3

We all know and love the fact that in most of cases $ f(z)=\sum_{n=0}^{\infty}\frac { f^{(n)}(0)z^n}{n!} $

In context of my question, what I need is general method of evaluation $\sum_{n=0}^{\infty}f^{(n)}(0)z^n$ as transformation of $f (z) $

Wreior
  • 396
  • 3
    Note that the first fact isn't always true, even if $f(x)$ is infinitely differentiable. – Greg Martin Jan 18 '23 at 16:49
  • To be honest I was thinking it is always true for infinitely differentiable functions (if you mean that series is divergent for me it doesn't matter becouse I'm working on divergent series). But even if not, theorem is so helpful in most of cases that it is worth to better know it than ignore even with few exceptions. So for that one reason I'm asking about this 'modified' taylor series. Becouse of utility in most of cases – Wreior Jan 22 '23 at 11:38
  • 1
    It is true for holomorphic functions (complex case), but in the real case there are infinitely differentiable which are not analytical functions. – Lelouch Jan 25 '23 at 20:39
  • 1
    @Wreior Isn't $f^{(n)}$ meant instead of $f^n$? Coud you please correct that? – IV_ Jan 28 '23 at 15:23

2 Answers2

4

First let's find transformation that $g (t^n) \rightarrow n!z^n $

$\displaystyle \int_0^{\infty} \frac {t^n e^{-\frac {t}{z}}}{z}dt =n!z^{n}$

By now, solution is trivial.

$\displaystyle \sum_{n=0}^{\infty}f^{(n )}(0)z^n= \sum_{n=0}^{\infty}\frac{f^{(n)}(0)z^n n! }{n!}=\sum_{n=0}^{\infty}\frac{f^{(n)} (0) \int_0^{\infty} \frac {t^n e^{-\frac {t}{z}}}{z}dt }{n!}=\int_0^{\infty} \frac { f (t) e^{-\frac {t}{z}}}{z} dt $

Hypothesis: If we just assume that integral at infinity is convergent to zero (Convergence of definite integral and its analytic continuation), we get

$\displaystyle \sum_{n=0}^{\infty}f^{(n)} (0)z^n=-\lim\limits_{t \rightarrow 0}\int \frac { f (t) e^{-\frac {t}{z}}}{z} dt $

Edit:I've just realise, that I have discovered Laplase transform XD

Wreior
  • 396
0

I've got $3$ examples for you in $\mathbb C$, but they only apply for $|z| < 1$:

$$\sum_{n=0}^{+\infty} z^n = \frac{1}{1-z} \tag 1$$ $$\sum_{n=0}^{+\infty} (-1)^n z^n = \frac{1}{1+z} \tag 2$$ $$\sum_{n=0}^{+\infty}(-1)^n z^{2n} = \frac{1}{1+z^2} \tag 3$$

Note that they also work in $\mathbb R$, but you didn't specify the nature of your "$x$".

  • $ \frac{1}{1-z}=\lim\limits_{t \rightarrow 0} \sum\limits_{n=0}^{\infty}\frac {\frac{d^n}{dt^n}\frac{1}{1-t} z^n}{n!}= $ $=\sum\limits_{n=0}^{\infty}\frac {n!  z^n}{n!}=  \sum\limits_{n=0}^{\infty}z^n $ , which implies that your anwser is invalid. It is still traditional Taylor series. In context of my question, what I need is $\lim\limits_{t \rightarrow 0} \sum\limits_{n=0}^{\infty}\frac{d^n}{dt^n}\frac{1}{1-t} z^n$ and what function that series would give. – Wreior Jan 22 '23 at 11:24
  • In that case indicate it explicitely in the question, it is hard to understand what you want overwise –  Jan 22 '23 at 11:28
  • Ok, so then could you edit my post to be more undstadable? I was thinking it is very clear what I mean, but if not, sorry – Wreior Jan 22 '23 at 11:33
  • Just copy the sentence "In context of my question, what I need is limt→0∑n=0∞dndtn11−tzn and what function that series would give" you gave in comment at the end of your question –  Jan 22 '23 at 11:38