Numerical Methods 1

Numerical Methods

N. B. Vyas

Department of Mathematics,Atmiya Institute of Tech. and Science,

Rajkot (Guj.)

N.B.V yas − Department of Mathematics, AITS − Rajkot

Introduction

There are two types of functions: (i) Algebraic function and(ii) Transcendental function

An algebraic function is informally a function thatsatisfies a polynomial equation

A function which is not algebraic is called a transcendentalfunction.

The values of x which satisfy the equation f(x) = 0 arecalled roots of f(x).

If f(x) is quadratic, cubic or bi-quadratic expression, thenalgebraic formulae are available for getting the solution.

If f(x) is a higher degree polynomial or transcendentalfunction then algebraic methods are not available.


Introduction








Introduction








Introduction








Introduction








Introduction








Errors

It is never possible to measure anything exactly.

So in order to make valid conclusions, it is good to make theerror as small as possible.

The result of any physical measurement has two essentialcomponents :

i) A numerical value

ii) A degree of uncertainty Or Errors.


Errors







Errors







Errors







Errors







Errors

Exact Numbers: There are the numbers in which there isno uncertainty and no approximation.

Approximate Numbers: These are the numbers whichrepresent a certain degree of accuracy but not the exactvalue.

These numbers cannot be represented in terms of finitenumber of digits.

Significant Digits: It refers to the number of digits in anumber excluding leading zeros.


Errors






Errors






Errors






Bisection Method

Consider a continuous function f(x).

Numbers a < b such that f(a) and f(b) have opposite signs.

Let f(a) be negative and f(b) be positive for [a, b].

Then there exists at least one point(root), say x, a < x < bsuch that f(x) = 0.

Now according to Bisection method, bisect the interval [a, b],

x1 =a + b

2(a < x1 < b).

If f(x1) = 0 then x1 be the root of the given equation.

Otherwise the root lies between x1 and b if f(x1) < 0.

OR the root lies between a and x1 if f(x1) > 0.

Then again bisect this interval to get next point x2.

Repeat the above procedure to generate x1, x2, . . . till theroot upto desired accuracy is obtained.


Bisection Method






x1 =a + b

2(a < x1 < b).







Bisection Method






x1 =a + b

2(a < x1 < b).







Bisection Method






x1 =a + b

2(a < x1 < b).







Bisection Method






x1 =a + b

2(a < x1 < b).







Bisection Method






x1 =a + b

2(a < x1 < b).







Bisection Method






x1 =a + b

2(a < x1 < b).







Bisection Method






x1 =a + b

2(a < x1 < b).







Bisection Method






x1 =a + b

2(a < x1 < b).







Bisection Method

Characteristics:

1 This method always slowly converge to a root.

2 It gives only one root at a time on the the selection of smallinterval near the root.

3 In case of the multiple roots of an equation, other initialinterval can be chosen.

4 Smallest interval must be selected to obtain immediateconvergence to the root, .


Bisection Method

Characteristics:






Bisection Method

Characteristics:






Bisection Method

Characteristics:






Bisection Method- Example

Ex. Find real root of x3 − x− 1 = 0 correct upto threedecimal places using Bisection method



Sol. Let f(x) = x3 − x− 1 = 0

f(0) = −1 < 0

f(1) = −1 < 0

f(2) = 5 > 0

∴ since f(x) is continuous function there must be a root in thelying in the interval (1, 2)

Now according to Bisection method, the next approximationis obtained by taking the midpoint of (1, 2)

c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Sol. Let f(x) = x3 − x− 1 = 0

f(0) =

−1 < 0

f(1) = −1 < 0

f(2) = 5 > 0



c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Sol. Let f(x) = x3 − x− 1 = 0

f(0) = −1 < 0

f(1) = −1 < 0

f(2) = 5 > 0



c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Sol. Let f(x) = x3 − x− 1 = 0

f(0) = −1 < 0

f(1) =

−1 < 0

f(2) = 5 > 0



c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Sol. Let f(x) = x3 − x− 1 = 0

f(0) = −1 < 0

f(1) = −1 < 0

f(2) = 5 > 0



c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Sol. Let f(x) = x3 − x− 1 = 0

f(0) = −1 < 0

f(1) = −1 < 0

f(2) =

5 > 0



c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Sol. Let f(x) = x3 − x− 1 = 0

