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does some one knows how to obtain $f(x)$ knowing that in x=0 they have the following value

$f^{n}(0)= \frac{1}{n-s}$ if $ n=1,3,5,\cdots$ and $f^{n}(0)=0$ otherwise

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    $\begingroup$ Are we talking about holomorphic functions here? Because otherwise you can't guarantee uniqueness of the solution. $\endgroup$ – Keen Jul 3 at 12:53
  • $\begingroup$ yes, is analytic in the in the interval $\endgroup$ – Utente Flow Jul 3 at 12:54
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    $\begingroup$ Taylor series around $0$ $\endgroup$ – Yuriy S Jul 3 at 12:58
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Using the derivatives to fill in a Taylor series expansion... $$ f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \\ \frac{x}{1! (1-s)} + \frac{x^3}{3! (3 - s)} + \frac{x^5}{5! (5 - s)} + \cdots \\ \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!(2n-1-s)} $$

This would be notoriously difficult to manually evaluate ot recognize as the expansion of a known function. Luckily, Wolfram Alpha gives us: $$ \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!(2n-1-s)}= \frac{_1F_2(\frac{1}{2}-\frac{s}{2};\frac{3}{2}, \frac{3}{2}-\frac{s}{2}; \frac{x^2}{4})x}{s-1} $$

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  • $\begingroup$ Generalized hypergeometric functions are easy enough to recognize from the series expansion and their parameters are found by writing down the ratio of terms in the form $$\frac{c_{n+1}}{c_n} = \frac{(n+a_1) \cdots (n+a_p) }{(n+b_1) \cdots (n+b_q) } \frac{z}{n+1}$$ Which gives us $$\sum_{n=0}^\infty c_n x^n =c_0 {_p F_q} (a_1, \ldots a_p; b_1, \ldots b_q; z)$$ en.wikipedia.org/wiki/Generalized_hypergeometric_function $\endgroup$ – Yuriy S Jul 3 at 13:08
  • $\begingroup$ Where shall I see for the method ? What theory is that? $\endgroup$ – Utente Flow Jul 3 at 13:10
  • $\begingroup$ In the last formula above $x$ not necessarily equal to $z$, there may be other factors $\endgroup$ – Yuriy S Jul 3 at 13:14
  • $\begingroup$ Does some know what kind of method is used to move from Taylor Series to Hypergeometric series? the method used in math wolfram? $\endgroup$ – Utente Flow Jul 3 at 17:42
  • $\begingroup$ @UtenteFlow, hypergeometric series is a particular case of Taylor series. I have made a short instruction on how to get the hypergeometric function from the series in my comment above, for more information please see the Wikipedia link or other sources $\endgroup$ – Yuriy S Jul 5 at 13:11

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