Chapter 23, Problem 8P

Compute the first-order central difference approximations of $O\left(h^4\right)$ for each of the following functions at the specified location and for the specified step size:

(a) $y=x^3+4x-15$ at $x=0,h=0.25$

(b) $y=x^2\cos x$ at $x=0.4,h=0.1$

(c) $y=\tan \left(x/3\right)$ at $x=3,h=0.5$

(d) $y=\sin \left(0.5\sqrt{x}\right)/x$ at $x=1,h=0.2$

(e) $y=e^x+x$ at $x=2,h=0.2$

Compare your results with the analytical solutions.

To determine

To calculate: The first order central difference approximations of $O\left(h^4\right)$ for $y=x^3+4x-15$ at $x=0$, where step size is $h=0.25$ also compare the result with the analytic solution.

Answer to Problem 8P

Solution:

Analytic value of the first derivative is $4$ and numeric value is $4$

Explanation of Solution

Given information:

The function, $y=x^3+4x-15$

The value of x, $x=0$

Step size $h=0.25$

Formula used:

Central difference approximation of $O\left(h^4\right)$ for first derivative is given by,

$f^{\text{'}}\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+8f\left(x_{i+1}\right)-8f\left(x_{i-1}\right)+f\left(x_{i-2}\right)}{12h}$

Here, h is step size and $x_{i+1}=x_i+h$

Calculation:

Consider the function,

$f\left(x\right)=x^3+4x-15$

Differentiate the function with respect to x,

$f^{\prime }\left(x\right)=3x^2+4$

Now substitute $x=0$ in the above equation,

$\begin{array}{c}{f}^{\prime }\left(0\right)=3{\left(0\right)}^2+4\\ =4\end{array}$

Thus, the analytic value of the first derivative of the function is $4$.

Again, consider the function,

$f\left(x\right)=x^3+4x-15$

Here, $x_i=0$ and $h=0.25$

The value of $x_{i+1}$ is,

$\begin{array}{c}{x}_{i+1}=x_i+h\\ =0+0.25\\ =0.25\end{array}$

The value of the function at $x_{i+1}=0.25$ is,

$\begin{array}{c}f\left(0.25\right)={\left(0.25\right)}^3+4\left(0.25\right)-15\\ =-13.984375\end{array}$

The value of $x_{i+2}$ is,

$\begin{array}{c}{x}_{i+2}=x_{i+1}+h\\ =0.25+0.25\\ =0.5\end{array}$

The value of the function at $x_{i+2}=0.5$ is,

$\begin{array}{c}f\left(0.5\right)={\left(0.5\right)}^3+4\left(0.5\right)-15\\ =-12.875\end{array}$

The value of $x_{i-1}$ is,

$\begin{array}{c}{x}_{i-1}=x_i-h\\ =0-0.25\\ =-0.25\end{array}$

The value of the function at $x_{i-1}=-0.25$ is,

$\begin{array}{c}f\left(-0.25\right)={\left(-0.25\right)}^3+4\left(-0.25\right)-15\\ =-16.015625\end{array}$

The value of $x_{i-2}$ is,

$\begin{array}{c}{x}_{i-2}=x_{i-1}-h\\ =-0.25-0.25\\ =-0.5\end{array}$

The value of the function at $x_{i-2}=-0.5$ is,

$\begin{array}{c}f\left(-0.5\right)={\left(-0.5\right)}^3+4\left(-0.5\right)-15\\ =-17.125\end{array}$

Now, central difference approximation of $O\left(h^4\right)$ for first derivative is given by,

$f^{\text{'}}\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+8f\left(x_{i+1}\right)-8f\left(x_{i-1}\right)+f\left(x_{i-2}\right)}{12h}$

Substitute the values of $f\left(x_{i+2}\right),f\left(x_{i+1}\right),f\left(x_{i-1}\right),f\left(x_{i-2}\right),x_i=0$ and $h=0.25$ in the above formula,

$\begin{array}{c}{f}^{\text{'}}\left(0\right)=\frac{-\left(-12.875\right)+8\left(-13.984375\right)-8\left(-16.015625\right)+\left(-17.125\right)}{12\left(0.25\right)}\\ =\frac{12}{3}\\ =4\end{array}$

Therefore, the analytic value of the first derivative at $x=0$ is $4$ and numerical value is also $4$.

