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I am working through some practice problems and I have this one which is stumping me:

Take $f$ and entire function such that $f(z) = f(z+1)=f(z+\sqrt{2})$, then it is constant.

I was thinking, given the function is entire, then it has a Taylor series expansion about the origin, with an infinite radius of convergence:

$$f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}z^n=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(z+1)^n=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(z+\sqrt{2})^n$$

Which we can also write:

$$\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}z^n = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\sum_{k=0}^{n}{n\choose k}z^k = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\sum_{k=0}^{n}{n\choose k}z^k\sqrt{2}^{n-k}$$

But I am not quite sure this is the correct place to go. Any ideas as to how I can move forward on this problem?

Thank you!

José Carlos Santos
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rannoudanames
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1 Answers1

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Note that the set $D$ of the numbers of the form $a+b\sqrt2$, with $a,b\in\mathbb Z$ form a dense subset of $\mathbb R$. Since $f$ is continuous and constant on $D$, $f|_{\mathbb R}$ is constant. It follows from the identity theorem that $f$ is constant.

José Carlos Santos
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  • I understand the use of the identity theorem assuming D is dense, However, do you mind further elaborating why D is dense? I am not quite familiar with the concept. In fact, when I learned the Identity theorem I learned it in the context of an accumulation point – rannoudanames Nov 04 '18 at 17:35
  • For every real numbers $x$ and $y$, with $x<y$, there is a number of the frm $a+b\sqrt2$ (with $a,b\in\mathbb Z$) such that $x<a+b\sqrt2<y$. In other words, the numbers of that form are spread everywhere (in $\mathbb R$). It follows from this that every continuous function from $\mathbb R$ into $\mathbb R$ whch is constant on $D$ is actually constant everywhere. – José Carlos Santos Nov 04 '18 at 17:41
  • That is interesting, is there a theorem or a property that I could look into in order to prove the claim that $a+b\sqrt{2}$ can be made to be between any two real numbers, taking $a$ and $b$ in $\mathbb{Z}$? – rannoudanames Nov 04 '18 at 17:44
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    You will find a proof here. – José Carlos Santos Nov 04 '18 at 17:47