Solution of Non Linear Equations

Numerical Solutions of Non- Linear equations

Raju Sharma Assistant Professor Department of Civil Engineering Chandigarh University

Introduction of polynomial

If f(x) is a quadratic or cubic equation then

we can find the root of the equation f(x)

=0

(1)2x2-8x+6=0

(2)2x3+x2-13x+5=0

f(x) =a0xn + a1xn-1 + …. + an-1x +an

• When f(x) is a polynomial of higher degree or an

expression involving transcendental functions

(equation contains trigonometric, logarithmic,

exponential) e.g. 1+cos-5x, xtanx-cosx, e-x – sin x

etc, algebraic methods are not available.

• To find the roots of such types of equations,

numerical methods are used.

Following are the Numerical Methods

1. Bisection Method2. Regula - falsi Method or method of false

position3. Secant Method4. Newton-Raphson Method5. Iteration Method

Bisection Method

• This method is based on the

theorem:

• If a function f(x) is continuous b/w a

& b, and f(a) , f(b) are of opposite

signs then there exist at least one

root b/w a & b.

• Suppose that a root of f(x)=0 lies in

the interval (a, b) i.e. f(a) f(b)<0. we

bisect this interval and obtain

• x1 = (a+b)/2

• If f(x1)=0, then x1 is a root of f(x)=0

• Otherwise root lies b/w a and x1 or

x1 & b

Y =

f (x

)

Y

a x2

b x1x3 x0

Then, we bisect the interval as

before and continue the

process until the root is found

to desired accuracy.

• Then the second approximation to the root is x2 =

(a+x1)/2

• We have f(x2)<0

• and f(x2) f(x1) <0

• Root lies in (x2, x1)

• If f(x2) is –ve, The root lies between x1 & x2 . Then

the third approximation to the root is

x3 = (x1+ x2 )/2

and so on…..

Q-1 Find a root of the equation x3-4x-9=0,using the bisection method correct the three decimal places• f(x) =x3-4x-9 ……(1)

• f(2)=23-4 x 2 -9=-9

• f(3)= 33-4 x 3 -9=6

• Hence root lies b/w 2 and 3

• First approximation to the

root is x1=(2+3)/2 = 2.5

• Then put x1 in eq. 1

• f(2.5)= 2.53-4 x 2.5-9=-

3.375< 0

• f(x1) f(3) < 0

• root lies b/w x1 and 3

• x2 = (2.5+3)/2 =2.75

• Then put x2 in eq. (1)

• f(2.75)= 2.753-4 x 2.75-

9=0.796 > 0

• f(x1) f(x2) < 0

• Root lies between x1 & x2

• Thus, the third approximation to the root is

• x 3 = (x1 + x2 )/2 = 2.625

• f(2.625)= 2.6253-4 x 2.625-9=-1.4121 <0

• And f(x2) f(x3) < 0

• Root lies b/w x2 & x3

Thus, the fourth approximation to the root is

x4 = (x2 + x3 )/2 = 2.6875

• By repeating the process, the successive approximation are

• x5 = 2.71875

• x6 = 2.70313

• x7 = 2.71094

• x8 = 2.70703

• x9 = 2.70508

• x10 = 2.70605

• x11 = 2.70654

• x12 = 2.70642

• Hence the required root is 2.706

Regula - falsi Method or method of false position

• Choose two points x0 & x1,

such that f(x0) & f(x1) are of

opposite signs i.e., f(x0) f(x1)

< 0

• The graph of y= f(x) crosses

the x-axis b/w these points.

• This method consist in

replacing the curve AB by the

chord AB and taking the point

of intersection of the chord

with the x-axis as an

approximation to the root

Y

XO

A[x0, f(x0)]

P(x)

x1x2x3

x0

B[x1, f(x1)]

• Equation of the chord joining the

points A[x0, f(x0)] and B[x1, f(x1)] is

• y-f(xo) =((f(x1)- f(xo)/(x1-xo))/ (x-xo)

• The abscissa of the point where the

chord cuts the x-axis (y=0) is given by

• x =x2=(xo – (x1-xo)/(f(x1)- f(xo)) f(xo)

---- (1)

• Which is the approximation to the root.

• Or x2 = (xo f(x1)-x1 f(xo))/( f(x1)- f(xo))

• If now f(xo) and f(x2) are of opposite

signs, then the root lies b/w xo and x2

• So replacing x1 by x2 and find the new

root x3

•Y

•X•O

•A[x0, f(x0)]

•P(x)

•x1

•x2

•x3

•x0

•B[x1, f(x1)]

• The procedure is repeated till the root is found to

desired accuracy. The iteration process based on

equation 1 is known as Regula Falsi Method.

• Note :- if the root lies initially in (xo, x1), then one

of the end points is fixed for all iterations.