To determine

To calculate: The first order central difference approximations of $O\left(h^4\right)$ for $y=x^2\cos x$ at $x=0.4$, where step size is $h=0.1$ also compare the result with the analytic solution.

Answer to Problem 8P

Solution:

Analytic value of the first derivative is $0.674542$ and numeric value is $0.674504$

Explanation of Solution

Given information:

The function, $y=x^2\cos x$

The value of x, $x=0.4$

Step size $h=0.1$

Formula used:

Central difference approximation of $O\left(h^4\right)$ for first derivative is given by,

$f^{\text{'}}\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+8f\left(x_{i+1}\right)-8f\left(x_{i-1}\right)+f\left(x_{i-2}\right)}{12h}$

Here, h is step size and $x_{i+1}=x_i+h$

Calculation:

Consider the function,

$f\left(x\right)=x^2\cos x$

Differentiate the function with respect to x,

$f^{\prime }\left(x\right)=2x\cos x-x^2\sin x$

Now substitute $x=0.4$ in the above equation,

$\begin{array}{c}{f}^{\prime }\left(0.4\right)=2\left(0.4\right)\cos \left(0.4\right)-{\left(0.4\right)}^2\sin \left(0.4\right)\\ =0.674542\end{array}$

Thus, the analytic value of the first derivative of the function is $4$.

Again, consider the function,

$f\left(x\right)=x^2\cos x$

Here, $x_i=0.4$ and $h=0.1$

The value of $x_{i+1}$ is,

$\begin{array}{c}{x}_{i+1}=x_i+h\\ =0.4+0.1\\ =0.5\end{array}$

The value of the function at $x_{i+1}=0.5$ is,

$\begin{array}{c}f\left(0.5\right)={\left(0.5\right)}^2\cos \left(0.5\right)\\ =0.219396\end{array}$

The value of $x_{i+2}$ is,

$\begin{array}{c}{x}_{i+2}=x_{i+1}+h\\ =0.5+0.1\\ =0.6\end{array}$

The value of the function at $x_{i+2}=0.6$ is,

$\begin{array}{c}f\left(0.6\right)={\left(0.6\right)}^2\cos \left(0.6\right)\\ =0.297121\end{array}$

The value of $x_{i-1}$ is,

$\begin{array}{c}{x}_{i-1}=x_i-h\\ =0.4-0.1\\ =0.3\end{array}$

The value of the function at $x_{i-1}=0.3$ is,

$\begin{array}{c}f\left(0.3\right)={\left(0.3\right)}^2\cos \left(0.3\right)\\ =0.0859803\end{array}$

The value of $x_{i-2}$ is,

$\begin{array}{c}{x}_{i-2}=x_{i-1}-h\\ =0.3-0.1\\ =0.2\end{array}$

The value of the function at $x_{i-2}=0.2$ is,

$\begin{array}{c}f\left(0.2\right)={\left(0.2\right)}^2\cos \left(0.2\right)\\ =0.0392027\end{array}$

Now, central difference approximation of $O\left(h^4\right)$ for first derivative is given by,

$f^{\text{'}}\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+8f\left(x_{i+1}\right)-8f\left(x_{i-1}\right)+f\left(x_{i-2}\right)}{12h}$

Substitute the values of $f\left(x_{i+2}\right),f\left(x_{i+1}\right),f\left(x_{i-1}\right),f\left(x_{i-2}\right),x_i=0.4$ and $h=0.1$ in the above formula,

$\begin{array}{c}{f}^{\text{'}}\left(0.4\right)=\frac{-\left(0.297121\right)+8\left(0.219396\right)-8\left(0.0859803\right)+\left(0.0392027\right)}{12\left(0.1\right)}\\ =\frac{0.80940730}{1.2}\\ =0.674504\end{array}$

Therefore, the analytic value of the first derivative at $x=0.4$ is $0.674542$ and numerical value is $0.674504$.

