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Numerical Solutions of Equations

Extracts from this document...

Introduction

        A2 Level Maths Coursework

A Level Coursework – Numerical Solutions of Equations

My coursework is based on finding roots of non-quadratic equations. I will be investigating 3 methods for finding these roots and comparing them.

Change Of Sign Method

image00.png

Y = x4+4x-6

This method simply uses Microsoft Excel to tabulate points on a function in different stages. It ‘narrows down’ the desired root by getting progressively closer to it. The root is always in the interval between the 2 x values where the y values change sign.

-5

599

-4

234

-3

63

-2

2

-1

-9

0

-6

1

-1

2

18

3

87

4

266

5

639

The roots of this equation lie in the intervals -2 < x < -1 & 1 < x < 2.

To find the lower root, I tabulated -2 < x < -1.image01.png

x

y

-2

2

-1.9

-0.5679

-1.8

-2.7024

-1.7

-4.4479

-1.6

-5.8464

-1.5

-6.9375

-1.4

-7.7584

-1.3

-8.3439

-1.2

-8.7264

-1.1

-8.9359

-1

-9

There is a change of sign between -2 & -1.9, therefore the boundaries of the root lie in this interval.

I tabulated in steps of 0.01, and again in steps of 0.001 & 0.0001

x

y

-1.924

0.007146

-1.9239

0.004697

-1.9238

0.002249

-1.9237

-0.0002

-1.9236

-0.00265

-1.9235

-0.00509

-1.9234

-0.00754

-1.9233

-0.00999

-1.9232

-0.01243

-1.9231

-0.01488

-1.923

-0.01732

x

y

-1.93

0.15488

-1.929

0.130146

-1.928

0.105457

-1.927

0.080812

-1.926

0.056212

-1.925

0.031657

-1.924

0.007146

-1.923

-0.01732

-1.922

-0.04174

-1.921

-0.06612

-1.92

-0.09046

x

y

-2

2

-1.99

1.722392

-1.98

1.449536

-1.97

1.181385

-1.96

0.917891

-1.95

0.659006

-1.94

0.404685

-1.93

0.15488

-1.92

-0.09046

-1.91

-0.33137

-1.9

-0.5679

...read more.

Middle

-1.92376

0.001269505

-1.92375

0.001024725

-1.92374

0.00077995

-1.92373

0.000535179

-1.92372

0.000290412

-1.92371

4.56502E-05

-1.9237

-0.000199108

image11.png

The other root of the equation can be found using this method. It is, to 5 significant figures, 1.1144.

Failure of the Change of Sign Method

The Change of Sign method only works for certain equations, e.g. where there is a change of sign or where a change of sign is shown in the first table of values. One equation where the Change of Sign method does not find the roots is y = 4x4-4x+1.

image12.png

x

y

-5

2521

-4

1041

-3

337

-2

73

-1

9

0

1

1

1

2

57

3

313

4

1009

5

2481

x

y

0

1

0.1

0.6004

0.2

0.2064

0.3

-0.1676

0.4

-0.4976

0.5

-0.75

0.6

-0.8816

0.7

-0.8396

0.8

-0.5616

0.9

0.0244

1

1


Newton Raphson Method

x3 – 10x + 6 = 0

image08.png

The Newton Raphson Iteration works as follows:

It is based on the formula

image13.png

F’(Xn) is F(Xn) differentiatied, so if F(Xn) = x3 – 10x + 6, than F’(xn) = 3x2 - 10. Iterations work by repeating a formula that eventually converges toward a root.

Xo is the original estimation of a root. Therefore we replace Xn with Xo.

My original estimate for the middle root is 3, so substitute 3 for Xo.

Therefore: f(Xn) = 33 – 10x3 + 6 = 3

And f’(Xn) = 3*32 – 10 = 17

X1 = 3 – (3/17) = 2.823529412 or 48/17

X2 = 2.803785064

X3 = 2.803542459

X4 = 2.803542496

X5 = 2.803542496

The other two roots of this equation are:

0.62434 & -3.4279, each to 5 significant figures.

This method can be illustrated graphically:

image14.png

image15.png

image16.png


Failure of the Newton-Raphson Method

However, the Newton-Raphson method does not work for all roots. One example of an equation where the Newton Raphson method does not work is y = 2/X3+6.

image17.png

x

y

-5

5.99936

-4

5.998047

-3

5.99177

-2

5.9375

-1

4

0

#DIV/0!

1

8

2

6.0625

3

6.00823

4

6.001953

5

6.00064

...read more.

Conclusion

However, the time it takes to set up and apply a method also affects it’s efficiency.

The Newton Raphson method took the least amount of steps to find the root, but also took the most time to substitute into the formula as the function had to be differentiated. Although this took a while, once the formula was in the calculator, it could easily be repeated for each Xn.

The Change of Sign method took 6 steps to converge to the root (up to 5 s.f.) but took by far the most time to work out. This was because the table on excel for each step had to be changed, while using the other methods a calculator could just re-iterate the chosen formula.

The Fixed Point iteration took more steps to converge to the root, twice that of Change of Sign and three times that of the Newton Raphson method. However, it was the quickest to find using hardware and software as the formula was easily re-arranged and could then be quickly written into and then re-iterated on a graphical calculator for each Xn.

...read more.

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