Chapter 18, Problem 43P

To determine

Investigate the meaning of numerical analysis and give its examples.

Explanation of Solution

In order to obtain the numerical solutions for the mathematical problems, numerical analysis is used. In the field of engineering and physics, numerical analysis plays a vital role.

Give the areas of study of numerical analysis as below.

• Value of functions
• Solving system of linear equations
• Differential equation
• Integral
• Eigen values and vectors

Value of functions:

The function of numerical analysis includes square root, cube root, logarithmic etc.

Example:

Let us consider the value,

$x=4$

Square root:

The square root of $x$ can be calculated as follows:

$\sqrt{x}=\sqrt{4}=2$

Cube root:

The cube root of $x$ can be calculated as follows:

$\sqrt[3]{x}=\sqrt[3]{4}=1.5874$

Logarithmic function:

The logarithmic value of $x$ can be calculated as follows:

$\log \left(x\right)=\log \left(4\right)=0.602$

Solving system of linear equations:

It is a collection of two or more linear equation involves same set of variables.

Consider the linear equation as follows:

$x_1+x_2+x_3=6$ (1)

$2x_1+5x_2+x_3=15$ (2)

$-3x_1+x_2+5x_3=14$ (3)

Solve equation (1) and (2)

Multiply equation (1) by $-2$

Therefore equation (2) becomes,

$-2x_1-2x_2-2x_3=-12$ (4)

Solve equation (2) and (4)

$\begin{array}{l}\underset{\_}{\begin{array}{c}-2x_1-2x_2-2x_3=-12\\ 2x_1+5x_2+x_3=15\end{array}}\\ 3x_2-x_3=3\end{array}$ (5)

Multiply equation (1) by $3$

Therefore equation (1) becomes,

$3x_1+3x_2+3x_3=18$ (6)

Solve equation (6) and (3)

$\begin{array}{l}\underset{\_}{\begin{array}{l}3x_1+3x_2+3x_3=18\\ -3x_1+x_2+5x_3=14\end{array}}\\ 4x_2+8x_3=32\\ x_2+2x_3=8\end{array}$ (7)

Solve equation (5) and (7)

Multiply equation (7) by $-3$

Therefore equation (7) becomes,

$-3x_2-6x_3=-24$ (8)

Solve equation (5) and (8)

$\begin{array}{l}\underset{\_}{\begin{array}{l}3x_2-x_3=3\\ -3x_2-6x_3-24\end{array}}\\ -7x_3=-21\\ x_3=3\end{array}$

Substitute $3$ for $x_3$ in equation (7)

$\begin{array}{l}{x}_2+2(3)=8\\ x_2=8-6\\ x_2=2\end{array}$

Substitute $3$ for $x_3$, and $2$ for $x_2$ in equation (1)

$\begin{array}{c}{x}_1+2+3=6\\ x_1=6-2-3\\ =1\end{array}$

Therefore, the solution is $x_1=1,x_2=2,x_3=3$.

Differential equation:

The differential equation is the equation which contains functions of derivatives represents their rate of change.

Let us consider the equation as follows:

$y=x^3+2x^2+3x+4$ (9)

Differentiate equation (9) with respect to $x$.

Therefore equation (9) becomes:

$\begin{array}{c}\frac{dy}{dx}=3x^2+2\left(2\right)x+3\left(1\right)+0\\ =3x^2+4x+3\end{array}$

Thus, the rate of change of equation (9) is $3x^2+4x+3$.

Integral:

Integral calculus is the development of differential calculus. It is used to find the displacement, moment of inertia, area and volume in the mathematical concepts.

The indefinite integral can be represented by,

$F\left(x\right)={\displaystyle \int f\left(x\right)dx}$ (10)

Let us consider the function,

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

Substitute $x^3+2x^2+3x+4$ for $f\left(x\right)$ in equation (10)

$\begin{array}{c}F\left(x\right)={\displaystyle \int \left(x^3+2x^2+3x+4\right)dx}\\ =\frac{x^4}{4}+2\frac{x^3}{3}+3\frac{x^2}{2}+4x\end{array}$

Eigen values and Eigen vectors:

The Eigen value is a non-zero vector that changes by a scalar factor when the linear transformation is applied.

Consider the Eigen value problem:

$Ax=\lambda x$ (11)

Here,

$A$ is the matrix,

$x$ is the vector,

$\lambda$ is the Eigen values.

