Fórum de Matemática
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Suppose that \(f(x)\) is a sum of power series for \(\mid x \mid <r\) and Let \(g(x) = f(x^2)\).

Then prove that for each \(n\;,\)

\(\hspace{-16}g^{(n)}(0)=\left\{\begin{matrix}%200&\;\;,%20\;\;if\;\; n=odd\\\\%20\displaystyle%20\frac{n!}{\left(\frac{n}{2}\right)!}f^{\left(\frac{n}{2}\right)}(0)&\;\;,%20\;\;if\; n = even\\\\%20\end{matrix}\right\)

There must be a mistake in the exercise... The proposed formula is not true. If \(f(x)\) has a power series representation for \(|x|< r\), then it necessarily coincides with its Taylor series:

\(f(x) = \sum_{n=0}^{+\infty} \frac{f^{(n)}(0)}{n!} x^n\).

Taking \(|x| < min \{r,1}, \quad g(x)=f(x^2)\) also has a power series representaion given by

\(g(x) = f(x^2) = \sum_{n=0}^{+\infty} \frac{f^{(n)}(0)}{n!} (x^2)^n = \sum_{n=0}^{+\infty} \frac{f^{(n)}(0)}{n!} x^{2n}\).

But, because g also coincides with its Taylor series, the coeficients are related to the derivatives of g. Since we only have even powers, all odd order derivatives of g will be zero (at x=0). On the other hand, the derivatives of even order 2n are given by \(f^{(n)}(0)/n!\). hence

\(g^{(n)}(0) = \left\{\begin{array}{ccc}0, & & n \textrm{ even}\\ \\ \frac{f^{(n/2)}(0)}{(n/2)!}, & &n \textrm{ Odd}\end{array}\right.\)