Dr. Nirav Vyas numerical method 1.pdf

8/20/2019 Dr. Nirav Vyas numerical method 1.pdf

1/92

Numerical Methods

N. B. Vyas

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

Rajkot (Guj.)

N.B.V yas−Department of M athematics, AIT S − Rajkot


2/92

Introduction

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



3/92

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



4/92

Introduction




A function which is not algebraic is called a transcendental

function.



5/92

Introduction





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



6/92

Introduction






If f (x) is quadratic, cubic or bi-quadratic expression, then

algebraic formulae are available for getting the solution.



7/92

Introduction






If f (x) is quadratic, cubic or bi-quadratic expression, then

algebraic formulae are available for getting the solution.If f (x) is a higher degree polynomial or transcendentalfunction then algebraic methods are not available.


E


8/92

Errors

It is never possible to measure anything exactly.



9/92

E


10/92

Errors


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 :


Errors


11/92

Errors




i) A numerical value


Errors


12/92

Errors




i) A numerical value

ii) A degree of uncertainty Or Errors.


Errors


13/92

Errors

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


Errors


14/92

Errors


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

value.


Errors


15/92

Errors



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


Errors


16/92

Errors



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

Significant Digits: It refers to the number of digits in a

number excluding leading zeros.


Bisection Method


17/92

Bisection Method

Consider a continuous function f (x).


Bisection Method


18/92

Bisection Method


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


Bisection Method


19/92

Bisection Method



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


Bisection Method


20/92

Bisection Method




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


Bisection Method


21/92

Bisection Method





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

x1 = a + b

2 (a < x1 < b).


Bisection Method


22/92

Bisection Method






x1 = a + b

2 (a < x1 < b).

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


Bisection Method


23/92






x1 = a + b

2 (a < x1 < b).


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


24/92






x1 = a + b

2 (a < x1 < b).


Otherwise the root lies between x1 and b if f (x1) 0.Then again bisect this interval to get next point x2.


Bisection Method


25/92






x1 = a + b

2 (a < x1 < b).


Otherwise the root lies between x1 and b 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


26/92

Characteristics:

1 This method always slowly converge to a root.


Bisection Method


27/92

Characteristics:

1 This method always slowly converge to a root.2 It gives only one root at a time on the the selection of small

interval near the root.


Bisection Method


28/92

Characteristics:


interval near the root.3 In case of the multiple roots of an equation, other initial

interval can be chosen.


Bisection Method


29/92

Characteristics:


interval near the root.3 In case of the multiple roots of an equation, other initial

interval can be chosen.4 Smallest interval must be selected to obtain immediate

convergence to the root, .



30/92

Bisection Method- Example


31/92

Sol. Let f (x) = x3 − x − 1 = 0

N.B.V yas−

Department of M athematics, AIT S−

Rajkot



32/92

Sol. Let f (x) = x3 − x − 1 = 0

f (0) =

N.B.V yas−


Rajkot



33/92

Sol. Let f (x) = x3 − x − 1 = 0

f (0) = −1


34/92

Sol. Let f (x) = x3 − x − 1 = 0

f (0) = −1


35/92

Sol. Let f (x) = x3 − x − 1 = 0

f (0) = −1


36/92

Sol. Let f (x) = x3 − x − 1 = 0

f (0) = −1


37/92

Sol. Let f (x) = x3 − x − 1 = 0

f (0) = −1


38/92

Sol. Let f (x) = x3 − x − 1 = 0

f (0) = −1


39/92

Sol. Let f (x) = x3 − x − 1 = 0

f (0) = −1


40/92

Sol. Let f (x) = x3 − x − 1 = 0

f (0) = −1


41/92

Sol. Let f (x) = x3 − x − 1 = 0

f (0) = −1


42/92

Ex. Find real root of x3 − 4x + 1 = 0 correct upto four

decimal places using Bisection method

N.B.V yas−


Rajkot



43/92

Ex. Find real root of x2 − lnx − 12 = 0 correct upto

three decimal places using Bisection method

N.B.V yas−


Rajkot

Newton-Raphson Method(N-R Method)


44/92

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.

N.B.V yas−


Rajkot


G hi l d i i f h M h d


45/92



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

N.B.V yas−


Rajkot


G hi l d i ti f th M th d


46/92




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

N.B.V yas−


Rajkot




47/92




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

N.B.V yas−


Rajkot




48/92





Now ∠ACB = α, tanα = AB

AC

N.B.V yas−


Rajkot




49/92






AC

∴ f (x0) = f (x0)

x0 − x1

N.B.V yas−


Rajkot




50/92






AC

∴ f (x0) = f (x0)

x0 − x1

∴ x0 − x1 = f (x0

f

(x0)

⇒ x0 − f (x0

f

(x0)

= x1

N.B.V yas−


Rajkot




51/92






AC

∴ f (x0) = f (x0)

x0 − x1

∴ x0 − x1 = f (x0

f

(x0)

⇒ x0 − f (x0

f

(x0)

= x1

∴ x1 = x0 − f (x0)

f (x0)

N.B.V yas−


Rajkot

Newton-Raphson Method(N-R Method)Graphical derivation of the Method:


52/92






AC

∴ f (x0) = f (x0)

x0 − x1

∴ x0 − x1 = f (x0

f

(x0)

⇒ x0 − f (x0

f

(x0)

= x1

∴ x1 = x0 − f (x0)

f (x0)

In general xn+1 = xn − f (xn)

f (xn); where n = 0, 1, 2, 3, . . .

