The document provides an overview of the Newton-Raphson method, an iterative technique used to find solutions to non-linear equations. It explains the derivation of the iteration formula, the convergence criteria, and detailing stopping criteria through examples. Additionally, it includes practical applications demonstrating how to apply the method to find approximated roots.

1. Overview of Newton Raphson Method
Numerical Analysis
Isaac Amornortey Yowetu
NIMS-Ghana
June 24, 2020

2. Background of Newton Raphson Iteration
1 Background of Newton Raphson Iteration
Graphical Example
2 Steps in Deriving Newton Raphson Iteration
Derivations of Newton Raphson Method
3 Performing Iteration to find approximate Solution
4 Application of Newton Raphson Iteration
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3. Background of Newton Raphson Iteration
Background of Newton Raphson Iteration
This iterative method Newton Raphson is said to be named after
Isaac Newton and Joseph Raphson. It is one of the opened Iterative
method that helps in a quick way of finding the solutions of
non-linear equations. This is achieved by producing successively
better approximations to the actual roots of a real-valued functions
f (x) = 0.
This iterative method is applicable when f (x) and f (x) are both
continuous and differentiable. The function f (x) is approximated by
a straight line tangential to function curve, starting with an initial
guess then, the x-intercept of the tangent line gives us the next
approximation.
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4. Background of Newton Raphson Iteration Graphical Example
Graphical Example
Figure: A Graphic Example of Newton Raphson’s Iterations
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5. Steps in Deriving Newton Raphson Iteration
Steps in Deriving Newton Raphson Iteration
Given f (x) = 0
tanθ = f (x) =
f (xn+1) − f (xn)
xn+1 − xn
(1)
From the graph, when we can consider these Cartesian coordinates:
xn+1, f (xn+1) and xn, f (xn) , but f (xn+1) = 0. Hence,
f (x) =
0 − f (xn)
xn+1 − xn
(2)
(xn+1 − xn)f (x) = −f (xn) (3)
(xn+1 − xn) = −
f (xn)
f (x)
(4)
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6. Steps in Deriving Newton Raphson Iteration Derivations of Newton Raphson Method
xn+1 = xn −
f (xn)
f (x)
(5)
Remarks
The xn+1 − value at which there is converges is the solution of
interest.
It takes an intial guess to start its iteration process
Second Derivation
Suppose x∗
is the solution of f (x) = 0 and let xn be the approximate
solution of x∗
s.t |x∗
− xn| = δ << 1.
By the use of Taylor Series approximation, we can have:
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7. Steps in Deriving Newton Raphson Iteration Derivations of Newton Raphson Method
0 = f (x∗
) = f (xn + δ) = f (xn+1) (6)
f (xn+1) = f (xn) + δf (xn) +
δ2
2!
f (xn) + ... = 0 (7)
But neglecting the powers of δ, n ≥ 2, we now have;
f (xn) + δf (xn) ≈ 0 (8)
But δ = xn+1 − xn. We can write eqn(8) as:
f (xn) = −(xn+1 − xn)f (xn) (9)
f (xn)
f (xn)
= xn − xn+1 (10)
xn+1 = xn −
f (xn)
f (xn)
(11)
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8. Performing Iteration to find approximate Solution
Performing Newton Raphson Iteration
We start our iteration with an initial point x0
x1 = x0 −
f (x0)
f (x0)
x2 = x1 −
f (x1)
f (x1)
x3 = x2 −
f (x2)
f (x2)
x4 = x3 −
f (x3)
f (x3)
...
xn+1 = xn −
f (xn)
f (xn)
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9. Performing Iteration to find approximate Solution
Convergence Criteria of Newton Raphson
Using the Newton Raphson Formula: xn+1 = xn − f (xn)
f (xn)
Φ(xn) = xn −
f (xn)
f (xn)
Φ (xn) =
d
dx
xn −
f (xn)
f (xn)
= 1 −
d
dx
f (xn)
f (xn)
= 1 −
d
dx
f (xn) · f (xn)−1
= 1 − f (xn) · f (xn)−1
− f (xn) · f (xn)−2
· f (xn)
= 1 −
f (xn)
f (xn)
−
f (xn) · f (xn)
f (xn)2
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10. Performing Iteration to find approximate Solution
Derivation continue....
Φ (xn) =
f (xn) · f (xn)
f (xn)2
For convergences with respect to the initial guess,
|Φ (xn)| < 1
We can also consider checking the convergence criterion as:
f (xn) · f (xn) < f (xn)2
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11. Performing Iteration to find approximate Solution
Stopping Criteria
Error Formula
Absolute Error = |x − x∗
|
Relative Error =
|x − x∗
|
|x|
Where x, x∗
are true-value and approximated-value respectively.
These error formula is not of direct use as the true value x is not
known.
Commonly Use Stopping Criteria
|xn+1 − xn| < ε
|xn+1−xn|
|xn+1|
< ε or |xn−xn+1|
|xn|
< ε
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12. Application of Newton Raphson Iteration
Application of Newton Raphson Iteration
Example 1
Find a root of xex
= 1 using Newton Raphson iteration. Take
x0 = 0.5
Considering our f (x) = xex
− 1 = 0
Solution
Checking the convergence at x0 = 0.5
f (x) = ex
+ xex
= ex
(1 + x) (12)
f (x) = ex
+ ex
+ xex
= ex
(2 + x) (13)
|Φ (0.5)| = (0.5·e0.5−1)·e0.5(2+0.5)
e0.5(1+0.5)
2 = 0.1184 < 1.
Hence, there will be convergence
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13. Application of Newton Raphson Iteration
Solution Continue...
x1 = x0 −
f (x0)
f (x0)
= 0.5 −
0.5e0.5
− 1
e0.5(1 + 0.5)
= 0.5710
x2 = x1 −
f (x1)
f (x1)
= 0.5672
x3 = x2 −
f (x2)
f (x2)
= 0.5671
...
x100 = x99 −
f (x99)
f (x99)
= 0.5671 −
0.5671e0.5671
− 1
e0.5671(1 + 0.5671)
= 0.5671
After successive iterations, the approximated value converges
x∗
= 0.5671
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14. Application of Newton Raphson Iteration
Example 2
Consider the nonlinear equation x3
= 2x + 1 with a solution with the
interval I = [1.5, 2.0] using Newton Raphson iteration with initial
guess x0 = 1.5, find the approximated root.
Considering our f (x) = x3
− 2x − 1 = 0
Solution
Checking the convergence at x0 = 1.5
f (x) = 3x2
− 2 (14)
f (x) = 6x (15)
|Φ (1.5)| =
(1.5)3−2(1.5)−1 · 6(1.5)
3(1.5)2−2
2 = 0.2493 < 1.
Hence, there will be convergence
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15. Application of Newton Raphson Iteration
Solution Continue...
x1 = x0 −
f (x0)
f (x0)
= 1.5 −
(1.5)3
− 2(1.5) − 1
3(1.5)2 − 2
= 1.6316
x2 = x1 −
f (x1)
f (x1)
= 1.6182
x3 = x2 −
f (x2)
f (x2)
= 1.6180
...
x100 = x99 −
f (x99)
f (x99)
= 1.6180 −
(1.6180)3
− 2(1.6180) − 1
3(1.6180)2 − 2
= 1.6180
After successive iterations, the approximated value converges
x∗
= 1.6180
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16. Application of Newton Raphson Iteration
End
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