2. Newton
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Transcript of 2. Newton
Newton-Raphson Method
Newton-Raphson Method
)(xf
)f(x - = xx
i
iii 1
f ( x )
f ( x i )
f ( x i - 1 )
x i + 2 x i + 1 x i X
ii xfx ,
Figure 1 Geometrical illustration of the Newton-Raphson method.2
Derivation
f(x)
f(xi)
xi+1 xi
X
B
C A
)(
)(1
i
iii xf
xfxx
1
)()('
ii
ii
xx
xfxf
AC
ABtan(
Figure 2 Derivation of the Newton-Raphson method.3
Algorithm for Newton-Raphson Method
4
Step 1
)(xf Evaluate symbolically.
5
Step 2
i
iii xf
xf - = xx
1
Use an initial guess of the root, , to estimate the new value of the root, , as
ix
1ix
6
Step 3
0101
1 x
- xx =
i
iia
Find the absolute relative approximate error asa
7
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
8
Convergence
1 1'
1
2
( )Comparing x with x ( )of the iteration method,
( )
( )( )
'( )
( )( )
'( )
( ) "( )'( )
'( )
Since iteration method converges if '( ) 1
Newton's method con
n n n n
nn n n
n
f xx x
f x
f xx x x
f x
f xIn general x x
f x
f x f xx
f x
x
2verges if ( ) "( ) '( )
in the interval considered.
f x f x f x
The rate at which the iteration method converges if the initial approximation to the root is sufficiently close to the desired root is called the rate of convergence.
Example 1
Q1. Using Newton-Raphson method find the real root of the equation 3x=cosx + 1 correct to four decimal places.
11
Solution
A root of f(x)=0 lies between 0 and 1.
( ) 3 cos 1
(0) 2( )
(1) 1.4597( )
f x x x
f ve
f ve
Cont… Also f’(x)=3+sin x
New Iteration formula gives,
1
( )
'( )
3 cos 1...............(1)
3 sin
nn n
n
nn
n
f xx x
f x
x xx
x
Cont… Put n=0 the first approximation is given by
0 0 01
0
sin cos 1
3 sin
0.6sin 0.6 cos 0.6 1
3 sin 0.60.6071
x x xx
x
Cont.. Put n=1,Then second approximation is
1 1 12
1
sin cos 1
3 sin
.6071sin(.6071) cos(.6071) 1
3 sin(.6071)
0.6071
x x xx
x
Cont..Clearly X1 = X2 .
Hence desired root is 0.6071 correct to 4 decimal places.
Example 2Q 2: Find to four places of decimal, the
smallest root of the equation
sinxe x
Solution
The given equation is
So that,
Take x0 = 0.6, then
( ) sin 0xf x e x
1
sin
cos
n
n
xn
n n xn
e xx x
e x
1
2
0.58848
0.588559
x
x
Cont..Hence desired value of the root is a
0.5885
http://numericalmethods.eng.usf.edu19
20
2
2
Example 3
Evaluate 12 to four decimal places by Newton's iteration method.
Sol. Let 12 so that 12 0 .........(1)
( ) 12, ' ,
n
x x
Take f x x Newton s iteration formula gives
x
2
1
0
1 00
( ) 12 1 12........(2)
'( ) 2 2
Now since (3) 3 and (4) 4
The root of (1) lies between 3 and 4.
take 3.5, from eqn.(2),
1 12 1 123.5 3.4643
2 2 3.5
n nn n n
n n n
f x xx x x
f x x x
f f
x
x xx
21
2 11
3 22
2 3
1 12 1 123.4643 3.4641
2 2 3.4643
1 123.4641
2
, 4 .
, 12 3.4641.
x xx
x xx
Since x x upto decimal places
So
Newton Iterative formulaThe reciprocal or inverse of s number ‘a’ can
be considered as a root of the equation which can be solve
by Newton’s method.
10a
x
2
1 1( ) , '( )f x a f x
x x
Cont… Newton’s Formula gives,
1
2
1
1
nn n
n
ax
x x
x
1 (2 )n n nx x ax
Square Root The square root of ‘a’ can be
considered as a root of equation solved by Newton’s method.
2 0x a
2( ) , ( ) 2f x x a f x x 2
1
1
2
1
2
nn n
n
n nn
x ax x
x
ax x
x
Inverse Square Root Equation is
Iterative Formula is
2
10a
x
21
1(3 )
2n n nx x ax
General Formula for pth root
1 1
( 1) pn
n pn
p x ax
px
Newton’s Iterative Formulae
Inverse:
Square Root:
Inverse Square Root:
For pth root :
27
1 (2 )n n nx x ax
1
1
2n nn
ax x
x
21
13
2n n nx x ax
1 1
( 1) pn
n pn
p x ax
px
Example 1 Cont.
