CHAPTER 3
Numerical Methods
Numerical methods are methods for solving problems numerically on computer or calculator (or some times by hand). Numerical methods are often divided into elementary ones such as finding the root of an equation, integrating a function or solving a linear system of equations to intensive ones like the finite element method.
A. Solution of algebraic and transcendental equation / Root Finding
1. Bisection Method2. Regula falsi method 3. Secant method
4. Newton- Raphson method 5. Horner’s method
6. Descarte’s rule
Consider an equation f(x) = 0 1. Bisection method
This method finds the root between points “a” and “b”.
If f(x) is continuous between a and b and f (a) and f (b) are of opposite sign then there is a root between a & b (Intermediate Value Theorem).
First approx. to the root is x1 =
.
If f(x1) = 0, then x1 is the root of f(x) = 0, otherwise root of f(x)=0 lies between a and
x1 or x1 and b.
Similarly, x2 and x3 . . . are determined.
a b
y = f(x) y
Note
Simplest iterative method
Bisection method always converge but often slowly. This method can’t be used for finding the complex roots. Rate of convergence is linear
Example
Find the roots of Solution
f(2) = -ve , f(3) = +ve a root in [2,3] x1 = = 2.5 f(x1) = -ve x2 = = 2.75 f(x2) = +ve x3 = = 2.625 f(x3) = -ve x4 = = 2.6875
2. Newton Raphson method (or successive substitution method or tangent method)
xn+1 = xn –
This method is commonly used for its simplicity and greater speed.
Here f x is assumed to have continuous derivative f’ x . f(x)
Y=f(x)
X2 X1 X0
This method fails if f’ x .
It has second order of convergence or quadratic convergence, i.e. the subsequent error at each step is proportional to the square of the error at previous step.
Sensitive to starting value, i.e. The Newton’s method converges provided the initial approximation is chosen sufficiently close to the root.
Rate of convergence is quadratic
Geometric Interpretation y – f(x0 f’ x0) (x-x0)
x1 = x0 -
Draw a tangent at f(x0). Tangent intersects X-axis at x1. Draw a tangent at f(x1 . And so on….
Example
Evaluate √ to four decimal places Solution X2 – 28 = 0 let x0 = 5 f’ x x x1 = x0 - x1= 5 + x1= 5.3 x2 = 5.3 – . . x2= 5.2915094 x3 = 5.2915094– . . x3 = 2.64574 3. Secant Method
Here, the formula is derived from Newton-Raphson by replacing the derivative by the difference co- efficient. =
–
Note
a) Here derivative is avoided. b) Convergence is not guaranteed.
c) If converges, Convergence super linear (more rapid than linear, almost quadratic
4. Regula falsi method or (method of false position)
Regula falsi method always converge. However, it converges slowly.
If converges, order of convergence is between 1 & 2 (closer to 1). It is superior to Bisection method.
Given, f(x) = 0
Select x0 and x1 such that f(x0) f(x1) < 0 (i.e. opposite sign)
Y – f(x0) = – (x- x0) = - – = – Check if f(x0) f(x2) < 0 or f(x1) f(x2) < 0 Compute ………
which is an approximation to the root.
Example
Find the roots of x3 +x-1 =0
Solution
f(x) = x3+ x-1=0
f(1) = 1+1-1= 1 (+ve)
f(0.5) = (0.5)3+(0.5)-1 = -0.375 (-ve)
Hence root lie between 0.5 & 1 x 0 =0.5 , x 1 = 1 x2 = – = 0.64 f(x) X 3 (X0, f(x0)) (X1, f(x1)) X1 X2 X0 X
f(x2) = -0.0979 (-ve)
Root lies between 0.64 & 1 x3 = . .
. = 0.672
f(x3)= -0.0245 (-ve)
Hence the root in between 0.672 & 1 X4 = 0.6778
B. Solution of linear system of equations
Gauss elimination method
Gauss jordan method
Do little method
Crout’s triangularization
Cholesky method
Jacobi iteration method
Gauss siedel iteration method
Relaxation method 1. Gauss Elimination Method
Here equations are converted into “upper triangular matrix” form, then solved by “back substitution” method.
