Taylor series operator is multiplicative

Statement

Suppose $f$ and $g$ are functions defined on subsets of the reals such that $x_0$ is a point in the interior of the domain of both, and both $f$ and $g$ are infinitely differentiable at $x_0$. Then, the pointwise product of functions $fg$ is also infinitely differentiable at $x_0$. Further, the Taylor series of $fg$ at $x_0$ is the product of the Taylor series of $f$ at $x_0$ and the Taylor series of $g$ at $x_0$.

Facts used

1. Product rule for higher derivatives: This states that $\! (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}g^{(k)}$ wherever the right side makes sense.

Proof

Given: $f$ and $g$ are functions defined on subsets of the reals such that $x_0$ is a point in the interior of the domain of both, and both $f$ and $g$ are infinitely differentiable at $x_0$.

To prove: The pointwise product $fg$ is infinitely differentiable at $x_0$ and its Taylor series at $x_0$ is the product of the Taylor series of $f$ at $x_0$ and the Taylor series of $g$ at $x_0$.

Proof: By Fact (1), we have:

$\! (fg)^{(n)}(x_0) = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x_0)g^{(k)}(x_0) \qquad (\dagger)$

with equality holding wherever the right side makes sense. Since the right side always makes sense, so does the left side, so $fg$ is infinitely differentiable.

We recall the expressions for the Taylor series:

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

$\mbox{Taylor series for } g \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{g^{(k)}(x_0)}{k!}(x - x_0)^k$

We now want to multiply these Taylor series. To do this, we need to determine the coefficient of $(x - x_0)^n$ in the product.

$(x - x_0)^n$ can arise in the product by picking a factor $(x - x_0)^k$ from the Taylor series of $f$ and a factor $(x - x_0)^{n-k}$ from the Taylor series of $g$. The coefficient arising from such a product is $\frac{f^{(k)}(x_0)g^{(n-k)}(x_0)}{k!(n - k)!}$. The overall coefficient is thus:

$\mbox{Coefficient of } (x - x_0)^n = \sum_{k=0}^n \frac{f^{(k)}(x_0)g^{(n-k)}(x_0)}{k!(n - k)!}$

Multiply and divide the right side by $n!$ to get:

$\mbox{Coefficient of } (x - x_0)^n = \frac{1}{n!} \sum_{k=0}^n \frac{n!}{k!(n- k)!} f^{(k)}(x_0)g^{(n-k)}(x_0)$

We have that $\frac{n!}{k!(n- k)!} = \binom{n}{k}$, and we get:

$\mbox{Coefficient of } (x - x_0)^n = \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} f^{(k)}(x_0)g^{(n-k)}(x_0)$

Now, using $(\dagger)$, we get that:

$\mbox{Coefficient of } (x - x_0)^n = \frac{1}{n!} (fg)^{(n)}(x_0)$

So the product of the Taylor series for $f$ and $g$ is:

$\sum_{n=0}^\infty \frac{(fg)^{(n)}(x_0)}{n!}(x - x_0)^n$

Since $n$ is a dummy variable, it can be replaced by the dummy variable $k$, giving:

$\sum_{k=0}^\infty \frac{(fg)^{(k)}(x_0)}{k!}(x - x_0)^k$

This is precisely the Taylor series for $fg$, and we have completed the proof.