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# Newton-Raphson Method: This is a fixed-point estimation method.

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Introduction

Method 2: Newton Raphson

Newton Raphson

Newton-Raphson Method:

This is a fixed-point estimation method. The estimate starts at x 1,for a root of f(x) = 0. A tangent is then draw to the curve y = f(x) at the point (x 1, f(x 1)). The point at which the tangent cuts the x-axis then gives the next approximation for the root, and the process is repeated.

I am going to use the equation y = x³ - 3x + 1.

***

As you can see there are three roots in this graph, they are in the interval [-2, -1]

[0, 1]

[1, 2]

The gradient for the tangent to the curve at (x 1, f(x 1)) is f’(x 1) (meaning dy/ dx for x). The equation of the tangent is: y-y1 = m(x-x1). Therefore y-f(x1) = f’(x1) [x-x1]. This tangent passes through the point (x2, 0). Carrying on with this process, this will get closer and closer to the tangent. But there is a general formula for this process:

x n+1 = x n – f(Xn)/ f’(Xn)

Returning to my function: f(x)

Middle

0.28

0.007317

-4.9216

0.28148678

0.281487

6.2E-07

-4.92077

0.28148691

0.281487

4.22E-15

-4.92077

0.28148691

Therefore,

x b = 0.28148691 (8 d.p.)

1. Root c:
 x n f(x)` x n+1 -4.5 15.25 -4.07540984 -4.07541 11.60897 -4.00772671 -4.00773 11.06187 -4.00604832 -4.00605 11.04842 -4.00604730 -4.00605 11.04841 -4.00604730

Therefore,

x c = -4.00604730 (8 d.p.)

NEWTON-RAPHSON METHOD DOESN’T ALWAYS WORK!

This method will not work if:

1. If the initial value is not close to the root, or is near a turning point, the iteration may diverge or converge to another root!
1. This method can break down when the equation is discontinuous.

I will demonstrate this using the equation g(x) = x 3 – 5x 2 + 8x -4.1

In this case, if we use the value x=2 as the starting point we get a “DIVERGENT” pattern. So the iteration cannot go to the upper or lower root.

At x=2 the trial will fail because the derivative there is 0; and in the equation x n+1 = x n – g(x n)/g`(x n), g`(x n) can’t equal 0 as it will lead to an undefined quantity for x n+1.

Conclusion

 x y 1.1 -0.0034 1.111 0.016937 1.112 0.037361 1.113 0.05782

This table now shows that the root is in the interval [1.11,1.111].

## Error Bounds

X=

Failure of the Change of Sign Method

Although the Change of Sign method has been proved to work above, there are many examples of cases or equations which would be wrongly represented by this technique. Examples of these are shown below:

1. Repeated Roots - in an initial search using the Change of Sign method, the table of values for an equation such as y = (x-1.26)2 (x+1.4) would overlook the second root. As you can see from the graph below, there are two roots; one in the interval [-2,-1] and one in the interval [-1,0]. However, the table only shows one change of sign in the interval [-2,-1].

2.   The Change of Sign Method would also fail with an equation where all of the roots fall within the same interval, such as in the equation y = x3-1.7x2+0.84x-0.108. The table below shows only one change of sign in the interval [0,1]. This would indicate that there is only one root, rather than three, in this interval.

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I have noticed that the gradient is always equal to 6x. I predict that when x = 5, the gradient will be equal to 30. 5 25 75 30 The gradient here is 30, so therefore I was correct. Now, to the binomial proof...

2. ## Numerical Method (Maths Investigation)

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1. ## The method I am going to use to solve x&amp;amp;#8722;3x-1=0 is the Change ...

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2. ## Change of Sign Method.

The iterative formula calculates these values as they converge. xr+1 = 1/4( x3r+2x2r-4.58) x1 = 1/4(( -1)3+2(-1)2-4.58) =1/4(-1+2-4.58) =1/4(-3.58) x1 = -0.895 The following table was obtained using a Microsoft Excel spreadsheet, with the formulae shown. The sequence works using this particular iterative formula because when the gradient of y=g(x)

1. ## Change of Sign Method

Therefore: x=-1.6511(5s.f.) -1.65115<x<-1.65105 Now I will find the other two roots of the equation. I will firstly find the root between x=1 and x=2 and I will use x0=1 as this is close to the root and is most likely to find that root.

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For example in the following case: Equation A: � The graph above shows us that the roots of the equation are very close. Zoom in on the x and y axes, we can see that they all lie between x=0 and x=1.

1. ## Change of sign method.

-0.0000916085 0.0000805507 -0.0000055299 1.1040115356 1.1040191650 1.1040153503 -0.0000055299 0.0000805507 0.0000375101 1.1040115356 1.1040153503 1.1040134430 -0.0000055299 0.0000375101 0.0000159901 .1040115356 1.1040134430 1.1040124893 -0.0000055299 0.0000159901 0.0000052301 1.1040115356 1.1040124893 1.1040120125 -0.0000055299 0.0000052301 -0.0000001499 1.1040120125 1.1040124893 1.1040122509 -0.0000001499 0.0000052301 0.0000025401 1.1040120125 1.1040122509 1.1040121317 -0.0000001499 0.0000025401 0.0000011951 Error bounds need to be established for the roots.

2. ## newton raphson

0,680760 -0,000056 0.67 -0,06939 0,687 0,040909 0,6807 -0,00045 0,680770 0,000010 0.68 -0,00501 0,688 0,047526 0,6808 0,000205 0,680780 0,000075 0.69 0,060803 0,689 0,054157 0,6809 0,000857 0,680790 0,000140 0.7 0,12807 0,69 0,060803 0,681 0,00151 0,680800 0,000205 I can express this interval using the Error bounds.

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