Every power series \(P(z)=\sum_{n=0}^{\infty} a_n(z-z_0)^n\) is differentiable (or holomorphic in the complex setting) on its radius of convergence. The derivative is given by
\begin{equation*} P'(z)=\sum_{n=1}^{\infty} na_n(z-z_0)^{n-1}. \end{equation*}The same is true for multidimensional power series [1, Proposition 2.2.3].
Proof
The statement is the consequence of this theorem
and by observing that \(\sqrt[n]{na_n}\) has the same limes superior like the sequence \(\sqrt{a_n}\) since \(\sqrt[n]{n}\to 1\) (see here
).
Remarks
- Therefore, \(P\in C^{\infty}\) and each derivative is a power series.
- This formula implies a representation of the coefficients which is determined by the derivatives of \(P\).
References Link to heading
- S. Krantz and H. Parks, A Primer of Real Analytic Functions. Boston, MA: Birkhäuser Boston, 2002. doi:10.1007/978-0-8176-8134-0