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

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 $x_{0}$.

Related facts

Similar facts

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

Facts used

1. Differentiation is linear, repeated differentiation is linear

Proof

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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 $g$ 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.