$\max\limits_{-1\le x\le 1}\Big\vert\frac{1-(-x^2)^{n+1}}{1+x^2}\Big\vert $?

Question

How to calculate the maximum of $$\max_{-1\le x\le 1}\Big\vert\frac{1-(-x^2)^{n+1}}{1+x^2}\Big\vert $$?

I think one solution is first to find the first derivative and then to find the critic points and then to calculate the second derivative respect to x and then evaluate the critic points in the second derivative and finally to check if it's positive or negative.

Since the first derivative it's already too complicated expression so the 2nd derivative will be worst.

Is there an easier way to proceed?

Answer 1

Change first $t=x^2$ to simplify $$ \max_{t\in[0,1]}\Big|\dfrac{1-(-1)^{n+1}t^{n+1}}{1+t}\Big|. $$ Then you have two cases:

1. $n+1$ is even. Then the expression is $$ \dfrac{1-t^{n+1}}{1+t}. $$ It is clear that the maximum is at $t=0$ since for larger $t$ we get the numerator smaller and the denominator larger, hence, the fraction becomes smaller.
2. $n+1$ is odd. Then the expression is $$ \dfrac{1+t^{n+1}}{1+t}=1-t+t^2-t^3+\ldots+t^n. $$ It is easy to differentiate, but it does not look like an easy task to solve for the critical point (somewhere in $(0,1)$). However, if you can prove that the function has only one critical point, which is minimum, and the end points have the same value, then the maximum is again at $t=0$ (or $t=1$).

Hint to continue: differentiate $$ \dfrac{1+t^{n+1}}{1+t} $$ and study what happens with the sign when $t$ increases from $0$ to $1$, conclude intervals of monotonicity and draw conclusion.

Answer 2

Consider the nominator for various $n$.

$n=0$: $\ 1+x^2$
$n=1$: $\ 1-x^4$
$n=2$: $\ 1+x^6\ $ etc.

Then apply some division of polynoms.
$(1+x^2):(1+x^2)=1$
$(1-x^4):(1+x^2)=1-x^2$
$(1+x^6):(1+x^2)=1-x^2+x^4\ $ etc.

Now you can differentiate, as you suggested already:
$0$
$-2x$
$-2x+4x^3=2x(-1+2x^2)\ $ etc.

You'll get that $x=0$ will always be a root. And, except for $n=0$, which is trivial, every second derivative would have the value $-2$ for $x=0$, i.e. it is a relative maximum indeed.

The function itself would evaluate to just $1$ for $x=0$.

But then you are looking for the global maximum within $-1\le x\le 1$.
$1\ $ is $1$ throughout.
$1-x^2\ $ runs down to $0$ at the borders.
$1-x^2+x^4\ $ gets back to $1$ there again.
Etc., always alternating between $0$ and $1$ there.

Thus your global maximum indeed is just $1$.