• For example, if xo is fixed, then the method is

of the form

• xk+1= xo f(xk)- xk f(xo)/ f(xk) - f(xo)

• K=1,2,3,………

• Find a real root of the equation x3-2x-5=0, by regula –falsi method correct to three decimal places •Solution:-

•A real root of the equation f(x) = x3-5x+1=0 lies in the interval (0, 1). Perform four iterations of regular-falsi method to obtain this root

Solution:-

Secant method • This method is an improvement over the method

of false position (or Regula-falsi method ) as it

does not require the condition

• f(xo) x f(x1) < 0

• In this method two most recent approximations

to the root are used to find the next

approximation

• Let xk-1, xk be two

approximations to the root

of f(x)=0. Then p (xk-1 , f(xk-

1)) Q (xk, f (xk)) are points

on the curve y=f(x)

• the equation of straight

line PQ is given by

• y- f(xk) = (f(xk-1)- f(xk)/(xk-1-xk)) (x-

xk)

Substitute y=o, and solving

for x, we get

p((xk-1, f(xk-

1))

a(xk , f(xk))

xk+1 x

yo

•x=xk-((xk-xk-1)/f(xk)-f(xk-1)) x f(xk)

•The next approximation, xk+1 , to the root

is written as

•xk+1=(xk-1 f(xk) – xk f(xk-1))/ f(xk) - f(xk-1)

•k = 1,2,3……

Find a root of the equation x3-2x-5=0 using secant method correct to three decimal places •Solution:-

Drawback of secant method

• If at any intersection f(xn)= f(xn-1), this

method fails and shows that it does not

converge necessarily. This is a drawback

of secant method over the method of false

position which always converge. But if the

secant method once converges, its rate of

convergence is 1.6 which is faster that

that of the method of false position.

Newton-Raphson Method

•Let x0 be an approximate root of the

equation f(x)=0. if x1 = x0+h be the exact

root then f(x1)=0

•Expanding f(x0+h) by Talor’s series

• f(x0) +h f ´(x0) +(h2 /2!) f ´´ (x0) +_ _ _ _=0

•Since h is very small, neglecting h2 & higher

power of h, we get

• f(x0) +h f ´(x0) =0 h = -(f(x0)/ f ´(x0))

•We have x1 = x0+h

• x1 = x0 -(f(x0)/ f ´(x0))

•Successive approximation are given by x2, x3, _ _ _ _, xn+1

•Where •xn+1 = x0 -(f(xn)/ f ´(xn))

•Which is known as Newton-Raphson formula

Find the positive root of x4-x=10 correct to three decimal places, using newton - raphson method •Solution:-

Some deduction from Newton-Rapson formula

• Iterative formula to find 1/N is xn+1 = xn (2-Nxn)

• Iterative formula to find √N is xn+1 = ½ (xn

+N/xn)

• Iterative formula to find 1/ √N is xn+1 = ½ (xn

+1/Nxn )

• Iterative formula to find 1/ √N is xn+1 =1/ k [(k-

1) xn+ N/xnk-1)

k

• Let x = 1/N or (1/x)-N =0• Take f(x) = (1/x)-N where f(x) = -(1/x2)• Newton’s formula gives

• xn+1 = xn – f(xn)/f’ (xn)

• = xn – ((1/ xn) –N)/ -(1/xn2 )

• = xn + ((1/ xn) –N) xn2

•

• = xn + xn –N xn2

• = 2xn –N xn2

•

• xn(2-N xn)

• Hence Proved

Evaluate 1/31 correct to four decimal places by newton/s iteration method

•Solution:-

Iteration Method

• To find the root of the equation f(x) =0 …..(i)

• By successive approximation, we rewrite (i) in

the form x=Ф (x).

x

y

Y =x

Y =Ф (x)

x0 x2 x3 x1o

• Let x = xo be an initial approximation of the

desired root α

• then the first approximation x1 is given by

• x1=Ф (x0)

• Now treating x1 as the initial value, the second

approximation is x2=Ф (x1)

• Processing in this way, the nth approximation is

xn=Ф (xn-1)

Sufficient condition for the convergence of iterations

• If

I. α is a root of f(x)=0 which is equivalent to x=Ф (x)

II. I, be any interval containing the point x= α

III. | Ф´ (x)|< 1 for all x lies b/w I

Then the sequence of approximations x0, x1, x2,_

_ _ _,xn all converge to the root α provided the

initial approximation x0 in chosen I

Find the real root of the equation Cosx=3x-1.correct to three decimal places using iteration method.•Solution

Rate of convergence • Let x0,x1,x2, _ _ _ _ be the value of root (α) of an

equation at the o , 1st ,2nd ……… iterations while

its actual value is 3.5567

• The value of this root calculated by three

different methods are as given below:-Root 1st Method 2nd Method 3rd method

x0 5 5 5

x1 5.6 3.8527 3.8327

x2 6.4 3.5693 3.56834

x3 8.3 3.55798 3.55743

x4 9.7 3.55687 3.55672

x5 10.6 3.55676

x6 11.9 3.55671

Rate of convergence

Bisection Method 1/2

Regula-falsi Method 1.62

Secent method 1.6

Newton Rapson 2

The fastness of convergence in any method is represented by its rate of convergence