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 M athematics, AIT S − Rajkot

Introduction

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


Introduction

An algebraic function is informally a function that
satisﬁes a polynomial equation


Introduction

A function which is not algebraic is called a transcendental
function.


Introduction

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


Introduction

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


Introduction

function.
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 transcendental
function then algebraic methods are not available.


Errors

It is never possible to measure anything exactly.


Errors

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


Errors

The result of any physical measurement has two essential
components :


Errors

components :
i) A numerical value


Errors

components :
i) A numerical value
ii) A degree of uncertainty Or Errors.


Errors

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


Errors

Approximate Numbers: These are the numbers which
represent a certain degree of accuracy but not the exact
value.


Errors

value.
These numbers cannot be represented in terms of ﬁnite
number of digits.


Errors

value.
These numbers cannot be represented in terms of ﬁnite
number of digits.
Signiﬁcant Digits: It refers to the number of digits in a
number excluding leading zeros.


Bisection Method

Consider a continuous function f (x).


Bisection Method

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


Bisection Method

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


Bisection Method

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


Bisection Method

Now according to Bisection method, bisect the interval [a, b],
a+b
x1 = (a < x1 < b).
2


Bisection Method

a+b
x1 = (a < x1 < b).
2
If f (x1 ) = 0 then x1 be the root of the given equation.


Bisection Method

a+b
x1 = (a < x1 < b).
2
Otherwise the root lies between x1 and b if f (x1 ) < 0.


Bisection Method

a+b
x1 = (a < x1 < b).
2
OR the root lies between a and x1 if f (x1 ) > 0.
Then again bisect this interval to get next point x2 .


Bisection Method

a+b
x1 = (a < x1 < b).
2
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 the
root upto desired accuracy is obtained.

Bisection Method

Characteristics:
1 This method always slowly converge to a root.


Bisection Method

Characteristics:

2 It gives only one root at a time on the the selection of small

interval near the root.


Bisection Method

Characteristics:


3 In case of the multiple roots of an equation, other initial

interval can be chosen.


Bisection Method

Characteristics:


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, .


Bisection Method- Example

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



Sol. Let f (x) = x3 − x − 1 = 0



Sol. Let f (x) = x3 − x − 1 = 0
f (0) =



Sol. Let f (x) = x3 − x − 1 = 0
f (0) = −1 < 0



Sol. Let f (x) = x3 − x − 1 = 0
f (0) = −1 < 0
f (1) =



Sol. Let f (x) = x3 − x − 1 = 0
f (0) = −1 < 0
f (1) = −1 < 0



Sol. Let f (x) = x3 − x − 1 = 0
f (0) = −1 < 0
f (1) = −1 < 0
f (2) =



Sol. Let f (x) = x3 − x − 1 = 0
f (0) = −1 < 0
f (1) = −1 < 0
f (2) = 5 > 0



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 the
lying in the interval (1, 2)



Sol. Let f (x) = x3 − x − 1 = 0
f (0) = −1 < 0
f (1) = −1 < 0
f (2) = 5 > 0
Now according to Bisection method, the next approximation
is obtained by taking the midpoint of (1, 2)



Sol. Let f (x) = x3 − x − 1 = 0
f (0) = −1 < 0
f (1) = −1 < 0
f (2) = 5 > 0
1+2
c= = 1.5, f (1.5) =
2



Sol. Let f (x) = x3 − x − 1 = 0
f (0) = −1 < 0
f (1) = −1 < 0
f (2) = 5 > 0
1+2
c= = 1.5, f (1.5) =
2
a+b
No. of a b c= f(c)
2
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 four
decimal places using Bisection method



Ex. Find real root of x2 − lnx − 12 = 0 correct upto
three 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 crosses
X − axis at R corresponding to the equation f (x) = 0.


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


guess x0 at A.
The tangent at B cuts the X − axis at C which gives ﬁrst
approximation x1 .


guess x0 at A.
approximation x1 . Thus AB = f (x0 ) and AC = x0 − x1


guess x0 at A.
AB
Now ∠ACB = α, tanα =
AC


guess x0 at A.
AB
AC
f (x0 )
∴ f (x0 ) =
x0 − x1


guess x0 at A.
AB
AC
f (x0 )
∴ f (x0 ) =
x0 − x1
f (x0 f (x0
∴ x0 − x1 = ⇒ x0 − = x1
f (x0 ) f (x0 )


guess x0 at A.
AB
AC
f (x0 )
∴ f (x0 ) =
x0 − x1
f (x0 f (x0
∴ x0 − x1 = ⇒ x0 − = x1
f (x0 ) f (x0 )
f (x0 )
∴ x1 = x0 −
f (x0 )


guess x0 at A.
AB
AC
f (x0 )
∴ f (x0 ) =
x0 − x1
f (x0 f (x0
∴ x0 − x1 = ⇒ x0 − = x1
f (x0 ) f (x0 )
f (x0 )
∴ x1 = x0 −
f (x0 )
f (xn )
In general xn+1 = xn − ; where n = 0, 1, 2, 3, . . .
f (xn )

