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

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. $$\frac{d}{dx}f(x) \neq \sum_{n=1}^{\infty} a_n n (x-x_0)^{n-1}$$?

• the same theorem (and proof) holds. Note more generally that for local properties (such as differentiability), it usually doesn't matter whether you consider real or complex analyticity, because given a real-analytic function, you can use the same coefficients, but replace real $x$ with complex $z$, and now you have in that region a complex analytic function. Oct 30, 2022 at 22:14

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.

• Sorry, but did you prove that there are or there aren't functions such that $\frac{d}{dx}f(x) \neq \sum_{n=1}^{\infty} a_n n (x-x_0)^{n-1}$ ? Oct 30, 2022 at 22:45
• Thank you for your answer:) Could you maybe quickly explain why the sequence of the $f_i’$ would always converge uniformly, is this just a theorem that I forgot? Oct 30, 2022 at 22:48
• @HenryT. Let $g(x) = \sum\limits_{n = 0}^\infty n a_n (z - z_0)^{n - 1}$. Within a radius $R’ < R$, we have $|g(x) - f_i’(x)| = |\sum\limits_{n = i + 1}^\infty n a_n (z - z_0)^{n - 1}| \leq \sum\limits_{n = i + 1}^\infty n |a_n| R’^{n - 1}$. So it suffices to show that $\sum\limits_{i = 0}^\infty n |a_n| R’^{n - 1}$ converges, which is true by the root test since $R’ < R$. Oct 30, 2022 at 23:11

If $$f(x) = \sum_{n = 0}^{\infty}{a_n (x - x_0)^n}$$ (with domain $$(a,b)$$) where the sum has radius of convergence $$R > 0$$, then you can define

$$\begin{split} g : \{ |z - x_0| < R \} &\to \mathbb{C} \\ z &\mapsto g(z) = \sum_{n=0}^{\infty}{a_n (z - x_0)^n } \end{split}$$

Whenever $$x \in (x_0 - R, x_0 + R) \cap (a,b)$$ you have $$g(x) = f(x)$$, also $$g$$ is holomorphic so for every $$n$$ $$g^{(n)}(x_0)$$ exists and clearly

$$f^{(n)}(x_0) = g^{(n)}(x_0) = a_n$$

Therefore if the theorem holds for complex analytic functions it holds also for real analytic functions!

• So you’re saying $g(z)$ also converges automatically on $\{|z-x_0| < R \}$ if it converges on $\{|z-x_0| < R \} \cap \mathbb{R}$ and if a power series converges, it is automatically holomorphic? Thank you for your answer:) Oct 30, 2022 at 22:59
• Yes. For any power series $\sum_{n=0}^{\infty}{a_n (z - z_0)^n}$ there is a number $R \in [0,+\infty]$ ($R$ can be recovered by the formula $1/R = \limsup_{n \to \infty}{\sqrt[n]{|a_n|} }$ ) such that the series converges when $|z - z_0| < R$ and doesn't converges ($a_n (z - z_0)^n \not\to 0$) when $|z - z_0| > R$, this implies that the radius of convergence is the same over $\mathbb{R}$ and over $\mathbb{C}$. Also a power series can be differentiated (in $\mathbb{C}$) term by term, so it is differentiable in $\mathbb{C}$ so it is by definition holomorphic
– Paul
Oct 30, 2022 at 23:21

Some calculus book and most books on real analysis should have a proof of termwise differentiation of power series. As evidence, here is where I found the property proved in some calculus and analysis books taken off my shelf.

Apostol, Calculus, Volume 1, first theorem of section 9.21.

Apostol, Mathematical Analysis, Theorem 13-24.

Bers, Calculus, Lemma 6 in Section 8.5.

Lang, A First Course in Calculus, 5th edition, Theorem 7.2.

Rudin, Principles of Mathematical Analysis, 2nd edition, Theorem 8.1.

Strichartz, The Way of Analysis, Theorem 7.4.2.

Kolmogorov and Sinai, Introductory Real Analysis: no treatment of power series.

I think the best way to see at a glance all the basic properties of real-analytic functions once you are familiar with complex-analytic functions is to take advantage of your knowledge of analytic functions in complex analysis. The starting point is that for a power series $$\sum_{n \geq 0} a_nx^n$$ with real coefficients (we may center it at the origin without loss of generality), Hadamard's radius-of-convergence formula for this series centered at $$0$$ on $$\mathbf R$$ and centered at $$0$$ on $$\mathbf C$$ are the same formula: the maximal open interval of convergence of that series in $$\mathbf R$$ is $$(-R,R)$$ and the maximal open disc of convergence of that series in $$\mathbf C$$ is $$\{z : |z| < R\}$$ for the same value of $$R$$, where $$1/R = \varlimsup \sqrt[n]{|a_n|}$$. Allowing a general center now, a function expressible as a power series on an open interval $$I$$ in $$\mathbf R$$ is an analytic function on the open disc in $$\mathbf C$$ having $$I$$ as a diameter. So if you know that termwise differentiation is valid for complex power series (perhaps proved using the Cauchy integral formula or some other technique of complex analysis), then you immediately get termwise differentiation of real power series since the real derivative (using a limit as $$h \to 0$$ in $$\mathbf R$$) is a restricted version of the complex derivative (using a limit as $$h \to 0$$ in $$\mathbf C$$) if the complex derivative exists.

• Oh, I should maybe buy some good analysis books then, I looked through the script form my analysis class last year and couldn’t find the theorem, that’s why I asked. But I just downloaded Rudin as a PDF! Oct 30, 2022 at 23:36