Q 75. - Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1. Given information.

It is given that function is an odd function.

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

Step 2. Prove the given statement.

It is given that the function $f\left(x\right)$ is odd then we have $f\left(x\right)=-f(-x)$.

So, evaluate $f(-x)$.

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

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

That is $a_k={(-1)}^{k+1}{a}_k$.

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