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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 term. It is possible to solve algebraically but it had certain level of difficulties because it involves imaginary number and this increase the difficulties of solving cubic equations algebraically. A problem also occur because  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, 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)

Middle

0.672982

After 6 converge we can see that the 1st target root is to , with an error bound and solution bound.

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 is to , with an error bound and solution bound.

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 isto , with an error bound and solution bound.

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

‘The equation  had 1 solution which comprising 3 roots,

, to. Their error bounds are  and their solution bound are, and  respectively.’

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

This is part of the equation, I will focus on region  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 then 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

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 5 2

After 5 steps the sequence becomes and it is clearly that this sequence is diverging to. 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. Rearrangeinto form

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

:

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  and  respectively. They are then shown by computer graphically:

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

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

-number of iteration done.

- value of x

- output of x when

- 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

Conclusion

converges 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  and the error bound is  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 and the other two methods states it is. But there is onlyerror 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 20Generally more than 15 6Generally less than 10 8Generally between 10 -20 Time consuming? very time consuming in generalCan be fast when starting point is near to the root. usually fastspeed of convergence depends on the gradientcan be speed up when starting point near to the root usually fastif restriction - achieve and also gradient is small, it will be speed up. Error / error bound small errorError can be build up during bisection. small errorDeveloped manually small errorDeveloped 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 1st.This method fails badly if. Degree of ease with computers. HighCan be done by excel, autograph HighCan be done by excel, autograph HighCan be done by excel, autograph Degree of ease with paper, calculator and pencil Too complicated, take too much step to doAlso very time consuming.Slow converges. Quicker than method of bisection in general.had to be worked out. Relatively quick.Iteration can be calculate with calculators.

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