Find the inverse of this function.

Question

Let define the function

$$f(x)=(a+b)\frac{16(\pi+x)x}{5\pi^2+4(\pi+x)x}-b\frac{(\pi^2-4x^2)x}{\pi^2+x^2}$$

for $x \in [\frac{-\pi}{2},0]$ such that $a,b \in \mathbb{R}$ and $0<a<b$.

What is the inverse of $f$ on $[\frac{-\pi}{2},0]$ ?

I tried to calculate it several times but I did not succeed.

Where did this problem come from? Was it inspired by this answer? — TonyK, Jul 27 '20 at 12:28
@TonyK this is a modified question from this question https://socratic.org/questions/5a08c0da7c0149261b6c1130 by taking the Mahabhaskariya of Bhaskara approxiamtion — Moon Estrella, Jul 27 '20 at 15:18
@MoonEstrella. The extra $x$ at the end of the equation makes a problem. Are you sure about it ? — Claude Leibovici, Jul 28 '20 at 01:19

score 3 · Answer 1 · answered Jul 27 '20 at 12:44

3

This is not an answer.

This is amazing since, for $-\frac \pi 2 <x<0$, $$\frac{16(\pi+x)x}{5\pi^2+4(\pi+x)x} \sim \sin(x)$$ $$\frac{(\pi^2-4x^2)}{\pi^2+x^2}\sim \cos(x)$$ These approximations are $1,400$ years old.

If the problem was (I removed the last $x$) $$f(x)=(a+b)\frac{16(\pi+x)x}{5\pi^2+4(\pi+x)x}-b\frac{(\pi^2-4x^2)}{\pi^2+x^2}$$ then an approximation would be $$f(x)=(a+b) \sin(x)-b \cos(x)$$ which is a basic trigonometric equation.

answered Jul 27 '20 at 12:44

Claude Leibovici

260,315

Claude you are too old for me (+1) for the wisdom – Miss and Mister cassoulet char Jul 27 '20 at 18:40

Find the inverse of this function.

1 Answers1