110 marksNumericalSecant method and ConvergenceAnswer

What is non-linear equation? Derive the required expression to calculate the root of non-linear equation using secant method. Using this expression find a root of following equation.

$x^2 + \cos(x) - e^{-x} - 2 = 0$

[10]

What is non-linear equation? Derive the required expression to calculate the root of non-linear equation using secant method. Using this expression find a root of following equation.

$$x^2 + \cos(x) - e^{-x} - 2 = 0$$

[10]

Non-Linear Equations and the Secant Method

1. Non-Linear Equation

A non-linear equation is an equation $f(x)=0$ in which the unknown appears in a degree higher than one, or inside transcendental functions (trigonometric, exponential, logarithmic). Its graph is not a straight line, and it generally cannot be solved by direct algebraic manipulation, so iterative numerical methods are used.

Examples: $x^2+\cos x-e^{-x}-2=0$ , $x^3-4x+1=0$ , $e^x-3x=0$ .

2. Derivation of the Secant Method

Start from the Newton-Raphson formula:

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

The Secant method replaces the analytic derivative with a finite-difference approximation using two known points:

$f'(x_n)\approx\frac{f(x_n)-f(x_{n-1})}{x_n-x_{n-1}}$

Substituting:

$\boxed{x_{n+1}=x_n-\frac{f(x_n)\,(x_n-x_{n-1})}{f(x_n)-f(x_{n-1})}}$

This needs two initial guesses $x_0,\,x_1$ .

3. Numerical Solution

$f(x)=x^2+\cos x-e^{-x}-2$

(radian mode)

Bracketing:

$x$	$f(x)$
1	$1+0.5403-0.3679-2=-0.8276$
2	$4-0.4161-0.1353-2=+1.4486$

Root lies in $(1,2)$ . Take $x_0=1,\ x_1=2$ .

Iteration 1

$x_2=2-\frac{1.4486(2-1)}{1.4486-(-0.8276)}=2-\frac{1.4486}{2.2762}=2-0.6364=1.3636$

$f(1.3636)=1.8594+0.2050-0.2557-2$

Check $\cos(1.3636)=0.2064$ (radians), $e^{-1.3636}=0.2557$ , $(1.3636)^2=1.8594$ : $f(1.3636)=1.8594+0.2064-0.2557-2=-0.1899$

Iteration 2

$x_3=1.3636-\frac{(-0.1899)(1.3636-2)}{-0.1899-1.4486}$ $=1.3636-\frac{(-0.1899)(-0.6364)}{-1.6385}=1.3636-\frac{0.1209}{-1.6385}$ $=1.3636+0.0738=1.4374$

$f(1.4374)=2.0661+\cos(1.4374)-e^{-1.4374}-2$ $\cos(1.4374)=0.1330,\ e^{-1.4374}=0.2375$ $=2.0661+0.1330-0.2375-2=-0.0384$

Iteration 3

$x_4=1.4374-\frac{(-0.0384)(1.4374-1.3636)}{-0.0384-(-0.1899)}$ $=1.4374-\frac{(-0.0384)(0.0738)}{0.1515}=1.4374-\frac{-0.002834}{0.1515}$ $=1.4374+0.0187=1.4561$

$f(1.4561)=2.1202+\cos(1.4561)-e^{-1.4561}-2$ $\cos(1.4561)=0.1145,\ e^{-1.4561}=0.2331$ $=2.1202+0.1145-0.2331-2=+0.0016$

Iteration 4

$x_5=1.4561-\frac{(0.0016)(1.4561-1.4374)}{0.0016-(-0.0384)}$ $=1.4561-\frac{0.0016\times0.0187}{0.0400}=1.4561-0.00075=1.4553$

$f(1.4553)\approx 0.0000$ .

Result

$\boxed{x\approx 1.4553}$

Watch the trigonometry: $\cos(1.3636) \approx 0.2064$ , not $0.2190$ , so $f(1.3636) \approx -0.19$ rather than $-0.1773$ . The slip does not move the converged root, which stays at $\approx 1.455$ .

Non-Linear Equations and the Secant Method

1. Non-Linear Equation

A non-linear equation is an equation $f(x)=0$ in which the unknown appears in a degree higher than one, or inside transcendental functions (trigonometric, exponential, logarithmic). Its graph is not a straight line, and it generally cannot be solved by direct algebraic manipulation, so iterative numerical methods are used.

Examples: $x^2+\cos x-e^{-x}-2=0$, $x^3-4x+1=0$, $e^x-3x=0$.

2. Derivation of the Secant Method

Start from the Newton-Raphson formula:

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

The Secant method replaces the analytic derivative with a finite-difference approximation using two known points:

$$f'(x_n)\approx\frac{f(x_n)-f(x_{n-1})}{x_n-x_{n-1}}$$

Substituting:

$$\boxed{x_{n+1}=x_n-\frac{f(x_n),(x_n-x_{n-1})}{f(x_n)-f(x_{n-1})}}$$

This needs two initial guesses $x_0,,x_1$.

