Chapter 23, Problem 20P

Use a Taylor series expansion to derive a centered finite-difference approximation to the third derivative that is second-order accurate. To do this, you will have to use four different expansions for the points $x_{i-2},x_{i-1},x_{i+1},$ and $x_{i+2}$. In each case, the expansion will be around the point $x_i$. The interval $\Delta x$ will be used in each case of $i-1$ and $i+1$, and $2\Delta x$ will be used in each case of $i-2$ and $i+2$. The four equations must then be combined in a way to eliminate the first and second derivatives. Carry enough terms along in each expansion to evaluate the first term that will be truncated to determine the order of the approximation.

To determine

A centered finite difference approximation to the third derivative with the help of Taylor series expansion.

Answer to Problem 20P

Solution:

$f\text{'}\text{'}\text{'}\left(x_i\right)=\frac{f\left(x_{i+2}\right)+f\left(x_{i-2}\right)-2f\left(x_{i+1}\right)+2f\left(x_{i-1}\right)}{2{\left(\Delta x\right)}^3}-\frac{1}{4}{\left(\Delta x\right)}^2{f}^5\left(x_i\right)+\mathrm{....}$

Explanation of Solution

Given information:

Step size $h=2\Delta x$ for $i-2$ and $i+2$

Step size $h=\Delta x$ for $i-1$ and $i+1$

Formula used:

The Taylor series expansion about a and x can be expressed as,

$\begin{array}{l}f\left(x\right)=f\left(a\right)+hf\text{'}\left(a\right)+\frac{1}{2}{h}^{2}f"\left(a\right)+\frac{1}{6}{h}^3{f}^{‴}\left(a\right)+\frac{1}{24}{h}^4{f}^4\left(a\right)\\ +\frac{1}{120}{h}^5{f}^5\left(a\right)+\cdot \cdot \cdot \end{array}$

Calculation:

The Taylor series expansion about a and x is,

$\begin{array}{l}f\left(x\right)=f\left(a\right)+hf\text{'}\left(a\right)+\frac{1}{2}{h}^{2}f"\left(a\right)+\frac{1}{6}{h}^3{f}^{‴}\left(a\right)+\frac{1}{24}{h}^4{f}^4\left(a\right)\\ +\frac{1}{120}{h}^5{f}^5\left(a\right)+\cdot \cdot \cdot \end{array}$

For forward expansion, substitute $a=x_i,h=2\Delta x$ and $x=x_{i+2}$ in Taylor series expansion,

$\begin{array}{c}f\left(x_{i+2}\right)=f\left(x_i\right)+\left(2\Delta x\right)f\text{'}\left(x_i\right)+\frac{1}{2}{\left(2\Delta x\right)}^{2}f"\left(x_i\right)+\frac{1}{6}{\left(2\Delta x\right)}^3{f}^{‴}\left(x_i\right)\\ +\frac{1}{24}{\left(2\Delta x\right)}^4{f}^4\left(x_i\right)+\frac{1}{120}{\left(2\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \\ f\left(x_{i+2}\right)=f\left(x_i\right)+2\left(\Delta x\right)f\text{'}\left(x_i\right)+2{\left(\Delta x\right)}^{2}f"\left(x_i\right)+\frac{8}{6}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{16}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)+\frac{32}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \end{array}$

Now, substitute $a=x_i,h=\Delta x$ and $x=x_{i+1}$ in Taylor series expansion,

$\begin{array}{c}f\left(x_{i+1}\right)=f\left(x_i\right)+\left(\Delta x\right)f\text{'}\left(x_i\right)+\frac{1}{2}{\left(\Delta x\right)}^{2}f"\left(x_i\right)+\frac{1}{6}{\left(\Delta x\right)}^3{f}^{‴}\left(x_i\right)\\ +\frac{1}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)+\frac{1}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \\ f\left(x_{i+1}\right)=f\left(x_i\right)+\left(\Delta x\right)f\text{'}\left(x_i\right)+\frac{1}{2}{\left(\Delta x\right)}^{2}f"\left(x_i\right)+\frac{1}{6}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{1}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)+\frac{1}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \end{array}$

