Taylor series operator commutes with differentiation

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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.

Related facts

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'.