f(0) = −1 < 0

f(1) = −1 < 0

f(2) = 5 > 0



c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Sol. Let f(x) = x3 − x− 1 = 0

f(0) = −1 < 0

f(1) = −1 < 0

f(2) = 5 > 0



c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Sol. Let f(x) = x3 − x− 1 = 0

f(0) = −1 < 0

f(1) = −1 < 0

f(2) = 5 > 0



c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Sol. Let f(x) = x3 − x− 1 = 0

f(0) = −1 < 0

f(1) = −1 < 0

f(2) = 5 > 0



c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Sol. Let f(x) = x3 − x− 1 = 0

f(0) = −1 < 0

f(1) = −1 < 0

f(2) = 5 > 0



c =1 + 2

2= 1.5, f(1.5) =

No. of a b c =a+ b

2f(c)

iterations (f(a) < 0) (f(b) > 0) (< 0, > 0)

1 1 2 1.5 -



Ex. Find real root of x3 − 4x + 1 = 0 correct upto fourdecimal places using Bisection method



Ex. Find real root of x2 − lnx− 12 = 0 correct uptothree decimal places using Bisection method


Newton-Raphson Method(N-R Method)

Graphical derivation of the Method:

Consider the portion of the graph y = f(x) which crossesX − axis at R corresponding to the equation f(x) = 0.

Let B be the point on the curve corresponding to the initialguess x0 at A.

The tangent at B cuts the X − axis at C which gives firstapproximation x1. Thus AB = f(x0) and AC = x0 − x1Now ∠ACB = α, tanα =

AB

AC

∴ f ′(x0) =f(x0)

x0 − x1∴ x0 − x1 =

f(x0f ′(x0)

⇒ x0 −f(x0f ′(x0)

= x1

∴ x1 = x0 −f(x0)

f ′(x0)

In general xn+1 = xn −f(xn)

f ′(xn); where n = 0, 1, 2, 3, . . .







AB

AC

∴ f ′(x0) =f(x0)

x0 − x1∴ x0 − x1 =

f(x0f ′(x0)

⇒ x0 −f(x0f ′(x0)

= x1

∴ x1 = x0 −f(x0)

f ′(x0)


f ′(xn); where n = 0, 1, 2, 3, . . .






The tangent at B cuts the X − axis at C which gives firstapproximation x1.

Thus AB = f(x0) and AC = x0 − x1Now ∠ACB = α, tanα =

AB

AC

∴ f ′(x0) =f(x0)

x0 − x1∴ x0 − x1 =

f(x0f ′(x0)

⇒ x0 −f(x0f ′(x0)

= x1

∴ x1 = x0 −f(x0)

f ′(x0)


f ′(xn); where n = 0, 1, 2, 3, . . .






The tangent at B cuts the X − axis at C which gives firstapproximation x1. Thus AB = f(x0) and AC = x0 − x1

Now ∠ACB = α, tanα =AB

AC

∴ f ′(x0) =f(x0)

x0 − x1∴ x0 − x1 =

f(x0f ′(x0)

⇒ x0 −f(x0f ′(x0)

= x1

∴ x1 = x0 −f(x0)

f ′(x0)


f ′(xn); where n = 0, 1, 2, 3, . . .







AB

AC

∴ f ′(x0) =f(x0)

x0 − x1∴ x0 − x1 =

f(x0f ′(x0)

⇒ x0 −f(x0f ′(x0)

= x1

∴ x1 = x0 −f(x0)

f ′(x0)


f ′(xn); where n = 0, 1, 2, 3, . . .







AB

AC

∴ f ′(x0) =f(x0)

x0 − x1

∴ x0 − x1 =f(x0f ′(x0)

⇒ x0 −f(x0f ′(x0)

= x1

∴ x1 = x0 −f(x0)

f ′(x0)


f ′(xn); where n = 0, 1, 2, 3, . . .







AB

AC

∴ f ′(x0) =f(x0)

x0 − x1∴ x0 − x1 =

f(x0f ′(x0)

⇒ x0 −f(x0f ′(x0)

= x1

∴ x1 = x0 −f(x0)

f ′(x0)


f ′(xn); where n = 0, 1, 2, 3, . . .







AB

AC

∴ f ′(x0) =f(x0)

x0 − x1∴ x0 − x1 =

f(x0f ′(x0)

⇒ x0 −f(x0f ′(x0)

= x1

∴ x1 = x0 −f(x0)

f ′(x0)


f ′(xn); where n = 0, 1, 2, 3, . . .







