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 $k$?

  • 3
    $\begingroup$ This has been asked at math.stackexchange.com/questions/83951/series-expansion too $\endgroup$ – Mariano Suárez-Álvarez Nov 21 '11 at 7:03
  • $\begingroup$ Did you really want the exponents to be strictly decreasing? It seems that you can replace that condition with convergence on $(0,1)$, with increasing exponents. $\endgroup$ – S. Carnahan Nov 21 '11 at 7:04
  • 3
    $\begingroup$ I don't think this is at the correct level for this site (without any motivation to say otherwise, it is clearly undergraduate-analysis-level material, which is considered off-topic here), but I was able to provide an answer (on MSE) assuming decreasing exponents. $\endgroup$ – B R Nov 21 '11 at 8:51

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.


I am not sure, my instinct is that the answer is no. However, I could imagine that maybe there is a solution where the $b_i$ are dense some neighborhood of $8$ (or maybe the entire real line.) However that would not be allowed by your condition that $b_k>b_{k+1}$ (I could also imagine that there is a simple reason that even that is impossible.)

You will probably want all the $b_i$ positive (or to restrict to $x \gt 0.)$

With your condition $b_k>b_{k+1}$ there are only finitely many $b_k \gt 8$ (or else only finitely many $b_k \lt 8.$ ) Resolve to avoid $x=0$ and divide through to get $1=\sum_{k=1}^{\infty}a_kx^{c_k}$ where $c_k=b_k-8$ and all but finitely many of the $c_k$ have the same sign. That seems like a problem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.