MHB Evaluate Inverse of Hi M.H.B.: Math Problem

[anemone]

Hi MHB,

The following problem has been a really vexing problem (for me) because I believe there would be a tricky way of approaching it but I could not solve it after working with it on and off for two days, it is an Olympiad math competition problem, and so far no one that I know of has solved it.

I think the time has come to ask for help at MHB. If anyone has ideas to solve it, I would appreciate the help.

Problem:

Let $f(x)=(x^{256}+1)(x^{64}+1)(x^{16}+1)(x^{4}+1)(x+1)$ for $0<x<1$.

Evaluate $f^{-1}\left(\dfrac{8}{5f\left(\dfrac{3}{8}\right)}\right)$.

My futile attempt is based on the core concept of utilizing the formula $f^{-1}(f(x))=x$ where I got

$f(x)=\dfrac{x^{512}-1}{(x-1)(x^2+1)(x^8+1)(x^{32}+1)(x^{128}+1)}$ that leads to $\dfrac{1}{(1-x)f(x)}=\dfrac{(x^2+1)(x^8+1)(x^{32}+1)(x^{128}+1)}{1-x^{512}}$, unfortunately all of these did not help to shed any insight for me to proceed.

 

[earboth]

...that leads to $\dfrac{1}{(1-x)f(x)}=\dfrac{(x^2+1)(x^8+1)(x^{32}+1)(x^{128}+1)}{1-x^{512}}$, unfortunately all of these did not help to shed any insight for me to proceed.

Good afternoon,

I don't know if this could be a step into the right direction, but you can factor the denominator into a lot of factors:

$$1-x^{512} = (x + 1)(1 - x)(x^2 + 1)(x^4 + 1)(x^8 + 1)(x^{16} + 1)(x^{32} + 1)(x^{64} + 1)(x^{128} + 1)(x^{256} + 1)$$

Now cancel as much factors as possible.

 

[anemone]

Good afternoon,

I don't know if this could be a step into the right direction, but you can factor the denominator into a lot of factors:

$$1-x^{512} = (x + 1)(1 - x)(x^2 + 1)(x^4 + 1)(x^8 + 1)(x^{16} + 1)(x^{32} + 1)(x^{64} + 1)(x^{128} + 1)(x^{256} + 1)$$

Now cancel as much factors as possible.

Hi earboth!

Thank you for the reply...but even after cancelling out the common factors, I could not see what I could do further to evaluate target expression...:(

$\dfrac{1}{(1-x)f(x)}=\dfrac{1}{(x^2-1)(x^4+1)(x^{16}+1)(x^{64}+1)(x^{256}+1)}$

 

[anemone]

Someone from France, an expert of solving Olympiad Mathematics problems has offered me a great insight which I thought to share it with members at MHB. But I have to say it out loud here that his approach led to the approximate but not exact value of $f^{-1}\left(\dfrac{8}{5f\left(\dfrac{3}{8}\right)}\right)$.

He mentioned about since $f(x)=(x^{256}+1)(x^{64}+1)(x^{16}+1)(x^{4}+1)(x+1)$, then we have $f(x^2)=(x^{512}+1)(x^{128}+1)(x^{32}+1)(x^{8}+1)(x^2+1)$ and note that

$$(x+1)(x^2 + 1)(x^4 + 1)(x^8 + 1)(x^{16} + 1)(x^{32} + 1)(x^{64} + 1)(x^{128} + 1)(x^{256} + 1)(x^{512}+1)=\dfrac{(x^{2014}+1)}{(x-1)}$$

We then obtained $f(x)f(x^2)=\dfrac{(x^{2014}-1)}{(x-1)}=\dfrac{(1-x^{2014})}{(1-x)}$.

At $x=\dfrac{3}{8}$, $f\left(\dfrac{3}{8}\right)f\left(\dfrac{3}{8}\right)^2=\dfrac{1-\left(\dfrac{3}{8}\right)^{2014}}{1-\left(\dfrac{3}{8}\right)}\approx \dfrac{8}{5}$.

Therefore, $f\left(\dfrac{3}{8}\right)^2\approx \dfrac{8}{5f\left(\dfrac{3}{8}\right)}$ and hence $f^{-1}\left(\dfrac{8}{5f\left(\dfrac{3}{8}\right)}\right)\approx \dfrac{9}{64}$.