# Transformation of a Taylor series: “doubling” the derivative order

Suppose a function $f(z)$ has a convergent Taylor expansion:

$$f(z)=\sum_{n=0}^{\infty} c_n \frac{z^n}{n!}$$

Are there general tools to compute $$g(z) = \sum_{n=0}^{\infty} c_{2n} \frac{z^n}{n!}=\sum_{n=0}^{\infty} \, f^{(2n)}(0)\frac{z^n}{n!} \text{ ?}$$

I came across this general problem in a more specific context explained in this question.

One possible way is to do an integral transform $$\dfrac{1}{s} \int_0^{\infty} e^{-z/s} f(z) dz$$ to remove the factorials, then replace $s \to \sqrt{s}$ and do the inverse transform. However, this easily leads to very complicated integrals, and does not seem to work for my particular problem.

-
 Don't you mean $\dfrac{d^{2n}f(0)}{dz^{2n}}$? – Peter Tamaroff Mar 8 '12 at 3:55 Yes, thanks! It looked embarrassingly inconsistent, just corrected. – Slaviks Mar 8 '12 at 6:35 I will put some thought on this problem. Give me some time. – Peter Tamaroff Mar 8 '12 at 6:40 @Slavilks I couldn't find the answer again. Now I did, so I've thought of something. Maybe it can help. – Peter Tamaroff Mar 25 '12 at 6:08

This seems rather complicated and awkward, but, FWIW, here it is:

You have that $f$ can be written as

$$f(z) =\sum\limits_{n=0}^\infty f^{(n)}(0)\frac{z^n}{n!}$$

I assume this series converges absolutely to $f$ in some domain $D$.

You need

$$g(z) =\sum\limits_{n=0}^\infty f^{(2n)}(0)\frac{z^n}{n!}$$

Put

$$g(z^2) =\sum\limits_{n=0}^\infty f^{(2n)}(0)\frac{z^{2n}}{n!}$$

Now,

$$\int \frac{g(z^2)}{z}dz=\sum\limits_{n=0}^\infty f^{(2n)}(0)\frac{z^{2n}}{(2n)n!}$$

Then

$$\int\frac{dz}{z^2}\int \frac{g(z^2)}{z}dz=\sum\limits_{n=0}^\infty f^{(2n)}(0)\frac{z^{2n-1}}{(2n)(2n-1)n!}$$

Continuing this will give

$$\int \frac{1}{z^k}\int \frac{1}{z^{k-1}}\cdots \int\frac{1}{z^2}\int \frac{g(z^2)}{z}dz^k=\sum\limits_{n=0}^\infty f^{(2n)}(0)\frac{z^{2n-k+1}}{(2n)(2n-1)\cdots(2n-k+1)n!}$$

Now choose $k=n$ and multiply by $z^{n-1}$ to get

$$j_n(z)=z^{n-1}\int \frac{1}{z^n}\int \frac{1}{z^{n-1}}\cdots \int\frac{1}{z^2}\int \frac{g(z^2)}{z}dz^n=\sum\limits_{n=0}^\infty f^{(2n)}(0)\frac{z^{2n}}{(2n)!}$$

I'm guessing the new $n$ will need to tend to $\infty$ for the result to hold. I'm still not confident about the above.

I'm thinking that it might be necessary to introduce the "odd" transformation

$$h(z) =\sum\limits_{n=0}^\infty f^{(2n+1)}(0)\frac{z^n}{n!}$$

In a similar manner you get

$$\int \frac{1}{z^k}\int \frac{1}{z^{k-1}}\cdots \int\frac{1}{z^2}\int {h(z^2)}dz^k=\sum\limits_{n=0}^\infty f^{(2n+1)}(0)\frac{z^{2n-k+2}}{(2n)(2n-1)\cdots(2n-k+2)n!}$$

Then $k=n+1$ and $z^n$

$$i_n(z) = z^n \int \frac{1}{z^{n+1}}\int \frac{1}{z^{n}}\cdots \int\frac{1}{z^2}\int {h(z^2)}dz^{n+1}=\sum\limits_{n=0}^\infty f^{(2n+1)}(0)\frac{z^{2n+1}}{(2n+1)!}$$

Then $i_n(z)+j_n(z)=f(z)$

So you are looking at some combination of the operations of multiplying by $z^n$ and $\frac{d^n}{dz^n}$ alternately. And clearly then putting $\sqrt{z}$ instead of $z$ as you suggest.

That's all I got for now.

-
 One cannot write things like $j_n(z)=\sum\limits_{n=0}^{\infty}A_n$ for some expressions $A_n$ depending on $n$. Remember that $\sum\limits_{n=0}^{\infty}A_n=\sum\limits_{s=0}^{\infty}A_s=$ $\sum\limits_{u=0}^{\infty}A_u=\ldots$ is a tautology. Until that point your solution is OK but from that point on, it becomes nonsense. Sorry. – Did May 4 '12 at 6:43