1. ## Maximum of function

For each integer$\displaystyle n\geq1$ find the maximum of fucntion
$\displaystyle f(x)=\frac{x+x^2+...+x^{2n-1}}{(1+x^n)^2}$ for $\displaystyle x>0$ and every $\displaystyle x>0$ to which it's taken

2. ## Re: Maximum of function

Originally Posted by geagry
For each integer$\displaystyle n\geq1$ find the maximum of fucntion
$\displaystyle f(x)=\frac{x+x^2+...+x^{2n-1}}{(1+x^n)^2}$ for $\displaystyle x>0$ and every $\displaystyle x>0$ to which it's taken
Dear geagry,

$\displaystyle f(x)=\frac{x+x^2+...+x^{2n-1}}{(1+x^n)^2}~;~n\geq 1\mbox{ and }x>0$

$\displaystyle f'(x)=\frac{(1+x^n)^2\left[1+2x+3x^2+.....+(2n-1)x^{2n-2}\right]-2(1+x^n)nx^{n-1}\left(x+x^2+....+x^{2n-1}\right)}{(1+x^n)^4}$

$\displaystyle f'(x)=\frac{(1+x^n)\left[1+2x+3x^2+.....+(2n-1)x^{2n-2}\right]-2nx^{n-1}\left(x+x^2+....+x^{2n-1}\right)}{(1+x^n)^3}$

Substituting x=1 you will see that, $\displaystyle f'(1)=0$

Also observe that, $\displaystyle f(1)=\frac{2n-1}{4}$

Therefore, x=1 is an inflection point and, $\displaystyle f(1)=\frac{2n-1}{4}$

It could be shown that at x=1 is a maximum by considering the second derivative (the calculation of $\displaystyle f''(1)$ is terrible ).

$\displaystyle f''(1)=-\frac{n}{24}(2n-1)(n+2)<0\mbox{ since }n\geq{1}$

Therefore, at x=1 the function has a maximum and $\displaystyle f(1)=\frac{2n-1}{4}\mbox{ where }n=1,2,3,......$