Taking the Third Order Taylor Polynomial is a Linear Transformation

Linear Transformation problems and solutions

Problem 675

The space $C^{\infty} (\mathbb{R})$ is the vector space of real functions which are infinitely differentiable. Let $T : C^{\infty} (\mathbb{R}) \rightarrow \mathrm{P}_3$ be the map which takes $f \in C^{\infty}(\mathbb{R})$ to its third order Taylor polynomial, specifically defined by
\[ T(f)(x) = f(0) + f'(0) x + \frac{f^{\prime\prime}(0)}{2} x^2 + \frac{f^{\prime \prime \prime}(0)}{6} x^3.\] Here, $f’, f^{\prime\prime}$ and $f^{\prime \prime \prime}$ denote the first, second, and third derivatives of $f$, respectively.

Prove that $T$ is a linear transformation.

 
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Proof.

To prove that $T$ is a linear transformation, we must check that it satisfies the two axioms for linear transformations. First, suppose $f, g \in C^{\infty} ( \mathbb{R} )$. Then basic properties of the derivative tell us that
\begin{align*}
(f + g)'(x) &= f'(x) + g'(x) \\
(f+g)^{\prime\prime}(x) &= f^{\prime\prime}(x) + g^{\prime\prime}(x)\\
(f+g)^{\prime \prime \prime}(x) &= f^{\prime \prime \prime}(x) + g^{\prime \prime \prime}(x).
\end{align*}

And so we can see
\begin{align*}
&T(f+g)(x)\\
&= (f+g)(0) + (f+g)'(0) x + \frac{(f+g)^{\prime\prime}(0)}{2} x^2 + \frac{(f+g)^{\prime \prime \prime}(0)}{6} x^3 \\[6pt] &= f(0) + g(0) + f'(0) x + g'(0) x + \frac{ f^{\prime\prime}(0) }{2} x^2 + \frac{g^{\prime\prime}(0)}{2} x^2 + \frac{f^{\prime \prime \prime}(0)}{6} x^3 + \frac{g^{\prime \prime \prime}(0)}{6} x^3 \\[6pt] &= f(0) + f'(0) x + \frac{ f^{\prime\prime}(0) }{2} x^2 + \frac{f^{\prime \prime \prime}(0)}{6} x^3 + g(0) + g'(0) x + \frac{g^{\prime\prime}(0)}{2} x^2 + \frac{g^{\prime \prime \prime}(0)}{6} x^3 \\[6pt] &= T(f)(x) + T(g)(x) .
\end{align*}


Now for a scalar $c \in \mathbb{R}$, the equality $(cf)'(x) = c f'(x)$ is a basic property of the derivative. Then,
\begin{align*}
T( cf )(x) &= c f(0) + (cf)'(0) x + \frac{(cf)^{\prime\prime}(0)}{2} x^2 + \frac{(cf)^{\prime \prime \prime}(0)}{6} x^3 \\
&= c f(0) + c f'(0) x + c \frac{ f^{\prime\prime}(0) }{2} x^2 + c \frac{ f^{\prime \prime \prime}(0) }{6} x^3 \\
&= c \left( f(0) + f'(0) x + \frac{ f^{\prime\prime}(0) }{2} x^2 + \frac{ f^{\prime \prime \prime}(0) }{6} x^3 \right) \\
&= c T(f)(x)
\end{align*}

Thus, $T$ is a linear transformation.


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