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# newton raphson

Extracts from this document...

Introduction

Numerical solutions of equations.

Decimal Search

##### The first method to be looked at is the Decimal Search.  This involves finding an interval of the x- axis in which a root of the equation lies.

Let's consider the function: y=x5+4x2-2. The region to scan to find the first root is -4<x<4.

 x f(x) -4 -962 -3 -209 -2 -18 -1 1 0 -2 1 3 2 46 3 277 4 1086

##### I am  going to concentrate my search on this interval [0,1]  and look at interval of width 0.1 unit.
 x f(x) 0.000000 -2.00027 0.100000 -1.95999 0.200000 -1.83968 0.300000 -1.63757 0.400000 -1.34976 0.500000 -0.96875 0.600000 -0.48224 0.700000 0.12807 0.800000 0.88768 0.900000 1.83049 1.000000 3

## The change of sign occurs in [0.6 , 0.7] which is where the root lies.

Middle

##### 1.
 Equation:  x5-2.5x3+1.7 = 0 Intervals: [-2,-1] [1,2] x f(x) -2 -10.3 -1 3.2 0 1.7 1 0.2 2 13.7
##### 2.
 Equation:   x³+0.7x²–2.1775x+0.845 = 0 We are looking for the change in signs between -4 and 4. x f(x) -4 -43.245 -3 -13.3225 -2 0 -1 2.7225 0 0,845 1 0,3675 2 7.29 3 27.6125 4 67.335
##### 3.
 Equation: 1/(x-2.5) = 0 We are looking for the change of sign between -4 and 4 for discounting type of function. x f(x) -4 -0.14925 -3 -0.17544 -2 -0.21277 -1 -0.27027 0 -0.37037 1 -0.58824 2 -1.42857 3 3.333333 4 0.769231

Conclusion

I will try two starting point, one at x=0, and next one at x=0.7. Look what I’ve done.

At x=0.  At x=0.7.

This is a little bit weird. Why they not come to this root which I want. Answer is very simple and depends of the gradient of g(x). Gradient is in this method responsible for finding the roots. A rearrangement will find only root at x if the gradient of g(x) at this point x is between -1 and 1. So algebraically, if -1<g(x),1.

g(x) =((6x3-1)/4)0.2

g’(x)=

We can not find the root in [0, 1] using this rearrangement. We can however use a different rearrangement to find this root.

2. Second rearrangement.

I will look on the graph of function y=(x) and y=((-4xn 5-1)/6)1/3. Below I show you that graph. I must find only one remaining root. I know that it lies in interval [0,1]. I know that gradient at point x=0 is less than 1 and bigger than -1. So I can clearly say that starting pint I’ve choose will go definitely towards the right root.

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