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 $f$.

Related facts

• Differentiation theorem for power series
• Taylor series operator is linear
• Taylor series operator is multiplicative

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 $k\geq 1$, set $l=k-1$ (note that $l\geq 0$ 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:

Failed to parse (syntax error): {\displaystyle \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 $f'$.