3. Numerical Solution

$$f(x)=x^2+\cos x-e^{-x}-2$$

(radian mode)

Bracketing:

$x$	$f(x)$
1	$1+0.5403-0.3679-2=-0.8276$
2	$4-0.4161-0.1353-2=+1.4486$

Root lies in $(1,2)$. Take $x_0=1,\ x_1=2$.

Iteration 1

$$x_2=2-\frac{1.4486(2-1)}{1.4486-(-0.8276)}=2-\frac{1.4486}{2.2762}=2-0.6364=1.3636$$

$$f(1.3636)=1.8594+0.2050-0.2557-2$$

Check $\cos(1.3636)=0.2064$ (radians), $e^{-1.3636}=0.2557$, $(1.3636)^2=1.8594$: $$f(1.3636)=1.8594+0.2064-0.2557-2=-0.1899$$

Iteration 2

$$x_3=1.3636-\frac{(-0.1899)(1.3636-2)}{-0.1899-1.4486}$$ $$=1.3636-\frac{(-0.1899)(-0.6364)}{-1.6385}=1.3636-\frac{0.1209}{-1.6385}$$ $$=1.3636+0.0738=1.4374$$

$$f(1.4374)=2.0661+\cos(1.4374)-e^{-1.4374}-2$$ $\cos(1.4374)=0.1330,\ e^{-1.4374}=0.2375$ $$=2.0661+0.1330-0.2375-2=-0.0384$$

Iteration 3

$$x_4=1.4374-\frac{(-0.0384)(1.4374-1.3636)}{-0.0384-(-0.1899)}$$ $$=1.4374-\frac{(-0.0384)(0.0738)}{0.1515}=1.4374-\frac{-0.002834}{0.1515}$$ $$=1.4374+0.0187=1.4561$$

$$f(1.4561)=2.1202+\cos(1.4561)-e^{-1.4561}-2$$ $\cos(1.4561)=0.1145,\ e^{-1.4561}=0.2331$ $$=2.1202+0.1145-0.2331-2=+0.0016$$

Iteration 4

$$x_5=1.4561-\frac{(0.0016)(1.4561-1.4374)}{0.0016-(-0.0384)}$$ $$=1.4561-\frac{0.0016\times0.0187}{0.0400}=1.4561-0.00075=1.4553$$

$f(1.4553)\approx 0.0000$.

Result

$$\boxed{x\approx 1.4553}$$

Watch the trigonometry: $\cos(1.3636) \approx 0.2064$, not $0.2190$, so $f(1.3636) \approx -0.19$ rather than $-0.1773$. The slip does not move the converged root, which stays at $\approx 1.455$.

210 marksNumericalMatrix factorization and Solving System ofAnswer

Matrix Factorization and LU Decomposition

Matrix factorization (decomposition) means expressing a matrix $A$ as a product of two or more matrices. In LU decomposition, $A = LU$ , where $L$ is lower triangular and $U$ is upper triangular. The Doolittle method fixes the diagonal of...

Matrix factorization (decomposition) means expressing a matrix $A$ as a product of two or more matrices. In LU decomposition, $A = LU$, where $L$ is lower triangular and $U$ is upper triangular. The Doolittle method fixes the diagonal of...

310 marksShooting method and its algorithmAnswer

What is initial value problem and boundary value problem? Write an algorithm and program to solve the boundary value problem using shooting method.[10]

An Initial Value Problem is a differential equation in which all the conditions (values of the dependent variable and its derivatives) are specified at a single point (the initial point). General Form: Here, the value of y is known at x ...

45 marksNumericalNewton's method and ConvergenceAnswer

Calculate a real negative root of following equation using Newton’s method for polynomial. $x^4 + 2x^3 + 3x^2 + 4x = 5$ . [5]

Calculate a real negative root of following equation using Newton’s method for polynomial. $x^4 + 2x^3 + 3x^2 + 4x = 5$. [5]

Equation: $x^4 + 2x^3 + 3x^2 + 4x = 5$ Rearranged: $f(x) = x^4 + 2x^3 + 3x^2 + 4x - 5 = 0$ $x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 4x^3 + 6x^2 + 6x + 4$ $f(-1) = 1 - 2 + 3 - 4 - 5 = -7 \;(<0)$ $$f(-2) = 16 - 16 + 12 -...

Equation: $$x^4 + 2x^3 + 3x^2 + 4x = 5$$ Rearranged: $$f(x) = x^4 + 2x^3 + 3x^2 + 4x - 5 = 0$$ $$x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 4x^3 + 6x^2 + 6x + 4$$ $$f(-1) = 1 - 2 + 3 - 4 - 5 = -7 ;(<0)$$ $$f(-2) = 16 - 16 + 12 -...

