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 $k$?
Best Answer
Of course, as you all say, it is undergraduate level analysis. But, since some undergraduate analysis courses are more equal than others, I'll post the full argument nevertheless.
Let's show the uniqueness part in the monotone case first.
I'll use the equivalent version with $x\in(0,1]$ and increasing $b_k$ in the proof. WLOG, $b_1=0$, $a_1\ne 0$. We know that the series converges for $x=1$. Hence, we know that $A_k=\sum_{j\le k}a_j$ are uniformly bounded and $A_1=a_1\ne 0$. Now, we have $$ 0=\sum_k A_k(x^{b_k}-x^{b_{k+1}}) $$ by Abel-Dirichlet. From here we conclude that $$ |A_1|(1-x^{b_2})\le (\sup_k|A_k|)x^{b_2} $$ for all $x>0$. Taking the limit as $x\to 0+$, we get $A_1=0$, which is impossible.
Let's show that we can have a non-trivial series like that converging uniformly on $[0,1]$ with pairwise distinct $b_k$ if no ordering is assumed.
The key point is that every continuous function that is $0$ at $0$ can be uniformly approximated by a polynomial without a free term on $[0,1]$. Now, if you have any polynomial without a free term consisting of monomials $P_k$, you can always go over all of them with some very small common coefficient $c$ first getting $cP_1+\dots+cP_n=cP$, then repeat, then repeat again, until you exhaust the whole $P$. This will allow you to keep the partial sums bounded by the norm of the total sum. We have a lot of repeating powers here, but perturbing each of them independently a tiny bit allows us to make them distinct without making our approximation any worse. Now just telescope, as usual, approximating minus the partial sum you have at each stage and adding this approximation to get the one for the next stage.