Q. 84 - Calculus

Question

Let,

$f\left(x\right)=\left\{\begin{array}{l}\frac{1-\cos x}{x} ,if x\ne 0\\ 0 ,if x=0 \end{array}\right.$

(a) Use the definition of the derivative to prove that $f$ is differentiable at $0$

(b) Use the Maclaurin series for $\cos x$ to find a Maclaurin series for $f$

Step-by-Step Solution

Answer

(a) The function is differentiable at $0$ is proved

(b) The required Maclaurin series for the function  $\frac{1-\cos x}{x}$ is $\frac{1-\cos x}{x}=\frac{1}{x}\left[1-\sum _{k=1}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}\right]$

Step 1 Given Information

Consider the function given as:

$f\left(x\right)=\left\{\begin{array}{l}\frac{1-\cos x}{x} ,if x\ne 0\\ 0 ,if x=0 \end{array}\right.$

Part (a) Step 1 Derivative test

Derivative test states that,
If $f$ is a function defined from $(a,b)\to R$ and $k\in (a,b)$, then $f$ is differentiable at $k$ if the limit $\underset{h\to \infty }{\lim }\frac{f(k+h)-f\left(h\right)}{h}$ exists and equals $f^{\text{'}}\left(k\right)$.

The objective is to use the derivative test to prove that the function $f$ is differentiable at $k=0$.

Part (a) Step 2 Proof

Prove the function $f$ is differentiable at $k=0$ as follows:

Consider the following expression and solve as,
$\begin{array}{rcl}\underset{h\to \infty }{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}& =& \underset{h\to \infty }{\lim }\left[\frac{\frac{1-\cos h}{h}-0}{h-0}\right]\\ & =& \underset{h\to \infty }{\lim }\left[\frac{{\displaystyle 1-\cos h}}{{\displaystyle h^2}}\right]\\ & =& 0\end{array}$

Hence, by the definition of derivative the function $f$ is differentiable at $k=0$.

Part(b) Step 1 Calculation

Tho objective is to use the Maclaurin series for the function $f\left(x\right)=\cos x$ to find the Maclaurin series for the function $\frac{1-\cos x}{x}$.
The Maclaurin series for the function $f\left(x\right)=\cos x$ is $\cos x=\sum _{k=1}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}$.
Therefore, the Maclaurin series for the function $\frac{1-\cos x}{x}$ is $\frac{1-\cos x}{x}=\frac{1}{x}\left[1-\sum _{k=1}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}\right]$.