdots a^{n-2}}{n-3}$$ seems to suggest that some results on mean-inequalities can be used but I can't figure out what that is.

The inequality holds trivially for $a = 1$, and it is invariant under the substitution $a \to 1/a$. Therefore it suffices to prove the inequality for $\mathbf{a > 1}$. The hyperbolic sine is a convex function on $[0, \infty)$, so for $a > 1$ the function $$ f(x) = 2 \sinh \bigl(\log a \cdot \frac x2 \bigr) = a^{x/2} - a^{-x/2} = \frac{a^x - 1}{a^{x/2}} \quad (x \ge 0) $$ is also convex, and therefore $$ \frac{f(x) - f(0)}{x-0} = \frac{a^x - 1}{x \, a^{x/2}} $$ is increasing. It follows that for $0 \le x \le y$ $$ \frac{a^y - 1}{y \, a^{y/2}} \ge \frac{a^x - 1}{x \, a^{x/2}} $$ which is equivalent to $$ \frac{a^y-1}{a^x-1} \ge a^{(y-x)/2} \frac yx \, . $$ The desired inequality follows as a special case for $x = n-3$ and $y = n+1$, but we have shown that the inequality can be generalized to arbitrary positive real numbers as exponents.

Indefinite integral $\int x\sqrt{1+x}\mathrm{d}x$ using integration by parts $$\int x\sqrt{1+x}\mathrm{d}x$$ $v'=\sqrt{1+x}$ $v=\frac{2}{3}(1+x)^{\frac{3}{2}}$ $u=x$ $u'=1$ $$\frac{2x}{3}(1+x)^{\frac{3}{2}}-\int\frac{2}{3}(1+x)^{\frac{3}{2}}=\frac{2x}{3}(1+x)^{\frac{3}{2}}-\frac{2}{3}*\frac{2}{5}(1+x)^{\frac{5}{2}}+c$$ result: $$\frac{2x}{3}(1+x)^{\frac{3}{2}}-\frac{4}{15}(1+x)^{\frac{5}{2}}+c$$ Deriving: $$x(1+x)^\frac{1}{2}-\frac{20}{30}(1+x)^{\frac{3}{2}}$$ Where did I get wrong?

You're correct. $u = x$, $dv=\sqrt{x+1}dx$ $\Rightarrow$ $v=\frac{2}{3}(x+1)^{3/2}$, $du=dx$. & $$\int udv=uv-\int vdu$$ $$\Downarrow$$ SOLUTION: $$\int x \sqrt{x+1}dx=x\frac{2}{3}(x+1)^{3/2}-\int\frac{2}{3}(x+1)^{3/2}dx=\bbox[5px,border:2px solid #F0A]{x\frac{2}{3}(x+1)^{3/2} -\frac{4}{15}(x+1)^{5/2}+C}$$ VERIFICATION: Do not forget applying the product rule when taking derivative. $$\frac{d}{dx}\bigg(x\frac{2}{3}(x+1)^{3/2} -\frac{4}{15}(x+1)^{5/2}+C\bigg)=\frac{2}{3}(x+1)^{3/2}+x\sqrt{x+1}-\frac{2}{3}(x+1)^{3/2}+0=\bbox[5px,border:2px solid #F0A]{x\sqrt{x+1}}$$

$AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$ $AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$.Find $a,b,c$. The point $(1,2)$ is inside the circle $x^2+y^2-6x-8y-11=0$.I let the points $A(x_1,y_1)$ and $B(x_2,y_2)$ are the end points of the chord $AB$.As $AB$ subtend $90^\circ$ at $(1,2)$ So $\frac{y_1-2}{x_1-1}\times \frac{y_2-2}{x_2-1}=-1$ But i do not know how to find the locus of mid point of chord $AB$ $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

Use polar coordinate. Let $(x_1,y_1)=(3+6\cos\theta_1,4+6\sin\theta_1), (x_2,y_2)=(3+6\cos\theta_2, 4+6\cos \theta_2)$. Then by your equation, we have $$(2+6\cos \theta_1)(2+6\cos \theta_2)+(2+6\sin\theta_1)(2+6\sin\theta_2)=0$$ Using sum and difference formula, this gives us $$18\cos(\theta_2-\theta_1)=-4-3(\cos\theta_1+\cos\theta_2)-3(\sin\theta_1+\sin\theta_2)$$ Now find the midpoint $x=3+3\cos\theta_1+3\cos\theta_2, y=4+3\sin\theta_1+3\sin\theta_2$. Compute: $$x^2+y^2=\dots=43+18(\cos\theta_1+\cos\theta_2)+12(\sin\theta_1+\sin\theta_2)+18\cos(\theta_2-\theta_1)$$ With the above derived formula, we get $$x^2+y^2=39+15(\cos\theta_1+\cos\theta_2)+9(\sin\theta_1+\sin\theta_2)$$ Now this is equal to $2ax+2by+c=2a(3+3(\cos\theta_1+\cos\theta_2))+2b(4+3(\sin\theta_1+\sin\theta_2))+c$ So you can find $a,b,c$.