To determine

To calculate: The first order central difference approximations of $O\left(h^4\right)$ for $y=\tan \left(\frac{x}{3}\right)$ at $x=3$, where step size is $h=0.5$ also compare the result with the analytic solution.

Answer to Problem 8P

Solution:

Analytic value of the first derivative is $1.14184$ and numeric value is $1.092486$

Explanation of Solution

Given information:

The function, $y=\tan \left(\frac{x}{3}\right)$

The value of x, $x=3$

Step size $h=0.5$

Formula used:

Central difference approximation of $O\left(h^4\right)$ for first derivative is given by,

$f^{\text{'}}\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+8f\left(x_{i+1}\right)-8f\left(x_{i-1}\right)+f\left(x_{i-2}\right)}{12h}$

Here, h is step size and $x_{i+1}=x_i+h$

Calculation:

Consider the function,

$f\left(x\right)=\tan \left(\frac{x}{3}\right)$

Differentiate the function with respect to x,

$f^{\prime }\left(x\right)=\left(\frac{1}{3}\right){\sec }^2\left(\frac{x}{3}\right)$

Now substitute $x=3$ in the above equation,

$\begin{array}{c}{f}^{\prime }\left(3\right)=\left(\frac{1}{3}\right){\sec }^2\left(\frac{3}{3}\right)\\ =1.14184\end{array}$

Thus, the analytic value of the first derivative of the function is $1.14184$.

Again, consider the function,

$f\left(x\right)=\tan \left(\frac{x}{3}\right)$

Here, $x_i=3$ and $h=0.5$

The value of $x_{i+1}$ is,

$\begin{array}{c}{x}_{i+1}=x_i+h\\ =3+0.5\\ =3.5\end{array}$

The value of the function at $x_{i+1}=3.5$ is,

$\begin{array}{c}f\left(3.5\right)=\tan \left(\frac{3.5}{3}\right)\\ =2.33825\end{array}$

The value of $x_{i+2}$ is,

$\begin{array}{c}{x}_{i+2}=x_{i+1}+h\\ =3.5+0.5\\ =4\end{array}$

The value of the function at $x_{i+2}=4$ is,

$\begin{array}{c}f\left(4\right)=\tan \left(\frac{4}{3}\right)\\ =4.131729\end{array}$

The value of $x_{i-1}$ is,

$\begin{array}{c}{x}_{i-1}=x_i-h\\ =3-0.5\\ =2.5\end{array}$

The value of the function at $x_{i-1}=2.5$ is,

$\begin{array}{c}f\left(2.5\right)=\tan \left(\frac{2.5}{3}\right)\\ =1.100778\end{array}$

The value of $x_{i-2}$ is,

$\begin{array}{c}{x}_{i-2}=x_{i-1}-h\\ =2.5-0.5\\ =2\end{array}$

The value of the function at $x_{i-2}=2$ is,

$\begin{array}{c}f\left(2\right)=\tan \left(\frac{2}{3}\right)\\ =0.786843\end{array}$

Now, central difference approximation of $O\left(h^4\right)$ for first derivative is given by,

$f^{\text{'}}\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+8f\left(x_{i+1}\right)-8f\left(x_{i-1}\right)+f\left(x_{i-2}\right)}{12h}$

Substitute the values of $f\left(x_{i+2}\right),f\left(x_{i+1}\right),f\left(x_{i-1}\right),f\left(x_{i-2}\right),x_i=3$ and $h=0.5$ in the above formula,

$\begin{array}{c}{f}^{\text{'}}\left(3\right)=\frac{-\left(4.131729\right)+8\left(2.338254\right)-8\left(1.100778\right)+0.786843}{12\left(0.5\right)}\\ =\frac{6.554922}{6}\\ =1.092486\end{array}$

Therefore, the analytic value of the first derivative at $x=3$ is $1.14184$ and numerical value is $1.092486$.

