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The problem is given as:

Show that there exists no power series $f(z)=\sum_{n=0}^{\infty}C_nz^n$ such that:$f(z)=1$ for $z=\frac12,\frac13,\frac14,...$ and $f'(0)>0$

My approach so far:

Let's assume that such a series exists and: $f(z)=C_0+C_1z+C_2z^2+...$

From the given conditions we have: $f(\frac12)=f(\frac14)$, thus:

$C_0+\frac{C_1}2+\frac{C_2}4+...=C_0+\frac{C_1}4+\frac{C_2}{16}+...\Rightarrow C_1(\frac12-\frac14)+C_2(\frac14-\frac1{16})+...=0$ $(*)$

Similar expressions can be formed by taking any pairs of $\frac1n,\frac1l$ simply by using the fact that all equal each other.

We also know that for the terms of the power series it holds: $C_n=\frac{f^{(n)}(0)}{n!} $ for every $n$.But since $f'(0)>0 \Rightarrow C_1=f'(0)>0$.

Furthermore,if a power series $f(z)=\sum_{n=0}^{\infty}C_nz^n$ converges for $|z|<R \Rightarrow f'(z)$ exists and is equal to $\sum_{n=0}^{\infty}nC_{n-1}z^{n-1}$. If we show that for this $f$,every derivative is positive in $z=0$,this amounts to every $C_j$ term being positive and from $(*)$ our work is done.I don't know how to go about doing this and whether it actually is a stronger condition which does not apply.(One can even speculate whether for a random $f$ it is true that: $f'(0)>0 \Rightarrow f^{(k)}(0)>0$).

The only way I can think off is using the Uniqueness Theorem for both $f^{n}$ and $f^{n+1}$ in $0$.But I am not sure on how to apply it effectively.

Is this a right path to the proof?Any ideas are welcome as always..

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The principle of isolated zeros is really the driving force to prove the non-existence of such power series. The condition of $f'(0)>0$ is there merely to insure that $f$ can't be a constant function. To see this, by means of contradiction if such power series exists, it defines a holomorphic function $f$ on some open disk $D:=D(0,\epsilon)$. Since $\{1/n\}_{n\in\mathbb{N}}$ is a sequence converging to zero, there are all but finitely many terms lying inside $D(0,\epsilon)$. $f$ being holomorphic function on $D$ is also continuous. And since $f(1/n)=1$ for all $n\geq 2$, it implies that $f(0)=1$. Define $g:=f-1$. Now note that $g$ is also a holomorphic function on $D$ and is zero on a set that has limit point. And therefore, $g\equiv 0$. Which means that $f\equiv 1$. But that's impossible because $f'(0)>0$ !

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  • $\begingroup$ Oh, I didn't realize that this series in $\mathbb{C}$, not in $\mathbb{R}$. It's very nice and clear solution. $\endgroup$ – Michael Galuza Jun 4 '15 at 5:19
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If this series is exist, it converges if $|z| < R$, and $f(z)$ is continuous; $f'(z)$ is also exist. Let's we have series $\{u_n\}$, $u_n\to 0$. In this case, $u_n = 1/n$. So, $f(u_n)= 1$ and $f$ is continuous; therefore $f(0) = 1$. But $f'(0) = \lim\limits_{x\to 0} \frac{f(x) - f(0)}{x}$. Now we use Heine's definition of limit: $f'(0) = \lim\limits_{n\to\infty} \frac{f(x_n) - f(0)}{x_n}$ for any $\{x_n\}\to\infty$, and take $u_n$ as as series; hence $f'(0) = 0$, if it exists, but $f'(0) > 0$.

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