Q 74. - Calculus

Question

Let $f$ be an even function with Maclaurin series representation $\sum _{k=0}^{\infty } a_k{x}^k$. Prove that  $a_{2k+1}=0$ for every nonnegative integer $k$.

Step-by-Step Solution

Answer

The solution is $a_{2k+1}=0$ for every nonnegative integer $k$.

Step 1. Given information.

The given even function is:

$f\left(x\right)=\sum _{k=0}^{\infty } a_k{x}^k$

Step 2. Prove the given statement.

It is given that $f\left(x\right)$ is an even function then $f\left(x\right)=f(-x)$.

Now evaluate $f(-x)$.

$\begin{array}{rcl}f(-x)& =& \sum _{k=0}^{\infty } a_k{\left(-x\right)}^k \\ & =& \sum _{k=0}^{\infty } (-1{)}^k{a}_k{x}^k\end{array}$

Since $f\left(x\right)=f(-x)$, implies that $\sum _{k=0}^{\infty } a_k{x}^k\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \sum _{k=0}^{\infty }\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & (\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 1\end{array}\begin{array}{rcl}& & {)}^k\end{array}\begin{array}{rcl}& & a_k\end{array}\begin{array}{rcl}& & x^k\end{array}$.

That is, $a_k={\left(-1\right)}^k{a}_k$

We know that it is possible only when $a_{2k+1}=0$, for every value of $k$.