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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

image00.png

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].

image01.pngimage12.png

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

-0.66796875

-0.8125

-0.123535156

-0.78125

0.119567871

-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) =

...read more.

Middle

image22.png

f(x)= x³+6x²+4x  

f’(x)= 3x²+12x+4

Using excel I have found the roots

By using the Newton-Raphson equation

x0

-2image03.png

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

image04.png

image05.pngimage06.png

.

        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

image07.pngimage08.png

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

-0.009985081

-0.762

-0.006586728

-0.763

-0.003180947

-0.764

0.000232256

-0.765

0.003652875

...read more.

Conclusion

G’(x)= √12x+2

image14.png

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.

image15.png

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

-1.027566442

-1.027566442

1.485825571

1.485825571

2.47808672

2.47808672

3.442647822

3.442647822

4.254251613

4.254251613

4.878406213

4.878406213

5.331530284

5.331530284

5.648456535

5.648456535

5.864822571

5.864822571

6.010227998

6.010227998

6.106948094

6.106948094

6.170855114

        The

 f(x)= 2x³–5x²–4x+1

f’(x)= 6x²–10x–4

-1

-0.833333333

-0.833333333

-0.798474946

-0.798474946

-0.796929982

-0.796929982

-0.796926979

-0.796926979

-0.796926979

-0.796926979

-0.796926979

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.

...read more.

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