## Solve fr x and $x^2 + y^2 = 1$ and $x \pm y = \frac \pi4$

Solve fr x and $$x^2+ y^2 = 1$$ and $$x \pm y = \frac \pi4$$

I tried solving this by substitute method. And using the quadratic formula, but that create lots of cases. The original problem was to solve $$\arcsin x + \arcsin y = \frac\pi2$$ and $$\sin 2x = \cos2y$$.

## Why is $u(z) = \frac {z^{1-p}}{1-p}$ taken as $log(z)$ when $p=1$?

Let $$u(z) = \frac {z^{1-p}}{1-p}$$ when $$p$$ is not 1, and $$u(z) = log(z)$$ when $$p=1$$.

Why do we take it as $$log(z)$$?

## Find limit for sequence $(\frac n{n+1})_{n=1}^{\infty}.$

Need help in getting limit of sequence. This question is taken from section 3.2.1 of Chapter 3 of the CRM series book by MAA: Exploratory Examples for Real Analysis, By Joanne E. Snow, Kirk E. Weller.

I request to vet attempt; as well help to find the solution.

It is not solvable as L’ Hopital’s rule applies for $$\frac 0 0$$ or $$\frac {\infty}{\infty}$$ cases alone, & the form is $$\frac 1{\infty}$$.

Q.3. Identify limit of Seq. 3 using calculus. $$(\frac n{n+1})_{n=1}^{\infty}$$

Let, $$f(n) = (\frac n{n+1})_{n=1}^{\infty}$$, need find $$f'(n)\approx\frac{\Delta y}{\Delta n}= \lim_{\Delta n \to0 } (\frac{f(n+\Delta n)-f(n)}{(n+\Delta n)-n})_{n=1}^{\infty}$$

$$= \lim_{h\to0 }(\frac{f(n+h)-f(n)}{h})_{n=1}^{\infty}$$

$$=\lim_{h\to0 }(\frac{\frac {n+h}{n+h+1}-\frac n{n+1}}{h})_{n=1}^{\infty}$$

$$=\lim_{h\to0 }(\frac{\frac {(n+h)(n+1) -(n)(n+h+1)}{(n+h+1)(n+1)}}{h})_{n=1}^{\infty}$$

$$=\lim_{h\to0 }(\frac{\frac {(n^2+n+nh+h) -(n^2+nh+n)}{(n^2+n+nh+h+n+1)}}{h})_{n=1}^{\infty}$$

$$=\lim_{h\to0 }(\frac{\frac {h}{(n^2+nh+2n+1)}}{h})_{n=1}^{\infty}$$ $$=(\frac {1}{n^2+2n+1})_{n=1}^{\infty}$$

## What will be $\lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$

Evaluate

$$\lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$$

I tried by taking log but it wouldn’t work because there are infinitely many points in any neighbourhood of $$0$$ where $$\ln \left ( \sin {\frac {1} {x}} \right )$$ doesn’t exist. How to overcome this situation?

Any help will be highly appreciated.Thank you very much for your valuable time.

## show this inequality to $\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3}$

Let $$a,b$$ and $$c$$ be positive real numbers. Prove that $$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3}$$

This problem is from Iran 3rd round-2017-Algebra final exam-P3,Now I can’t find this inequality have solve it,maybe it seem can use integral to solve it?

## Solving $u_{\epsilon \eta} = \frac{ u_\epsilon – u_eta}{4 (\epsilon – \eta) } .$

I’m trying to solve the following PDE

$$y^2 u_{xx} – u_{yy} = 0.$$

I’ve found its canonical form as $$u_{\epsilon \eta} = \frac{ u_\epsilon – u_eta}{4 (\epsilon – \eta) } .$$

However, I’m having trouble in solving this PDE. I’ve tried solving with mathematica, and that gave me

$$u = \frac{ x + y^3}{ 3} ,$$ but, I need to know how to obtain such a result from the canonical form of the given PDE.

## Estimate $\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$

Suppose $$a>1,b>0$$ are real numbers. Consider the summation of the infinite series: $$S=\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$$ How can I give a tight estimation on the summation? Apparently, one can get the upper bound: $$S\le \sum_{k=1} ^\infty \frac 1 {(a+b)^k}=\frac 1 {a+b-1}$$ But it is not tight enough. For example, fix $$a\rightarrow 1$$, and $$b=0.001$$, then $$S=38.969939$$, it seems that $$S=O(\sqrt{1/b})$$. Another example: $$a=1$$, and $$b=0.00001$$,$$S=395.039235$$.

## Compute $\frac{x}{y-z} +\frac {y}{z-x}+\frac {z}{x-y}$

Compute $$\frac{x}{y-z} +\frac {y}{z-x}+\frac {z}{x-y}$$ if $$\frac{x^2}{ (y-z)^2} +\frac {y^2}{ (z-x)^2}+\frac {z^2}{ (x-y)^2}=11.$$

I have no idea how can I start.

## On the determinant $\det[(\frac{i^2+dj^2}p)]_{0\le i,j\le(p-1)/2}$ with $(\frac dp)=-1$

Let $$p$$ be an odd prime. For $$d\in\mathbb Z$$ we define $$T(d,p):=\det\left[\left(\frac{i^2+dj^2}p\right)\right]_{0\le i,j\le(p-1)/2},$$ where $$(\frac{\cdot}p)$$ is the Legendre symbol. By (1.17) of my paper arXiv:1308.2900, if$$(\frac dp)=-1$$ then $$(\frac{T(d,p)}p)=1$$.

Suppose that $$p\equiv3\pmod4$$. Then $$T(d,p)=T(-1,p)$$ for any $$d\in\mathbb Z$$ with $$(\frac dp)=-1$$. As $$T(-1,p)$$ is a skew-symmetric determiant of even order, $$T(-1,p)$$ is always a square.

In the case $$p\equiv1\pmod4$$, if $$d$$ and $$d’$$ are both quadratic nonresidues modulo $$p$$, then we clearly have $$T(d,p)=\pm T(d’,p)$$.

I have the following conjecture which seems quite challenging.

Conjecture. Let $$p\equiv1\pmod4$$ be a prime and write $$p=x^2+4y^2$$ with $$x$$ and $$y$$ positive integers. Then, for any integer $$d\in\mathbb Z$$ with $$(\frac dp)=-1$$, there is a positive integer $$t(p)$$ (not depending on $$d$$) such that $$|T(d,p)|=2^{(p-1)/2}t(p)^2y.$$

Via Mathematica, I find that $$\begin{gather}t(5)=1,\ t(13)=3,\ t(17)=4,\ t(29)=91,\ t(37)=81,\ t(41)=180, \t(53)=1703,\ t(61)=87120,\ t(73)=16104096,\ t(89)=3947892146, \ t(97)=19299520512,\ t(101)=885623936875,\ t(109)=36548185365.\end{gather}$$