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 $x_{0}$.

Related facts

• Taylor series operator is linear
• Taylor series operator commutes with differentiation

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 $g$ and a factor $(x-x_{0})^{n-k}$ from the Taylor series of $f$. The coefficient arising from such a product is ${\frac {f^{(n-k)}(x_{0})g^{(k)}(x_{0})}{(n-k)!k!}}$. The overall coefficient is thus:

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

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

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

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

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

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.