# How to obtain the function knowing its higher derivatives at $0$

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

• Are we talking about holomorphic functions here? Because otherwise you can't guarantee uniqueness of the solution. – Keen Jul 3 at 12:53
• yes, is analytic in the in the interval – Utente Flow Jul 3 at 12:54
• Taylor series around $0$ – Yuriy S Jul 3 at 12:58

## 1 Answer

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: $$\sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!(2n-1-s)}= \frac{_1F_2(\frac{1}{2}-\frac{s}{2};\frac{3}{2}, \frac{3}{2}-\frac{s}{2}; \frac{x^2}{4})x}{s-1}$$

• 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 – Yuriy S Jul 3 at 13:08
• Where shall I see for the method ? What theory is that? – Utente Flow Jul 3 at 13:10
• In the last formula above $x$ not necessarily equal to $z$, there may be other factors – Yuriy S Jul 3 at 13:14
• Does some know what kind of method is used to move from Taylor Series to Hypergeometric series? the method used in math wolfram? – Utente Flow Jul 3 at 17:42
• @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 – Yuriy S Jul 5 at 13:11