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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 $g(x)=2$.

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$$f(x)=\frac{\color{red}{(x-1)}(x+1)(x^2+1)(x^4+1)\ldots\left({x^2}^{2007}+1\right)}{\color{red}{x-1}}$$ 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 $x=2$, $$g(2)\cdot f(2)=f(2)-1$$ $$g(2)=\frac{f(2)-1}{f(2)}\implies g(2)=1-\frac{1}{f(2)}$$ $$\therefore g(2)=1-\frac{1}{4^{2008}-1}$$ $$\fbox{$g(2)=\frac{1}{4^{2008}-1}$}$$

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    $\begingroup$ I think the numerator of $f(x)$ gives $x^{2^{2008}}-1$, thus we have $$f(x)=\frac{x^{2^{2008}}-1}{x-1}$$. $\endgroup$
    – Alessio K
    Commented Sep 29, 2020 at 9:19
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    $\begingroup$ Oh yes spotted the error. Thanks! Will edit the post @Äres $\endgroup$
    – rash
    Commented Sep 29, 2020 at 9:22
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    $\begingroup$ Won't $f(2)=4^{2008}-1$? $\endgroup$
    – Amadeus
    Commented Sep 29, 2020 at 11:15
  • $\begingroup$ @l1mbo yes, have edited it. Thank you! $\endgroup$
    – rash
    Commented Sep 29, 2020 at 11:36
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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 $g(2)$.

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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 $$f(2)=2^{2^{2008}}-1$$ $$g(2)=1-\frac{1}{2^{2^{2008}}-1}$$

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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 $\epsilon=\frac{1} {{\left( {2 + 1} \right)\left( {2^2 + 1} \right) \ldots \left( {2^{2^{2007} } + 1} \right)}}=\frac{1}{4^{2008}-1}$.

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  • $\begingroup$ Perhaps you should substitute x=2 in the expression of e. $\endgroup$
    – cgss
    Commented Sep 29, 2020 at 9:13

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