Consider a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Step 1. To eliminate x from second and third equation (we do this by subtracting suitable multiple of first equation from second and third equation)
a1x + b1y + c1z = d1 pivotal equation, a1 pivot point.
b ’y + c ’ z = d ’
b3’y + c3’ z = d3’
Step 2. Eliminate y from third equation a1x + b1y + c1z = d1
b ’y + 4c z = d ’ pivotal equation, b ’ is pivot point.
c3’’z = d3”
Step 3. The value of x , y and z can be found by back substitution. Note: Number of operations: N = + -
2. Gauss Jordon method
- Used to find inverse of the matrix and solving linear equations.
- Here back substitution is avoided by additional computations that reduce the matrix to “diagonal from”, instead to triangular form in Gauss elimination method.
- Number of operations is more than Gauss elimination as the effort of back substitution is saved at the cost of additional computation.
Step 1: Eliminate x from 2nd and 3rd
Step 3: Eliminate z from 1st and 2nd
3. L U decomposition
- It is modification of the Gauss eliminiation method. - Also used for finding the inverse of the matrix. [A]n x n = [ L ] n x n [U] n x n
a11 a12 a13 1 0 0 U11 U12 U13
a21 b22 c23 = L21 1 0 0 U22 U23
a31 b32 c33 L31 L32 1 0 0 U31
Ax = LUX = b can be written as a) LY=b, where b) UX=Y
Solve for Y from a then solve for X from b . This method is known as Dolittle’s method. Similar methods are Crout’s method and Cholesky methods.
4. Iterative Method
a. Jacobi iteration method a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
If a1, b2 , c3 are large compared to other coefficients, then solving these for x, y, z
respectively
x = k1 – l1y – m1z
y = k2 – l2x – m2z
z = k3 – l3x – m3y
Let us start with initial approximation x0 , y0 , z0
x1= k1 – l1y0 – m1z0
y1= k2 – l2x0 – m2z0
z1= k3 – l3x0 – m3y0
Note: No of component of x(k) is used in computation unless y(k) and z(k) is computed.
The process is repeated till the difference between two consecutive approximation is negligible. In generalized form: x(k+1) = k1 – l1 y(k) – m1z(k) y(k+1) = k 2 – l2 x(k) – m2z(k) z(k+1) = k3 – l3 x(k) – m3y(k)
b. Gauss Siedel iteration method
Modification of the Jacobi’s iteration method
Start with (x0, y0, z0)= (0, 0, 0) or anything [No specific condition]
In first equation, put y = y0 z = z0 which will give x1
In second equation, put x = x1 z = z0 which will give y1
In third equation, put x = x1 y = y1 which will give z1
Note: To compute any variable, use the latest available value. In generalized form:
x(k+1) = k1 – l1y(k) – m1z(k)
y(k+1) = k2 – l2x(k+1) – m2z(k)
z(k+1) = k3 – l3x(k+1) – m3y(k+1)
C. Solution of the non – linear simultaneous equation f(x,y) = 0 ; g(x, y)=0; Assume, x1 = x0 + h , y1 = y0 + k f( x0 + h , y0 + k)=0; g( x0 + h , y0 + k) = 0 f(x0 , y0 ) + h , , + k , , = 0 g(x0 , y0 ) + h , , + k , , = 0
Solving for h and k , we get a new approximation to the root as x1 = x0 + h , y1 = y0 + k
D. Numerical Integration
The process of evaluation of definite Y X0+h X0 X0+2 h X0+n h X Y0 Y1 Y 2
integeral ∫ from a set of tabulated value of the integrand f(x) is called numerical integration.
This process when applied to a function of a single variable is known as quadrature. Trapezoidal formula:
∫
{ }
Error = Exact – approx
The error in approximating an integral using Trapezoidal rule is bounded b
max ,
Simpson’s one third rule Simpson’s rule
∫ { } this is known as Simpsons one third rule or Simpsons rule.
