# Taylor series operator commutes with differentiation

## Statement

Suppose $f$ if a function defined on a subset of the reals that is infinitely differentiable at a point $x_0$ in its domain. Then, the derivative $f'$ is also defined and infinitely differentiable at $x_0$, and the Taylor series for $f'$ is the derivative (in the sense of derivative of power series) of the Taylor series for $f$.

## Proof

Given: $f$ is a function defined on a subset of the reals and it is infinitely differentiable at a point $x_0$ in its domain.

To prove: The derivative $f'$ is defined and infinitely differentiable at $x_0$ and the Taylor series for $f'$ is the derivative of the Taylor series for $f$.

Proof: We have the following relation between the derivatives of $f$ and $f'$: $(f')^{(k)}(x_0) = f^{(k+1)}(x_0) \ \forall \ k \in \{ 0,1,2,\dots,\} \qquad (\dagger)$

In particular, this means that, since $f$ is infinitely differentiable at $x_0$, so is $f'$, with the derivatives given by the above relationship. Thus, it makes sense to take the Taylor series of $f'$.

### Derivative of the Taylor series

We note that: $\mbox{Taylor series for } f \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!}(x - x_0)^k$

Let's try differentiating this. Differentiation of power series is term-wise, so we compute the derivative for each term, i.e., we try to calculate: $\frac{d}{dx}\left[\frac{f^{(k)}(x_0)}{k!}(x - x_0)^k\right]$

The expression $\frac{f^{(k)}(x_0)}{k!}$ is constant, so we use the differentiation rule for power functions on $(x - x_0)^k$ (with the chain rule) and obtain: $\frac{d}{dx}\left[\frac{f^{(k)}(x_0)}{k!}(x - x_0)^k\right] = \frac{kf^{(k)}(x_0)}{k!}(x - x_0)^{k-1}$

For $k = 0$, this becomes 0. For $k \ge 1$, set $l= k - 1$ (note that $l \ge 0$ now) and this becomes: $\frac{f^{(l + 1)}(x_0)}{l!}(x - x_0)^l$

Now adding together all the terms, we get: $\mbox{Derivative of Taylor series for } f \mbox{ at } x_0 = \sum_{l=0}^\infty \frac{f^{(l + 1)}(x_0)}{l!}(x - x_0)^l$

Since $l$ is a dummy variable, we can use the dummy variable $k$ instead, and get: $\mbox{Derivative of Taylor series for } f \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{f^{(k + 1)}(x_0)}{k!}(x - x_0)^k$

### Taylor series of the derivative

We note that: $\mbox{Taylor series for } f' \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{(f')^{(k)}(x_0)}{k!}(x - x_0)^k$

By $(\dagger)$, we can rewrite $(f')^{(k)}(x_0)$ on the right side as $f^{(k+1)}(x_0)$, and get: $\mbox{Taylor series for } f' \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{f^{(k+1)}(x_0)}{k!}(x - x_0)^k$

### Checking equality

It is now easy to verify that we have precisely the same expressions for the derivative of the Taylor series of $f$ and the Taylor series of $f'$.