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Newton-Raphson MethodMajor: All Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
05/04/23 1http://
numericalmethods.eng.usf.edu
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Newton-Raphson Method
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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. http://numericalmethods.eng.usf.edu3
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Derivation
f(x)
f(xi)
xi+1 xi X
B
C A
)()(
1i
iii xf
xfxx
1
)()('
ii
ii
xxxfxf
ACAB
tan(
Figure 2 Derivation of the Newton-Raphson method.4 http://numericalmethods.eng.usf.edu
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Algorithm for Newton-Raphson Method
5 http://numericalmethods.eng.usf.edu
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Step 1
)(xf Evaluate symbolically.
http://numericalmethods.eng.usf.edu6
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Step 2
i
iii xf
xf - = xx1
Use an initial guess of the root, , to estimate the new value of the root, , as
ix1ix
http://numericalmethods.eng.usf.edu7
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Step 3
0101
1 x
- xx = i
iia
Find the absolute relative approximate error asa
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Step 4Compare 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
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Example 1You are working for ‘DOWN THE TOILET COMPANY’ that makes floats for ABC commodes. The floating ball has a specific gravity of 0.6 and has a radius of 5.5 cm. You are asked to find the depth to which the ball is submerged when floating in water.
Figure 3 Floating ball problem. http://numericalmethods.eng.usf.edu10
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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.
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Figure 3 Floating ball problem.
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Example 1 Cont.
423 1099331650 -.+x.-xxf
12 http://numericalmethods.eng.usf.edu
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)
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Example 1 Cont.
13 http://numericalmethods.eng.usf.edu
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 '
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Example 1 Cont.
06242.0
01242.00.05 10910118.10.05
05.033.005.0310.993305.0165.005.005.0
'
3
4
2
423
0
001
xfxfxx
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Iteration 1The estimate of the root is
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Example 1 Cont.
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Figure 5 Estimate of the root for the first iteration.
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Example 1 Cont.
%90.19
10006242.0
05.006242.0
1001
01
x
xxa
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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.
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Example 1 Cont.
06238.0
104646.406242.0 1090973.81097781.306242.0
06242.033.006242.0310.993306242.0165.006242.006242.0
'
5
3
7
2
423
1
112
xfxfxx
17 http://numericalmethods.eng.usf.edu
Iteration 2The estimate of the root is
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Example 1 Cont.
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Figure 6 Estimate of the root for the Iteration 2.
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Example 1 Cont.
%0716.0
10006238.0
06242.006238.0
1002
12
x
xxa
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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
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Example 1 Cont.
06238.0
109822.406238.0 1091171.8
1044.406238.0
06238.033.006238.0310.993306238.0165.006238.006238.0
'
9
3
11
2
423
2
223
xfxfxx
20 http://numericalmethods.eng.usf.edu
Iteration 3The estimate of the root is
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Example 1 Cont.
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Figure 7 Estimate of the root for the Iteration 3.
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Example 1 Cont.
%0
10006238.0
06238.006238.0
1002
12
x
xxa
22 http://numericalmethods.eng.usf.edu
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.
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Advantages and Drawbacks of Newton
Raphson Method
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Advantages Converges fast (quadratic
convergence), if it converges. Requires only one guess
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Drawbacks
25 http://numericalmethods.eng.usf.edu
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 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 13512.01
i
iii x
xxx
1x
.2.0x
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Drawbacks – Inflection Points
Iteration Number
xi
0 5.00001 3.65602 2.74653 2.10844 1.60005 0.925896 −30.1197 −19.74618 0.2000 0512.01 3 xxf
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Figure 8 Divergence at inflection point for
Table 1 Divergence near inflection point.
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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
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ii
iiii xx
xxxx06.03
104.203.02
623
1
02.0or 0 00 xx Figure 9 Pitfall of division by zero or near a zero number
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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
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3. Oscillations near local maximum and minimum
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Drawbacks – Oscillations near local maximum and minimum
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-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
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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
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xf
539822.74.20 x
0x
2831853.62 x Figure 11 Root jumping from intended location of root for
. 0 sin xxf
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Additional ResourcesFor all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit
http://numericalmethods.eng.usf.edu/topics/newton_raphson.html