The equation that gives the depth x in meters to which the ball is submerged under water is given by
423 1099331650 -.+x.-xxf
Use the Newton’s method of finding roots of equations to find a)the depth ‘x’ to which the ball is submerged under water. Conduct three iterations to estimate the root of the above equation. b)The absolute relative approximate error at the end of each iteration, andc)The number of significant digits at least correct at the end of each iteration.
28
Figure 3 Floating ball problem.
Example 1 Cont.
423 1099331650 -.+x.-xxf
29
To aid in the understanding of how this method works to find the root of an equation, the graph of f(x) is shown to the right,
where
Solution
Figure 4 Graph of the function f(x)
Example 1 Cont.
30
x-xxf
.+x.-xxf -
33.03'
10993316502
423
Let us assume the initial guess of the root of is . This is a reasonable guess (discuss why and are not good choices) as the extreme values of the depth x would be 0 and the diameter (0.11 m) of the ball.
0xfm05.00 x
0x m11.0x
Solve for xf '
Example 1 Cont.
06242.0
01242.00.05 109
10118.10.05
05.033.005.03
10.993305.0165.005.005.0
'
3
4
2
423
0
001
xf
xfxx
31
Iteration 1The estimate of the root is
Example 1 Cont.
32
Figure 5 Estimate of the root for the first iteration.
Example 1 Cont.
%90.19
10006242.0
05.006242.0
1001
01
x
xxa
33
The absolute relative approximate error at the end of Iteration 1 is
a
The number of significant digits at least correct is 0, as you need an absolute relative approximate error of 5% or less for at least one significant digits to be correct in your result.
Example 1 Cont.
06238.0
104646.406242.0
1090973.8
1097781.306242.0
06242.033.006242.03
10.993306242.0165.006242.006242.0
'
5
3
7
2
423
1
112
xf
xfxx
34
Iteration 2The estimate of the root is
Example 1 Cont.
35
Figure 6 Estimate of the root for the Iteration 2.
Example 1 Cont.
%0716.0
10006238.0
06242.006238.0
1002
12
x
xxa
36
The absolute relative approximate error at the end of Iteration 2 is
a
The maximum value of m for which is 2.844. Hence, the number of significant digits at least correct in the answer is 2.
ma
2105.0
Example 1 Cont.
06238.0
109822.406238.0
1091171.8
1044.406238.0
06238.033.006238.03
10.993306238.0165.006238.006238.0
'
9
3
11
2
423
2
223
xf
xfxx
37
Iteration 3The estimate of the root is
Example 1 Cont.
38
Figure 7 Estimate of the root for the Iteration 3.
Example 1 Cont.
%0
10006238.0
06238.006238.0
1002
12
x
xxa
39
The absolute relative approximate error at the end of Iteration 3 is
a
The number of significant digits at least correct is 4, as only 4 significant digits are carried through all the calculations.
Advantages and Drawbacks of Newton
Raphson Method
40
Advantages
Converges fast (quadratic convergence), if it converges.
Requires only one guess
41
Drawbacks
42
1. Divergence at inflection pointsSelection 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 their 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
113
512.01
i
iii
x
xxx
1x
.2.0x
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 0512.01 3 xxf
43
Figure 8 Divergence at inflection point for
Table 1 Divergence near inflection point.
2. Division by zeroFor the equation
the Newton-Raphson method reduces to
For , the denominator will equal zero.
Drawbacks – Division by Zero
0104.203.0 623 xxxf
44
ii
iiii xx
xxxx
06.03
104.203.02
623
1
02.0or 0 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
02 2 xxf
45
3. Oscillations near local maximum and minimum
Drawbacks – Oscillations near local maximum and minimum
46
-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 .
2 2 xxf
Iteration Number
0123456789
–1.0000 0.5–1.75–0.30357 3.1423 1.2529–0.17166 5.7395 2.6955 0.97678
3.002.255.063 2.09211.8743.5702.02934.9429.2662.954
300.00128.571 476.47109.66150.80829.88102.99112.93175.96
Table 3 Oscillations near local maxima and mimima in Newton-Raphson method.
ix ixf %a
4. Root JumpingIn 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
0 sin xxf
47
xf
539822.74.20 x
0x
2831853.62 x Figure 11 Root jumping from intended location of root for
. 0 sin xxf
THE END