CSC212 · TU past paper

Numerical Method 2078 question paper

The complete TU 2078 exam paper for Numerical Method (CSC212), all 12 questions with solved model answers written to the mark scheme.

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110 marksNumericalHorner's methodHideAnswer
How can Horner’s rule be used to evaluate the f(x) and f(x) of a polynomial at a given point? Explain. Write an algorithm and program to calculate a real root of a polynomial using Horner’s rule.[10]
Horner's rule is an efficient scheme for evaluating a polynomial and its derivative at a point by rewriting the polynomial in nested form. It is closely related to synthetic division and the Remainder Theorem. For a polynomial of degree ...
210 marksNumericalMatrix factorization and Solving System ofHideAnswer
Matrix Factorization and Solving Linear Systems
Matrix factorization expresses a square matrix $A$ as a product of two triangular matrices: $A = LU$ where $L$ is lower triangular (with unit diagonal in Doolittle's method) and $U$ is upper triangular. This is called LU decomposition....
Matrix factorization expresses a square matrix $A$ as a product of two triangular matrices: $$A = LU$$ where $L$ is lower triangular (with unit diagonal in Doolittle's method) and $U$ is upper triangular. This is called LU decomposition....
310 marksNumericalSolution of the higher order equationsHideAnswer
Higher-Order Differential Equations
A higher-order differential equation is a differential equation in which the highest derivative of the dependent variable is of order two or more. The given equation $\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 5y = 0$ is a second-order ODE (...
A higher-order differential equation is a differential equation in which the highest derivative of the dependent variable is of order two or more. The given equation $$\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 5y = 0$$ is a second-order ODE (...
45 marksNumericalHalf-Interval method and ConvergenceHideAnswer
How the half-interval method can be estimate a root of a non-linear equation? Find a real root of the following equation using the half-interval method to correct up to two decimal places. $x^2 - e^{-x} - x = 1$ [5]
How the half-interval method can be estimate a root of a non-linear equation? Find a real root of the following equation using the half-interval method to correct up to two decimal places. $x^2 - e^{-x} - x = 1$ [5]
Equation: $x^2 - e^{-x} - x = 1$ - Rearranged: $f(x) = x^2 - e^{-x} - x - 1 = 0$ - Required accuracy: two decimal places The half-interval (bisection) method is based on the Intermediate Value Theorem: if $f(x)$ is continuous on

Equation: $x^2 - e^{-x} - x = 1$ - Rearranged: $f(x) = x^2 - e^{-x} - x - 1 = 0$ - Required accuracy: two decimal places The half-interval (bisection) method is based on the Intermediate Value Theorem: if $f(x)$ is continuous on
55 marksNumericalFixed point iteration and its convergenceHideAnswer
Calculate the real root of the given equation using fixed point iteration correct up to 3 significant figures. $2x^3 - 2x = 5$ . [5]
Calculate the real root of the given equation using fixed point iteration correct up to 3 significant figures. $2x^3 - 2x = 5$. [5]
Equation: $2x^3 - 2x = 5$ , i.e. $f(x) = 2x^3 - 2x - 5 = 0$ - Method: Fixed point iteration - Required accuracy: 3 significant figures No initial guess given; must be determined by root location. Rearrange to $x = g(x)$ : $$2x^3 = 5 + 2x...

Equation: $2x^3 - 2x = 5$, i.e. $f(x) = 2x^3 - 2x - 5 = 0$ - Method: Fixed point iteration - Required accuracy: 3 significant figures No initial guess given; must be determined by root location. Rearrange to $x = g(x)$: $$2x^3 = 5 + 2x...
65 marksNumericalNewton's Interpolation using divided diffeHideAnswer
Newton's Interpolation

What is Newton's interpolation? Obtain the divided difference table from the following data set and estimate the f(x) at x = 2 and x = 5.

$\begin{array}{c|ccccc} x & 3.2 & 2.7 & 1.0 & 4.8 & 5.6 \\ \hline f(x) & 22.0 & 17.8 & 14.2 & 38.3 & 51.7 \end{array}$

[5]
Newton's Interpolation

What is Newton's interpolation? Obtain the divided difference table from the following data set and estimate the f(x) at x = 2 and x = 5.

$$\begin{array}{c|ccccc} x & 3.2 & 2.7 & 1.0 & 4.8 & 5.6 \ \hline f(x) & 22.0 & 17.8 & 14.2 & 38.3 & 51.7 \end{array}$$

[5]
Newton's interpolation constructs an interpolating polynomial through unequally (or equally) spaced data points using divided differences. Its form is: $Pn(x) = f[x0] + f[x0,x1](x-x0) + f[x0,x1,x2](x-x0)(x-x1) + \cdots$ $i$ $xi$
Newton's interpolation constructs an interpolating polynomial through unequally (or equally) spaced data points using divided differences. Its form is: $$Pn(x) = f[x0] + fx0,x1 + fx0,x1,x2(x-x1) + \cdots$$ $i$ $xi$
75 marksNumericalLinear RegressionHideAnswer
Linear Regression Question

What is linear regression? Fit the linear function to the following data.

