Solving by Lagrange’s auxiliary equations $ frac {dx}{z} =frac {dy}{z} =frac {(3z+1)(dz)}{x+y} $

BY taking 1st and 2nd equations 1st solution is $$ x=y+c_{0}$$
but i am getting two different solutions slightly differing :

Solution 1 : using lagrange multiplier 1,1,0 we get
$$ frac{dx+dy}{2z}=frac{(3z+1)dz}{x+y}$$
$$ frac {(x+y)d(x+y)}{2z}= {(3z+1)(dz)}$$
$$ frac {d(x+y)^{2}}{2} = (2z)(3z+1)(dz) =(6z^{2}+2z)dz$$
$$ frac {(x+y)^2}{2} = {2z^3}+(z^2)+c_{1} $$
$$ (x+y)^{2}= 4z^{3}+2z^{2}+c_{1′}$$

Solution 2 : taking x,y,-z as multipliers

$$ frac {xdx+ydy-(zdz)(3z+1)}{xz+yz-z(x+y)} =frac {xdx+ydy-(zdz)(3z+1)}{0}$$

$$ xdx+ydy= (3z+1)(zdz)=(3z^{2}+z)dz$$
$$ frac{x^2}{2} +frac{y^2}{2}= z^{3}+ frac{z^2}{2}+c_{2} $$
$$ x^{2}+y^{2}=2z^{3}+z^{2}+c_{2′}$$
$$ 2(x^{2}+y^{2})=4z^{3}+2z^{2}+c_{2”}$$

here in the first solution the term of $$xy$$ is extra .. i believe i have done mistake in my first solution but i am unable to find out where?

solution 1 is my solution and solution 2 is the solution from the book

Proving $lim_{xto 0} xsin {frac 1x} = 0$

Prove $$lim_{xto 0} xsin {frac 1x} = 0$$ with squeeze theorem.

I know there’s a way of proving this by using $$lim_{xto 0} x = 0 \ |sin x| le 1$$ but how can I prove it with squeeze theorem?

inequality – Prove that $frac{ 1}{a^2+b^2+ab}+frac{ 1}{b^2+c^2+bc}+ frac{ 1}{c^2+a^2+ca}ge frac{ 9}{left(a+b+cright)^2}$.

Let $a,b,c$ be positive real numbers. Prove that $$frac{ 1}{a^2+b^2+ab}+frac{ 1}{b^2+c^2+bc}+ frac{ 1}{c^2+a^2+ca}ge frac{ 9}{left(a+b+cright)^2}.$$

I want to prove the inequality with elementary inequalities ( Mean inequalities, Cauchy-Schwarz etc.). I tried to use Cauchy-Schwarz, but I got $$frac{ 1}{a^2+b^2+ab}+frac{ 1}{b^2+c^2+bc}+ frac{ 1}{c^2+a^2+ca}gefrac{9}{2(a^2+b^2+c^2)+ab+bc+ca}.$$
Then we have to show that $$2(a^2+b^2+c^2)+ab+bc+caleq (a+b+c)^2 $$
which is not true.
Multiplying the numerators of the fractions in LHS by $a^2,b^2,c^2$ also makes the problem more difficult.

So, how to solve the problem with elementary inequalities?

partial differential equations – When is $frac {mathrm d y}{mathrm d x}=frac {mathrm d r}{mathrm d theta}?$

I came up with this idea, which is when will $frac {mathrm d y}{mathrm d x}=frac {mathrm d r}{mathrm d theta}$?

