# Taylor series operator is multiplicative

## Contents

## Statement

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

## Related facts

## Facts used

- Product rule for higher derivatives: This states that wherever the right side makes sense.

## Proof

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

**To prove**: The pointwise product is infinitely differentiable at and its Taylor series at is the product of the Taylor series of at and the Taylor series of at .

**Proof**: By Fact (1), we have:

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

We recall the expressions for the Taylor series:

We now want to multiply these Taylor series. To do this, we need to determine the coefficient of in the product.

can arise in the product by picking a factor from the Taylor series of and a factor from the Taylor series of . The coefficient arising from such a product is . The overall coefficient is thus:

Multiply and divide the right side by to get:

We have that , and we get:

Now, using , we get that:

So the product of the Taylor series for and is:

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

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