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GOVERNMENT ENGINEERING
COLLEGE,GANDHI NAGAR
Subject:Numerical Techniques &
Statistical Methode
 Made By:-
 130130103006 Priyank
 130130103013 Ashish
 130130103014 Shirish
 Guide By:-
 M.I.Bosmia
NEWTON-RAPHSON
METHOD
NEWTON-RAPHSON METHOD
)(xf
)f(x
-= xx
i
i
ii
1
f(x)
f(xi)
f(xi-1)
xi+2 xi+1 xi
X

  ii xfx ,
Figure 1 Geometrical illustration of the Newton-Raphson method.
http://numericalmethods.eng.usf.edu3
DERIVATION
4
f(x)
f(xi)
xi+1 xi
X
B
C A
)(
)(
1
i
i
ii
xf
xf
xx


1
)(
)('


ii
i
i
xx
xf
xf
AC
AB
tan(
Figure 2 Derivation of the Newton-Raphson method.
http://numericalmethods.eng.usf.edu
ALGORITHM FOR NEWTON-RAPHSON
METHOD
5http://numericalmethods.eng.usf.edu
STEP 1
)(xf Evaluate symbolically.
http://numericalmethods.eng.usf.edu6
STEP 2
 
 i
i
ii
xf
xf
-= xx
1
Use an initial guess of the root, , to estimate the new
value of the root, , as
ix
1ix
http://numericalmethods.eng.usf.edu7
STEP 3
010
1
1
x
- xx
=
i
ii
a 


Find the absolute relative approximate error asa
http://numericalmethods.eng.usf.edu8
STEP 4
Compare the absolute relative approximate error
with the pre-specified relative error tolerance .
Also, check if the number of iterations has exceeded
the maximum number of iterations allowed. If so,
one needs to terminate the algorithm and notify the
user.
s
Is ?
Yes
No
Go to Step 2 using new
estimate of the root.
Stop the algorithm
sa 
http://numericalmethods.eng.usf.edu9
EXAMPLE 1
gecg10
xxxxxv 018507.010x66722.010x13533.010x42493.0)( 4-65-83-4

0)( 
dx
dv
xf
EXAMPLE 1 CONT.
Let us take the initial guess of the root of as .
11
 
  01025496010800670107066.2
00185070101274801026689010676650
32538
233548


x.x.xxf'
.x.x.x.xf
---
---
Solution
  0xf 100 x
Iteration 1
The estimate of the root is
 
 
     
     
3
3
32538
233548
0
0
01
107219.1
104956.8
10
101025496.0101080067.010107066.2
018507.0101012748.0101026689.0101067665.0
10
'











xf
xf
xx
EXAMPLE 1 CONT.
12
Figure 4 Graph of the estimate of the root after Iteration 1.
 
934.14
9339.410


Entered function along given interval with current and next root and the
tangent line of the curve at the current root
0 5 10 15 20 25 30
0.019
0.00831
0
0.013
0.024
f(x)
prev. guess
new guess
tangent line
0.02422
0.02571
0
f x( )
f x( )
f x( )
tan x( )
301 x x 0 x 1 x
EXAMPLE 1 CONT.
The absolute relative approximate error at the end of Iteration 1 is
13
a
%038.33
100
934.14
10934.14
100
1
01







x
xx
a
The number of significant digits at least correct is 0, as you need an
absolute relative approximate error of less than 5% for one significant
digit to be correct in your result.
EXAMPLE 1 CONT.
14
Iteration 2
The estimate of the root is
 
 
   
 
   
 
 
572.14
36149.0934.14
109317.1
109829.6
934.14
934.141025496.0
934.141080067.0934.14107066.2
018507.0934.141012748.0
934.141026689.0934.141067665.0
934.14
'
3
4
3
2538
23
3548
1
1
12

































xf
xf
xx
EXAMPLE 1 CONT.
The absolute relative approximate error at the end of Iteration 2 is
15
a
%4806.2
100
572.14
934.14572.14
100
2
12







x
xx
a
The number of significant digits at least correct is 1, because the
absolute relative approximate error is less than 5%.
EXAMPLE 1 CONT.
16
Iteration 3
The estimate of the root is
 