To determine

To calculate: The first order central difference approximations of $O\left(h^4\right)$ for $y=\frac{\sin \left(0.5\sqrt{x}\right)}{x}$ at $x=1$, where step size is $h=0.2$ also compare the result with the analytic solution.

Answer to Problem 8P

Solution:

Analytic value of the first derivative is $-0.26003$ and numeric value is $-0.25908$

Explanation of Solution

Given information:

The function, $y=\frac{\sin \left(0.5\sqrt{x}\right)}{x}$

The value of x, $x=1$

Step size $h=0.2$

Formula used:

Central difference approximation of $O\left(h^4\right)$ for first derivative is given by,

$f^{\text{'}}\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+8f\left(x_{i+1}\right)-8f\left(x_{i-1}\right)+f\left(x_{i-2}\right)}{12h}$

Here, h is step size and $x_{i+1}=x_i+h$

Calculation:

Consider the function,

$f\left(x\right)=\frac{\sin \left(0.5\sqrt{x}\right)}{x}$

Differentiate the function with respect to x,

$f^{\prime }\left(x\right)=\frac{0.25\cos \left(0.5\sqrt{x}\right)}{x^{\frac{3}{2}}}-\frac{\sin \left(0.5\sqrt{x}\right)}{x^2}$

Now substitute $x=1$ in the above equation,

$\begin{array}{c}{f}^{\prime }\left(1\right)=\frac{0.25\cos \left(0.5\sqrt{1}\right)}{{\left(1\right)}^{\frac{3}{2}}}-\frac{\sin \left(0.5\sqrt{1}\right)}{{\left(1\right)}^2}\\ =-0.26003\end{array}$

Thus, the analytic value of the first derivative of the function is $-0.26003$.

Again, consider the function,

$f\left(x\right)=\frac{\sin \left(0.5\sqrt{x}\right)}{x}$

Here, $x_i=1$ and $h=0.2$

The value of $x_{i+1}$ is,

$\begin{array}{c}{x}_{i+1}=x_i+h\\ =1+0.2\\ =1.2\end{array}$

The value of the function at $x_{i+1}=1.2$ is,

$\begin{array}{c}f\left(1.2\right)=\frac{\sin \left(0.5\sqrt{1.2}\right)}{1.2}\\ =0.433954\end{array}$

The value of $x_{i+2}$ is,

$\begin{array}{c}{x}_{i+2}=x_{i+1}+h\\ =1.2+0.2\\ =1.4\end{array}$

The value of the function at $x_{i+2}=1.4$ is,

$\begin{array}{c}f\left(1.4\right)=\frac{\sin \left(0.5\sqrt{1.4}\right)}{1.4}\\ =0.398355\end{array}$

The value of $x_{i-1}$ is,

$\begin{array}{c}{x}_{i-1}=x_i-h\\ =1-0.2\\ =0.8\end{array}$

The value of the function at $x_{i-1}=0.8$ is,

$\begin{array}{c}f\left(0.8\right)=\frac{\sin \left(0.5\sqrt{0.8}\right)}{0.8}\\ =0.540569\end{array}$

The value of $x_{i-2}$ is,

$\begin{array}{c}{x}_{i-2}=x_{i-1}-h\\ =0.8-0.2\\ =0.6\end{array}$

The value of the function at $x_{i-2}=0.6$ is,

$\begin{array}{c}f\left(0.6\right)=\frac{\sin \left(0.5\sqrt{0.6}\right)}{0.6}\\ =0.629480\end{array}$

Now, central difference approximation of $O\left(h^4\right)$ for first derivative is given by,

$f^{\text{'}}\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+8f\left(x_{i+1}\right)-8f\left(x_{i-1}\right)+f\left(x_{i-2}\right)}{12h}$

Substitute the values of $f\left(x_{i+2}\right),f\left(x_{i+1}\right),f\left(x_{i-1}\right),f\left(x_{i-2}\right),x_i=3$ and $h=0.5$ in the above formula,

$\begin{array}{c}{f}^{\text{'}}\left(x_i\right)=\frac{-\left(0.398355\right)+8\left(0.433954\right)-8\left(0.540569\right)+0.629480}{12\left(0.2\right)}\\ =-\frac{0.621795}{2.4}\\ =-0.25908\end{array}$

Therefore, the analytic value of the first derivative at $x=1$ is $-0.26003$ and numerical value is $-0.25908$.

