Chapter 2, Problem 4RWDT

To determine

To check : If $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a-h\right)}{2h}$ exists, will it always equal to $f\text{'}\left(a\right)$ .

Answer to Problem 4RWDT

No.

Explanation of Solution

Given information :

The numerical derivative uses the symmetric difference quotient $\frac{f\left(a+h\right)-f\left(a-h\right)}{2h}$ with a small value of $h$ , to approximate $f\text{'}\left(a\right)$ .

Also, $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a-h\right)}{2h}$ exists.

Formula used :

A function $y=f\left(x\right)$ is differentiable on a closed interval $\left[a,b\right]$ if it has a derivative at every interior point of the interval, and if the limits

  $\begin{array}{l}\underset{h\to 0^{+}}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ \underset{h\to 0^{-}}{\lim }\frac{f\left(a+h\right)-f\left(b\right)}{h}\\ \end{array}$

Exist at the end point.

Proof:

No, by taking the example of $y=f(x)=\left|x\right|$ .

Check the numerical derivative of $\left|x\right|$ at $x=0$ .

  

Here, $y=\left|x\right|$ is not differentiable at $x=0$ , since R.H.L and L.H.L derivatives at $x=0$ are not same.

  $\begin{array}{c}NDER\left(\left|x\right|,0\right)=\frac{\left|0+0.001\right|-\left|0-0.001\right|}{2\left(0.001\right)}\\ =\frac{0.001-0.001}{2\left(0.001\right)}\\ =0\end{array}$

For the limit,

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{\left|0+h\right|-\left|0-h\right|}{2h}=\underset{h\to 0}{\lim }\frac{0}{2h}\\ =0\end{array}$

By this example, the derivative cannot be defined as the limit of the symmetric difference quotient.