 
   
 
   
 
 
572.14
104375.2572.14
109314.1
107078.4
572.14
572.141025496.0
572.141080067.0572.14107066.2
018507.0572.141012748.0
572.141026689.0572.141067665.0
572.14
'
6
3
9
3
2538
23
3548
2
2
23


































xf
xf
xx
EXAMPLE 1 CONT.
The absolute relative approximate error at the end of Iteration 3 is
17
a
%101.6727
100
572.14
572.14572.14
100
5
2
12








x
xx
a
EXAMPLE 1 CONT.
18
Hence the number if significant digits at least correct is given by the
largest value of m for which
 
  4756.6103454.3log2
2103454.3log
10103454.3
105.0106727.1
105.0
5
5
25
25
2










m
m
m
m
m
a
So
6m
The number of significant digits at least correct in the estimated
root 14.572 is 6.
ADVANTAGES
 Converges fast (quadratic convergence), if
it converges.
 Requires only one guess
19
DRAWBACKS
20
1. Divergence at inflection points
Selection of the initial guess or an iteration value of the root that
is close to the inflection point of the function may start
diverging away from the root in ther Newton-Raphson method.
For example, to find the root of the equation .
The Newton-Raphson method reduces to .
Table 1 shows the iterated values of the root of the equation.
The root starts to diverge at Iteration 6 because the previous estimate
of 0.92589 is close to the inflection point of .
Eventually after 12 more iterations the root converges to the exact
value of
 xf
    0512.01
3
 xxf
 
 2
33
1
13
512.01



i
i
ii
x
x
xx
1x
.2.0x
DRAWBACKS – INFLECTION POINTS
Iteration
Number
xi
0 5.0000
1 3.6560
2 2.7465
3 2.1084
4 1.6000
5 0.92589
6 −30.119
7 −19.746
18 0.2000
21
    0512.01
3
 xxf
Figure 8 Divergence at inflection point for
Table 1 Divergence near inflection point.
2. Division by zero
For the equation
the Newton-Raphson method
reduces to
For , the
denominator will equal zero.
DRAWBACKS – DIVISION BY ZERO
22
  0104.203.0 623
 
xxxf
ii
ii
ii
xx
xx
xx
06.03
104.203.0
2
623
1





02.0or0 00  xx Figure 9 Pitfall of division by zero
or near a zero number
Results obtained from the Newton-Raphson method may
oscillate about the local maximum or minimum without
converging on a root but converging on the local maximum or
minimum.
Eventually, it may lead to division by a number close to zero
and may diverge.
For example for the equation has no real
roots.
DRAWBACKS – OSCILLATIONS NEAR
LOCAL MAXIMUM AND MINIMUM
23
  022
 xxf
3. Oscillations near local maximum and minimum
DRAWBACKS – OSCILLATIONS NEAR
LOCAL MAXIMUM AND MINIMUM
24
-1
0
1
2
3
4
5
6
-2 -1 0 1 2 3
f(x)
x
3
4
2
1
-1.75 -0.3040 0.5 3.142
Figure 10 Oscillations around local
minima for .  22
 xxf
Iteration
Number
0
1
2
3
4
5
6
7
8
9
–1.0000
0.5
–1.75
–0.30357
3.1423
1.2529
–0.17166
5.7395
2.6955
0.97678
3.00
2.25
5.063
2.092
11.874
3.570
2.029
34.942
9.266
2.954
300.00
128.571
476.47
109.66
150.80
829.88
102.99
112.93
175.96
Table 3 Oscillations near local maxima
and mimima in Newton-Raphson method.
ix  ixf %a
4. Root Jumping
In some cases where the function is oscillating and has a number
of roots, one may choose an initial guess close to a root. However, the
guesses may jump and converge to some other root.
For example
Choose
It will converge to
instead of
-1.5
-1
-0.5
0
0.5
1
1.5
-2 0 2 4 6 8 10
x
f(x)
-0.06307 0.5499 4.461 7.539822
DRAWBACKS – ROOT JUMPING
25
  0sin  xxf
 xf
539822.74.20  x
0x
2831853.62  x Figure 11 Root jumping from intended
location of root for
.  0sin  xxf

