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Goal: If $f(z)^8$ is analytic on some domain D and if $f(z)$ is continuous on domain $D$ with $f(0) = 0$, then the power series $f(z)^8 = \sum a_nz^n$ will begin with $n$ divisible by $8$

My attempt:

Allow $g(z) = f(z)^8$ which is analytic on some domain D. Then, we use Cauchy integral to calculate our $a_n$.

$$g^{(n)}(z) = a_n = \frac{n!}{2\pi i}\int_{D} \frac{g(w)}{(w-z)^{n+1}} dw$$

Now, I notice that we can split the integrand up, but I'm not quite sure if it's helpful

$$\frac{n!}{2\pi i} \int_D \frac{1}{(w-z)^{j}} \frac{f(w)^8}{(w-z)^{8k+1}} dw$$.

Where $j + 8k + 1 = n + 1$

Can anyone tell me if this is going in the right direction or any hints?

Good Morning Captain
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  • @BrevanEllefsen Sadly, we have not covered the argument principle. – Good Morning Captain Oct 31 '18 at 23:07
  • Take $f(z):=z+z^2$, then $f(z)^8=z^8+8z^9+\cdots$. – JRen Oct 31 '18 at 23:08
  • @JRen My bad, I believe I phrased my question incorrectly. I meant more that the first term will be some factor of $8$! Allow me to edit – Good Morning Captain Oct 31 '18 at 23:09
  • @BrevanEllefsen that's very interesting. I appreciate the response. – Good Morning Captain Nov 01 '18 at 06:19
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    Slight typo in the formula I gave earlier. I'll perhaps look it up later and fix. That being said, I'm not sure I understand what this post is saying... for example, if $f(x) = \cos(x)$ then $f^8$ is definitely analytic and $f$ is definitely continuous, but $f(x)^8 = 1-4x^2+\cdots$ – Brevan Ellefsen Nov 01 '18 at 06:40
  • @BrevanEllefsen what if we make another restriction that $f(z_0) = 0$, where $z_0$ is the point we're making our taylor series around. WLOG, we can just say it's $z_0 = 0$. For example $\sin(z)^8 = x^8 - \frac{4}{3}x^{10} + ...$ – Good Morning Captain Nov 01 '18 at 18:58
  • I think you will have an easier time discussing the Cauchy product. Start with $f(z), f^2(z)$ and by induction what must $f^8(z)$ look like. – Doug M Nov 01 '18 at 19:23
  • @OP - the condition $f(0)=0$ should definitely help... The idea being that $f$ is analytic then $f(x) = cx+\mathcal{O}(x^2)$ so that $f(x)^8 = (cx)^8 + \mathcal{O}(x^9)$ – Brevan Ellefsen Nov 01 '18 at 23:15
  • To finish the proof, we use this post, which says that if $f$ is continuous and $f^2$ is analytic then $f$ is analytic. Merely note $f$ being continuous means $f^2$ and $f^4$ are, so we apply the lemma to show $f^4$ and then $f^2$ and then $f$ are analytic. We then use my prior comment. I'll write these comments up as an answer when I get a computer later. – Brevan Ellefsen Nov 01 '18 at 23:40
  • @BrevanEllefsen awesome, awesome, awesome! – Good Morning Captain Nov 02 '18 at 07:12
  • @GoodMorningCaptain I have written up a more complete answer now. – Brevan Ellefsen Nov 02 '18 at 20:51

1 Answers1

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Your assumption of continuity of $f$ along with the analyticity of $f^8$ is enough to conclude $f$ is analytic. For example, see the following:

The former gives the result immediately but is a tad tougher to prove, while the latter gives the result after noting $f$ being continuous implies $f^2$ and $f^4$ are continuous, at which point we apply the latter result three times.

We thus know $f$ is analytic, so $$f(z) = \sum_{k=0}^\infty c_kz^k$$
Applying your edit that $f(0)=0,$ we must have $c_0 = 0.$ Let us first assume $c_1 \neq 0,$ so $$f(z) = \sum_{k=1}^\infty c_kz^k = c_1 z + \mathcal{O}(z^2)$$ Thus, by the Binomial Theorem, we have that $$f(z)^8 = (c_1 z + \mathcal{O}(z^2))^8 = (c_1 z)^8 + \mathcal{O}(z^9)$$ Which proves the claim whenever $c_1 \neq 0.$

We now use this idea to prove the general case. Suppose $c_k = 0$ for $0 \le k \le n-1,$ but $c_n \neq 0.$ Then: $$f(z)^8 = \left(\sum_{k=n}^\infty c_kz^k\right)^8 = (c_n z^n + \mathcal{O}(z^{n+1}))^8 = (c_n z^n)^8 + \mathcal{O}(z^{8(n+1)})$$ Which proves that the leading nonzero term will have a power divisible by $8.$ This of course extends readily to other powers of $f.$

Brevan Ellefsen
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