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

Additivity and scalar multiples

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 $(x - x_{0})^{k}$ .
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 $λ$ is a real number. Then, $λ f$ is infinitely differentiable at $x_{0}$ and the Taylor series of the function $λ f$ at $x_{0}$ is $λ$ times the Taylor series of $f$ at $x_{0}$ .

Related facts

Similar facts

Facts used

Differentiation is linear, repeated differentiation is linear

Proof

Proof of additivity

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 math>g</math> at $x_{0}$ .

Proof: By Fact (1), we have:

$(f + g)^{(k)} (x_{0}) = f^{(k)} (x_{0}) + g^{(k)} (x_{0}) \forall k \in {0, 1, 2, \dots} (†)$

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:

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

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

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:

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

By $(†)$ , 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. $λ$ is a real number.

To prove: $λ f$ is infinitely differentiable at $x_{0}$ and the Taylor series of the function $λ f$ at $x_{0}$ is $λ$ times the Taylor series of $f$ at $x_{0}$ .

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

$(λ f)^{(k)} (x_{0}) = λ f^{(k)} (x_{0}) \forall k \in {0, 1, 2, \dots} († †)$

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 $λ f$ is infinitely differentiable and it makes sense to takes its Taylor series.

We note that:

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

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

The Taylor series for $λ f$ is:

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

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