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Important Questions

Important Questions

Solution of Nonlinear Equations

Asked in 2082Long Question2+4+4 Marks
1.
Explain how bisection method differ from secant. Derive the formula for Newton Raphson. Use Newton Raphson method to solve the equation f(x)=x3+2xβˆ’2f(x) = x^3 + 2x - 2 correct upto three decimal places. [2+4+4]
Asked in 2080Short Question5 Marks
2.
What are the sources of errors? Discuss various types of errors encounters in numerical computation. [5]
Asked in 2080Short Question5 Marks
3.
Use secant method to estimate the root of the equation x2βˆ’5x+6=0x^2-5x+6=0, with initial estimate x1=4x_1 = 4 and x2=2x_2 = 2 (EPS=0.05). [5]
Asked in 2080Long Question10 Marks
4.
Write an algorithm and a C-Program to obtain roots of non-linear equation using Newton Raphson Method. [10]
Asked in 2079Short Question5 Marks
5.
Write an algorithm for Honer's method. Evaluate the polynomial f(x)=x4+3x3+5x2+7x+9f(x) = x^4 + 3x^3 + 5x^2 + 7^x + 9 at x = 2 by using Honer's method. [5]
Asked in 2079Long Question10 Marks
6.
Define true error and relative error. Derive the bisection method for solving non-linear equation and using this method solve 2x3βˆ’2xβˆ’52x^3-2x-5 with initial x0=1x_0=1 and x1=2x_1=2. Calculate upto 10th iteration. [10]
Asked in 2078Short Question5 Marks
7.
Show that the rate of convergence of Newtons Raphson method is quadratic. [5]
Asked in 2078Long Question10 Marks
8.
Derive the formula for integration using simpsons 3/8 rule. Use Secant Method to estimate the root of equation with initial estimate x₁ = 4 and xβ‚‚ = 2, x2βˆ’4xβˆ’10=0x^2 - 4x - 10 = 0. [10]
ModelShort Question5 Marks
9.
Define the terms true error and relative error? Write down algorithm for Horner' method to evaluate polynomial and use the method to evaluate the polynomial 2x3βˆ’3x2+5xβˆ’22x^3-3x^2+5x-2 at x=3. [5]
ModelLong Question10 Marks
10.
How Secant methods differs from Newton Raphson method? Derive the formula for Secant Method. Solve the equation cosx+2sinxβˆ’x2=0cosx+2sinx-x^2=0 using Secant method. Assume error precision is 0.01. [10]