Uniqueness of power series Is there two sequences of real numbers $a_i$ and $b_i\neq 8$, not depending on $x$, such that $x^8=\sum_{k=1}^{\infty}a_kx^{b_k}$ for all $x$?
If $\displaystyle\sum_{k=1}^{\infty}a_kx^{b_k}=\sum_{k=1}^{\infty}c_kx^{d_k}$ for all $x>1$, and all coefficients are real and $b_k>b_{k+1}$, and $d_k>d_{k+1}$, is there a way to prove that $a_k=c_k$ and $b_k=d_k$ for all $k$?

If $\displaystyle\sum_{k=1}^{\infty}a_kx^{b_k}=0$ for all $x>1$ and $b_i\neq b_j$, and all coefficients are real, must $a_k=0$?
 A: Note that these questions are the same as asking whether non-zero $a_i$, $b_i$ can satisfy $\sum_{k=1}^\infty a_kx^{b_k}=0$ for all $x$ (or for $x>1$). This assumes the series converge absolutely, so that we can combine them. If we're not assuming absolute convergence, then can prove absolute convergence assuming that the $b_i$ are positive (this can be relaxed a bit, but not done away with). 
Assuming the $b_i$ are strictly decreasing, we can answer it as follows. Assume the series converges absolutely for $x\ge A$ (that such an $A> 1$ exists will be shown at the end). Without loss of generality, assume $a_1\ne 0$. For $x\ge A$, we have
$$0=a_1 x^{b_1}+\sum_{n=2}^\infty a_nx^{b_n}$$
So we solve for $a_1$ and take absolute values
$$|a_1|\le x^{-b_1}\sum_{n=2}^\infty |a_n|x^{b_n}=x^{-b_1+b_2}\sum_{n=2}^\infty |a_n|x^{b_n-b_2}$$
Since the $b_i$ are strictly decreasing, both $-b_1+b_2$ and $b_n-b_2$ are negative (or zero, for $n=2$), thus for $x\ge A$, $x^{b_n-b_2}\le A^{b_n-b_2}$. Therefore, for $x\ge A$,
$$|a_1|\le x^{-b_1+b_2}\sum_{n=2}^\infty |a_n|A^{b_n-b_2}$$
Since $x$ can be arbitrarily large, this implies that $a_1=0$. Contradiction.
We now prove the existence of such an $A$. Assume the series converges at $x_0$. 
Fix $\epsilon>0$ and set 
$$x_n=x_0e^{\log(n^{-(1+\epsilon)})/b_n}$$
Then
$$|a_n|x_n^{b_n}=|a_n|x_0^{b_n}(x_n/x_0)^{b_n}\le (|a_n|x_0^{b_n})n^{-(1+\epsilon)}$$
Set $A=\lim\sup_n x_n$. Since the series converges at $x_0$, the terms $|a_n|x_0^{b_n}$ are bounded, hence the series will converge absolutely for $x\ge A$. A priori, $A$ could be infinite. But this won't happen as the $b_n$ are decreasing and positive. 
A: Asked and answered at MO:
https://mathoverflow.net/questions/81482/uniqueness-of-power-series