55 marksLeast squares methodAnswer

What is least squares approximation of fitting a function? How does it differ with polynomial interpolation? Explain with suitable example. [5]

Least squares approximation is a method of fitting a function (usually a polynomial) to a set of data points such that the sum of the squares of the errors (residuals) between the actual data values and the approximated function values i...

65 marksNumericalLagrange's InterpolationAnswer

Find the lowest degree polynomial, which passes through the following points. Using this polynomial estimate f(x) at x = 0.

$\begin{array}{c|cccccc} X & -2 & -1 & 1 & 2 & 3 & 4 \\ \hline F(x) & -19 & 0 & 2 & -3 & -4 & 5 \end{array}$

[5]

Find the lowest degree polynomial, which passes through the following points. Using this polynomial estimate f(x) at x = 0.

$$\begin{array}{c|cccccc} X & -2 & -1 & 1 & 2 & 3 & 4 \ \hline F(x) & -19 & 0 & 2 & -3 & -4 & 5 \end{array}$$

[5]

X -2 -1 1 2 3 4 ----------------------- F(x) -19 0 2 -3 -4 5 Find the lowest degree polynomial through these 6 points, then estimate $f(0)$ . Using Newton's Divided Difference Interpolation. First differences: $$f[x0,x1]=\frac{0-(-19)}{-1...

X -2 -1 1 2 3 4 ----------------------- F(x) -19 0 2 -3 -4 5 Find the lowest degree polynomial through these 6 points, then estimate $f(0)$. Using Newton's Divided Difference Interpolation. First differences: $$f[x0,x1]=\frac{0-(-19)}{-1...

75 marksNumericalLinear RegressionAnswer

The fit function of type $y = a + bx$ for the following points using the least square method.

$\begin{array}{c|cccccc} X & -1 & 1.2 & 2 & 2.7 & 3.6 & 4 \\ \hline F(x) & 1 & 20 & 27 & 33 & 41 & 45 \end{array}$

[5]

The fit function of type $y = a + bx$ for the following points using the least square method.

$$\begin{array}{c|cccccc} X & -1 & 1.2 & 2 & 2.7 & 3.6 & 4 \ \hline F(x) & 1 & 20 & 27 & 33 & 41 & 45 \end{array}$$

[5]

X -1 1.2 2 2.7 3.6 4 --------------------------- F(x) 1 20 27 33 41 45 Number of points: $n = 6$ Normal equations for $y = a + bx$ : $\sum y = na + b\sum x$ $\sum xy = a\sum x + b\sum x^2$ x y x² xy --------------------------- -1 1 1....

X -1 1.2 2 2.7 3.6 4 --------------------------- F(x) 1 20 27 33 41 45 Number of points: $n = 6$ Normal equations for $y = a + bx$: $$\sum y = na + b\sum x$$ $$\sum xy = a\sum x + b\sum x^2$$ x y x² xy --------------------------- -1 1 1....

85 marksNumericalSimpson's 1/3 ruleAnswer

Calculate the integral value of the function given below from $x = 1.8$ to $x = 3.4$ using Simpson's 1/3 rule.

$\begin{array}{c|cccccccc} x & 1.8 & 2.0 & 2.2 & 2.4 & 2.6 & 2.8 & 3.0 & 3.4 \\ \hline f(x) & 0.003 & 0.778 & 1.632 & 2.566 & 3.579 & 4.672 & 7.097 & 8.429 \end{array}$

[5]

Calculate the integral value of the function given below from $x = 1.8$ to $x = 3.4$ using Simpson's 1/3 rule.

$$\begin{array}{c|cccccccc} x & 1.8 & 2.0 & 2.2 & 2.4 & 2.6 & 2.8 & 3.0 & 3.4 \ \hline f(x) & 0.003 & 0.778 & 1.632 & 2.566 & 3.579 & 4.672 & 7.097 & 8.429 \end{array}$$

[5]

X 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.4 -------------------------------------------------------------- F(x) 0.003 0.778 1.632 2.566 3.579 4.672 7.097 8.429 Limits: $x = 1.8$ to $x = 3.4$ Spacing check: - 1.8 to 3.0: step $h = 0.2$ (equally spa...

95 marksNumericalRomberg integrationAnswer

Evaluate the following integration using Romberg integration. $\int_{0}^{1} \frac{\sin x}{x} dx$ [5]

Evaluate the following integration using Romberg integration. $$\int_{0}^{1} \frac{\sin x}{x} dx$$ [5]

Integrand: $f(x) = \dfrac{\sin x}{x}$ - Limits: $a = 0$ , $b = 1$ - Special value at $x=0$ : $\lim{x\to 0}\dfrac{\sin x}{x} = 1$ , so $f(0)=1$ - Method: Romberg integration - Interval width: $b-a = 1$ I compute the required function value...

