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Newton-Raphson Method
𝑥 𝑛+1 = 𝑥 𝑛 −

𝑓(𝑥 𝑛 )
𝑓 ′ (𝑥 𝑛 )

The Newton-Raphson method is a fixed point estimation method, i
...
it
estimates the actual value of a root as opposed to the interval in which it lies
...

It is necessary to find f’(x) to use in the formula:
f(x) = 𝑥 3 + 3𝑥 2 − 𝑥 − 5
f’(x) = 3𝑥 2 − 6𝑥 − 1
This gives the iterative formula:
𝑥 𝑛+1 = 𝑥 𝑛 −

𝑥 3 + 3𝑥 2 − 𝑥 𝑛 − 5
𝑛
𝑛
3𝑥 𝑛 2 − 6𝑥 𝑛 − 1

If the numbers in the sequence approach a limiting value then they are said to
‘converge’ to this value
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Where this
tangent crosses the x-axis
will be the next prediction x1
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25
x

y

This graph shows the
prediction x1 and the creation
of a tangent
...

This can be seen in greater
detail on the graph below:

Tangent x=1
...
25
on y=f(x)

x2

y=f(x)

x1

y

x2 at 1
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f
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21508

x3

Tangent at
x=1
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21432 (to 6 s
...
)

Scale is too small to
show, this should read:
1
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2143 (to 5 s
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) and so the root lies in the interval [1
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2144]
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2143 (to 5 s
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Using this technique all 3 roots can be found:
The roots lie in the intervals: [-3,-2], [-2,-1] and [1, 2]

Root: 1
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5392

Root: -2
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An equation that is proof of this is:
Here is the curve: 𝑦 =

𝑥3
3

𝑥3
3

+ 0
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22953𝑥 + 1
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618𝑥 2 − 2
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1 represented graphically:
y

y=f(x)

x

To verify the roots, look for change in sign:
f(-4)= -1
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35059

So root in interval [-4,-3]

f(0)= 1
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1782

So root in interval [0,1]

f(1)= -0
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77961

So root in interval [1, 2]

It is necessary to find f’(x) to use in the formula:
𝑥3

f(x) = 3 + 0
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22953𝑥 + 1
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236𝑥 − 2
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618𝑥 𝑛 − 2
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1
= 𝑥𝑛−
𝑥 𝑛 2 + 1
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22953

Since f(1) = -0
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1 it seems reasonable to assume that the
root is closer to 1 than 0, so start with x1 = 1
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Instead
it converged on another root in
the interval [1, 2]
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e
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54199 – this is already seen to create a significant problem for the method
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54199
x

Tangent at x=1 i
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x0

y

x
Tangent at x = 28
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e
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87877 which
gives the new prediction:
x2 = 18
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y

y=f(x)

x
Tangent at x = 1
...
e
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319581 which
gives the new prediction:
x13 = 1
...


y
Scale is too small to
show, this should read:
1
...
319592
New
prediction, x13

x13 = 1
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f
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3196 (5 s
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So in this instance the method failed to find the root in the
interval [0, 1]

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