To determine

To calculate: The first order central difference approximations of $O\left(h^4\right)$ for $y=e^x+x$ at $x=2$, where step size is $h=0.2$ also compare the result with the analytic solution.

Answer to Problem 8P

Solution:

Analytic value of the first derivative is $8.389056$ and numeric value is $8.38866$

Explanation of Solution

Given information:

The function, $y=e^x+x$

The value of x, $x=2$

Step size $h=0.2$

Formula used:

Central difference approximation of $O\left(h^4\right)$ for first derivative is given by,

$f^{\text{'}}\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+8f\left(x_{i+1}\right)-8f\left(x_{i-1}\right)+f\left(x_{i-2}\right)}{12h}$

Here, h is step size and $x_{i+1}=x_i+h$

Calculation:

Consider the function,

$f\left(x\right)=e^x+x$

Differentiate the function with respect to x,

$f^{\prime }\left(x\right)=e^x+1$

Now substitute $x=2$ in the above equation,

$\begin{array}{c}{f}^{\prime }\left(2\right)=e^{\left(2\right)}+1\\ =8.389056\end{array}$

Thus, the analytic value of the first derivative of the function is $8.389056$.

Again, consider the function,

$f\left(x\right)=e^x+x$

Here, $x_i=2$ and $h=0.2$

The value of $x_{i+1}$ is,

$\begin{array}{c}{x}_{i+1}=x_i+h\\ =2+0.2\\ =2.2\end{array}$

The value of the function at $x_{i+1}=2.2$ is,

$\begin{array}{c}f\left(2.2\right)=e^{\left(2.2\right)}+\left(2.2\right)\\ =11.2250\end{array}$

The value of $x_{i+2}$ is,

$\begin{array}{c}{x}_{i+2}=x_{i+1}+h\\ =2.2+0.2\\ =2.4\end{array}$

The value of the function at $x_{i+2}=2.4$ is,

$\begin{array}{c}f\left(2.4\right)=e^{\left(2.4\right)}+\left(2.4\right)\\ =13.4232\end{array}$

The value of $x_{i-1}$ is,

$\begin{array}{c}{x}_{i-1}=x_i-h\\ =2-0.2\\ =1.8\end{array}$

The value of the function at $x_{i-1}=1.8$ is,

$\begin{array}{c}f\left(1.8\right)=e^{\left(1.8\right)}+\left(1.8\right)\\ =7.84965\end{array}$

The value of $x_{i-2}$ is,

$\begin{array}{c}{x}_{i-2}=x_{i-1}-h\\ =1.8-0.2\\ =1.6\end{array}$

The value of the function at $x_{i-2}=1.6$ is,

$\begin{array}{c}f\left(1.6\right)=e^{\left(1.6\right)}+\left(1.6\right)\\ =6.55303\end{array}$

Now, central difference approximation of $O\left(h^4\right)$ for first derivative is given by,

$f^{\text{'}}\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+8f\left(x_{i+1}\right)-8f\left(x_{i-1}\right)+f\left(x_{i-2}\right)}{12h}$

Substitute the values of $f\left(x_{i+2}\right),f\left(x_{i+1}\right),f\left(x_{i-1}\right),f\left(x_{i-2}\right),x_i=3$ and $h=0.5$ in the above formula,

$\begin{array}{c}{f}^{\text{'}}\left(x_i\right)=\frac{-13.4232+8\left(11.2250\right)-8\left(7.84965\right)+6.55303}{12\left(0.2\right)}\\ =\frac{20.13263}{2.4}\\ =8.38866\end{array}$

Therefore, the analytic value of the first derivative at $x=2$ is $8.389056$ and numerical value is $8.38866$.