PRMO level question of functions Consider the functions $f(x)$ and $g(x)$ which are defined as $f(x)=(x+1)(x^2+1)(x^4+1)\ldots\left({x^2}^{2007}+1\right)$ and $g(x)\left({x^2}^{2008}-1\right)= f(x)-1$.
Find $g(2)$
This is a PRMO level question of functions and I tried it with substituting values also but to no avail and the solution Of this question is also not available thought answer is given as $g(x)=2$.
 A: $$f(x)=\frac{\color{red}{(x-1)}(x+1)(x^2+1)(x^4+1)\ldots\left({x^2}^{2007}+1\right)}{\color{red}{x-1}}$$
The $f(x)$ then becomes
$$f(x)=\frac{x^{2^{2008}}-1}{x-1}\implies f(2)=4^{2008}-1$$
The given condition then becomes,
$$g(x)\left({x^2}^{2008}-1\right)= f(x)-1 \implies g(x)\cdot f(x)(x-1)=f(x)-1$$
Substituting $x=2$,
$$g(2)\cdot f(2)=f(2)-1$$
$$g(2)=\frac{f(2)-1}{f(2)}\implies g(2)=1-\frac{1}{f(2)}$$
$$\therefore g(2)=1-\frac{1}{4^{2008}-1}$$
$$\fbox{$g(2)=\frac{1}{4^{2008}-1}$}$$
A: Some hints:
Note that
$$f(x)(x-1) = x^{2^{2008}}-1$$
This gives $f(2)$.
Now, use this in
$$g(x)\left({x^2}^{2008}-1\right) = g(x)f(x)(x-1)= f(x)-1$$
to evaluate $g(2)$.
A: We can write
$$
g\left( x \right) = \frac{{\left( {x + 1} \right)\left( {x^2  + 1} \right) \ldots \left( {x^{2^{2007} }  + 1} \right) - 1}}
{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {x^2  + 1} \right) \ldots \left( {x^{2^{2007} }  + 1} \right)}}$$
and then get
$$
g\left( x \right) = \frac{1}
{{x - 1}} - \frac{1}
{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {x^2  + 1} \right) \ldots \left( {x^{2^{2007} }  + 1} \right)}}$$
So $g(2) = 1-\epsilon$
where $\epsilon=\frac{1}
{{\left( {2 + 1} \right)\left( {2^2  + 1} \right) \ldots \left( {2^{2^{2007} }  + 1} \right)}}=\frac{1}{4^{2008}-1}$.
A: Every integer admits a unique binary expansion. Since the exponent of $x$'s in $f$ are powers of 2, we have
$$f(x)=1+x+x^2+\cdots +x^{2^{2007}+2^{2006}+\cdots+1}$$
Using the formula for the sum of geometric sequence:
$$2^{2007}+2^{2006}+\cdots+1=\frac{1-2^{2008}}{1-2}=2^{2008}-1$$
Hence,
$$f(x)=1+x+x^2+\cdots +x^{2^{2008}-1}=\frac{1-x^{2^{2008}}}{1-x}$$
so
$$f(2)=2^{2^{2008}}-1$$
$$g(2)=1-\frac{1}{2^{2^{2008}}-1}$$