Prove a function is uniformly continuous Prove the function $f(x)=\sqrt{x^2+1}$ $ (x\in\mathbb{R})$ is uniformly continuous. Now I understand the definition, I am just struggling on what to assign $x$ and $x_0$ Let $\epsilon>0$ we want $|x-x_0|<\delta$ so that $|f(x)-f(x_0)|<\epsilon$ Could anyone help fill in the missing bits? Thanks

You have $$\begin{aligned}\vert f(x)-f(y) \vert &= \left\vert \sqrt{x^2+1}-\sqrt{y^2+1} \right\vert \\ &= \left\vert (\sqrt{x^2+1}-\sqrt{y^2+1}) \frac{\sqrt{x^2+1}+\sqrt{y^2+1}}{\sqrt{x^2+1}+\sqrt{y^2+1}} \right\vert \\ &= \left\vert \frac{x^2-y^2}{\sqrt{x^2+1}+\sqrt{y^2+1}} \right\vert \\ &\le \frac{\vert x-y \vert (\vert x \vert + \vert y \vert )}{ \sqrt{x^2+1}+\sqrt{y^2+1}} \\ &\le \vert x-y \vert \end{aligned}$$ hence choosing $\delta = \epsilon$ will work.

integrate $\int \cos^{4}x\sin^{4}xdx$ $$\int \cos^4x\sin^4xdx$$ How should I approach this? I know that $\sin^2x={1-\cos2x\over 2}$ and $\cos^2x={1+\cos2x\over 2}$

Continuing from where you left off: $$\sin^2x\cos^2x=\left({1-\cos2x\over 2}\right)\left({1+\cos2x\over 2}\right)$$ $$\sin^2x\cos^2x=\frac{1}{4}(1 - \cos^2(2x))$$ $$\sin^2x\cos^2x=\frac{1}{4}(\sin^2(2x))$$ Squaring both the sides: $$\sin^4x\cos^4x=\frac{1}{16}(\sin^4(2x))$$ This can be integrated in two ways: Method 1: $$\sin^4x\cos^4x=\frac{1}{16}(\sin^4(2x))$$ $$=\frac{1}{16}(\sin^4(2x)) = \frac{1}{16} \sin^2{(2x)} (1 - \cos^2{(2x)})$$ $$=\frac{1}{16} \left[\sin^2{(2x)} - \sin^2{(2x)} \cos^2{(2x)}\right]$$ $$=\frac{1}{16} \left[\sin^2{(2x)} - \frac{1}{4}(1 - \cos^2{(4x)})\right]$$ $$=\frac{1}{16} \left[\sin^2{(2x)} - \frac{1}{4}\sin^2{(4x)}\right] $$ $$=\frac{1}{32}[1 - \cos{(4x)}] - \frac{1}{128}[1 - \cos{(8x)}]$$ Which is easy to integrate Method 2: $$\:\frac{1}{16}\left[\sin^2(2x)\right]^2\;\;$$ $$=\frac{1}{16}\left[\frac{1\,-\,\cos(4x)}{2}\right]^2\:$$ $$=\:\frac{1}{64}\left[1\,-\,2\cdot\cos(4x) \,+\,\cos^2(4x)\right]$$ $$=\:\frac{1}{64}\left[1\,-\,2\cdot\cos(4x) \,+\,\frac{1+\cos(8x)}{2}\right]$$ $$=\:\frac{1}{128}\left[3\,-\,4\cdot\cos(4x)\,+\,\cos(8x)\right]$$ Which is again easy to integrate

Calculate $\lim_{n\to \infty} \frac{\frac{2}{1}+\frac{3^2}{2}+\frac{4^3}{3^2}+...+\frac{(n+1)^n}{n^{n-1}}}{n^2}$ Calculate $$\lim_{n\to \infty} \frac{\displaystyle \frac{2}{1}+\frac{3^2}{2}+\frac{4^3}{3^2}+...+\frac{(n+1)^n}{n^{n-1}}}{n^2}$$ I have messed around with this task for quite a while now, but I haven't succeeded to find the solution yet. Help is appreciated!

Combining $$\frac{1}{1-1/n}>1+\frac1n \ \ \ n>1 $$ with the fact that for all positive $n$ we have $$ \left(1+\frac1n\right)^n<e<\left(1+\frac{1}{n}\right)^{n+1}, $$ and after rewriting your sequence $a_n$ as $\frac{1}{n^2}\sum_{k=1}^n k(1+1/k)^k$, we find the general term $s_k$ of the sum satisfies $$k\left(e-\frac ek\right)<s_k<ke$$. Thus $$\frac{e}{2} \leftarrow \frac{e}{2}\frac{n(n+1)}{n^2} - \frac{e}{n} =\frac{e}{n^2} \sum_{k=1}^n(k-1)<a_n< \frac{e}{n^2} \sum_{k=1}^n k=\frac e2 \frac{n(n+1)}{n^2} \to \frac e2,$$ and by the squeeze theorem we conclude $\lim\limits_{n\to\infty} a_n=\displaystyle\frac e2$.

