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NUMERICAL METHODS

Solutions of Transcendental Equations

US03CMTH21

By: Dr. Snehal Shukla

BASIC CONCEPT OF � NUMERICAL METHODS:

• Numerical methods are concerned with all aspects of the numerical solution of a problem, from the theoretical development and understanding of numerical methods to their practical implementation. Most numerical analysts specialize in small subfields, but they share some common concerns, perspectives, and mathematical methods of analysis.

What are numerical methods?

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is a root finding method which repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing.

It is very simple and robust method.

Draw Back of the method:

It is a relatively slow and lengthy process

The Bi-section method in

numerical methods

Exa. Find the missing term of the following data.

(1)

S0l. Here we have four values. So, we need to go

up-to ∆4 in Binomial expansion .

(2)

(3)

S0l. Here we have five values of y. So, the fourth difference vanish, i.e. ∆4=0 for all values of x.

Bi-section Method

(1) Consider the transcendental equation

f(x)=0 --------(1)

which has zero in interval [a, b] and f(a)*f(b)<0.

Suppose f(a)<0 and f(b)>0

(2) Find first approximate

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Explanation of Method

x₁=a+b/2

Calculate f(x₁) and examine its sign.

(2.1) If f(x₁)<0 implies root lies between x₁ and b,

than second approximate is given by

x₂= x₁ +b/2

(2.2) If f(x₁)>0 implies root lies between x₁ and a,

than second approximate is given by

x₂= x₁ +a/2

Calculate f(x₂) and repeat steps (2.1) and (2.2).

Continuous this process until one can achieve the desire root of the equation up to required decimal places.

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Using Bi-section method

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find a root of the equation

correct up to three decimal places.

Sol.- First of all we have to find two nearest values for which function keeps opposite sign for respectively value.

Here f(0)=-1 and f(1)=1. (opposite sign)

Therefore, root of the equation lies in (0,1)

Compare [0,1] with [a,b], we have a=0 and b=1.

From the table it clear that has

same values after three decimal places.

Our desire root is 0.754

FALSE POSITION

METHOD

Object: To find real root of an equation.

Let f(x)=0 ---------------(1)

be a continuous function on [a,b], and f(a) and f(b) have opposite signs.

Therefore, there must be a real root of the equation in [a,b].

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a

b

B(b, f(b))

A(a, f(a))

C(c,0)

A better approximation is find to point (c,0) on X-axis.

Find the slope of =

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Similarly, find the slope of

=

In this way we have

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=

c can be treated as an approximation of actual real root.

Now, we have

(1) f(a) and f(c) have opposite sign than root lies in [a,c].

(2) f(c) and f(b) have opposite sign than root lies in [c,b].

Once more we obtain next approximation x-coordinate of the point and joining the chord and do as above.

Continuing similarly we get a sequence of approximations x1,x2,---------

In this way one can find the desire root of the equation of (1) up to given (m) decimal places.

Example:

Find the real root of the equation

By method of False position or Regular falsi method

Up to three decimal places.

Sol.- First of all we have to find two nearest values for which function keeps opposite sign for respectively value.

Here f(2)=-9 and f(3)=6. (opposite sign)

Therefore, root of the equation lies in (2,3)

Compare [2,3] with [a,b], we have a=2 and b=3.

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From the table it clear that and

has same values after three decimal places.

Our desire root is 2.70 6

ITERATION

METHOD

Object: To find real root of an equation.

Statement:

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Let be a root of f(x)=0 and let I be an

interval containing . Let and

Be continuous in I, where is defined by

which is equivalent to f(x)=0. Then if

For all x € I, the sequence of approximations

defined by

Converges to , provided that the initial approximation is chosen I.

Proof:

f(x)=0 is a real valued function, and be a root.

If we express the equation equivalently as

Then we have

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Now, let be the initial approximation chosen in some interval I.

