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# In my coursework I will be using three equations to investigate their solutions using three numerical methods which are: change of sign using Newton-Raphson by finding the fixed point iteration fixed point iteration after rearranging the equa

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Introduction

Numerical Methods

In my coursework I will be using three equations to investigate there solutions using three numerical methods which are:

1. change of sign using
2. Newton-Raphson by finding the fixed point iteration
3. fixed point iteration after rearranging the equation f(x) = 0 into the form x = g(x)

Change of sign

This method involves finding an interval in which f(x) changes sign.

I am using the function f(x) = 2x³ – 5x² – 4x + 1 for this method I have used autograph to show the graph of y= 2x³ – 5x² – 4x + 1

Here is the table of values

 -2 -27 -1 -2 0 1 1 -6 2 -11 3 -2 4 33

This shows that there are three intervals containing roots:

[–1,0], [0,1] and [3,4].  It can be seen that f(-1) > 0 and f(0) < 0 so there is a root in the interval [-1,0].

Bisection

In this method the interval is successively halved by looking at the value of f(x) at its

mid-point. For the root in the interval [-1,0], you next try an x-value of-0.5.

f(-0.5) = 1.5

 -0.75 0.34375 -0.875 -0.667969 -0.8125 -0.123535 -0.78125 0.119568 -0.79688 0.000365464

Since this is positive the root lies in the interval [-1,-0.5]. The next value I’ll try is the midpoint of the new interval, -0.75

f(-0.75) = 0.34375

f(-0.875) =

Middle f(x)= x³+6x²+4x

f’(x)= 3x²+12x+4

Using excel I have found the roots

By using the Newton-Raphson equation

 x0 -2 x1 -1 x2 -0.8 x3 -0.765217391 x4 -0.76393381 x5 -0.763932023 x6 -0.763932023 x7 -0.763932023

This is fixed point iteration   .

I have used autograph to both draw my graphs and display

a table of results to show where the root is

By using Newton Raphson method  So for my example Xo= -2

So (-2)³+6*(-2)²+4*(-2) = 8

3*(-2)²+12*(-2)+4 = -8

Then (-2) – 8/-8 = -1

And then carrying on the principle x would be replaced with -1

by looking for a change of sign I can identify my error bounds and the sign changes between -1 and -0.5 so -1= 1 and -0.5= -0.625

so by using excel I have imputed the x values into the equation y = x³+6x²+4x  and here are the results:

x        y

 -0.5 -0.625 -0.6 -0.456 -0.7 -0.203 -0.8 0.128

I can see there is a sign change

between -0.7 and -0.8
so I use 2 d.p

x        y

 -0.75 -0.046875 -0.76 -0.013376 -0.77 0.020867 -0.78 0.055848 -0.79 0.091561

I can see a sign change

Between -0.76 and -0.77

 -0.761 -0.00998508 -0.762 -0.00658673 -0.763 -0.00318095 -0.764 0.000232256 -0.765 0.00365287

Conclusion

G’(x)= √12x+2 From the graph above you can see where both the rearrangement is successful and where it fails this is because of the gradient of y=³√6x²+2x-1 at where the graph y=x interact near (-0.5,-0.5) is more than 1. the reason for this fail is because the gradient of y=³√6x²+2x-1 is much larger than one meaning that section neither converges or diverges correctly so it was impossible to use both staircase diagram and cobweb diagram to determine the root.

x        y

 -0.05 -1.02757 -1.02757 1.48583 1.48583 2.47809 2.47809 3.44265 3.44265 4.25425 4.25425 4.87841 4.87841 5.33153 5.33153 5.64846 5.64846 5.86482 5.86482 6.01023 6.01023 6.10695 6.10695 6.17086

The

f(x)= 2x³–5x²–4x+1

f’(x)= 6x²–10x–4

 -1 -0.833333 -0.833333 -0.798475 -0.798475 -0.79693 -0.79693 -0.796927 -0.796927 -0.796927 -0.796927 -0.796927

x0-f(x)/f’(x)= 1-

I found that each method had its own advantages and disadvantages for me I found Newton raphson method easy to use on the computer because autograph has a special tool that works everything out for you.

One of the equations used above is selected and the

Other two methods are applied successfully to find the Same root. There is a sensible comparison of the relative merits of The three methods in terms of speed of convergence. There is a sensible comparison of the relative merits of the three methods in terms of ease of use with available hardware and software.

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