Formulae for Catalan's constant. Some years ago, someone had shown me the formula (1). I have searched for its origin and for a proof. I wasn't able to get true origin of this formula but I was able to find out an elementary proof for it. Since then, I'm interested in different approaches to find more formulae as (1). What other formulas similar to ($1$) are known? Two days ago, reading the book of Lewin "Polylogarithms and Associated Functions" I was able to find out formula (2). $\displaystyle \dfrac{1}{3}C=\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{2-x}\right)dx\tag1$ $\displaystyle \dfrac{2}{5}C=\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{\sqrt{5}x(1-x)}{1+\sqrt{5}-\sqrt{5}x}\right)dx-\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{3+\sqrt{5}-x}\right)dx\tag2$ $C$ being the Catalan's constant. I have a proof for both of these formulae. My approach relies on the following identity: For all real $x>1$, $\displaystyle \int_0^1 \dfrac{1}{t} \arctan \left (\dfrac{t(1-t)}{\frac{x+1}{2}-t}\right) dt=\int_1^{\frac{\sqrt{x}+1}{\sqrt{x}-1}}\dfrac{\log(t)}{1+t^2}dt$

A more natural proof. \begin{align} \beta&=\sqrt{3}-1\\ J&=\int_0^1 \frac{\arctan\left(\frac{x(1-x)}{2-x}\right)}{x}dx\\ &\overset{\text{IBP}}=\left[\arctan\left(\frac{x(1-x)}{2-x}\right)\ln x\right]_0^1-\int_0^1 \frac{(x^2-4x+2)\ln x}{(x^2+\beta x+2)(x^2-(\beta+2)x+2)}dx\\ &=-\int_0^1 \frac{(x^2-4x+2)\ln x}{(x^2+\beta x+2)(x^2-(\beta+2)x+2)}dx\\ &=\int_0^1 \frac{\beta\ln x}{2(x^2-(2+\beta) x+2)}-\int_0^1 \frac{(2+\beta)\ln x}{2(x^2+\beta x+2)}\\ &=\underbrace{\int_0^1 \frac{2\ln x}{\beta\left(\left(\frac{2x-2-\beta}{\beta}\right)^2+1\right)}dx}_{y=\frac{\beta}{2+\beta-2x}}-\underbrace{\int_0^1 \frac{2\ln x}{(2+\beta)\left(\left(\frac{2x+\beta}{2+\beta}\right)^2+1\right)}dx}_{y=\frac{2x+\beta}{2+\beta}}\\ &=-\int_{\frac{\beta}{\beta+2}}^1 \frac{\ln y}{1+y^2}dy\\ &\overset{y=\tan \theta}=-\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\ln\left(\tan \theta\right)d\theta\\ &=-\int_{0}^{\frac{\pi}{4}}\ln\left(\tan \theta\right)d\theta+\int_0^{\frac{\pi}{12}}\ln\left(\tan \theta\right)d\theta\\ &=\text{G}-\frac{2}{3}\text{G}\\ &=\boxed{\dfrac{1}{3}\text{G}} \end{align} NB: For the latter integral see Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$

Extreme of $\cos A\cos B\cos C$ in a triangle without calculus. If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$. I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing helps. On this topic we learned also about Cauchy inequality, but I have no experience with it. The answer according to Mathematica is when $A=B=C=60$. Any ideas?

If $y=\cos A\cos B\cos C,$ $2y=\cos C[2\cos A\cos B]=\cos C\{\cos(A-B)+\cos(A+B)\}$ As $A+B=\pi-C,\cos(A+B)=-\cos C$ On rearrangement we have $$\cos^2C-\cos C\cos(A-B)+2y=0$$ As $C$ is real, so will be $\cos C$ $\implies$ the discriminant $$\cos^2(A-B)-8y\ge0\iff y\le\dfrac{\cos^2(A-B)}8\le\dfrac18$$ The equality occurs if $\cos^2(A-B)=1\iff\sin^2(A-B)=0$ $\implies A-B=n\pi$ where $n$ is any integer As $0<A,B<\pi, n=0\iff A=B$ and consequently $$\cos^2C-\cos C+2\cdot\dfrac18=0\implies \cos C=\dfrac12\implies C=\dfrac\pi3$$ $\implies A=B=\dfrac{A+B}2=\dfrac{\pi-C}2=\dfrac\pi3=C$

Evaluation of Real-Valued Integrals (Complex Analysis) How to get calculate the integration of follwing: $$\int_{0}^{2\pi} \frac{dt}{a + cos t} (a>1)$$ My attempt: let, $z=e^{it}$ $\implies dt = \frac{dz}{it}$ and $$\cos t = \frac{z + \frac{1}{z}}{2}$$ On substituting everything in the integral I got: $$\frac{2}{i}\int_{c} \frac{dz}{z^2+2az+1}$$ Now how do I decompose this fraction so that I can use the Residue Theorem? Or is there anyother way to solve this?? Thanks for the help.