$\begin{array}{c|cccccccc} x & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 & 2.2 & 2.4 \\ \hline f(x) & 2.0 & 2.6 & 3.9 & 6.0 & 9.3 & 15.0 & 20.6 & 30.4 \end{array}$

[5]
Linear Regression Question

What is linear regression? Fit the linear function to the following data.

$$\begin{array}{c|cccccccc} x & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 & 2.2 & 2.4 \ \hline f(x) & 2.0 & 2.6 & 3.9 & 6.0 & 9.3 & 15.0 & 20.6 & 30.4 \end{array}$$

[5]
Linear regression is a curve-fitting technique that determines the best-fit straight line through a set of data points using the least squares principle, which minimizes the sum of squares of the residuals (vertical deviations) between o...
85 marksCubic spline interpolationHideAnswer
What are the problems with polynomial interpolation for a large number of data set? How such problems are addressed? Explain with an example. [5]
When the number of data points is large, a single polynomial of degree n-1 must be fitted through all n points. This approach suffers from several serious problems: For a large number of data points, a high-degree interpolating polynomia...
95 marksNumericalRomberg integrationHideAnswer
Evaluate the following integration using Romberg integration. $\int_{0}^{1} \frac{\sin^2 x}{x} dx$ [5]
Evaluate the following integration using Romberg integration. $$\int_{0}^{1} \frac{\sin^2 x}{x} dx$$ [5]
Integrand: $f(x) = \dfrac{\sin^2 x}{x}$ - Limits: $a = 0$ , $b = 1$ - Interval length: $b - a = 1$ Removable singularity at $x=0$ : $\lim{x\to 0}\frac{\sin^2 x}{x} = \lim{x\to0}\frac{x^2}{x} = 0$ , so $f(0)=0$ . $x$ $\sin x$ $\sin^2 x$

Integrand: $f(x) = \dfrac{\sin^2 x}{x}$ - Limits: $a = 0$, $b = 1$ - Interval length: $b - a = 1$ Removable singularity at $x=0$: $\lim{x\to 0}\frac{\sin^2 x}{x} = \lim{x\to0}\frac{x^2}{x} = 0$, so $f(0)=0$. $x$ $\sin x$ $\sin^2 x$
105 marksNumericalGauss-Jordan methodHideAnswer
Solve the following set of linear equations using the Gauss-Jordan method.

$x_2 + 2x_3 + 3x_4 = 9$ $7x_1 + 6x_2 + 5x_3 + 4x_4 = 33$ $8x_1 + 9x_2 + x_4 = 27$ $2x_1 + 5x_2 + 4x_3 + 3x_4 = 23$

[5]
Solve the following set of linear equations using the Gauss-Jordan method.

$$x_2 + 2x_3 + 3x_4 = 9$$ $$7x_1 + 6x_2 + 5x_3 + 4x_4 = 33$$ $$8x_1 + 9x_2 + x_4 = 27$$ $$2x_1 + 5x_2 + 4x_3 + 3x_4 = 23$$

[5]
$\begin{aligned} x2 + 2x3 + 3x4 &= 9 \quad (1)\\ 7x1 + 6x2 + 5x3 + 4x4 &= 33 \quad (2)\\ 8x1 + 9x2 + 0x3 + x4 &= 27 \quad (3)\\ 2x1 + 5x2 + 4x3 + 3x4 &= 23 \quad (4) \end{aligned}$ $$ \left[\begin{array}{ccccc} 0 & 1 & 2 & 3 & 9 \ 7...
$$ \begin{aligned} x2 + 2x3 + 3x4 &= 9 \quad (1)\ 7x1 + 6x2 + 5x3 + 4x4 &= 33 \quad (2)\ 8x1 + 9x2 + 0x3 + x4 &= 27 \quad (3)\ 2x1 + 5x2 + 4x3 + 3x4 &= 23 \quad (4) \end{aligned} $$ $$ \left[\begin{array}{ccccc} 0 & 1 & 2 & 3 & 9 \ 7...
115 marksNumericalHeun's methodHideAnswer
Solve the following differential equation for $1 \leq x \leq 2$ , taking $h = 0.25$ using Heun’s method. $y'(x) + x^2y = 3x$ , with $y(1) = 1$ . [5]
Solve the following differential equation for $1 \leq x \leq 2$, taking $h = 0.25$ using Heun’s method. $y'(x) + x^2y = 3x$, with $y(1) = 1$. [5]
ODE: $y'(x) + x^2 y = 3x$ , so $y'(x) = 3x - x^2 y = f(x,y)$ - Initial condition: $y(1) = 1$ - Step size: $h = 0.25$ - Interval: $1 \le x \le 2$ - Number of steps: $(2-1)/0.25 = 4$ $y{i+1} = yi + \frac{h}{2}(m1 + m2)$ where

ODE: $y'(x) + x^2 y = 3x$, so $y'(x) = 3x - x^2 y = f(x,y)$ - Initial condition: $y(1) = 1$ - Step size: $h = 0.25$ - Interval: $1 \le x \le 2$ - Number of steps: $(2-1)/0.25 = 4$ $$y{i+1} = yi + \frac{h}{2}(m1 + m2)$$ where
125 marksNumericalLaplacian equation and Poisson's equationHideAnswer
Consider a metallic plate of size 90cm by 90cm. The two adjacent sides of the plate are maintained at a temperature of $100^\circ C$ and the remaining two adjacent sides are held at $200^\circ C$ . Calculate the steady-state temperature at interior points assuming a grid size of 30 cm by 30 cm. [5]
Consider a metallic plate of size 90cm by 90cm. The two adjacent sides of the plate are maintained at a temperature of $100^\circ C$ and the remaining two adjacent sides are held at $200^\circ C$. Calculate the steady-state temperature at interior points assuming a grid size of 30 cm by 30 cm. [5]
Plate: 90 cm × 90 cm - Grid spacing: $h = 30$ cm → interior nodes form a 2×2 arrangement (4 unknowns) - Two adjacent sides at $100^\circ C$ - Remaining two adjacent sides at $200^\circ C$ Governing equation: $$\frac{\partial^2 T}{\part...

Plate: 90 cm × 90 cm - Grid spacing: $h = 30$ cm → interior nodes form a 2×2 arrangement (4 unknowns) - Two adjacent sides at $100^\circ C$ - Remaining two adjacent sides at $200^\circ C$ Governing equation: $$\frac{\partial^2 T}{\part...

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