Are there real analytic functions whose derivative is not the sum of the derivatives of the terms in its Taylor expansion

Question

In complex analysis we know that for every complex analytic function $f(z)$, that $f(z)$ has (by definition) a Taylor expansion around every point $z_0$: $\sum_{n=0}^{\infty} a_n (z-z_0)^{n}$ with radius of convergence $R>0$. Furthermore we know that $\frac{d}{dz}f(z)=\sum_{n=1}^{\infty} a_n n (z-z_0)^{n-1}$, and this series has again radius of convergence $R$.

I couldn’t, however, find a similar theorem for analytic functions on $\mathbb{R}$. Clarification: For $a,b \in \mathbb{R}$ and $f \in C^{\infty}((a,b))$, we say that $f$ is analytic if for every point $x_0 \in (a,b) \ \exists R >0$ s.t. f has a Taylor series around $x_0$ with radius of convergence $R$ that agrees with $f$ on $(x_0-R,x_0+R) \cap (a,b)$.

So my question is, are there real analytic functions s.t. $\frac{d}{dx}f(x) \neq \sum_{n=1}^{\infty} a_n n (x-x_0)^{n-1} $?

Answer 1

You can always differentiate a power series termwise. The proof is fairly straightforward.

Recall the following theorem.

Let $f_1, f_2, \ldots$ be a sequence of differentiable functions $[a, b] \to \mathbb{R}$. Suppose that $f_1', f_2', \ldots$ converges uniformly, and suppose there is an $x_0 \in [a, b]$ such that $\lim\limits_{n \to \infty} f_n(x)$ exists. Then $f_1, f_2, \ldots$ converges uniformly to some $f$, and $f'$ is the limit of the $f_i'$s.

With this theorem in hand, consider $f_i = \sum\limits_{n = 0}^i a_n (z - z_0)^n$. Note that the radius of convergence for the differential partial sums $f_i' = \sum\limits_{n = 0}^i n a_n (z - z_0)^{n - 1}$ is also $R$ by the root test. In fact, the sequence of $f_i'$s converges uniformly for any radius $< R$. So we can apply the above theorem.