# Taylor series operator is linear

This article gives a statement of the form that a certain operator from a space of functions to another space of functions is a linear operator, i.e., applying the operator to the sum of two functions gives the sum of the applications to each function, and applying it to a scalar multiple of a function gives the same scalar multiple of its application to the function.

## Statement

1. Additivity: Suppose $f$ and $g$ are functions defined on subsets of reals that are both infinitely differentiable functions at a point $x_0$ which is in the domain of both functions. Then, $f + g$ is infinitely differentiable at $x_0$ and the Taylor series of $f + g$ about $x_0$ is the sum of the Taylor series of $f$ at $x_0$ and the Taylor series of $g$ at $x_0$. Note that we add Taylor series formally by adding the coefficients for each power $(x - x_0)^k$.
2. Scalar multiples: Suppose $f$ is a function defined on a subset of the reals and it is infinitely differentiable at a point $x_0$ in its domain. Suppose $\lambda$ is a real number. Then, $\lambda f$ is infinitely differentiable at $x_0$ and the Taylor series of the function $\lambda f$ at $x_0$ is $\lambda$ times the Taylor series of $f$ at $x_0$.

## Facts used

1. Differentiation is linear, repeated differentiation is linear

## Proof

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

To prove: $f + g$ is infinitely differentiable at $x_0$ and the Taylor series of $f + g$ about $x_0$ is the sum of the Taylor series of $f$ at $x_0$ and the Taylor series of $g$ at $x_0$.

Proof: By Fact (1), we have:

$(f + g)^{(k)}(x_0) = f^{(k)}(x_0) + g^{(k)}(x_0)\ \forall \ k \in \{ 0,1,2,\dots \} \qquad (\dagger)$

with equality holding whenever the right side makes sense. Since by assumption, the right side always makes sense, the left side always makes sense, so $f + g$ is infinitely differentiable at $x_0$ and it makes sense to take its Taylor series.

We recall the expressions for the Taylor series:

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

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

Thus, the sum of these Taylor series is:

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

The Taylor series for $f + g$ is:

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

By $(\dagger)$, we obtain that these two Taylor series are equal coefficient-wise, hence equal.

### Proof of scalar multiples

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. $\lambda$ is a real number.

To prove: $\lambda f$ is infinitely differentiable at $x_0$ and the Taylor series of the function $\lambda f$ at $x_0$ is $\lambda$ times the Taylor series of $f$ at $x_0$.

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

$(\lambda f)^{(k)}(x_0) = \lambda f^{(k)}(x_0) \ \forall \ k \in \{0,1,2,\dots \} \qquad (\dagger\dagger)$

with equality holding whenever the right side makes sense. Since by assumption, the right side always makes sense, the left side always makes sense, so $\lambda f$ is infinitely differentiable and it makes sense to takes its Taylor series.

We note that:

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

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

The Taylor series for $\lambda f$ is:

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

By $(\dagger\dagger)$, these two Taylor series are equal coefficient-wise, hence equal.