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Newton Raphson Of Root Equation

  • 1. GOVERNMENT ENGINEERING COLLEGE,GANDHI NAGAR Subject:Numerical Techniques & Statistical Methode  Made By:-  130130103006 Priyank  130130103013 Ashish  130130103014 Shirish  Guide By:-  M.I.Bosmia
  • 3. NEWTON-RAPHSON METHOD )(xf )f(x -= xx i i ii 1 f(x) f(xi) f(xi-1) xi+2 xi+1 xi X    ii xfx , Figure 1 Geometrical illustration of the Newton-Raphson method. http://numericalmethods.eng.usf.edu3
  • 6. STEP 1 )(xf Evaluate symbolically. http://numericalmethods.eng.usf.edu6
  • 7. STEP 2    i i ii xf xf -= xx 1 Use an initial guess of the root, , to estimate the new value of the root, , as ix 1ix http://numericalmethods.eng.usf.edu7
  • 8. STEP 3 010 1 1 x - xx = i ii a    Find the absolute relative approximate error asa http://numericalmethods.eng.usf.edu8
  • 9. STEP 4 Compare the absolute relative approximate error with the pre-specified relative error tolerance . Also, check if the number of iterations has exceeded the maximum number of iterations allowed. If so, one needs to terminate the algorithm and notify the user. s Is ? Yes No Go to Step 2 using new estimate of the root. Stop the algorithm sa  http://numericalmethods.eng.usf.edu9
  • 10. EXAMPLE 1 gecg10 xxxxxv 018507.010x66722.010x13533.010x42493.0)( 4-65-83-4  0)(  dx dv xf
  • 11. EXAMPLE 1 CONT. Let us take the initial guess of the root of as . 11     01025496010800670107066.2 00185070101274801026689010676650 32538 233548   x.x.xxf' .x.x.x.xf --- --- Solution   0xf 100 x Iteration 1 The estimate of the root is                 3 3 32538 233548 0 0 01 107219.1 104956.8 10 101025496.0101080067.010107066.2 018507.0101012748.0101026689.0101067665.0 10 '            xf xf xx
  • 12. EXAMPLE 1 CONT. 12 Figure 4 Graph of the estimate of the root after Iteration 1.   934.14 9339.410   Entered function along given interval with current and next root and the tangent line of the curve at the current root 0 5 10 15 20 25 30 0.019 0.00831 0 0.013 0.024 f(x) prev. guess new guess tangent line 0.02422 0.02571 0 f x( ) f x( ) f x( ) tan x( ) 301 x x 0 x 1 x
  • 13. EXAMPLE 1 CONT. The absolute relative approximate error at the end of Iteration 1 is 13 a %038.33 100 934.14 10934.14 100 1 01        x xx a The number of significant digits at least correct is 0, as you need an absolute relative approximate error of less than 5% for one significant digit to be correct in your result.
  • 14. EXAMPLE 1 CONT. 14 Iteration 2 The estimate of the root is                   572.14 36149.0934.14 109317.1 109829.6 934.14 934.141025496.0 934.141080067.0934.14107066.2 018507.0934.141012748.0 934.141026689.0934.141067665.0 934.14 ' 3 4 3 2538 23 3548 1 1 12                                  xf xf xx
  • 15. EXAMPLE 1 CONT. The absolute relative approximate error at the end of Iteration 2 is 15 a %4806.2 100 572.14 934.14572.14 100 2 12        x xx a The number of significant digits at least correct is 1, because the absolute relative approximate error is less than 5%.
  • 16. EXAMPLE 1 CONT. 16 Iteration 3 The estimate of the root is                   572.14 104375.2572.14 109314.1 107078.4 572.14 572.141025496.0 572.141080067.0572.14107066.2 018507.0572.141012748.0 572.141026689.0572.141067665.0 572.14 ' 6 3 9 3 2538 23 3548 2 2 23                                   xf xf xx
