In Exercises 1 through 28, determine whether each of the given sets is a real linear space, if addition and multiplication by real scalars are defined in the usual way. For those that are not, tell which axioms fail to hold. The functions in Exercises 1 through 17 are real-valued. In Exercises 3, 4, and 5, each function has domain containing 0 and 1. In Exercises 7 through 12, each domain contains all real numbers.
16. All Taylor polynomials of degree $\leq n$ for a fixed $n$ (including the zero polynomial).
Recall Theorem 7.1 (Volume 1, Section 7.2):
$\text{Theorem 7.1. $\quad$}$ Let $f$ be a function with derivatives of order $n$ at the point $x = 0.$ Then there exists one and only one polynomial $P$ of degree $\leq n$ which satisfies the $n + 1$ conditions
\begin{align*}
P(0) &= f(0),
\quad
P'(0) = f'(0),
\quad
...,
\quad
P^{(n)}(0) = f^{(n)}(0).
\end{align*}This polynomial is given by the formula
\begin{align*}
P(x) &= \sum_{k = 0}^n\frac{f^{(k)}(0)}{k!}x^k
\end{align*}In the same way, we may show that there is one and only one polynomial of degree $\leq n$ which agrees with $f$ and its first $n$ derivatives at a point $x = a.$ If we evaluate the derivatives at $x = a$ in place of $x = 0,$ we find
\begin{align*}
P(x) &= \sum_{k = 0}^n\frac{f^{(k)}(a)}{k!}(x - a)^k
\end{align*}This is the one and only one polynomial $P$ of degree $\leq n$ which satisfies the $n + 1$ conditions
\begin{align*}
P(a) &= f(a),
\quad
P'(a) = f'(a),
\quad
...,
\quad
P^{(n)}(a) = f^{(n)}(a).
\end{align*}and it is referred to as a Taylor polynomial of degree $n$ generated by $f$ at the point $a.$ or in short, a $\textit{Taylor polynomial.}$
Let $P$ and $Q$ be Taylor polynomials of degree $\leq n$ that agree with $f,$ $g,$ and their first $n$ derivatives, respectively, at a given point $x = a$. In other words, define $P$ and $Q$ as
\begin{align*}
P(x) &= \sum_{k = 0}^n\frac{f^{(k)}(a)}{k!}(x - a)^k
\\
\\
Q(x) &= \sum_{k = 0}^n\frac{g^{(k)}(a)}{k!}(x - a)^k
\end{align*}Then, if we define function addition in the usual way, we find that
\begin{align*}
(P + Q)(x) &= P(x) + Q(x)
\\
\\
&= \sum_{k = 0}^n\frac{f^{(k)}(a)}{k!}(x - a)^k
\\
\\
&+ \sum_{k = 0}^n\frac{g^{(k)}(a)}{k!}(x - a)^k
\\
\\
&= \sum_{k = 0}^n\frac{[f^{(k)}(a) + g^{(k)}(a)]}{k!}(x - a)^k
\end{align*}But from Theorem 4.1 (i) (Volume 1, Section 4.5) we can see that the function $f^{(k)}(a) + g^{(k)}(a)$ can be written as $[f(a) + g(a)]^{(k)} = [(f + g)(a)]^{(k)}$ $[$That is, the $k^{th}$ derivative of $(f + g)(a)].$ As such, we can see that $(P + Q)(x)$ is a Taylor polynomial of degree $\leq n$ generated by $f + g$ at $a,$ thus verifying closure under addition.
To verify closure under multiplication by real numbers, we can use a specific case of Theorem 4.1 (iii):
$$(f \cdot g)' = f'\cdot g + g'\cdot f$$where $g(x) = b$ for some real constant $b.$ Then, using the zero-derivative theorem in (Theorem 5.2, Volume 1, Section 5.2) in conjunction with Theorem 4.1 (iii), we can see that $bP(x)$ is given by
\begin{align*}
bP(x) &= b\sum_{k = 0}^n\frac{f^{(k)}(a)}{k!}(x - a)^k
\\
\\
&= \sum_{k = 0}^n\frac{bf^{(k)}(a)}{k!}(x - a)^k
\end{align*}Then, if we set $h(x) = bf(x),$ we can see that $bP(x)$ is a Taylor polynomial of degree $\leq n$ generated by $h$ at $a,$ which verifies closure under multiplication by real numbers. If $b = -1,$ it also verifies the existence of negatives. The explicit inclusion of the zero polynomial $(b = 0)$ verifies the existence of the zero element. And since every $P$ in the set is a real polynomial in $x$ for all real $x,$ the remaining axioms can be verified by treating $P(x)$ as a real number for any given $x.$
As such, we have shown that the set of all Taylor polynomials of degree $\leq n$ for a fixed $n$ (including the zero polynomial) is a real linear space. $\,\blacksquare$
