<s>[INST] Prove the root is less than $2^n$ 
A polynomial $f(x)$ of degree $n$ such that coefficient of $x^k$ is $a_k$. Another constructed polynomial $g(x)$ of degree $n$ is present such that the coefficeint of $x^k$ is $\frac{a_k}{2^k-1}$. If $1$ and $2^{n+1}$ are roots of $g(x)$, show that $f(x)$ has a positive root less than $2^n$.

Personally a tough problem. Hints Only Please!
I got that: $f(x) = g(2x) - g(x)$
Also,
$f(x) = a_0 + \sum_{k=1}^{n} a_k x^k$ and $g(x) = \sum_{k=1}^{n} \frac{a_k x^k}{2^k - 1}$ and $h(x) = g(2x) = \sum_{k=1}^{n} \frac{2^k a_k x^k}{2^k - 1}$
$g(1) = \sum_{k=1}^{n} \frac{a_k}{2^k - 1} = 0$
$g(2^{n+1}) = \sum_{k=1}^{n} \frac{2^k a_k 2^{nk}}{2^k - 1} = 0$
Realize that: $g(2^{n + 1}) = g(2 \cdot 2^{n}) = h(2^n). $ Hence the root of $h(x) = g(2x) \implies x =  2^n$. 
Since $h(1/2) = g(1) = 0$, it follows $x= \frac{1}{2}$ is a root for $h(x) = g(2x)$. 
So I have:
$g(x) = 0 \implies x = \{1, 2^{n+1} \}$ 
$g(2x) = 0 \implies x = \{\frac{1}{2}, 2^{n}\}$.
Now suppose $x = 2^{n} + i$. Then, 
$g(2x) = g(2^{n+1} + 2i) > g(2^{n+1}) = 0$. 
$g(x) = g(2^n + i)$
But I'm not sure what to do next?
 [/INST] Warning: This is just a wordy hint... : 
Assuming $a_0=0$, notice that $g(2x)$ is a horizontally scaled version of $g(x)$, which means that both functions have the same sign in $(\dfrac 12,2^n)$ and $(1,2^{n+1})$ respectively, so they have the same sign in $(1,2^n)$. On the left-most side of the interval, around $x=1$ (the root of $g(x)$), one function must be "higher than" the other, so that $g(2x)-g(x)>0$ or $g(2x)-g(x)<0$. Conversely, on the right-most side, around $x=2^n$, the inverse is true. Hence the difference, in either case, changes sign somewhere in the interval $(1,2^n)$ which means that...
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