HINT: The denominator: $z^2+2az+1$ can be easily factorize using quadratic formula as follows $$z^2+2az+1=(z+a-\sqrt{a^2-1})(z+a+\sqrt{a^2-1})$$ hence, $$\frac{1}{z^2+2az+1}=\frac{A}{z+a-\sqrt{a^2-1}}+\frac{B}{z+a+\sqrt{a^2-1}}$$ $$=\frac{1}{2\sqrt{a^2-1}}\left(\frac{1}{z+a-\sqrt{a^2-1}}-\frac{1}{z+a+\sqrt{a^2-1}}\right)$$

Probability that $2^a+3^b+5^c$ is divisible by 4 If $a,b,c\in{1,2,3,4,5}$, find the probability that $2^a+3^b+5^c$ is divisible by 4. For a number to be divisible by $4$, the last two digits have to be divisible by $4$ $5^c= \_~\_25$ if $c>1$ $3^1=3,~3^2=9,~3^3=27,~3^4=81,~ 3^5=243$ $2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=32$ Should I add all possibilities? Is there a simpler method?

Observe that $$2^a+3^b+5^c \equiv 2^a+(-1)^b+1 \pmod{4}$$ So for this to be $0 \pmod 4$, we have the following scenarios * *$a \geq 2$, $b$ is odd and $c$ is any number. *$a=1$, $b$ is even and $c$ is any number. The number of three tuples $(a,b,c)$ that satisfy the first case =$(4)(3)(5)=60$ and the number of three tuples $(a,b,c)$ that satisfy the second case =$(1)(2)(5)=10.$ Probability is $\frac{70}{125}$.

Find prob. that only select red balls from $n$ (red+blue) balls There are 4 blue balls and 6 red balls(total 10 balls). $X$ is a random variable of the number of selected balls(without replacement), in which $$P(X=1)=0.1$$ $$P(X=2)=0.5$$ $$P(X=3)=0.2$$ $$P(X=4)=0.1$$ $$P(X=10)=0.1$$ Then, what is probability of only selecting red balls? This is what I have tried: The (conditional) probability that all $r$ of the balls are selected from the red is just: ${6\choose r}\big/{10\choose r}$, for $0\leq r\leq 6$ , and $0$ elsewhere. That is, let $N_R$ be the number of red balls selected, and $N_r$ the total number of balls selected, then: $$\mathsf P(N_r=N_R\mid N_r=r) = \frac{6!/(6-r)!}{10!/(10-r)!} \mathbf 1_{r\in\{1\ldots 6\}}$$ As the number of balls selected is a random variable with the specified distribution, then the probability that all balls selected are red is: $$\begin{align} \mathsf P(N_R=N_r) & =\frac{1}{10}\frac{6!\,(10-1)!}{(6-1)!\,10!}+\frac 5{10}\frac{6!\,(10-2)!}{(6-2)!\,10!}+\frac{1}{5}\frac{6!\,(10-3)!}{(6-3)!\,10!}+... \\[1ex] & \end{align}$$

Since there is $6$ red balls, if $10$ balls are selected, probability of selecting only red balls is $0$ and we only have to consider selecting $1,2,3,4$ balls. Let $R$ be number of red balls selected. $$\begin{align}P(R=i|X=i)&=\frac{_6P_i}{_{10}P_i}\\ P(R=X)&=\sum\limits_{i=1}^4P(X=i)\cdot P(R=i|X=i)\\ &=0.1\cdot \frac 6{10}+0.5\cdot \frac {6\cdot5}{10\cdot9}+0.2\cdot \frac {6\cdot5\cdot4}{10\cdot9\cdot8}+0.1\cdot \frac {6\cdot5\cdot4\cdot3}{10\cdot9\cdot8\cdot7}\\ &=\frac{187}{700}\end{align}$$

General solution for the series $a_n = \sqrt{(a_{n-1} \cdot a_{n-2})}$ Hey I'm searching a general solution for this recursive series: $a_n = \sqrt{(a_{n-1}\cdot a_{n-2})}$ $\forall n \geq 2$ $a_0 = 1$, $a_1 = 2$

Elaborating on what Wojowu has mentioned, $$a_n^2=a_{n-1}\cdot a_{n-2}$$ $$a_n^2\cdot a_{n-1}=a_{n-1}^2\cdot a_{n-2}$$ That is, $a_n^2\cdot a_{n-1}=$ constant is invariant. Hence $$a_n^2\cdot a_{n-1}=a_{n-1}^2\cdot a_{n-2}= \ldots = a_1^2a_0 = 4$$ or, $$a_n^2=\frac{4}{a_{n-1}}=\frac{4}{\frac{2}{\sqrt{a_{n-2}}}}=2\sqrt{a_{n-2}}$$ or, $$a_n=\sqrt{2\sqrt{a_{n-2}}}=\sqrt{2\sqrt{\sqrt{2\sqrt{a_{n-4}}}}}$$ or,$$a_n=\sqrt[2]{2\sqrt[4]{2\sqrt{a_{n-4}}}}$$ or,$$a_n=\sqrt[2]{2\sqrt[4]{2{\sqrt[4]{2\sqrt{a_{n-6}}}}}}$$ Therefore, $$a_n=\begin{cases}\sqrt[2]{2\sqrt[4]{2{\sqrt[4]{2\sqrt[4]{\ldots \sqrt{a_1}}}}}} & \text{if n is odd} \\\sqrt[2]{2\sqrt[4]{2{\sqrt[4]{2\sqrt[4]{\ldots \sqrt{a_0}}}}}} & \text{if n is even} \end{cases}$$

Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}.$ Prove that for all positive real numbers $a,b,$ and $c$, we have $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}.$$ What I tried is saying $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} = \dfrac{a^2c+b^2a+c^2b}{abc} \geq \dfrac{3abc}{abc} = 3$. Then how can I use this to prove that $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}$?