Derivation of the Newton-
Raphson method.
f(x)

y= f(x)
AB
tan(α ) =
f(x0 ) B AC

f ( x0 )
f '( x 0) =
x0 − x1

f ( x0 )
R C α A x1 = x0 −
X f ′( x0 )
x1 x0

4


Newton-Raphson method

f(x)

f(x0 ) f(xn )
 x0, f ( x0 ) xn +1 = xn -
  f ′ (xn )

f(x 1)
α
x2 x1 x0 X

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 that
is close to the inﬂection point of the function f (x) may start
diverging 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 about
the local maximum or minimum without converging on a
root 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 q th root:
1
xq − N = 0 i.e. x = N q


N-R Method

1
xq − N = 0 i.e. x = N q
Let f (x) = xq − N


N-R Method

1
xq − N = 0 i.e. x = N q
∴ f (x) = qxq−1


N-R Method

1
xq − N = 0 i.e. x = N q
∴ f (x) = qxq−1
f (xn ) (xn )q − N
Now by N-R method, xn+1 = xn − = xn −
f (xn ) q(xn )q−1


N-R Method

1
xq − N = 0 i.e. x = N q
∴ f (x) = qxq−1
f (xn ) (xn )q − N
f (xn ) q(xn )q−1
1 N
= xn − xn +
q q(xn )q−1


N-R Method

1
xq − N = 0 i.e. x = N q
∴ f (x) = qxq−1
f (xn ) (xn )q − N
f (xn ) q(xn )q−1
1 N
= xn − xn +
q q(xn )q−1
1 N
xn+1 = (q − 1)xn + ; n = 0, 1, 2, . . .
q (xn )q−1


N-R Method

Iterative formula for ﬁnding reciprocal of a positive number
N:
1 1
x= i.e. − N = 0
N x
1
Let f (x) = − N
x


Secant Method

In N-R method two functions f and f are required to be
evaluate per step.


Secant Method

evaluate per step.
Also it requires to evaluate derivative of f and sometimes it is
very complicated to evaluate f .


Secant Method

evaluate per step.
Often it requires a very good initial guess.


Secant Method

evaluate per step.
To overcome these drawbacks, the derivative of f of the function
f (xn−1 ) − f (xn )
f is approximated as f (xn ) =
xn−1 − xn


Secant Method

evaluate per step.
f (xn−1 ) − f (xn )
xn−1 − xn
Therefore formula of N-R method becomes


Secant Method

evaluate per step.
f (xn−1 ) − f (xn )
xn−1 − xn
f (xn )
xn+1 = xn −
f (xn−1 )−f (xn )
xn−1 −xn


Secant Method

evaluate per step.
f (xn−1 ) − f (xn )
xn−1 − xn
f (xn )
xn+1 = xn −
f (xn−1 )−f (xn )
xn−1 −xn
xn−1 − xn
∴ xn+1 = xn − f (xn )
f (xn−1 ) − f (xn )
where n = 1, 2, 3, . . ., f (xn−1 ) = f (xn )


Secant Method

This method requires two initial guesses


Secant Method

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

Geometrical interpretation of Secant Method:
Consider a continuous function y = f (x)


Secant Method

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


Secant Method

(xn−1 , f (xn−1 ))
Take the x−coordinate of intersection with X−axis as xn+1


Secant Method

(xn−1 , f (xn−1 ))
From the ﬁgure ABE and DCE are similar triangles.


Secant Method

(xn−1 , f (xn−1 ))
AB AE f (xn ) xn − xn+1
Hence = ⇒ =
DC DE f (xn−1 ) xn−1 − xn+1


Secant Method

(xn−1 , f (xn−1 ))
AB AE f (xn ) xn − xn+1
Hence = ⇒ =
DC DE f (xn−1 ) xn−1 − xn+1
xn−1 − xn
∴ xn+1 = xn − f (xn )
f (xn−1 ) − f (xn )


Secant Method

NOTE:
Do not combine the secant formula and write it in the form as
follows because it has enormous loss of signiﬁcance errors
xn f (xn−1 ) − xn−1 f (xn )
xn+1 =
f (xn−1 ) − f (xn )


Secant Method

General Features:
The secant method is an open method and may not converge.


Secant Method

General Features:
It requires fewer function evaluations.


Secant Method

General Features:
In some problems the secant method will work when Newton’s
method does not and vice-versa.


Secant Method

General Features:
The method is usually a bit slower than Newton’s method. It is
more rapidly convergent than the bisection method.


Secant Method

General Features:
This method does not require use of the derivative of the
function.


Secant Method

General Features:
This method does not require use of the derivative of the
function.
This method requires only one function evaluation per iteration.


Secant Method

Disadvantages:
There is no guaranteed error bound for the computed value.


Secant Method

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

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.


Numerical Methods 1

More Related Content

What's hot (20)

Viewers also liked (16)

Similar to Numerical Methods 1 (20)

More from Dr. Nirav Vyas (20)

Recently uploaded (20)

Numerical Methods 1