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 $f^{\prime }$ is also defined and infinitely differentiable at $x_0$, and the Taylor series for $f^{\prime }$ 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 $f^{\prime }$ is defined and infinitely differentiable at $x_0$ and the Taylor series for $f^{\prime }$ is the derivative of the Taylor series for $f$.

Proof: We have the following relation between the derivatives of $f$ and $f^{\prime }$:

$(f^{\prime }{)}^{(k)}(x_0)=f^{(k+1)}(x_0)\mathrm{\forall }k\in \{0,1,2,\dots ,\}(†)$

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

Derivative of the Taylor series

We note that:

$\text{Taylor series for }f\text{ at }{x}_0=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$

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:

$\frac{d}{dx}[\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k]$

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:

$\frac{d}{dx}[\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k]=\frac{k{f}^{(k)}(x_0)}{k!}(x-x_0{)}^{k-1}$

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:

$\text{Derivative of Taylor series for }f\text{ at }{x}_0=\sum _{l=0}^{\mathrm{\infty }}\frac{f^{(l+1)}(x_0)}{l!}(x-x_0{)}^l$

Since $l$ is a dummy variable, we can use the dummy variable $k$ instead, and get:

$\text{Derivative of Taylor series for }f\text{ at }{x}_0=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k+1)}(x_0)}{k!}(x-x_0{)}^k$

Taylor series of the derivative

We note that:

$\text{Taylor series for }{f}^{\prime }\text{ at }{x}_0=\sum _{k=0}^{\mathrm{\infty }}\frac{(f^{\prime }{)}^{(k)}(x_0)}{k!}(x-x_0{)}^k$

By $(†)$, we can rewrite $(f^{\prime }{)}^{(k)}(x_0)$ on the right side as $f^{(k+1)}(x_0)$, and get:

$\text{Taylor series for }{f}^{\prime }\text{ at }{x}_0=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k+1)}(x_0)}{k!}(x-x_0{)}^k$

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 $f^{\prime }$.