Without Loss of Generality, Let us assume that $c$ is the maximum of $a,b,c$ Notice that $$\sum _{ cyc }^{ }{ \frac { a }{ b } } -3=\frac { a }{ b } +\frac { b }{ a } -2+\frac { b }{ c } +\frac { c }{ a } +\frac { b }{ a } -1=\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}$$ However, since $(a-b)^2,(c-a)(c-b)\ge 0$, $$\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac} \ge \frac{(a-b)^2}{(c+a)(c+b)}+\frac{(c-a)(c-b)}{(a+c)(b+a)}=\sum _{ cyc }^{ }{ \frac { a+b }{ c+a } }-3$$ Our proof is done.

How to find Laurent expansion I have been presented with the function $g(z) = \frac{2z}{z^2 + z^3}$ and asked to find the Laurent expansion around the point $z=0$. I split the function into partial fractions to obtain $g(z) = \frac{2}{z} - \frac{2}{1+z}$, but do not know where to go from here.

The function $$g(z)=\frac{2}{z}-\frac{2}{1+z}$$ has two simple poles at $0$ and $-1$. We observe the fraction $\frac{2}{z}$ is already the principal part of the Laurent expansion at $z=0$. We can keep the focus on the other fraction. Since we want to find a Laurent expansion with center $0$, we look at the other pole $-1$ and have to distuinguish two regions. \begin{align*} |z|<1,\qquad\quad 1<|z| \end{align*} * *The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $-1$ at the boundary of the disc. In the interior of this disc the fraction with pole $-1$ admits a representation as power series at $z=0$. *The second region $1<|z|$ containing all points outside the disc with center $0$ and radius $1$ admits for all fractions a representation as principal part of a Laurent series at $z=0$. A power series expansion of $\frac{1}{z+a}$ at $z=0$ is \begin{align*} \frac{1}{z+a} &=\frac{1}{a}\frac{1}{1+\frac{z}{a}} =\frac{1}{a}\sum_{n=0}^{\infty}\left(-\frac{1}{a}\right)^nz^n\\ &=-\sum_{n=0}^{\infty}\left(-\frac{1}{a}\right)^{n+1}z^n\\ \end{align*} The principal part of $\frac{1}{z+a}$ at $z=0$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{z}\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}(-a)^n\frac{1}{z^n}\\ &=\sum_{n=1}^{\infty}(-a)^{n-1}\frac{1}{z^n}\\ \end{align*} We can now obtain the Laurent expansion of $g(x)$ at $z=0$ for both regions * *Region 1: $|z|<1$ \begin{align*} g(z)&=\frac{2}{z}-2\sum_{n=0}^{\infty}(-1)^nz^n=2\sum_{n=-1}^{\infty}(-1)^{n+1}z^n\\ \end{align*} * *Region 2: $1<|z|$ \begin{align*} g(z)&=\frac{2}{z}-2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{z^n}=2\sum_{n=2}^{\infty}(-1)^{n}\frac{1}{z^n}\\ \end{align*}

How to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ $B = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 4 & 1 \end{bmatrix}$ and I need to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ . I know that $AB = A\cdot\begin{bmatrix} 1\\1\\1 \end{bmatrix} \mid A\cdot\begin{bmatrix} 2\\3\\4 \end{bmatrix} \mid A\cdot\begin{bmatrix} 1\\1\\1 \end{bmatrix} = 0 \mid 0 \mid 0$ Then I can conclude that (assume $A_1,...,A_n$ are columns of $A$): 1) $A_1 + A_2 + A_3 = 0$ 2) $2A_1 + 3A_2 +4 A_3 = 0$ Meaning: $A_1 + 2A_2 +3A_3 = 0$ But now I got stuck... How should I continue from here?

