Taylor series operator is linear

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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

Facts used

  1. 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 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.