N.B.V yas−


Rajkot


53/92

Derivation of the Newton-

Raphson method. f(x)

f x0 B

y= f(x)

AC

AB=)α tan(

x1 x0

XC AαR

01 0

0

( )

( )

f x x x

f x

= −

′

00

0 1

( )'( )

f x f x

x x=

−

4

N.B.V yas−


Rajkot


54/92

Newton-Raphson method

x

f(x)

1n nn x = x - f (x )+ ′

x0

f(x1)

x2 x1 x0 X

( )0, 0 x f x

α

3

N.B.V yas−


Rajkot

N-R Method:


55/92

Advantages:

Converges Fast (if it converges).

Requires only one guess.

N.B.V yas−


Rajkot

Drawbacks:Divergence at inflection point.

S l ti f th i iti l it ti l f th t th t


56/92

Selection of the initial guess or an iteration value of the root thatis close to the inflection point of the function f (x) may start

diverging away from the root in the Newton-Raphson method.

N.B.V yas−


Rajkot

N-R Method:(Drawbacks)


57/92

Division by zero

N.B.V yas−


Rajkot


Results obtained from the N-R method may oscillate about


58/92

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.

N.B.V yas−


Rajkot



59/92

Root Jumping

N.B.V yas−


Rajkot

N-R Method


60/92

Iterative formula for finding q th root:

xq −N = 0 i.e. x = N 1q

N.B.V yas−


Rajkot

N-R Method


61/92


xq −N = 0 i.e. x = N 1q

Let f (x) = xq −N

N.B.V yas−


Rajkot


62/92

N-R Method


63/92


xq −N = 0 i.e. x = N 1q

Let f (x) = xq −N

∴ f (x) = qxq−1

Now by N-R method, xn+1 = xn − f (xn)

f (xn) = xn −

(xn)q −N

q (xn)q−1

N.B.V yas−


Rajkot

N-R Method


64/92


xq −N = 0 i.e. x = N 1q

Let f (x) = xq −N

∴ f (x) = qxq−1


f (xn) = xn −

(xn)q −N

q (xn)q−1

= xn − 1

q xn +

N

q (xn)q−1

N.B.V yas−


Rajkot

N-R Method


65/92


xq −N = 0 i.e. x = N 1q

Let f (x) = xq −N

∴ f (x) = qxq−1


f (xn) = xn −

(xn)q −N

q (xn)q−1

= xn − 1

q xn +

N

q (xn)q−1

xn+1 =

1

q

(q − 1)

xn +

N

(xn)q−1

; n

= 0,

1,

2, . . .

N.B.V yas−


Rajkot

N-R Method


66/92


xq −N = 0 i.e. x = N 1q

Let f (x) = xq −N

∴ f (x) = qxq−1


f (xn) = xn −

(xn)q −N

q (xn)q−1

= xn − 1

q xn +

N

q (xn)q−1

xn+1 =

1

q

(q −

1)xn +

N

(xn)q−1

; n

= 0,

1,

2, . . .

N.B.V yas−


Rajkot

N-R Method


67/92

Iterative formula for finding reciprocal of a positive number

N :

x = 1

N

i.e. 1

x

−N = 0

Let f (x) = 1

x −N

N.B.V yas−


Rajkot

Secant Method

In N-R method two functions f and f are required to be


68/92

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

N.B.V yas−


Rajkot

Secant Method



69/92

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 .

N.B.V yas−


Rajkot

Secant Method



70/92

R f f qevaluate per step.


Often it requires a very good initial guess.


Secant Method



71/92

f f qevaluate per step.



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

f is approximated as f

(xn) = f (xn−1) − f (xn)

xn−1 − xn


Secant Method



72/92

f f qevaluate per step.





(xn) = f (xn−1) − f (xn)

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


Secant Method



73/92

f f

evaluate per step.





(xn) = f (xn−1) − f (xn)


xn+1 = xn − f (xn)

f (xn−1)−f (xn)

xn−1−xn


Secant Method



74/92

evaluate per step.





(xn) = f (xn−1) − f (xn)


xn+1 = xn − f (xn)

f (xn−1)−f (xn)

xn−1−xn ∴ xn+1 = xn − f (xn)

xn−1 − xn

f (xn−1) − f (xn)

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


Secant Method


75/92

This method requires two initial guesses


Secant Method


76/92

This method requires two initial guesses

The two initial guesses do not need to bracket the root of the

equation, so it is not classified as a bracketing method.


Secant Method


77/92

Geometrical interpretation of Secant Method:

Consider a continuous function y = f (x)


Secant Method


78/92



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


Secant Method


79/92




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


Secant Method


80/92





From the figure ABE and DCE are similar triangles.


Secant Method


81/92






Hence AB

DC =

AE

DE ⇒

f (xn)

f (xn−1) =

xn − xn+1

xn−1 − xn+1


Secant Method


82/92






Hence AB

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


83/92

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


84/92

General Features:

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


Secant Method


85/92

General Features:


It requires fewer function evaluations.


Secant Method


86/92

General Features:



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


Secant Method


87/92

General Features:




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


Secant Method

G l F


88/92

General Features:





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


Secant Method

G l F t


89/92

General Features:





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

This method requires only one function evaluation per iteration.


Secant Method


90/92

Disadvantages:

There is no guaranteed error bound for the computed value.


Secant Method


91/92

Disadvantages:


It is likely to difficulty of f (x) = 0. This means X −axis is

tangent to the graph of y = f (x)


Secant Method


92/92

Disadvantages:


It is likely to difficulty of f (x) = 0. This means X −axis is

tangent to the graph of y = f (x)Method may converge very slowly or not at all.


Dr. Nirav Vyas numerical method 1.pdf

Documents

Transcript of Dr. Nirav Vyas numerical method 1.pdf