Then I can conclude that (assume $A_1,...,A_n$ are columns of $A$): 1) $A_1 + A_2 + A_3 = 0$ 2) $2A_1 + 3A_2 +4 A_3 = 0$ Close! Instead of rows they should be columns because matrix multiplication would take the dot products of the row vectors of the first matrix with the column vectors of the second. By elementary row operations, $(-2 eq. 1 + eq. 2)$ $A_2+2A_3=0 \Rightarrow A_2=-2A_3$ Substituting this relation into $eq. 1$ gives us $A_1-A_3=0 \Rightarrow A_1=A_3$. So, we can create a matrix that respects the previous conditions: $A = \begin{bmatrix} a&b &c \\ -2a&-2b &-2c \\ a&b &c \end{bmatrix} $ and we want to find such $a,b,c$ that $AB=0$ $\begin{bmatrix} a&b &c \\ -2a&-2b &-2c \\ a&b &c \end{bmatrix}\cdot \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 4 & 1 \end{bmatrix}=\begin{bmatrix} a+b+c&2a+3b+4c&a+b+c\\ -2(a+b+c)&-2(2a+3b+4c)&-2(a+b+c)\\ a+b+c&2a+3b+4c&a+b+c \end{bmatrix}$ See how the system we are trying to solve: $a+b+c=0, \quad\!\! 2a+3b+4c=0$ is similar to the one above with $A_1,A_2,A_3$. So, $-b=2a=2c$. The original matrix $A$ can now be expressed as $A = \begin{bmatrix} t&-2t &t \\ -2t&4t &-2t \\ t&-2t &t \end{bmatrix} $ where $t\in \mathbb{R}$ To construct a basis, notice that the matrix function $A(t)$ is of single variable. Thus, any multiple of \begin{bmatrix} 1&-2 &1 \\ -2&4 &-2 \\ 1&-2 &1 \end{bmatrix} forms a basis for $W$. $$W=span\begin{bmatrix} 1&-2 &1 \\ -2&4 &-2 \\ 1&-2 &1 \\ \end{bmatrix}=t\begin{bmatrix} 1&-2 &1 \\ -2&4 &-2 \\ 1&-2 &1 \\ \end{bmatrix}{\Huge{|}}t\in \mathbb{R}$$

What is a proof of this limit of this nested radical? It seems as if $$\lim_{x\to 0^+} \sqrt{x+\sqrt[3]{x+\sqrt[4]{\cdots}}}=1$$ I really am at a loss at a proof here. This doesn't come from anywhere, but just out of curiosity. Graphing proves this result fairly well.

For any $2 \le n \le m$, let $\phi_{n,m}(x) = \sqrt[n]{x + \sqrt[n+1]{x + \sqrt[n+2]{x + \cdots \sqrt[m]{x}}}}$. I will interpret the expression we have as following limit. $$\sqrt{x + \sqrt[3]{x + \sqrt[4]{x + \cdots }}}\; = \phi_{2,\infty}(x) \stackrel{def}{=}\;\lim_{m\to\infty} \phi_{2,m}(x)$$ For any $x \in (0,1)$, we have $\lim\limits_{m\to\infty}(1-x)^m = 0$. This implies the existence of an $N$ so that for all $m > N$, we have $$(1-x)^m < x \implies 1 - x < \sqrt[m]{x} \implies \phi_{m-1,m}(x) = \sqrt[m-1]{x + \sqrt[m]{x}} > 1$$ It is clear for such $m$, we will have $\phi_{2,m}(x) \ge 1$. Recall for any $k > 1$ and $t > 0$, $\sqrt[k]{1 + t} < 1 + \frac{t}{k}$. Start from $\phi_{m,m}(x) = \sqrt[m]{x} \le 1$, we have $$\begin{align} & \phi_{m-1,m}(x) = \sqrt[m-1]{x + \phi_{m,m}(x)} \le \sqrt[m-1]{x + 1} \le 1 + \frac{x}{m-1}\\ \implies & \phi_{m-2,m}(x) = \sqrt[m-2]{x + \phi_{m-1,m}(x)} \le \sqrt[m-2]{x + 1 + \frac{x}{m-1}} \le 1 + \frac{1}{m-2}\left(1 + \frac{1}{m-1}\right)x\\ \implies & \phi_{m-3,m}(x) = \sqrt[m-3]{x + \phi_{m-2,m}(x)} \le 1 + \frac{1}{m-3}\left(1 + \frac{1}{m-2}\left(1 + \frac{1}{m-1}\right)\right)x\\ & \vdots\\ \implies & \phi_{2,m}(x) \le 1 + \frac12\left( 1 + \frac13\left(1 + \cdots \left(1 + \frac{1}{m-1}\right)\right)\right)x \le 1 + (e-2)x \end{align} $$ Notice for fixed $x$ and as a sequence of $m$, $\phi_{2,m}(x)$ is monotonic increasing. By arguments above, this sequence is ultimately sandwiched between $1$ and $1 + (e-2)x$. As a result, $\phi_{2,\infty}(x)$ is defined for this $x$ and satisfies $$1 \le \phi_{2,\infty}(x) \le 1 + (e-2) x$$ Taking $x \to 0^{+}$, we get $$1 \le \liminf_{x\to 0^+} \phi_{2,\infty}(x) \le \limsup_{x\to 0^+}\phi_{2,\infty}(x) \le \limsup_{x\to 0^+}(1 + (e-2)x) = 1$$ This implies $\lim\limits_{x\to 0^+} \phi_{2,\infty}(x)$ exists and equal to $1$.

