Mathematical Immaturity

1.10 Exercises

Let $P_n$ denote the linear space of all real polynomials of degree $\leq n,$ where $n$ is fixed. In each of Exercises 11 through 20, let $S$ denote the set of all polynomials $f$ in $P_n$ satisfying the condition given. Determine whether or not $S$ is a subspace of $P_n.$ If $S$ is a subspace, compute $\dim S.$

• $f$ is odd.
• Recall from Section 1.6, Theorem 1.4, that a nonempty subset $S$ of a linear space $V$ is a subspace if and only if it satisfies the closure axioms:
  
  $\text{Axiom 1.}\quad$Closure under addition. $\quad$ For every pair of elements $x$ and $y$ in $V$ there corresponds a unique element in $V$ called the sum of $x$ and $y,$ denoted by $x + y.$
  $\text{Axiom 2.}\quad$Closure under multiplication by real numbers. $\quad$ For every $x$ in $V$ and every real number $a$ there corresponds an element in $V$ called the product of $a$ and $x,$ denoted by $ax.$
  
  The dimension of a basis for a linear space is the number of elements in the basis.
• Let $f$ and $g$ be odd polynomials. Then, by definition, $f(-x) = -f(x)$ and $g(-x) = -g(x).$ If we define the sum of two functions $f$ and $g$ as $(f + g)(x) = f(x) + g(x),$ we find the following: \begin{align*} (f + g)(-x) &= f(-x) + g(-x) \\ &= -f(x) - g(x) \\ &= -(f + g)(x). \end{align*} As we can see, $f + g$ is also odd, thus the set of odd polynomials in $P_n$ satisfies closure under addition. If we let $c$ be some real scalar, then $cf(-x) = -cf(x),$ satisfying closure under multiplication by real numbers. Hence, the set of odd polynomials of degree $\leq n$ is a subspace of $P_n.$
  
  To compute the dimension of $S,$ we revisit the condition of all $f$ in $S$ being odd functions. If $f$ is an odd polynomial of degree $\leq n,$ then for all real $x$ we have $f(-x) = -f(x)$ or \begin{align*} a_0 + a_1(-x) + \cdots + a_n(-x)^n &= -a_0 - a_1x - \cdots - a_nx^n \end{align*} But since $(-x)^k = x^k$ for all even $k,$ and $(-x)^k = -x^k$ for all odd $k,$ if $n$ is odd, we can simplify the above equation to \begin{align*} a_0 + a_2x^2 + \cdots + a_{n-1}x^{n-1} &= -a_0 - a_2x^2 - \cdots - a_{n-1}x^{n-1} \end{align*} Rearranging terms, we find that \begin{align*} 2a_0 + 2a_2x^2 + \cdots + 2a_{n-1}x^{n-1} &= 0 \end{align*} But if this is true for all real $x,$ it implies that the even coefficients $a_0, \cdots, a_{n-1}$ must all be zero. Thus, the odd polynomial $f,$ given by \begin{align*} f(x) &= a_1x + a_3x^3 + \cdots + a_{n}x^{n} \end{align*} is a linear combination of the $\frac{n+1}{2}$ independent elements $\{x, x^3, ..., x^{n}\},$ giving $\dim S = \frac{n+1}{2}$ for $n \geq 1.$ In the case where $n$ is even, then every $f$ in $S$ is a linear combination of the $\frac{n}{2}$ independent elements $\{x, ..., x^{n-1}\},$ making $\dim S = \frac{n}{2}$ for $n \geq 2. \quad \blacksquare$