Suppose,

Apply Langrage’s mean value therefore

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--------(1)

For some between and

Using we obtain the sequence

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And corresponding inequalities given below

For some between and

Same we apply for the next we have

Continuous the process we have

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Now, we have assume for some k

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Then above inequalities(1), (2), (n+1) give

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This implies that if for all i , then all

Successive approximations remain in I provided the initial approximation is chosen in I

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Finally, we show that the sequence of successive approximations converges to .

Multiplying the inequalities(1), (2), (n+1) and then simplifying we get,

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Since

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Also since k<1, the R.H.S. tends to zero as

Therefore the sequence of successive approximations will converges to

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If the condition in the theorem are satisfied then using iteration formula

We can find successive approximations .

Find a real root of

by Iteration method correct up to three decimal places.

Sol:

For f(x)=sinx-10(x-1) ----------(1)

f(1)=sin(1)-10(1-1)

= 0.841470984808>0

f(2) = sin(2)-10(2-1)

=-9.09070257317<0

There fore, root lies between 1 and 2.

We obtain initial approximation as a mid-value

By equ. (1)

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According to theorem

Now working with the formula

As the condition

Which satisfies for all x between 1 and 2.

From the table it is clear and has

Same value after three decimal places.

Our desire root is 1.088

AITKEN’S

-PROCESS

Object: To find real root of an equation.

Let be a real root of f(x)=0. So if we express the equation equivalently as

then we have

Now, let be the initial approximation chosen in some interval I.

Suppose, using Iteration method we have three successive approximations

to

Then for some k such that we have

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PROCESS:

Dividing we can eliminate k and obtain

Simplifying

On the right hand side adding and subtracting

In terms of forward difference notation we have

Therefore above formula can be expressed as

As it gives next approximation to we have the

Aitken’s process approximation formula as given

EXAMPLE:

Solve by Aitken’s Process correct up to three decimal places.

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Sol.

Root of the equation lies between 0 and 1.

Now, we obtain initial approximation as mid-value of an interval,

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Now, let us express the given equation as subject

to x we

We get following successive approximation.

Finding first three values we can apply Iteration method.

Formulae

In (*) i=2

In this way

For i=3

In this way,

NEWTON-RAPHSON METHOD

N-R METHOD

Object: To find real root of an equation.

Let f(x)=0 be an equation and be initial

approximation to be a root, say , of the

Equation.

Suppose for some h.

As is a root of the equation , we have

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Therefore

Initial x0

First

Approximation

If f satisfies all the conditions for Taylor’s

Series expansion then

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Neglecting the terms with second and higher order derivatives in above expansion we have

If f satisfies all the conditions for Taylor’s

Series expansion then

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Neglecting the terms with second and higher order derivatives in above expansion we have

Substituting in we get

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Using similar arguments we can obtain a

generalized formula called Newton-Raphson approximation formula as

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EXAMPLE:

Find a real root of , correct up to three decimal places, by using Newton-Raphson method.

Solution:

The initial approximation is

APPLICATION OF

N-R METHOD

APP-1: Formula for finding cube root of a equation

If is an approximation to the root of an equation f(x)=0, where f is differentiable on an interval containing , then by N-R formula

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Now, suppose x is a real cube root of N.

Then we have

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Now, to approximate the cube root, let us define

Here,

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Therefore, if is an approximation to x then using N-R approximation formula

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We get the next approximation by

Thus, the iterative formula for finding successive

Successive approximations of cube root of a

number N is

Exa.: Find the cube root of 12

APP-I1: Formula for finding square root of a equation

If is an approximation to the root of an equation f(x)=0, where f is differentiable on an interval containing , then by N-R formula

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Now, suppose x is a real square root of N.

We get the next approximation by

Thus, the iterative formula for finding successive

approximations of square root of a number N is

Exa.: Find the square root of 8

APP-I1: Formula for finding square root of a equation

If is an approximation to the root of an equation f(x)=0, where f is differentiable on an interval containing , then by N-R formula

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Now, suppose x is a real square root of N.

We get the next approximation by

Thus, the iterative formula for finding successive

approximations of square root of a number N is

Exa.: Find the square root of 8