By total differential, we have
$$mathrm d r=frac {partial r}{partial x}mathrm dx+frac {partial r}{partial y}mathrm dy=frac {xmathrm dx+ymathrm dy}{sqrt{x^2+y^2}}$$
$$mathrm d theta=frac {partial theta}{partial x}mathrm dx+frac {partial theta}{partial y}mathrm dy=frac {ymathrm dx+xmathrm dy}{x^2+y^2}$$
Thus $frac {mathrm dr}{mathrm dtheta}=frac {sqrt{x^2+y^2}(xmathrm dx+ymathrm dy)}{ymathrm dx+xmathrm dy}$
And notice that if we divide the numerator and denominator by $mathrm dx$ we will have a quadratic equation in terms of $frac {mathrm dy}{mathrm dx}$, how can we solve that?

calculus – The convergence of $sum_{n=1}^{infty} frac {(-1)^{E(sqrt n)}}{n}$

sum_{n=1}^{infty} frac{(-1)^{E(sqrt n)}}{n}

where $E(x)$ denote the integer part of $x$.

I try my best to show that it is convergent, but I failed. I can the the first three terms is negative , then the next five terms is positive. I think it is can be solved by the Leibniz’s series theorem. But I don’t know how to apply it to the problem. If this is proved , then can I guess that it is also convergent for
sum_{n=1}^{infty} frac {(-1)^{Eleft(n^{frac 13}right)}}{n}

or more generally?

limits – Does $lim {n to infty} a_n = alpha$ imply $lim left(1+ frac {a_n}{n} right) =e^alpha$ for real value

Does $lim_{n to infty} a_n = alpha$ imply $lim_{n to infty} left(1+ frac {a_n}{n} right)^n =e^alpha$?

I’m only considering real value
and also, does the value of $alpha$ makes a difference?

I tried considering $$|left(1+ frac {a_n}{n} right)^n-e^alpha|leq|left(1+ frac {a_n}{n} right)^n-left(1+ frac {alpha}{n} right)^n|+|left(1+ frac {alpha}{n} right)^n-e^alpha|$$
I know that for any fixed $alpha$, $|left(1+ frac {alpha}{n} right)^n-e^alpha|$ tends to zero
but I don’t know how to deal with $$|left(1+ frac {a_n}{n} right)^n-left(1+ frac {alpha}{n} right)^n|$$, I tried binomial expansion but didn’t get anywhere. Is this where the value of $alpha$ come to matter´╝č

sequences and series – Does there exists an upper bound on the Fourier coefficients of the reciprocal theta function $frac {1}{theta}$?

Define the theta function as
theta(x) = sum_{n=-infty}^infty e^{-gamma(x+n)^2}

where $gamma>0$. Clearly, $theta$ is 1-periodic, non-zero and smooth. Therefore, the reciprocal map $x mapsto frac{1}{theta(x)}$ is 1-periodic and smooth as well and can be expanded as a Fourier series:
frac{1}{theta(x)} = sum_{n=-infty}^infty a_ne^{2pi i nx}.

I’m interested in upper bounding the Fourier coefficients, i.e. I want to find a non-trivial map $f$ (with a fast decay) such that
|a_n| leq f(n).

Are there an results in this direction? Thank you!

real analysis – Any hints on how to prove that $;cos (ax-bx)+2cos (ax+bx);$ is positive over $frac12$ of its domain if $frac ab notinmathbb{Q}$?

We know that the periodic function $;cos x;$ is positive over $;frac12;$ of its domain. Now, I have this function

$$f(x)=cos (ax-bx)+2cos (ax+bx)\ qquadqquad =3cos (ax)cos (bx)-sin (ax)sin(bx)$$

with $x>0$, and $frac {a}{b}notinmathbb{Q}$, therefore, the function is not periodic.

How can I prove that the function $f(x)$ is also positive on $frac12$ of its domain?

Any hints and suggestions are appreciated.

taylor ‘s series of $sin frac { 1 }{x }$

Is it possible to built the Taylor ‘s series of the function $sin frac {1}{x}$ in $x=0$ also if $f (x )$ is not defined in this point?

How to prove $frac a{b + 2c} + frac b{c + 2a} + frac c{a + 2b} ge 1$?

$$frac a{b + 2c} + frac b{c + 2a} + frac c{a + 2b} ge 1$$

How can I prove this inequality?
Given that a, b, c are strictly positive reals