Root is in between [0.5179, 0.5180]
Failure Case
There are some cases where this method cannot be used to solve the equation.
Newton-Raphson Method
f `(x) = ∫f(x)
= ∫ (x5 – 4x +1)
= 5x4 -4
Roots are in between [0,1] and [1, 2].
We look for the roots in [0,1]
xn+1 = xn – f (xn)
f ` (xn)
xn+1 = xn - (xn5 – 4xn+1)
(5xn4 -4)
When x1= 0
x2 = 0 – ((0)5 - 4(0) + 1)
5 (0)4 - 4
= 0 - 1
-4
= 0.25
x3 = 0.25 - ((0.25) 5- 4(0.25) + 1)
5(0.25) 4
= 0.25024533856722
x4 = 0.20524 - ((0.20524) 5- 4(0.20524) + 1)
5(0.20524) 4
= 0.25024534093233
Root will be in between 0.250235 and 0.250245
Failure Case
Rearranging the Equation in the form x=g(x)
We start this method by putting f(x) = 0.
x4-4x + 2
We rearrange this equation to make x the subject
x4- 4x + 2 = 0
4x = x4+ 2
x = x4+ 2
4
Roots in [0, 1] Roots in [1, 2]
x1 = 1 x1= 2
x2 = 0.75 x2 = 4.5
x3 = 0.5791 x3 = 103.1
x4 = 0.5281 x4 = 28154798.49
x5 = 0.5194 x5 = 1.57 * 1029
x6 = 0.5182 x6 = ∞
x7 = 0.5180
x8 = 0.5180 x → ∞
x9 = 0.5179
x10 = 0.5179 So it is a failure case.
x11 = 0.5179
We can also rearrange the equation in another way
x4-4x + 2
x4 = 4x - 2
x = (4x - 2)1/4
or
x2 = (4x - 2)
x2
x = (4x-2)1/2
x
xn+1 = (4xn - 2)1/2
xn
x1 = 1 x1 = 2
x2 = 1.414 x2 = 1.2247
x3 = 1.352 x3 = 1.3901
x4 = 1.3654 x4 = 1.3573
x5 = 1.3626 x5 = 1.3643
x6= 1.3632 x6 = 1.3628
x7 = 1.3630 x7 = 1.3631
x8 = 1.3631 x8 = 1.3631
x9 = 1.3631
x = x4 + 2
4
is equivalent to solving
y = x
y = x4 + 2 is y = g(x)
4
The solution is where the line crosses the graph
Rearrangement will work if its gradient g’(x) is less than the gradient of y = x otherwise it fails. The gradient of y = x is 1.
| g’(x) | < 1 method works
| g’(x) | > 1 method fails
g(x) = x4 + 2
4 4
g’(x) = 4x3
4
g’(x) = x3
In [0, 1] , say x= 0.5
g’(x) = (0.5)3
= 0.625 < 1 . The method works
In [1, 2] , say x= 1.5
g’(x) = (1.5) 3
= 5.0625 > 1. The method fails.
