Infinitely differentiable functions and sequences

Asked Aug 22 '22 at 11:54

Active Aug 22 '22 at 11:54

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I have been completely stuck in proving the following statement:

Let $\{c_n\}_{n \in \Bbb N}$ be a complex-valued sequence, then there exists a function $f: \Bbb R \to \Bbb C$ such that $\forall n \in \Bbb N, f^{(n)}(0) = c_n n!$

As always any comment or answer is welcome and let me know if I can explain myself clearer.

asked Aug 22 '22 at 11:54

Matteo Menghini

1,159

@Lorago But what if $c_n=n^n$? Then $f$ is not defined on a interval containing 0, but just at 0. Hence it is not differentiable. – Matteo Menghini Aug 22 '22 at 13:31
@Lorago That need not converge – David C. Ullrich Aug 22 '22 at 14:23
This is an exercise somewhere in Rudin Real and Complex Analysis; I suspect there's some sort of hint or outline given there... – David C. Ullrich Aug 22 '22 at 14:24

Infinitely differentiable functions and sequences

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