Integrand: $f(x) = \dfrac{\sin x}{x}$ - Limits: $a = 0$, $b = 1$ - Special value at $x=0$: $\lim{x\to 0}\dfrac{\sin x}{x} = 1$, so $f(0)=1$ - Method: Romberg integration - Interval width: $b-a = 1$ I compute the required function value...

105 marksNumericalGauss-Seidal MethodAnswer

Solve the following set of equations using Gauss Seidel method.

\begin{aligned} x + 2y + 3z &= 4 \\ 6x - 4y + 5z &= 10 \\ 5x + 2y + 2z &= 25 \end{aligned}

[5]

Solve the following set of equations using Gauss Seidel method.

$$ \begin{aligned} x + 2y + 3z &= 4 \ 6x - 4y + 5z &= 10 \ 5x + 2y + 2z &= 25 \end{aligned} $$

[5]

System of equations: $x + 2y + 3z = 4 \quad \cdots (1)$ $6x - 4y + 5z = 10 \quad \cdots (2)$ $5x + 2y + 2z = 25 \quad \cdots (3)$ Initial guess (standard): $x^{(0)} = y^{(0)} = z^{(0)} = 0$ . To apply Gauss-Seidel we solve equation ...

System of equations: $$x + 2y + 3z = 4 \quad \cdots (1)$$ $$6x - 4y + 5z = 10 \quad \cdots (2)$$ $$5x + 2y + 2z = 25 \quad \cdots (3)$$ Initial guess (standard): $x^{(0)} = y^{(0)} = z^{(0)} = 0$. To apply Gauss-Seidel we solve equation ...

115 marksNumericalRunge-Kutta methodsAnswer

From the following differential equation estimate y(1) using RK 4th order method.

$\frac{dy}{dx} + 2x^2y = 4 \quad \text{with} \quad y(0) = 1$

[Take $h = 0.5$ ] [5]

From the following differential equation estimate y(1) using RK 4th order method.

$$\frac{dy}{dx} + 2x^2y = 4 \quad \text{with} \quad y(0) = 1$$

[Take $h = 0.5$] [5]

ODE: $\dfrac{dy}{dx} + 2x^2 y = 4 \Rightarrow \dfrac{dy}{dx} = 4 - 2x^2 y = f(x,y)$ - Initial condition: $x0 = 0,\ y0 = 1$ - Step size: $h = 0.5$ - Target: $y(1)$ → requires 2 steps ( $0 \to 0.5 \to 1.0$ ) RK4 formulas: $$y{n+1} = yn + ...

ODE: $\dfrac{dy}{dx} + 2x^2 y = 4 \Rightarrow \dfrac{dy}{dx} = 4 - 2x^2 y = f(x,y)$ - Initial condition: $x0 = 0,\ y0 = 1$ - Step size: $h = 0.5$ - Target: $y(1)$ → requires 2 steps ($0 \to 0.5 \to 1.0$) RK4 formulas: $$y{n+1} = yn + ...

125 marksNumericalLaplacian equation and Poisson's equationAnswer

Solve the Poisson's equation over the square domain $0 \leq x \leq 1.5$ , $0 \leq y \leq 1.5$ with $f = 0$ on the boundary and $h = 0.5$ . [5]

Solve the Poisson's equation over the square domain $0 \leq x \leq 1.5$, $0 \leq y \leq 1.5$ with $f = 0$ on the boundary and $h = 0.5$. [5]

Domain: $0 \le x \le 1.5$ , $0 \le y \le 1.5$ - Boundary condition: $f = 0$ on all boundaries - Step size: $h = 0.5$ - Poisson equation right-hand side: NOT LEGIBLE in the question. The function $f$ (source term) appears garbled. This s...

Domain: $0 \le x \le 1.5$, $0 \le y \le 1.5$ - Boundary condition: $f = 0$ on all boundaries - Step size: $h = 0.5$ - Poisson equation right-hand side: NOT LEGIBLE in the question. The function $f$ (source term) appears garbled. This s...

Numerical Method 2075 question paper

Non-Linear Equations and the Secant Method

1. Non-Linear Equation

2. Derivation of the Secant Method

3. Numerical Solution

Iteration 1

Iteration 2

Iteration 3

Iteration 4

Result

Non-Linear Equations and the Secant Method

1. Non-Linear Equation

2. Derivation of the Secant Method

3. Numerical Solution

Iteration 1

Iteration 2

Iteration 3

Iteration 4

Result

Non-Linear Equations and the Secant Method

1. Non-Linear Equation

2. Derivation of the Secant Method

3. Numerical Solution

Iteration 1

Iteration 2

Iteration 3

Iteration 4

Result