Solution of algebraic & transdental equation with Newton Raphson method & Secant

method

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Presented by: 130490106065 – Modi Nagma

130490106085 – Patel Pinal

140493106015 – Padhiyar Sagar

140493106025 – Taylor Kishan

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Two Fundamental Approaches

1. Bracketing or Closed Methods

- Bisection Method

- False-position Method (Regula falsi).

2. Open Methods

- Newton-Raphson Method

- Secant Method

- Fixed point Methods

Roots of Equations

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Newton Raphson’s Method The equation for

Newton’s Method can be determined graphically!

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Continue.....

The equation for Newton’s Method can be determined graphically!

From the diagram tan Ө = ƒ'(x0) = ƒ(x0)/(x0 – x1)

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Continue....

The equation for Newton’s Method can be determined graphically!

From the diagram tan Ө = ƒ'(x0) = ƒ(x0)/(x0 – x1)

Thus, x1=x0 -ƒ(x0)/ƒ'(x0).

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In general, if the nth approximation is xn and f’(xn) ≠0, then the next approximation is given by:

3

1

( )12

'( )

nn n

n

f xx x

f x

Equation/Formula 2

Continue…

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Starting with x1 = 2, find the third approximation x3 to the root of the equation

x3 – 2x – 5 = 0

Example Of Newton Raphson’s method

Example 1

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We apply Newton’s method with f(x) = x3 – 2x – 5 and f’(x) = 3x2 – 2

Newton himself used this equation to illustrate his method.

He chose x1 = 2 after some experimentation because f(1) = -6, f(2) = -1 and f(3) = 16

Continue…

Example 1

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Equation 2 becomes:

Continue…

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3

2

2 51

3 2

n nn n

n

x xx x

x

With n = 1, we have:

Continue…..

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3

1 12 1 2

1

3

2

2 5

3 2

2 2(2) 52

3(2) 2

2.1

x xx x

x

With n = 2, we obtain:

It turns out that this third approximation x3 ≈ 2.0946is accurate to four decimal places.

Continue….

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3

2 23 2 2

2

3

2

2 5

3 2

2.1 2(2.1) 52.1

3(2.1) 2

2.0946

x xx x

x

Use Newton’s method to find correct to eight decimal places.

First, we observe that finding is equivalent to finding the positive root of the equation x6 – 2 = 0

So, we take f(x) = x6 – 2

Then, f’(x) = 6x5

Example 2

Newton Raphson’s method6 2

6 2

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So, Formula 2 (Newton’s method) becomes:

Continue…

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6

1 5

2

6

nn n

n

xx x

x

Choosing x1 = 1 as the initial approximation, we obtain:

As x5 and x6 agree to eight decimal places, we conclude that to eight decimal places.

Continue…

2

3

4

5

6

1.16666667

1.12644368

1.12249707

1.12246205

1.12246205

x

x

x

x

x

6 2 1.12246205

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Newton Raphson’s method Use Newton’s method to find correct to four

decimal places.

Where, N=12

q=3

= 1

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1 1

11n n q

n

Nx x q

q x

0x

3 12

continue….

So, Formula 2 (Newton’s method) for find out qthrootbecomes:

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1 3 1

1 123 1 1

3n n

n

x xx

1 4.6667x

Continue…. Choosing x1 = 1 as the initial approximation, we obtain:

As x5 and x6 agree to four decimal places, we conclude that to four decimal places.

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3 12 2.2894

2

3

4

5

6

7

3.2948

2.5650

2.3180

2.2898

2.2894

2.2894

x

x

x

x

x

x

Secant Method

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Continue....

First we find two points(x0,x1), which are hopefully near the root (we may use the bisection method).

A line is then drawn through the two points and we find where the line intercepts the x-axis, x2.

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Continue....

If f(x) were truly linear, the straight line would intercept the x-axis at the root.

However since it is not linear, the intercept is not at the root but it should be close to it.

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Continue....

From similar triangles we can write that,

10

10

1

21

xfxf

xx

xf

xx

1x 0x2x

1xf

0xf

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Continue....

From similar triangles we can write that,

Solving for x2 we get:

10

10

1

21

xfxf

xx

xf

xx

10

10112

xfxf

xxxfxx

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Continue....

Iteratively this is written as:

nn

nnnnn

xfxf

xxxfxx

1

11

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Approx. f '(x) with backward FDD:

Substitute this into the N-R equation:

to obtain the iterative expression:

i 1 i

i 1 i

f (x ) f (x )f '(x)

x x

ii 1 i

i

f (x )x x

f '(x )

i i 1 ii 1 i

i 1 i

f (x )(x x )x x

f (x ) f (x )

Continue….

x

f(x)

1

2

new est.

x

f(x)

1

new est.

2

FALSE POSITION

SECANT METHOD

The new estimateis selected from theintersection with thex-axis

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Example of Secant Method

The floating ball has a specific gravity of 0.6 and has aradius of 5.5cm.You are asked to find the depth to which the ball issubmerged when floating in water.

The equation that gives the depth x to which the ball is

submerged under water is given by

010993.3165.0 423 xx

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Use the secant method of finding roots ofequations to find the depth to which the ball issubmerged under water. Conduct three iterations toestimate the root of the above equation. Find theabsolute relative approximate error at the end of eachiteration, and the number of significant digits at leastcorrect at the converged iteration.

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Secant Method

From the physics of the problem

11.00

)055.0(20

20

x

x

Rx

x

water

Figure 2 :Floating ballproblem

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Let us assume

11.0,0 UL xx

4423

4423

10662.210993.311.0165.011.011.0

10993.310993.30165.000

fxf

fxf

U

L

Hence,

010662.210993.311.00 44 ffxfxf UL

31

Iteration 1

0660.0

10662.210993.3

10662.2010993.311.044

44

UL

ULLUm

xfxf

xfxxfxx

5

423

101944.3

10993.30660.0165.00660.00660.0

fxf m

00660.00 ffxfxf mL

0660.0,0 UL xxS.N.P.I.T. & R.C.

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Iteration

1 0.0000 0.1100 0.0660 N/A -3.1944x10-5

2 0.0000 0.0660 0.0611 8.00 1.1320x10-5

3 0.0611 0.0660 0.0624 2.05 -1.1313x10-7

4 0.0611 0.0624 0.0632377619 0.02 -3.3471x10-10

LxUx mx %a mxf

010993.3165.0 423 xxxfTable 1: Root of

for secant Method.

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