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Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and also disadvantage in these methods and they will be discussing throughout this investigation.

Extracts from this document...

Introduction

Numerical Solution of Equation

INTRODUCTION

In many situations, equation can be solving by algebra and by graphical analysis. However when equation involves terms which had a higher order than two then it became very difficult to solve by algebra. In this investigation we are focus of cubic functions – functions which involves termimage65.png. It is possible to solve algebraically but it had certain level of difficulties because it involves imaginary number image66.pngand this increase the difficulties of solving cubic equations algebraically. A problem also occur because image66.png was not introduce before early 16 century therefore mathematicians use numerical analysis instead of solving cubic equation algebraically. Three numerical methods are discussed in this investigation. There are advantages and also disadvantage in these methods and they will be discussing throughout this investigation.

In cubic functions, there is 1 solution which constitute by 3 or less distinct roots. All roots of the equations are laid on the x axis, in other words it is intercepting linear equation x = 0. Three methods we are discussing in this investigation is base on this concept and derive to find one target root in an equation.

  1. Method of Bisection

 This is an example of method of interval estimation; it was used in continuous function. In continuous function f(x) = 0, if it intercepts liner equation x = 0 then this implies that f(x) =y,image131.png must have at least one real root for solution. One observation can be made is variable y will have a ‘sign change’ on either side of roots, it means when the y value on the left hand side of root is negative then the right hand side of the root must be positive.                

 It can be illustrate graphically:

Example 1) image139.png

image144.png

image127.png

image00.png

...read more.

Middle

0.672982

After 6 converge we can see that the 1st target root isimage146.png to image86.png, with an error bound image87.pngand solution boundimage147.png.

Other 2 root are also found:

Target root 2:

n

xn

f(xn)

f'(x)

f'(xn)

f(xn)/f'(xn)

xn+1

1

-1

-0.5

3x2+6x+0.5

-2.5

0.2

-1.2

2

-1.2

-0.008

3x2+6x+0.5

-2.38

0.003361345

-1.203361345

3

-1.203361

-6.81716E-06

3x2+6x+0.5

-2.37593

2.86926E-06

-1.203364214

4

-1.203364

-5.02265E-12

3x2+6x+0.5

-2.37593

2.11397E-12

-1.203364

5

-1.203364

0

3x2+6x+0.5

-2.37593

0

-1.203364

6

-1.203364

0

3x2+6x+0.5

-2.37593

0

-1.203364

After 5 converge we can see that the 2nd target root isimage148.png to image86.png, with an error bound image87.pngand solution boundimage149.png.

Target root 3

n

xn

f(xn)

f'(x)

f'(xn)

f(xn)/f'(xn)

xn+1

1

-3

-3.5

3x2+6x+0.5

9.5

-0.368421053

-2.631578

2

-2.631578

-0.764397142

3x2+6x+0.5

5.48615

-0.139332173

-2.492246

3

-2.492246

-0.092318832

3x2+6x+0.5

4.180401

-0.022083725

-2.47016

4

-2.47016

-0.002172495

3x2+6x+0.5

3.984138

-0.000545286

-2.469617

5

-2.469617

-1.31124E-06

3x2+6x+0.5

3.979329

-3.29513E-07

-2.469617

6

-2.469617

-4.77618E-13

3x2+6x+0.5

3.979326

-1.20025E-13

-2.469617

7

-2.469617

0

3x2+6x+0.5

3.979326

0

-2.469617

After 7 converge we can see that the 3rd target root isimage150.pngto image151.png, with an error bound image87.pngand solution boundimage152.png.

After three solutions were found then we can now create a statement:

‘The equation image153.png had 1 solution which comprising 3 roots,

image146.png,image148.pngimage150.png toimage86.png. Their error bounds are image155.png and their solution bound areimage147.png,image149.png and image152.png respectively.’

Newton Raphson’s method can also be shown graphically by computers:

image156.png

image157.png

image20.png

This is part of the equationimage158.png, I will focus on region image159.png because the root is between this region. ‘Autograph’ is also helpful to find a root by Newton Raphson’s method. Gradient can be drawn by autograph easily. We have to use the function – ‘Newton Raphson iteration’ and choose the value of image75.pngthen a gradient will be drawn by this software automatically. This is efficient way to find the target root without too much process.

However there are situations which this method fails:

Discontinuous function

image160.pngimage21.png

image24.pngimage161.pngimage23.png

image25.png

image04.pngimage26.pngimage27.png

A spread sheet had also been set up for help:

n

xn

f(xn)

f'(xn)

f(xn)/f'(xn)

xn+1

1

-2

1.8

-0.48

-3.75

1.75

2

1.75

2.119626

-0.131476985

-16.12165179

17.87165179

3

17.87165

2.000175

-2.93769E-05

-68086.64236

68104.51401

4

68104.51

2

image162.png

image163.png

image164.png

5

image165.png

2

image166.png

image167.png

image168.png

After 5 steps the sequence becomesimage168.png and it is clearly that this sequence is diverging toimage170.png. This is an example of failure case to Newton Raphson’s method.

