$\frac{1}{4rs}[(r+s)^2\cos \frac{\pi}{2}(r+s)-(s-r)^2\cos \frac{\pi}{2}(s-r)]$ near 0

Question

How to show there exists $|f(x)|>1$ near $x=0$, where $$f(x)=\frac{1}{4rs}[(r+s)^2\cos \frac{\pi}{2}(r+s)-(r-s)^2\cos \frac{\pi}{2}(r-s)]$$ $r=\sqrt{a+2x}$

$s=\sqrt{a-2x}$

for $a=k^2$, $k\in \mathbb{N}$.

I try taylor series but I have to calculate up to $x^4$ order which is a heavy calcluation.

Is there an easier way ?

Answer 1

Here is what Mathematica did for me. An important point was writing $$r(y):=k\sqrt{1+2y},\quad s(y):=k\sqrt{1-2y},\qquad y(x)={x\over k^2}\ .$$ It turns out that the claim is true, but the cases $k$ even and $k$ odd look different.