To find the sum of the series $\,1+ \frac{1}{3\cdot4}+\frac{1}{5\cdot4^2}+\frac{1}{7\cdot4^3}+\ldots$ The answer given is $\log 3$. Now looking at the series \begin{align} 1+ \dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot4^2}+\dfrac{1}{7\cdot4^3}+\ldots &= \sum\limits_{i=0}^\infty \dfrac{1}{\left(2n-1\right)\cdot4^n} \\ \log 3 &=\sum\limits_{i=1}^\infty \dfrac{\left(-1\right)^{n+1}\,2^n}{n} \end{align} How do I relate these two series?

HINT... consider the series for $\ln(1+x)$ and $\ln(1-x)$

Real values of $x$ satisfying the equation $x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$ Real values of $x$ satisfying the equation $$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$$ We can write it as $$512x^9+576x^6+216x^3-512x+219=0$$ I did not understand how can i factorise it. Help me

If this problem can be solved without computation it is reducible; we assume this. $f(x)=512x^9+576x^+216x^3-512x+219=0$ has two change-sign and $f(-x)$ has three ones so $f(x)$ has at least $9-5=4$ non real roots. We try to find a quadratic factor using the fact that $219=3\cdot73$ and $512=2^9$; this factor could correspond to real or non-real roots. Trying with $4x^2+ax\pm 3$ we find at once that $a=2$ and the sign minus fits; furthermore $4x^2+2x-3=0$ has two real roots because $\Delta=1+12>0$. The quotient gives $$128x^7-64x^6+128x^5+32x^4+80x^3-16x^2+122x-73=0$$ and this equation has necessarily a real root because $7$ is odd; assuming this 7-degree polynomial is reducible and noticing that: $$\begin{cases}(2x)^7-(2x)^6+4(2x)5+2(2x)^4+10(2x)^3-4(2x)^2+61(2x)-73=0\\1-1+4+2+10-4+61-73=0\end{cases}$$ it follows at once that $2x=1$ gives a third real root. Dividing again, now by $(2x)-1$ one gets $$g(x)=(2x)^6+4(2x)^4+6 (2x)^3+16(2x)^2+12(2x)+73=0$$ It is obvious that $g(x)>0$ for $x>0$ and it is easy to show that for $X<0$ $$X^6+4X^4+16X^2+73>-(6X^3+12X)$$ hence $g(x)$ is always positive so $g(x)=0$ has six non-real roots. Thus $f(x)=0$ has only three real roots, given by $$\color{red}{4x^2+2x-3=0}\space \text {and}\space\space \color{red}{ 2x-1=0}$$

The number of ordered pairs $(x,y)$ satisfying the equation The number of ordered pairs $(x,y)$ satisfying the equation $\lfloor\frac{x}{2}\rfloor+\lfloor\frac{2x}{3}\rfloor+\lfloor\frac{y}{4}\rfloor+\lfloor\frac{4y}{5}\rfloor=\frac{7x}{6}+\frac{21y}{20}$,where $0<x,y<30$ It appears that $\frac{x}{2}+\frac{2x}{3}=\frac{7x}{6}$ and $\frac{y}{4}+\frac{4y}{5}=\frac{21y}{20}$ but $\frac{x}{2}$ and $\frac{2x}{3}$ being inside the floor function,i can not add them up directly,same problem here,i cannot add $\frac{y}{4}$ and $\frac{4y}{5}$ because they are inside the floor function.What should i do to solve it?

Note that $$\Bigl\lfloor\frac x2\Bigr\rfloor\le\frac x2\ ,$$ and likewise for the other terms. So we always have $LHS\le RHS$, and the only way they can be equal is if $$\Bigl\lfloor\frac x2\Bigr\rfloor=\frac x2\ ,$$ and likewise for the other terms. So $$\frac x2\ ,\quad \frac{2x}3\ ,\quad \frac y4\ ,\quad\frac{4y}5$$ must all be integers. Can you take it from here?

Show that $2^n-(n-1)2^{n-2}+\frac{(n-2)(n-3)}{2!}2^{n-4}-...=n+1$ If n is a positive integer I need to show that $2^n-(n-1)2^{n-2}+\frac{(n-2)(n-3)}{2!}2^{n-4}-...=n+1$ My guess: Somehow I need two equivalent binomial expression whose coefficients I need to compare.But which two binomial expressions? I know not! P.S:Don't use Sterling Numbers or very high level maths...

