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Investigate the solution of equations, comparing the following methods, Systematic search for change of sign using a decimal search, Fixed point iteration using the Newton-Raphson method, Fixed point iteration after rearranging the equation f(x)=0 into th

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Introduction

Pure Mathematics 2: Component 02 (Coursework)

During this coursework, I intend to investigate the solution of equations, comparing the following methods.

1. Systematic search for change of sign using a decimal search
2. Fixed point iteration using the Newton-Raphson method
3. Fixed point iteration after rearranging the equation f(x)=0 into the form x=g(x)

After implementing these three equations to find the same root of the equation, I will compare the methods in terms of speed of convergence and whether available hardware/software simplify the problem.

Change of Sign Method

The equation I intend to use is…

f(x) = x³ - 4x² - 11x +10

When using this equation I am assuming that f(x) = y = 0

This equation cannot be solved by normal algebraical methods, such as factorising, which is why these numerical methods must be used.

Here is an unaltered graph of the line of the equation above.

If y changes sign in an interval, a root lies in between that interval.  I know for a fact that there is a root between 0 and 1 as the lines cuts the x-axis (where y=0) somewhere between those two points.

f(0) = 10

f(1) = -4

There is a change of sign between these points, which further confirms my theories.

Middle

0.683

0.71

0.531511

0.72

0.379648

0.73

0.227417

0.74

0.074824

0.75

-0.07813

0.76

-0.23142

0.77

-0.38507

0.78

-0.53905

0.79

-0.69336

0.80

-0.848

The table showing the change of sign to 4 decimal places shows that the change of sign occurs between 0.7448 and 0.7449.  .

It is positive when 0.7448 is inserted in to the equation.  f(0.7448) = 0.001453

It is negative when 0.7449 is inserted in to the equation.  f(0.7449) = -0.0000077

So, the root is 0.74485 plus or minus 0.00005.  This is true to 4 decimal places.  I therefore need to show this graphically.  I have demonstrated this in the graph below.   It clearly shows that the line cuts throught the x-axis in between 0.7448 and 0.7449.

Where this method fails:  Although this was successful, this cannot be said for some other examples, one of which I will demonstrate.  I will use a parabola as an example.  If I take the equation y=(x-3)², it is obvious that this is a parabola as a minimum.  This means that it only touches the x-axis at one point.  It is a repeated root.  If the change of sign method is used at any point, the sign will never change as the entire parabola is positive – never crossing the x-axis.

Newton-Raphson Method

When using this method, I will gind all of the roots of a second equation and I will illustrate one of these graphically.  I will be using this equation…, where f(x) = 0

f(x)

Conclusion

Comparison of Methods

I will now compare how the three methods manage to find the same root, the speed of convergence to find that root and how easy each method is to implicate with the software and hardware present.  Earlier, I found that a root of y=x³-4x²-11x+10, is 0.74485 plus or minus 0.00005.  I will now find the same root using the Newton-Raphson method (using 0 as the starting point).

x2 = x1 - x1³-4x1²-11x1+10

3x1²-8x1-11

x2 = 0.90909

x3 = 0.74735

x4 = 0.74490

x5 = 0.74489

x6 = 0.74489

So, there has been repetition to four significant figures and these four figures are the same as that of the earlier method (0.7448).  This has been successful.  I will now employ the rearrangement method, using the same starting point (0).

Making x the subject of y=x³-4x²-11x+10

xn+1=x1³-4x1²+10

11

x2 = 0.90909

x3 = 0.67687

x4 = 0.77068

x5 = 0.73472

x6 = 0.74885

x7 = 0.74335

x8 = 0.74550

x9 = 0.74466

x10 = 0.74499

x11 = 0.74486

x12  = 0.74491

x13  = 0.74489

x14  = 0.74490

x15  = 0.74489

x16  = 0.74490

x17  = 0.74489

x18  = 0.74490

x19  = 0.74489

x20  = 0.74489

So, it took 20 iterations for this oscillating method to get the root it was converging upon to 4 significant figures repeating twice to show it was finally there.  This method, therefore is very slow in finding the root, it takes 20 steps to converge, although there is no need to do the change of sign to check for error bounds as it continually oscillates anyway.  (That is an added bonus!)

This has created a cobweb diagram.

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