Make a copy of the original code and put it in another cell. Modify the code to find and graph backward, forward, and centered difference absolute true

errors for the derivative of $f(x)=si{n}^2(x)$ for $x=\frac{}{4}.$ Use the following values of $x$:$3E-1,3E-2,3E-3,$dots$3E-9.$ The $x$ values on the $x-$axis should

use reverse log scale $($as with the original code$).$ The expression E$-$n means ${10}^{-n}.$ You may enter it exaclty as it appears here, and Python will be able

to interpret it correctly. Notice that the relation between the accuracy for the three methods is different than in the original code. What is the difference?

Compute $($exactly$)$ the second derivatives $f^{\text{'}\text{'}}(x)$ for the original code and your code for part $1.$ How do you suppose the values of the second derivative

are related to the relative accuracy of the three methods? Put your comments in the code.

We have remarked that the centered difference approximation is the equal average of the forward $($FW$)$ and the backward $($BW$)$ difference

approximations: $(0\mathrm{.}5FW+0\mathrm{.}5BW).$ It is possible to take other $($unequal$)$ averages, such as $(0\mathrm{.}4FW+0\mathrm{.}6BW,$ and so on$.$ A general

expression for such an average is: $(FW+BW),$ where $+=1.$ Make a copy of the original code and put it in a new cell. In the new cell, add three

new variables and put the results for the following three combinations: $($i$)=0\mathrm{.}4=0\mathrm{.}6$; $($ii$)=0\mathrm{.}49=0\mathrm{.}51$; $($iii$)=0\mathrm{.}499,=0\mathrm{.}501.$ Add the

errors for these three cases to the existing graph. Describe and interpret the trends that you see in the graph. Pay particular attention to the slopes. If the

graphs for two different approximation methods have nearly the same slope, then what does that tell you about the accuracy of the corresponding

methods? Put your comments after the code.

In Chapter $2\mathrm{.}02$ page $10,$ we find the forward difference approximation for the second derivative of $f(x)$ at the point $x_{z}i$ :

$f^{\text{'}\text{'}}(x_i)$~~$\frac{f(x_{i+2})-2f(x_{i+1})+f(x_i)}{(x{)}^2}+O(x)$

On page $12,$ we find the centered difference approximation:

$f^{\text{'}\text{'}}(x_i)$~~$\frac{f(x_{i+1})-2f(x_i)+f(x_{i-1})}{(x{)}^2}+O(x{)}^2$

From these two formulas, you may infer the backward difference approximation formula.

Make a copy of the original code and put it in a new cell. Modify the code to compute the forward, backward, and centered difference approximations for

the second derivative of $f(x)=x^3$ at the point $x=1.$ Use the same values of $x$ as in the original code. Plot the errors for the different approximations.

What value of $x$ is required to obtain a true accuracy of $3$ decimal digits for the three methods? Put your answer as a comment below the code.

In a new cell, modify your code so that you can plot the errors for the forward, backwards, and centered difference approximation of the first derivative

obtained in the original alongside the error for forward, backward, and centered difference found in part $3.$ Compare the slopes of these lines. What does

this tell you about the accuracy of the second derivative approximations as compared to the accuracy of the first derivative approximations? Put your

comments after the code.

# The following code finds the forward, backward, and centered difference

# derivative approximations for a given function for a range of delta

# values, then plots the absolute errors on a log$-$log scale plot.

# Necessary imports

import numpy as np

import matplotlib.pyplot as plt

# "Magic" statement for jupyter notebook to display plots

$\%$matplotlib inline

## Initialization

# Point where derivative is evaluated

x$=0\mathrm{.}5$

#function to be differentiated

f$=$np$.$exp$($x$)$

#Derivative of function $($theo$.$ calculation$)$

fPrime$=$np$.$exp$($x$)$

#get vector of deltax values

deltax $=2.**$np$.$arange$(-1,-14,-1)$

## Calculation

#Compute forward approx.

fPlus $=$ np$.$exp$($x$+$deltax$)$

forward $=($fPlus$-$f$)/$deltax

#Compute backward approx

fMinus $=$ np$.$exp$($x$-$deltax$)$

backward $=($f$-$fMinus$)/$deltax

#use average of forward and backward for centered

centered $=($forward $+$ backward$)/2$

#compute absolute errors

forwardError $=$ abs$($fPrime $-$ forward$)$

backwardError $=$ abs$($fPrime $-$ backward$)$

centerError $=$ abs$($fPrime $-$ centered$)$

## Display section $--$ create figure

plt$.$figure$(1)$

#plot forward, backward, and cente