## 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？
Thanks

## 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