Here is a generating function approach $$ \begin{align} \sum_{n=0}^\infty a_nx^n &=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{n-k}{k}2^{n-2k}x^n\tag{1}\\ &=\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^k\binom{n-k}{k}2^{n-2k}x^n\tag{2}\\ &=\sum_{k=0}^\infty\left(-\frac14\right)^k\sum_{n=k}^\infty\binom{n-k}{k}(2x)^n\tag{3}\\ &=\sum_{k=0}^\infty\left(-\frac14\right)^k\sum_{n=0}^\infty\binom{n}{k}(2x)^{n+k}\tag{4}\\ &=\sum_{k=0}^\infty\left(-\frac x2\right)^k\sum_{n=0}^\infty\binom{n}{k}(2x)^n\tag{5}\\ &=\sum_{k=0}^\infty\left(-\frac x2\right)^k\sum_{n=0}^\infty(-1)^{n-k}\binom{-k-1}{n-k}(2x)^n\tag{6}\\ &=\sum_{k=0}^\infty\left(-\frac x2\right)^k\sum_{n=0}^\infty(-1)^n\binom{-k-1}{n}(2x)^{n+k}\tag{7}\\ &=\sum_{k=0}^\infty\left(-x^2\right)^k\sum_{n=0}^\infty(-1)^n\binom{-k-1}{n}(2x)^n\tag{8}\\ &=\sum_{k=0}^\infty\left(-x^2\right)^k\frac1{(1-2x)^{k+1}}\tag{9}\\ &=\frac1{1-2x}\frac1{1+\frac{x^2}{1-2x}}\tag{10}\\ &=\frac1{(1-x)^2}\tag{11}\\ &=\sum_{k=0}^\infty(-1)^k\binom{-2}{k}x^k\tag{12}\\ &=\sum_{k=0}^\infty(k+1)x^k\tag{13}\\ \end{align} $$ Explanation: $\phantom{0}(2)$: change order of summation $\phantom{0}(3)$: move $(-1)^k2^{-2k}=\left(-\frac14\right)^k$ out front $\phantom{0}(4)$: substitute $n\mapsto n+k$ $\phantom{0}(5)$: move $(2x)^k$ out front $\phantom{0}(6)$: $\binom{n}{k}=\binom{n}{n-k}=(-1)^{n-k}\binom{-k-1}{n-k}$ (see this answer) $\phantom{0}(7)$: substitute $n\mapsto n+k$ $\phantom{0}(8)$: move $(2x)^k$ out front $\phantom{0}(9)$: Binomial Theorem $(10)$: sum of a geometric series $(11)$: simplification $(12)$: Binomial Theorem $(13)$: $(-1)^k\binom{-2}{k}=\binom{k+1}{k}=\binom{k+1}{1}=k+1$ Equating the coefficients of $x^k$, we get $a_n=n+1$.

Find general solution to the equation $xy^{'}=2x^2\sqrt{y}+4y$ This is Bernoulli's differential equation: $$xy^{'}-4y=2x^2y^{\frac{1}{2}}$$ Substitution $y=z^{\frac{1}{1-\alpha}},\alpha=\frac{1}{2},y^{'}=z^{'2}$ gives $$xz^{'2}-4z^2=2x^2z$$ Is this correct? What is the method for solving this equation?

$$xy'(x)=2x^2\sqrt{y(x)}+4y(x)\Longleftrightarrow$$ $$xy'(x)-4y(x)=2x^2\sqrt{y(x)}-4y(x)\Longleftrightarrow$$ $$\frac{y'(x)}{2\sqrt{y(x)}}-\frac{2\sqrt{y(x)}}{x}=x\Longleftrightarrow$$ Let $v(x)=\sqrt{y(x)}$; which gives $v'(x)=\frac{y'(x)}{2\sqrt{y(x)}}$: $$v'(x)-\frac{2v(x)}{x}=x\Longleftrightarrow$$ Let $\mu(x)=e^{\int-\frac{2}{x}\space\text{d}x}=\frac{1}{x^2}$. Multiply both sides by $\mu(x)$: $$\frac{v'(x)}{x^2}-\frac{2v(x)}{x^3}=\frac{1}{x}\Longleftrightarrow$$ Substitute $-\frac{2}{x^3}=\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^2}\right)$: $$\frac{v'(x)}{x^2}+\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^2}\right)v(x)=\frac{1}{x}\Longleftrightarrow$$ Apply the reverse product rule $g\frac{\text{d}f}{\text{d}x}+f\frac{\text{d}g}{\text{d}x}=\frac{\text{d}}{\text{d}x}(fg)$ to the left-hand side: $$\frac{\text{d}}{\text{d}x}\left(\frac{v(x)}{x^2}\right)=\frac{1}{x}\Longleftrightarrow$$ $$\int\frac{\text{d}}{\text{d}x}\left(\frac{v(x)}{x^2}\right)\space\text{d}x=\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$ $$\frac{v(x)}{x^2}=\ln\left|x\right|+\text{C}\Longleftrightarrow$$ $$v(x)=x^2\left(\ln\left|x\right|+\text{C}\right)\Longleftrightarrow$$ $$y(x)=\left(x^2\left(\ln\left|x\right|+\text{C}\right)\right)^2\Longleftrightarrow$$ $$y(x)=x^4\left(\ln\left|x\right|+\text{C}\right)^2$$

How to prove that all odd powers of two add one are multiples of three For example \begin{align} 2^5 + 1 &= 33\\ 2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)} \end{align} I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?

$2^2=4\equiv1\pmod 3$, so $4^k\equiv1\pmod3$ for all integers $k$. And so for any odd number $2k+1$, we get $2^{2k+1}+1 = 4^k\cdot 2+1\equiv 2+1\equiv0\pmod3$.