Euler's original proof that $\sum_n 1/n^2 = \pi^2/6$ seemed to implicitly rely on an assumption that the series $1 - x/3! + x^2/5! + \ldots$ was the only series that had $\{(n \pi)^2\}_{n=1}^{\infty}$ as the roots. Is this true in general? I.e. if I have countable infinite many non-zero roots $x_n$ of $f$ such that $\sum_n 1/x_n$ converges absolutely, and $0$ is not a root of $f$, then if $f$ has a Taylor series at $0$ that converges everywhere, is $f$ (i.e. the Taylor series of $f$ at $0$) uniquely determined up to a scalar multiple, just by knowing the roots? If $f(0) = 1$, is the Taylor series of $f$ at $0$ always given by the usual formulas for polynomial coefficients in terms of reciprocals of the roots, assuming that the constant term of the polynomial is $1$? What can be said about a possible Taylor series for $f$ at $0$, if $f(0) = 1$ and $\sum_n 1/x_n$ diverges and/or does not converge absolutely (or is there no possible Taylor series with non-zero radius of convergence)?
-
Consider $e^x\sin x$. – Cameron Williams Aug 30 '13 at 15:48
-
@Cameron: Well put, and an excellent example. Make it (part of) an answer? – Cameron Buie Aug 30 '13 at 15:49
-
In fact, Euler does explicitly discuss why $e^x\sin x$ does not have the same product formula as $\sin x$, even though it has the same roots (the old masters are usually not stupid) – user8268 Aug 30 '13 at 15:52
-
@Cameron Thanks and done! – Cameron Williams Aug 30 '13 at 15:52
-
Suppose the infinite product is periodic. Is that sufficient for uniqueness? – Bill Kleinhans Aug 30 '13 at 16:51
2 Answers
Maybe Weierstraß-Factorization-Theorem is something for you. It says that for every sequence $a_n$ with $|a_n|\to \infty$ you find an entire function which do have a zero at every $a_n$ and are zero only there.
So in fact after knowing every zero (with multiplicity) two entire functions only differ by a $\exp(g(z))$ term, where $g$ is entire.
It says: (taken from Wikipedia)
Let $f$ be an entire function, and let $\{a_n\}$ be the non-zero zeros of $f$ repeated according to multiplicity; suppose also that $f$ has a zero at $z = 0$ of order $m \geq 0$ (a zero of order $m = 0$ at $z = 0$ means $f(0) \neq 0$). Then there exists an entire function $g$ and a sequence of integers $\{p_n\}$ such that $$f(z)=z^m e^{g(z)} \prod_{n=1}^\infty E_{p_n}\!\!\left(\frac{z}{a_n}\right)$$ where $$E_n(z) = \begin{cases} (1 -z) & \text{if }n=0, \\ (1-z)\exp \left( \frac{z^1}{1}+\frac{z^2}{2}+\cdots+\frac{z^n}{n} \right) & \text{otherwise}. \end{cases}$$
- 19,935
It's not true that you can uniquely extract a function from its zeroes when it is not a polynomial. E.g. $\sin x$ and $e^x \sin x$ have identical zeroes.
- 29,432
-
Thanks, I should have seen that and then rephrased the question as to why Euler's original proof could still be considered valid. Maybe I'll post another question about that, unless someone can explain here. – user2566092 Aug 30 '13 at 16:08
-
@user2566092 it was a flaw in his proof which he later improved on – Dominic Michaelis Aug 30 '13 at 19:18