1. $f(x) = a_0 + a_1(x) + a_2(x^2) + ... \;=\;$ an infinite power series with real coefficients
2. $g(x) = b_0 + b_1(x) + b_2(x^2) + ... \;=\;$ an infinite power series with real coefficients
3. There exist an infinite number of discrete real numbers $\;x_1, x_2, x_3, ...\;$ such that $\;f(x_i)=g(x_i)\;$ for each of these numbers.
4. Note: $\;0\;$ is not necessarily in $\;\{x_1, x_2, x_3, ...\}.$

To Prove: $\;\forall i\in\{0,1,2,...\},\;a_i=b_i.$

My Research:

The closest that I've come is theorem 11.10, p434, in "Calculus Volume 1, 2nd Edition", 1966, by Apostol: if two power series $\;\sum a_n(x-a)^n\;$ and $\;\sum b_n(x-a)^n\;$ have the same function $f$ in some neighborhood of the point $a,$ then the two series are equal term by term.

I don't see how I can use that theorem for my problem. I am aware that if $\;f(x)\;$ and $\;g(x)\;$ were each of finite degree, then the problem would be solved because $\;[f(x)-g(x)]\;$ has an infinite number of roots. However, I don't see how to use this insight.

Is there a relevant theorem? If not, how do I attack this problem?

If I'm understanding my own question correctly, it can be re-stated as: can a non-trivial (infinite) power series have an infinite number of roots?


It's not true. For example, let $f(x)$ be any power series with a discrete infinite set of zeros (e.g., the power series for $\sin(x)$, centered at $x=0$), and let $g(x) = 2f(x)$.

  • $\begingroup$ interesting, thanks. I was reading a paper where my (false) conjecture was apparently assumed. Based on your answer, I'll have to make a (separate) follow-up posting. $\endgroup$ – user2661923 Nov 1 '18 at 2:46

The two functions $$ f(x)=\sin x$$ and $$ g(x)=-\sin x$$ have the same values at infinitely many points and the power series are not the same at all.

  • $\begingroup$ Thanks. If I hadn't already accepted quasi's answer I would have accepted yours. $\endgroup$ – user2661923 Nov 1 '18 at 2:48
  • $\begingroup$ @user2661923 thanks for your attention $\endgroup$ – Mohammad Riazi-Kermani Nov 1 '18 at 3:09

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.