AB

AC

∴ f ′(x0) =f(x0)

x0 − x1∴ x0 − x1 =

f(x0f ′(x0)

⇒ x0 −f(x0f ′(x0)

= x1

∴ x1 = x0 −f(x0)

f ′(x0)


f ′(xn); where n = 0, 1, 2, 3, . . .


Derivation of the Newton-Raphson method.

f(x)

f(x0) B

y= f(x)

AC

AB=)αtan(

f(x0)

x1 x0

X

B

C A α R 0

1 0

0

( )

( )

f xx x

f x= −

′

00

0 1

( )'( )

f xf x

x x=

−

AC

4


Newton-Raphson method

nf(x )

x = x -

f(x)

f(x )

( ) 1

n

n n

n

f(x )x = x -

f (x )+

′ f(x0)

f(x1)

x2 x1 x0 X

( )0, 0x f x

α

3


N-R Method:

Advantages:

Converges Fast (if it converges).

Requires only one guess.


Drawbacks:

Divergence at inflection point.

Selection of the initial guess or an iteration value of the root thatis close to the inflection point of the function f(x) may startdiverging away from the root in the Newton-Raphson method.


N-R Method:(Drawbacks)

Division by zero



Results obtained from the N-R method may oscillate aboutthe local maximum or minimum without converging on aroot but converging on the local maximum or minimum.For example for f(x) = x2 + 2 = 0 the equation has no real roots.



Root Jumping


N-R Method

Iterative formula for finding qth root:

xq −N = 0 i.e. x = N1q

Let f(x) = xq −N∴ f ′(x) = qxq−1

Now by N-R method, xn+1 = xn −f(xn)

f ′(xn)= xn −

(xn)q −Nq(xn)q−1

= xn −1

qxn +

N

q(xn)q−1

xn+1 =1

q

[(q − 1)xn +

N

(xn)q−1

]; n = 0, 1, 2, . . .


N-R Method


xq −N = 0 i.e. x = N1q

Let f(x) = xq −N

∴ f ′(x) = qxq−1


f ′(xn)= xn −


= xn −1

qxn +

N

q(xn)q−1

xn+1 =1

q

[(q − 1)xn +

N

(xn)q−1

]; n = 0, 1, 2, . . .


N-R Method


xq −N = 0 i.e. x = N1q



f ′(xn)= xn −


= xn −1

qxn +

N

q(xn)q−1

xn+1 =1

q

[(q − 1)xn +

N

(xn)q−1

]; n = 0, 1, 2, . . .


N-R Method


xq −N = 0 i.e. x = N1q



f ′(xn)= xn −


= xn −1

qxn +

N

q(xn)q−1

xn+1 =1

q

[(q − 1)xn +

N

(xn)q−1

]; n = 0, 1, 2, . . .


N-R Method


xq −N = 0 i.e. x = N1q



f ′(xn)= xn −


= xn −1

qxn +

N

q(xn)q−1

xn+1 =1

q

[(q − 1)xn +

N

(xn)q−1

]; n = 0, 1, 2, . . .


N-R Method


xq −N = 0 i.e. x = N1q



f ′(xn)= xn −


= xn −1

qxn +

N

q(xn)q−1

xn+1 =1

q

[(q − 1)xn +

N

(xn)q−1

]; n = 0, 1, 2, . . .


N-R Method


xq −N = 0 i.e. x = N1q



f ′(xn)= xn −


= xn −1

qxn +

N

q(xn)q−1

xn+1 =1

q

[(q − 1)xn +

N

(xn)q−1

]; n = 0, 1, 2, . . .


N-R Method

Iterative formula for finding reciprocal of a positive numberN :

x =1

Ni.e.

1

x−N = 0

Let f(x) =1

x−N


Secant Method

In N-R method two functions f and f ′ are required to beevaluate per step.

Also it requires to evaluate derivative of f and sometimes it isvery complicated to evaluate f ′.

Often it requires a very good initial guess.

