Newton’s Method
Roots of Equation
Asifur Rahman
CEN 205 Numerical Methods Asifur Rahman January 22, 2017 1/5
Newton’s Method Newton’s Method
Slope of the tangent is,
f(xi )
f0 (xi ) =
xi − xi+1
Rearranging we find,
f(xi )
xi+1 = xi −
f0 (xi )
Steps
1. Find initial estimate of root, xi .
2. Then determine f0 (x) and evaulate f0 (xi )
3. Determine next root, xi+1
4. Goto step 2
CEN 205 Numerical Methods Asifur Rahman January 22, 2017 2/5
Example 1 Newton’s Method
Ques. Solve the following equation using Newton’s First Iteration
method. Approximate error should be less than 1 xo = 2.5
percent. 2 x
f(x) = 3x + 2e − 12 2 2.5
f(xo ) = 3(2.5) + 2e − 12 = 31.1150
0 2.5
Solution. f (xo ) = 6(2.5) + 2e = 39.3650
f(xo ) 31.1150
xn = xo − = 2.5 − = 1.7096
f0 (xo ) 39.3650
1.7096 − 2.5
a = = 46.23
1.7096
Second Iteration
xo = 1.7096
2 1.7096
f(xo ) = 3(1.7096) + 2e − 12 = 7.8212
0 1.7096
f (xo ) = 6(1.7096) + 2e = 21.3107
f(xo ) 7.8212
xn = xo − = 1.7096 − = 1.3426
f0 (xo ) 21.3107
Determine the first derivative of the function.
1.3426 − 1.7096
0 x a = = 27.33
f (x) = 6x + 2e
1.3426
Newton’s Method
i xo f(xo ) f0 (xo ) xn a
1 2.5000 31.1150 39.3650 1.7096 46.2350
2 1.7096 7.8212 21.3107 1.3426 27.3363
3 1.3426 1.0652 15.7131 1.2748 5.3178
4 1.2748 0.0310 14.8045 1.2727 0.1645
CEN 205 Numerical Methods Asifur Rahman January 22, 2017 3/5
Example 2 Newton’s Method
Ques. The following equation expresses oxygen level c Solution.
(mg/L) in a river downstream from a sewage discharge site, Since oxygen level c = 5 mg/L, substituing into equation, we find,
where x is the distance downstream in kilometers. −0.2x −0.75x
5 = 10 − 20(e −e )
−0.2x −0.75x
c = 10 − 20(e −e )
−0.2x −0.75x
1−4 e −e =0
Determine the distance downstream where the oxygen level
first falls to a reading of 5 mg/L. Your answer should be
The derivative of the funciton is,
accurate upto a 1% approximate error.
−0.2x −0.75x
0.8e − 3e
First Iteration
xo = 0.25
−0.2(0.25) −0.75(0.25)
f(xo ) = 1 − 4 e −e = 0.5112
0 −0.2(0.25) −0.75(0.25)
f (xo ) = 0.8e − 3e = −1.7261
f(xo ) 0.5112
xn = xo − = 0.25 − = 0.5462
f0 (xo ) −1.7261
0.5462 − 0.25
a = = 54.22
0.5462
Newton’s Method
i xo f(xo ) f0 (xo ) xn a
1 0.2500 0.5112 −1.7261 0.5462 54.2257
2 0.5462 0.0695 −1.2745 0.6007 9.0809
3 0.6007 0.0020 −1.2024 0.6024 0.2734
Ans. x = 0.6024 km
CEN 205 Numerical Methods Asifur Rahman January 22, 2017 4/5
Example 3 Newton’s Method
First Iteration
Ques. Solve the following equation using Newton’s xo = 2.5
method. Approximate error should be less than 1 percent.
f(xo ) = 2.5 sin 2.5 + cos 2.5 = 0.6950
f(x) = x sin x + cos x 0
f (xo ) = 2.5 cos 2.5 = −2.0029
Solution. f(xo ) 0.6950
xn = xo − = 2.5 − = 2.8470
The derivative of the funciton is, f0 (xo ) −2.0029
0
f (x) = x cos x + sin x − sin x 2.8470 − 2.5
a = = 12.1890
2.8470
0
f (x) = x cos x
Newton’s Method
i xo f(xo ) f0 (xo ) xn a
1 2.5000 0.6950 −2.0029 2.8470 12.1890
2 2.8470 −0.1304 −2.7244 2.7992 1.7093
3 2.7992 −0.0021 −2.6367 2.7984 0.0282
CEN 205 Numerical Methods Asifur Rahman January 22, 2017 5/5