The reason is because the gradient –f'(xn), the gradient is so small and on the process f(xn)/f'(xn) it becomes infinitely large and consequence xn+1 is very large and this value diverges to infinity.

  1. Rearrangeimage115.pnginto formimage111.png

 This is also a method by using fix point iteration and algebra was also involved. The basic concept is arrange image115.png into a new form - image111.png and we can then derive two equations – image69.pngandimage171.png. We can then able to solve image115.pngby determine the intersections of two equations and also with the help of fix point iteration.

image67.png:

image93.pngimage68.png

This equation had infinity number of arrangement, therefore one of them will be chosen by myself and I choose two new equations and they are image69.png and image70.png respectively. They are then shown by computer graphically:

image28.png

image72.pngimage71.png

image29.png

image30.pngimage22.png

image32.png

image33.png

image34.png

image73.png

image74.png

Then we set up a fixed point -image75.png and it can be any value but near to the target root and we are able to find a better approximation callimage76.png, this sequence can repeat and eventually estimations approach to the target root will be found.

image77.pngimage35.png

image36.pngimage79.png

image37.png

image38.png

image80.png


Microsoft Excel can also helpful and therefore a spread sheet is set up:

n

xn

g(xn)

g'(x)

1

0.5

=0.5*B2^3-B2^2+1

=(3/2)*B2^2-2*B2

2

=C2

=0.5*B3^3-B3^2+1

=(3/2)*B3^2-2*B3

3

=C3

=0.5*B4^3-B4^2+1

=(3/2)*B4^2-2*B4

image78.png-number of iteration done.

image81.png- value of x

image82.png- output of x when image83.png

image84.png- gradient of xn

n

xn

g(xn)

g'(xn)

1

0.500000

0.812500

-0.625

2

0.812500

0.608032

-0.63477

3

0.608032

0.742693

-0.66151

4

0.742693

0.653240

-0.658

5

0.653240

0.712654

-0.6664

6

0.712654

0.673094

-0.66349

7

0.673094

0.699419

-0.6666

8

0.699419

0.681886

-0.66506

9

0.681886

0.693559

-0.66632

10

0.693559

0.685785

-0.66558

11

0.685785

0.690961

-0.66612

12

0.690961

0.687514

-0.66578

13

0.687514

0.689810

-0.66601

14

0.689810

0.688281

-0.66586

15

0.688281

0.689299

-0.66597

16

0.689299

0.688621

-0.6659

17

0.688621

0.689073

-0.66594

18

0.689073

0.688772

-0.66591

...read more.

Conclusion

image116.pngconverges and unlike other two methods they take at least 15 steps to converges to the root..

Error is very small in all 3 situations because it is correct to image86.png and the error bound is image87.png therefore the error is relatively small. There is a slight disagreement of the root between rearranging method and other two methods. The rearranging method states that the root is image85.pngand the other two methods states it isimage113.png. But there is onlyimage118.pngerror between two roots therefore the error is relatively low in this case.


Table compares and contrasts between three methods

Bisection method

Newton Raphson

Rearrangement

Steps

20

Generally more than 15

6

Generally less than 10

8

Generally between 10 -20

Time consuming?

  • very time consuming in general
  • Can be fast when starting point is near to the root.
  • usually fast
  • speed of convergence depends on the gradient
  • can be speed up when starting point near to the root
  • usually fast

if restriction - image103.pngachieve and also gradient is small, it will be speed up.

Error / error bound

  • small error
  • Error can be build up during bisection.
  • small error
  • Developed manually
  • small error
  • Developed manually

Restriction/

disadvantages

  • It is not working in discontinuous equations.
  • Roots cannot be close together, otherwise target root may not be found.
  • The starting value cannot be a turning point, maximum and minimum.
  • It is also not work for discontinuous equations.
  • Function had to be rearranging into image111.png1st.
  • This method fails badly ifimage119.png.

Degree of ease with computers.

  • High
  • Can be done by excel, autograph
  • High
  • Can be done by excel, autograph
  • High
  • Can be done by excel, autograph

Degree of ease with paper, calculator and pencil

  • Too complicated, take too much step to do
  • Also very time consuming.
  • Slow converges.
  • Quicker than method of bisection in general.
  • image120.pnghad to be worked out.
  • Relatively quick.
  • Iteration can be calculate with calculators.

...read more.

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