To overcome these drawbacks, the derivative of f ′ of the function

f is approximated as f ′(xn) =f(xn−1)− f(xn)

xn−1 − xnTherefore formula of N-R method becomes

xn+1 = xn −f(xn)(

f(xn−1)−f(xn)xn−1−xn

)∴ xn+1 = xn − f(xn)

xn−1 − xnf(xn−1)− f(xn)

where n = 1, 2, 3, . . ., f(xn−1) 6= f(xn)


Secant Method







xn+1 = xn −f(xn)(


)∴ xn+1 = xn − f(xn)


where n = 1, 2, 3, . . ., f(xn−1) 6= f(xn)


Secant Method







xn+1 = xn −f(xn)(


)∴ xn+1 = xn − f(xn)


where n = 1, 2, 3, . . ., f(xn−1) 6= f(xn)


Secant Method






xn−1 − xn

Therefore formula of N-R method becomes

xn+1 = xn −f(xn)(


)∴ xn+1 = xn − f(xn)


where n = 1, 2, 3, . . ., f(xn−1) 6= f(xn)


Secant Method







xn+1 = xn −f(xn)(


)∴ xn+1 = xn − f(xn)


where n = 1, 2, 3, . . ., f(xn−1) 6= f(xn)


Secant Method







xn+1 = xn −f(xn)(


)

∴ xn+1 = xn − f(xn)xn−1 − xn

f(xn−1)− f(xn)

where n = 1, 2, 3, . . ., f(xn−1) 6= f(xn)


Secant Method







xn+1 = xn −f(xn)(


)∴ xn+1 = xn − f(xn)


where n = 1, 2, 3, . . ., f(xn−1) 6= f(xn)


Secant Method

This method requires two initial guesses

The two initial guesses do not need to bracket the root of theequation, so it is not classified as a bracketing method.


Secant Method

This method requires two initial guesses

The two initial guesses do not need to bracket the root of theequation, so it is not classified as a bracketing method.


Secant Method

Geometrical interpretation of Secant Method:

Consider a continuous function y = f(x)

Draw the straight line through the points (xn, f(xn)) and(xn−1, f(xn−1))

Take the x−coordinate of intersection with X−axis as xn+1

From the figure ABE and DCE are similar triangles.

HenceAB

DC=AE

DE⇒ f(xn)

f(xn−1)=

xn − xn+1

xn−1 − xn+1

∴ xn+1 = xn − f(xn)xn−1 − xn

f(xn−1)− f(xn)


Secant Method






HenceAB

DC=AE

DE⇒ f(xn)

f(xn−1)=

xn − xn+1

xn−1 − xn+1

∴ xn+1 = xn − f(xn)xn−1 − xn

f(xn−1)− f(xn)


Secant Method






HenceAB

DC=AE

DE⇒ f(xn)

f(xn−1)=

xn − xn+1

xn−1 − xn+1

∴ xn+1 = xn − f(xn)xn−1 − xn

f(xn−1)− f(xn)


Secant Method






HenceAB

DC=AE

DE⇒ f(xn)

f(xn−1)=

xn − xn+1

xn−1 − xn+1

∴ xn+1 = xn − f(xn)xn−1 − xn

f(xn−1)− f(xn)


Secant Method






HenceAB

DC=AE

DE⇒ f(xn)

f(xn−1)=

xn − xn+1

xn−1 − xn+1

∴ xn+1 = xn − f(xn)xn−1 − xn

f(xn−1)− f(xn)


Secant Method






HenceAB

DC=AE

DE⇒ f(xn)

f(xn−1)=

xn − xn+1

xn−1 − xn+1

∴ xn+1 = xn − f(xn)xn−1 − xn

f(xn−1)− f(xn)


Secant Method

NOTE:

Do not combine the secant formula and write it in the form asfollows because it has enormous loss of significance errors

xn+1 =xnf(xn−1)− xn−1f(xn)

f(xn−1)− f(xn)


Secant Method

General Features:

The secant method is an open method and may not converge.

It requires fewer function evaluations.

In some problems the secant method will work when Newton’smethod does not and vice-versa.

The method is usually a bit slower than Newton’s method. It ismore rapidly convergent than the bisection method.

This method does not require use of the derivative of thefunction.

This method requires only one function evaluation per iteration.


Secant Method

General Features:








Secant Method

General Features:








Secant Method

General Features:








Secant Method

General Features:








Secant Method

General Features:








Secant Method

Disadvantages:

There is no guaranteed error bound for the computed value.

It is likely to difficulty of f ′(x) = 0. This means X−axis istangent to the graph of y = f(x)

Method may converge very slowly or not at all.


Secant Method

Disadvantages:





Secant Method

Disadvantages:





Education

Numerical Methods 1