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Solving Equations Numerically

Extracts from this document...

Introduction

Josh Wakeford 12B1

05/04/03

Pure 2 Coursework: Solving Equations Numerically

Many cubic equations can be solved algebraically; however, many cannot, which means we have to give approximate answers.  Graphs, like the one seen below, cannot be solved algebraically and have to be solved some other way.

image00.pngimage01.png

So, how can we solve these graphs to any degree of accuracy, and in the most efficient way possible?  This problem comes up a lot in real life, as many equations in real life will not be able to be solved algebraically, and so other methods have to be used.

There are a number of methods that can be used here, and in this coursework, I am going to describe a trio of good methods that all have their advantages and disadvantages.


Interval Estimation – Change of Sign Method

Now, as are looking for the values of x when the graph crosses the x-axis, we know that when there is a root, f(x) will change sign, so assuming that the graph does not have any asymptotes or other breaks in it, we know that once we have located a interval with a root in it, f(x) will change sign.image08.png

For example, in the diagram to the left, we know that there is a root between 1 and 2.

Decimal Search

...read more.

Middle

1.  We then draw a tangent of the curve.  I.e. we draw the tangent of the curve at the point (x1,f(x1)).

As the equation of a straight line can be written

y-y1 = m(x-x1);

the equation of the tangent must be

y-f(x1) = f’(x1)(x-x1).

Where the tangent cuts the x axis, we will call the point (x2,0), so:

0-f(x1) = f’(x1)(x2-x1).

We can rearrange this to give:

x2 = x1-(f(x1)/f’(x1))

Which gives us the Newton-Raphson iteration formula, which is:

xn+1 = xn-(f(xn)/f’(xn))

When we use our cubic equation with this formula, we can values which will start to converge at a very quick rate, this makes using this formula a lot quicker than using the decimal search with the change of sign method which was explained earlier.

The method with which this is done is illustrated graphically on the next page.  Also, on the next page, is the way in which it is possible to find all three roots to the equation.  


If we plot this method of obtaining the roots in a table, we get the following:image06.png

x

f(x)

f'(x)

x'

-2

6.8

20

-2.34000000000000

-2.34000000000000

-2.58359200000000

32.94040000000000

-2.26156767980960

-2.26156767980960

-0.01675654070199

29.77062765342460

-2.26100482500865

-2.26100482500865

0.00075737613753

29.74828054340280

-2.26103028450156

-2.26103028450156

-0.00003455148215

29.74929124239710

-2.26102912307953

-2.26102912307953

0.00000157557904

29.74924513564450

-2.26102917604151

-2.26102917604151

-0.00000007184920

29.74924723815710

-2.26102917362635

-2.26102917362635

0.00000000327645

29.74924714227880

-2.26102917373649

-2.26102917373649

-0.00000000014942

29.74924714665100

-2.26102917373146

-2.26102917373146

0.00000000000682

29.74924714645160

-2.26102917373169

-2.26102917373169

-0.00000000000033

29.74924714646070

-2.26102917373168

-2.26102917373168

0.00000000000003

29.74924714646030

-2.26102917373168

-2.26102917373168

-0.00000000000001

29.74924714646030

-2.26102917373168

Where f(x) is applying the function to the value just found.  Where f(x) = 0 is where the difference between the tangent crossing the x-axis and the graph crossing the x-axis are the same.  The closest we managed to get in this case was 0.00000000000001, which is probably close enough.

So the root is -2.26102917373168 ± 0.000000000000005.


The other two roots are found in these following tables:

x

f(x)

f'(x)

x'

0

2

-14

0.14285714285714

0.14285714285714

0.01282798833819

-13.67346938775510

0.14379530916844

0.14379530916844

-0.00007910411767

-13.67011087238190

0.14378952252060

0.14378952252060

0.00000049930572

-13.67013163639910

0.14378955904590

0.14378955904590

-0.00000000315117

-13.67013150533860

0.14378955881538

0.14378955881538

0.00000000001989

-13.67013150616570

0.14378955881684

0.14378955881684

-0.00000000000013

-13.67013150616050

0.14378955881683

0.14378955881683

0.00000000000000

-13.67013150616050

0.14378955881683

The root found in this table is 0.143789558815385 ± 0.000000000000005.

x

f(x)

f'(x)

x'

5

312

216

3.55555555555556

3.55555555555556

89.59835390946500

103.33333333333300

2.68847471127041

2.68847471127041

24.10298233491410

53.73954116953500

2.23995985962932

2.23995985962932

5.36050535730701

33.39674141438490

2.07945002752342

2.07945002752342

0.72784915149579

26.99646178022800

2.05248911844500

2.05248911844500

0.04732869754988

25.96689335046130

2.05066646307722

2.05066646307722

0.00230660109554

25.89776294818390

2.05057739742703

2.05057739742703

0.00010973435526

25.89438636297440

2.05057315966083

2.05057315966083

0.00000521425557

25.89422570775330

2.05057295829333

2.05057295829333

0.00000024775204

25.89421807384790

2.05057294872547

2.05057294872547

0.00000001177175

25.89421771112760

2.05057294827086

2.05057294827086

0.00000000055932

25.89421769389320

2.05057294824926

2.05057294824926

0.00000000002657

25.89421769307440

2.05057294824824

2.05057294824824

0.00000000000126

25.89421769303550

2.05057294824819

2.05057294824819

0.00000000000006

25.89421769303360

2.05057294824819

2.05057294824818

-0.00000000000017

25.89421769303330

2.05057294824819

...read more.

Conclusion

However, to do this at any real speed, you need to have access to a powerful computer program, or be able to do large sums quite fast.  It is easy to do if you have a computer program like Excel or Autograph, but otherwise one of the other methods might be easier.

The Change of Sign method is not particularly fast.  It does get to a root, but you need to produce whole tables of data for each decimal place you obtain, and this is really quite a slow process.  On the plus side, it is possible to do this without any special hardware or software, a calculator and a piece of paper are enough to get to a few decimal places with this method, and if you have the time and patience, you can find many more decimal places.

The x = g(x) method is also slower than Newton Raphson in most cases.  It can be quite fast, but it often depends on where you start from, however, assuming you choose a reasonable value, you get quite a good rate of convergance.  This method is possible without a computer, but would be very time consuming, and it might be quicker to use the Change of Sign Method.

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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