6-)Find f(x)=(x-2)^3(x-4),
accelerate newton iteration numerically
sand show that R=2
For the given function
f(x) = (x - 2)^3 *(x - 4)
so,
f'(x) = 3(x - 2)^2(x - 4) + (x - 2)^3
now to find roots we have to put f(x) = 0
from there we can easily see x = 2 and x = 4 are solutions and R =
2 is repeated root
now, from newtons method, let xo = 1
then x1 = xo - f(xo)/f'(xo)
x1 = 1 - (-1)(-3)/(3(-3) + 1) = 1 + 3/8 = 11/8
the following excel sheet has values till x18
x0 | 1.00000 |
x1 | 1.30000 |
x2 | 1.51477 |
x3 | 1.66663 |
x4 | 1.77270 |
x5 | 1.84598 |
x6 | 1.89612 |
x7 | 1.93019 |
x8 | 1.95320 |
x9 | 1.96868 |
x10 | 1.97907 |
x11 | 1.98602 |
x12 | 1.99067 |
x13 | 1.99378 |
x14 | 1.99585 |
x15 | 1.99723 |
x16 | 1.99815 |
x17 | 1.99877 |
x18 |
1.99918 |
taking initial value 3 gives us
x0 | 3.00000 |
x1 | 2.50000 |
x2 | 2.31250 |
x3 | 2.20148 |
x4 | 2.13172 |
x5 | 2.08675 |
x6 | 2.05739 |
x7 | 2.03807 |
x8 | 2.02530 |
x9 | 2.01683 |
x10 | 2.01120 |
x11 | 2.00746 |
x12 | 2.00497 |
x13 | 2.00331 |
x14 | 2.00221 |
x15 | 2.00147 |
x16 | 2.00098 |
x17 | 2.00065 |
x18 | 2.00044 |
using initial value 3.5 gives us division by 0 error, using 3.6 gives
x0 | 3.60000 |
x1 | 5.20000 |
x2 | 4.63529 |
x3 | 4.26663 |
x4 | 4.06955 |
x5 | 4.00637 |
x6 | 4.00006 |
x7 | 4.00000 |
x8 | 4.00000 |
x9 | 4.00000 |
x10 | 4.00000 |
x11 | 4.00000 |
x12 | 4.00000 |
x13 | 4.00000 |
x14 | 4.00000 |
x15 | 4.00000 |
x16 | 4.00000 |
x17 | 4.00000 |
x18 | 4.00000 |
hence we can see, R = 2, is repeated root and E = 4 is single root
6-)Find f(x)=(x-2)^3(x-4), accelerate newton iteration numerically sand show that R=2
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