Multiply above equation by $2$,

$\begin{array}{c}2f\left(x_{i+1}\right)=2f\left(x_i\right)+2\left(\Delta x\right)f\text{'}\left(x_i\right)+\frac{2}{2}{\left(\Delta x\right)}^{2}f"\left(x_i\right)+\frac{2}{6}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{2}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)+\frac{2}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \\ 2f\left(x_{i+1}\right)=2f\left(x_i\right)+2\left(\Delta x\right)f\text{'}\left(x_i\right)+{\left(\Delta x\right)}^{2}f"\left(x_i\right)+\frac{1}{3}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{1}{12}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)+\frac{1}{60}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \end{array}$

Thus, the value of $f\left(x_{i+2}\right)-2f\left(x_{i+1}\right)$ is,

$\begin{array}{c}f\left(x_{i+2}\right)-2f\left(x_{i+1}\right)=f\left(x_i\right)-2f\left(x_i\right)+2\left(\Delta x\right)f\text{'}\left(x_i\right)-2\left(\Delta x\right)f\text{'}\left(x_i\right)\\ +2{\left(\Delta x\right)}^{2}f"\left(x_i\right)-{\left(\Delta x\right)}^{2}f"\left(x_i\right)+\frac{8}{6}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ -\frac{1}{3}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)+\frac{16}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)-\frac{1}{12}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)\\ +\frac{32}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)-\frac{1}{60}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \end{array}$

$\begin{array}{c}f\left(x_{i+2}\right)-2f\left(x_{i+1}\right)=-f\left(x_i\right)+{\left(\Delta x\right)}^{2}f"\left(x_i\right)+{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)+\frac{7}{12}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)\\ +\frac{1}{4}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\mathrm{.....}\end{array}$ …… (1)

For backward expansion, substitute $a=x_i,h=-2\Delta x$ and $x=x_{i-2}$ in Taylor series expansion,

$\begin{array}{c}f\left(x_{i-2}\right)=f\left(x_i\right)+\left(-2\Delta x\right)f\text{'}\left(x_i\right)+\frac{1}{2}{\left(-2\Delta x\right)}^{2}f"\left(x_i\right)+\frac{1}{6}{\left(-2\Delta x\right)}^3{f}^{‴}\left(x_i\right)\\ +\frac{1}{24}{\left(-2\Delta x\right)}^4{f}^4\left(x_i\right)+\frac{1}{120}{\left(-2\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \\ f\left(x_{i-2}\right)=f\left(x_i\right)-2\left(\Delta x\right)f\text{'}\left(x_i\right)+2{\left(\Delta x\right)}^{2}f"\left(x_i\right)-\frac{8}{6}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{16}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)-\frac{32}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \end{array}$

Now, substitute $a=x_i,h=-\Delta x$ and $x=x_{i-1}$ in Taylor series expansion,

$\begin{array}{c}f\left(x_{i-1}\right)=f\left(x_i\right)+\left(-\Delta x\right)f\text{'}\left(x_i\right)+\frac{1}{2}{\left(-\Delta x\right)}^{2}f"\left(x_i\right)+\frac{1}{6}{\left(-\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{1}{24}{\left(-\Delta x\right)}^4{f}^4\left(x_i\right)+\frac{1}{120}{\left(-\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \\ f\left(x_{i-1}\right)=f\left(x_i\right)-\left(\Delta x\right)f\text{'}\left(x_i\right)+\frac{1}{2}{\left(\Delta x\right)}^{2}f"\left(x_i\right)-\frac{1}{6}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{1}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)-\frac{1}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \end{array}$

Multiply above equation by $2$,

$\begin{array}{c}2f\left(x_{i-1}\right)=2f\left(x_i\right)-2\left(\Delta x\right)f\text{'}\left(x_i\right)+\frac{2}{2}{\left(\Delta x\right)}^{2}f"\left(x_i\right)-\frac{2}{6}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{2}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)-\frac{2}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \\ 2f\left(x_{i-1}\right)=2f\left(x_i\right)-2\left(\Delta x\right)f\text{'}\left(x_i\right)+{\left(\Delta x\right)}^{2}f"\left(x_i\right)-\frac{1}{3}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{1}{12}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)-\frac{1}{60}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \end{array}$