  • 17. EXAMPLE 1 CONT. The absolute relative approximate error at the end of Iteration 3 is 17 a %101.6727 100 572.14 572.14572.14 100 5 2 12         x xx a
  • 18. EXAMPLE 1 CONT. 18 Hence the number if significant digits at least correct is given by the largest value of m for which     4756.6103454.3log2 2103454.3log 10103454.3 105.0106727.1 105.0 5 5 25 25 2           m m m m m a So 6m The number of significant digits at least correct in the estimated root 14.572 is 6.
  • 19. ADVANTAGES  Converges fast (quadratic convergence), if it converges.  Requires only one guess 19
  • 20. DRAWBACKS 20 1. Divergence at inflection points Selection of the initial guess or an iteration value of the root that is close to the inflection point of the function may start diverging away from the root in ther Newton-Raphson method. For example, to find the root of the equation . The Newton-Raphson method reduces to . Table 1 shows the iterated values of the root of the equation. The root starts to diverge at Iteration 6 because the previous estimate of 0.92589 is close to the inflection point of . Eventually after 12 more iterations the root converges to the exact value of  xf     0512.01 3  xxf    2 33 1 13 512.01    i i ii x x xx 1x .2.0x
  • 21. DRAWBACKS – INFLECTION POINTS Iteration Number xi 0 5.0000 1 3.6560 2 2.7465 3 2.1084 4 1.6000 5 0.92589 6 −30.119 7 −19.746 18 0.2000 21     0512.01 3  xxf Figure 8 Divergence at inflection point for Table 1 Divergence near inflection point.
  • 22. 2. Division by zero For the equation the Newton-Raphson method reduces to For , the denominator will equal zero. DRAWBACKS – DIVISION BY ZERO 22   0104.203.0 623   xxxf ii ii ii xx xx xx 06.03 104.203.0 2 623 1      02.0or0 00  xx Figure 9 Pitfall of division by zero or near a zero number
  • 23. Results obtained from the Newton-Raphson method may oscillate about the local maximum or minimum without converging on a root but converging on the local maximum or minimum. Eventually, it may lead to division by a number close to zero and may diverge. For example for the equation has no real roots. DRAWBACKS – OSCILLATIONS NEAR LOCAL MAXIMUM AND MINIMUM 23   022  xxf 3. Oscillations near local maximum and minimum
  • 24. DRAWBACKS – OSCILLATIONS NEAR LOCAL MAXIMUM AND MINIMUM 24 -1 0 1 2 3 4 5 6 -2 -1 0 1 2 3 f(x) x 3 4 2 1 -1.75 -0.3040 0.5 3.142 Figure 10 Oscillations around local minima for .  22  xxf Iteration Number 0 1 2 3 4 5 6 7 8 9 –1.0000 0.5 –1.75 –0.30357 3.1423 1.2529 –0.17166 5.7395 2.6955 0.97678 3.00 2.25 5.063 2.092 11.874 3.570 2.029 34.942 9.266 2.954 300.00 128.571 476.47 109.66 150.80 829.88 102.99 112.93 175.96 Table 3 Oscillations near local maxima and mimima in Newton-Raphson method. ix  ixf %a
  • 25. 4. Root Jumping In some cases where the function is oscillating and has a number of roots, one may choose an initial guess close to a root. However, the guesses may jump and converge to some other root. For example Choose It will converge to instead of -1.5 -1 -0.5 0 0.5 1 1.5 -2 0 2 4 6 8 10 x f(x) -0.06307 0.5499 4.461 7.539822 DRAWBACKS – ROOT JUMPING 25   0sin  xxf  xf 539822.74.20  x 0x 2831853.62  x Figure 11 Root jumping from intended location of root for .  0sin  xxf