The error in approximating an integral using Simpson’s one third rule is max ,
Simpson’s three eighth rule
∫ { 3 3 }
The error in approximating an integral using Simpson’s 3/ rule is max ,
Example Evaluate = ∫ 1. Trapezoidal rule 2. Simpson’s rule 3. Simpson’s rule Solution f(x) =
We choose number of sub division to be six i.e. n = 6 Each sub division h = 1
x = 0 f(x) = 1 = y0
x = 1 f(x) = 0.5 = y1
x = 3 f(x) = 0.1 = y3 x = 4 f(x) = 0.0588= y4 x = 5 f(x) = 0.0385= y5 x = 6 f(x) = 0.027 = y6 1. By Trapezoidal rule ∫ { } = 1.4108 2. By Simpson’s rule ∫ { y y y y … … … y y y y } = 1.3662 3. By Simpson’s rule ∫ { 3 … … … … } = 1.3735 Example
Solve by Gauss Seidal iteration method x1 + x2 + 4x3 = 9
8x1 – 3x2 + 2x3 =20
4x1 + 11x2 – x3 =33
Solution
Arrange in proper order |8| > |-3| + |2 | |11| > |4| + |-1| |4| > |1| |+ |1| 8x1 – 3x2 + 2x3 = 20 4x1 + 11x2 –x3 = 33 X1 + X2 +4x3 = 9 = { 20 + 3 - } = { 33 - + } = { 9 - - } Initial guess = = = 0 Table K 0 0 0 0 1 2.5 2.0909 1.1023 2 3.0085 2.0062 0.9963 3 3.0032 1.9985 0.9996 4 2.9995 2.0001 1.0001 5 3.000
Hence, X1 = 3.00
X2 = 2.00
X3 = 1.00
E. Solving Differential Equation
Euler’s method
Runge kutta method
Picard’s method
Taylor’s series method
Adam bashford method
Milne’s method
1. Euler method (for first order differential equation ) Given equation is y’ f x, y ; y x0) = y0
Solution is given by, Yn+1 = yn + h f(xn,yn)
Example
Solve the following differential equation by Euler’s method y’ x y , y(0) =0,
Solution f(x, y) = x + y yn+1 = yn + 0.2{ xn + yn} n xn Yn 0.2(xn+yn) 0 0.0 0.000 0.000 1 0.2 0.000 0.040 2 0.4 0.040 0.088 3 0.6 0.128 0.146 4 0.8 0.274 0.215 5 1.0 0.485
2. Runge Kutta Method
Used for finding the y at a particular x without solving the 1st order differential equation
, K1 = h f(x0, y0) K2 = h f(x0 + y0 + ) K3 = h f(x0 + y0 + ) K4 = h f(x0 +h, y0 + k3) K = (k1 + 2k2 + 2k3 + k4) Y(x0+h) = y0 + k Example
= with y(0) =1 Solution f(x, y) = , x0 = 0, y0 = 1 , h = 0.2 K1 = h f(x0, y0) = 0.2 f(0,1) =0.2 K2 = h f(x0 + , y0 + ) = 0.2 f(0.1 ,1.1) = 0.19672 K3 = h f(x0 + , y0 + ) = 0.2 f(0.1 ,1.09836) = 0.1967 K4 = hf(x0 +h, y0 +k3) = 0.2 f(0.2 ,1.1967) = 0.1891 K = (k1 + 2k2 + 2k3 +k4) = [0.2+2(0.19672) +2(0.1967) +0.1891] = 0.19599 Y(0.2) = y0 +k = 1.196 Y(0.4) = ?, x1 =0.2 , y1 =1.196 , h = 0.2 K1 = 0.1891 K2 = 0.1795 K3 = 0.1793 K4 = 0.1688 K = ( 0.1892 +2 * 0.1795 + 2 * 0.1793 + 0.1688 ) =0.1792 Y(0.4) = y1 + K = 1.196 + 0.1792 = 1.3752
In numerical integration methods to find integration we make equal slides, in a given period and then sum of all the slides, its how we find numerical integration.
For trapezoidal formula and Simpsons one third rule Simpsons three eight rule. We find some error but in exact method we find integration by formula =∫ f x dx So it is exact method.
Important Points
- In trapezoidal formula error is zero, if the function is constant or linear then exact value is equals to approximate value.
- Simpson’s one third and three eights rule, is error is zero for. o Constant
o Linear - Quadratic - Cubical functions