Thus, the value of $2f\left(x_{i-1}\right)-f\left(x_{i-2}\right)$ is,

$\begin{array}{c}2f\left(x_{i-1}\right)-f\left(x_{i-2}\right)=2f\left(x_i\right)-f\left(x_i\right)-2\left(\Delta x\right)f\text{'}\left(x_i\right)+2\left(\Delta x\right)f\text{'}\left(x_i\right)\\ +{\left(\Delta x\right)}^{2}f"\left(x_i\right)-2{\left(\Delta x\right)}^{2}f"\left(x_i\right)-\frac{1}{3}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{8}{6}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)+\frac{1}{12}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)-\frac{16}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)\\ -\frac{1}{60}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\frac{32}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)+\cdot \cdot \cdot \end{array}$

$\begin{array}{c}2f\left(x_{i-1}\right)-f\left(x_{i-2}\right)=f\left(x_i\right)-{\left(\Delta x\right)}^{2}f"\left(x_i\right)+\frac{6}{6}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)-\frac{14}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)\\ +\frac{30}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)\end{array}$ …… (2)

Now, add equation (1) and equation (2),

$\begin{array}{c}f\left(x_{i+2}\right)+f\left(x_{i-2}\right)-2f\left(x_{i+1}\right)+2f\left(x_{i-1}\right)=f\left(x_i\right)-{\left(\Delta x\right)}^{2}f"\left(x_i\right)+\frac{6}{6}{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ -\frac{14}{24}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)+\frac{30}{120}{\left(\Delta x\right)}^5{f}^5\left(x_i\right)\\ -f\left(x_i\right)+{\left(\Delta x\right)}^{2}f"\left(x_i\right)+{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)\\ +\frac{7}{12}{\left(\Delta x\right)}^4{f}^4\left(x_i\right)+\frac{{\left(\Delta x\right)}^5}{4}{f}^5\left(x_i\right)+\mathrm{.....}\\ f\left(x_{i+2}\right)+f\left(x_{i-2}\right)-2f\left(x_{i+1}\right)+2f\left(x_{i-1}\right)=2{\left(\Delta x\right)}^{3}f\text{'}\text{'}\text{'}\left(x_i\right)+\frac{{\left(\Delta x\right)}^5}{2}{f}^5\left(x_i\right)\end{array}$

Rearrange the above equation as,

$\begin{array}{c}f\text{'}\text{'}\text{'}\left(x_i\right)=\frac{f\left(x_{i+2}\right)+f\left(x_{i-2}\right)-2f\left(x_{i+1}\right)+2f\left(x_{i-1}\right)}{2{\left(\Delta x\right)}^3}-\frac{{\left(\Delta x\right)}^5}{4}\frac{f^5\left(x_i\right)}{{\left(\Delta x\right)}^3}+\mathrm{....}\\ f\text{'}\text{'}\text{'}\left(x_i\right)=\frac{f\left(x_{i+2}\right)+f\left(x_{i-2}\right)-2f\left(x_{i+1}\right)+2f\left(x_{i-1}\right)}{2{\left(\Delta x\right)}^3}-\frac{{\left(\Delta x\right)}^2}{4}{f}^5\left(x_i\right)+\mathrm{....}\end{array}$

Therefore, the centered finite-difference approximation to the third derivative which is second order correct is,

$f\text{'}\text{'}\text{'}\left(x_i\right)=\frac{f\left(x_{i+2}\right)+f\left(x_{i-2}\right)-2f\left(x_{i+1}\right)+2f\left(x_{i-1}\right)}{2{\left(\Delta x\right)}^3}-\frac{{\left(\Delta x\right)}^2}{4}{f}^5\left(x_i\right)+\mathrm{....}$