Consider the matrix A as follows:

$A=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]$

The Eigen value of the matrix can be calculated by,

$\left|A-\lambda I\right|=0$ . (12)

Here,

$I$ is the identity matrix.

Substitute $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for $I$, and $\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]$ for $A$ in equation (12)

$\begin{array}{c}\left|\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0\\ \left|\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right|=0\\ \left|\left[\begin{array}{cc}-\lambda & 1\\ -2& -3-\lambda \end{array}\right]\right|=0\\ \left(-\lambda \right)\left(-3-\lambda \right)-\left(-2\right)\left(1\right)=0\\ {\lambda }^2+3\lambda +2=0\end{array}$ (13)

Solve the equation (13)

$\begin{array}{c}{\lambda }^2+3\lambda +2=0\\ {\lambda }^2+2\lambda +\lambda +2=0\\ \lambda \left(\lambda +2\right)+1\left(\lambda +2\right)=0\\ \left(\lambda +1\right)\left(\lambda +2\right)=0\end{array}$ . (14)

Reduce equation (14) as follows,

$\begin{array}{c}\lambda +1=0\lambda +2=0\\ \lambda =-1\lambda =-2\end{array}$

The Eigen values are,

$\begin{array}{l}{\lambda }_1=-1,\\ {\lambda }_2=-2\end{array}$.

Eigen vectors for the Eigen value can be calculated as follows:

$\begin{array}{c}A{x}_1={\lambda }_1{x}_1\\ \left(A-{\lambda }_1\right).x_1=0\end{array}$ (15)

Substitute $\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]$ for $A$, and $-1$ for ${\lambda }_1$ in equation (15)

$\begin{array}{c}\left(\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]-\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_1\end{array}\right]\right).x_1=0\\ \left[\begin{array}{cc}-{\lambda }_1& 1\\ -2& -{\lambda }_1-3\end{array}\right]x_1=0\end{array}$ (16)

Reduce the equation (16) as follows,

$\begin{array}{c}\left[\begin{array}{cc}-\left(-1\right)& 1\\ -2& -\left(-1\right)-3\end{array}\right]x_1=0\\ \left[\begin{array}{cc}1& 1\\ -2& -2\end{array}\right]x_1=0\\ \left[\begin{array}{cc}1& 1\\ -2& -2\end{array}\right]\left[\begin{array}{c}{x}_{1,1}\\ x_{1,2}\end{array}\right]=0\end{array}$ (17)

Write the matrix form of equation (17) into linear equation as follows,

$x_{1,1}+x_{1,2}=0$ (18)

$-2x_{1,1}-2x_{1,2}=0$ (19)

Solving equation (18) and (19)

$x_{1,1}=-x_{1,2}$

Therefore, the Eigen vector of $x_1$ can be written as

$x_1=\left[\begin{array}{c}1\\ -1\end{array}\right]$

Eigen vectors for the Eigen value can be calculated as follows:

$\begin{array}{c}A{x}_2={\lambda }_2{x}_2\\ \left(A-{\lambda }_2\right).x_2=0\end{array}$ (20)

Substitute $\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]$ for $A$, and $-1$ for ${\lambda }_1$ in equation (20)

$\begin{array}{c}\left(\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]-\left[\begin{array}{cc}{\lambda }_2& 0\\ 0& {\lambda }_2\end{array}\right]\right).x_2=0\\ \left[\begin{array}{cc}-{\lambda }_2& 1\\ -{\lambda }_2-2& -{\lambda }_2-3\end{array}\right]x_2=0\end{array}$ (21)

Reduce the equation (16) as follows,

$\begin{array}{c}\left[\begin{array}{cc}-\left(-2\right)& 1\\ -2& -\left(-2\right)-3\end{array}\right]x_2=0\\ \left[\begin{array}{cc}2& 1\\ 0& -1\end{array}\right]x_2=0\\ \left[\begin{array}{cc}2& 1\\ -2& -1\end{array}\right]\left[\begin{array}{c}{x}_{2,1}\\ x_{2,2}\end{array}\right]=0\end{array}$ (22)

Write the matrix form of equation (22) into linear equation as follows,

$2x_{2,1}+x_{2,2}=0$ (23)

$-2x_{2,1}-x_{2,2}=0$ (24)

Solving equation (23) and (24)

$x_{2,2}=-2x_{2,1}$

Therefore, the Eigen vector of $x_2$ can be written as

$x_2=\left[\begin{array}{c}1\\ -2\end{array}\right]$

Conclusion:

Thus, the numerical analysis and its examples are explained.