 Level: AS and A Level
 Subject: Maths
 Word count: 4262
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.
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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.
 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  6.81716E06  3x2+6x+0.5  2.37593  2.86926E06  1.203364214 
4  1.203364  5.02265E12  3x2+6x+0.5  2.37593  2.11397E12  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  1.31124E06  3x2+6x+0.5  3.979329  3.29513E07  2.469617 
6  2.469617  4.77618E13  3x2+6x+0.5  3.979326  1.20025E13  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  2.93769E05  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.
 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*B2^3B2^2+1  =(3/2)*B2^22*B2 
2  =C2  =0.5*B3^3B3^2+1  =(3/2)*B3^22*B3 
3  =C3  =0.5*B4^3B4^2+1  =(3/2)*B4^22*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
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  20 Generally more than 15  6 Generally less than 10  8 Generally between 10 20 
Time consuming? 


if restriction  achieve and also gradient is small, it will be speed up. 
Error / error bound 



Restriction/ disadvantages 



Degree of ease with computers. 



Degree of ease with paper, calculator and pencil 



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