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Pure Mathematics - In this coursework, I will be investigating and comparing the use of 3 different numerical methods for finding roots of equations.

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Introduction

Azeem Mir

Pure Mathematics 2 Coursework

Introduction

In this coursework, I will be investigating and comparing the use of 3 different numerical methods for finding roots of equations. The numerical methods that I will be using for this coursework are as follows:

  1. Change of sign: Decimal search
  2. Fixed point iteration: Newton-Raphson Method
  3. Rearranging f(x)=0 to the form x=g(x)

Change of Sign: Decimal Search

image00.png

For this method I will be using the above equation:  x³-4x²-3x+5=0. The first step for this method is to construct a table of values that range the values of the x-axis that the graph passes:

x

F(x)

  -5

-205

-4

-111

-3

-49

-2

-13

-1

3

0

5

1

-1

2

-9

3

-13

4

-7

5

15

F(x) changes sign between [-2,-1], [0,1] and [4,5]; indicating that a root lies between each of these x co-ordinates. The next few steps can now be repeated to obtain a more accurate value, taking x1 and x2 to be the lower and upper possible bounds of x. I shall be testing x between 4 and 5.

The table below gives the corresponding y-values for the f(x) values between x1=4 and x2=5. From this, it can be seen that a root lies between x=4 and x=5.

x

F(x)

4.0

-7

4.1

-5.61967

4.2

-4.07234

4.3

-2.35343

4.4

-0.45665

4.5

1.62554

4.6

3.89644

4.7

6.36388

4.8

9.03272

4.9

11.90922

5.0

15

The new values of x would be x1=4.4 and x2=4.5. These steps can be repeated to a necessary level of accuracy; each time gaining an extra decimal place. So far, x can be expressed as x=4.45 ±0.05, or 4.4<x<4.5.

...read more.

Middle

x

F(x)

1.0

2.49245

1.1

1.81617

1.2

1.20565

1.3

0.68818

1.4

0.29363

1.5

0.05452

1.6

0.00565

1.7

0.18412

1.8

0.62966

1.9

1.38411

2.0

2.49256

Fixed point iteration: Newton-Raphson Method

image02.png

To investigate this method, I shall use the function x³-3x²-3x+4=0. The equation is shown above. This method involves taking a rough approximation of x, and then improving this each time to get a more accurate value for x. Each successive iteration is calculated from the last using the formula: x1=x0-(F(x0))/(F’(x0)).

The differential of my formula is 3x²-6x-3=0. There is a root between x=0 and x=1 as indicated by the graph so I shall take my first value to be x=1. Therefore, the next value of x is calculated as follows:

                  x1=1-(F(1)/(F’(1))

                  x1=1-(-1/-6)

                  x1=0.833333333

Taking the x1=1, the new approximation of x is calculated as follows:

                  X2=0.833333333-(F(0.833333333)/(F’(0.833333333))

                  X2=0.833333333-(-1.963*10^-3/-5.916)

                  X2=0.832668188

And the next value of x:

                  X2=0.832668188-(F(0.832668188)/(F’(0.832668188))

                  X2=0.832668188-(-6.94*10-04/-5.92)

                  X2=0.832550958

Any other number of repetitions of this process past this point gives the same value of x of 0.83255, correct to 5 s.f.; i.e. ±0.000005.

To test that 0.83255 is a root, f(0.832545) and f(0.832555) are calculated. These equal

3.44*10-05 and -2.48*10-04 respectively. As there is a change of sign ±0.000005 either side of 0.83255, it can be concluded that this is a root.

This method can quickly find a very accurate estimate of the root in very few steps, with the accuracy limited only by the device used to calculate x.

...read more.

Conclusion

-0.00104

0.65669

-0.000257851

0.656629

-3.1761E-05

0.69

-0.12149

0.66

-0.0125

0.657

-0.00141

0.6567

-0.000294914

0.65663

-3.5468E-05

0.7

-0.157

So 0.65662 < x < 0.656621. By contrast, the equivalent result of ±0.00005 using the Newton-Raphson method takes only three iterations:

x

Δx

0.5

0.5

0.6471

0.1471

0.6566

0.009514

0.6566

4.76E-05

0.6566

1.20E-09

0.6566

2.22E-16

0.6566

1.11E-16

0.6566

0

image08.png

The G(x) iteration method is also faster than the decimal search method with 7 iterations:

G(x)

Δx

0.8

0.2

0.7024

0.0976

0.6693

0.03309

0.66

0.009342

0.6575

0.002476

0.6568

0.0006446

0.6567

0.000167

0.6566

4.32E-05

0.6566

1.12E-05

image09.png

So in speeds of convergence, the Change of Sign method takes 24 steps, x=g(x) method takes 7 steps and the Newton-Raphson method takes 3 for this equation. This order of speed is a good representation of most equations that I have encountered.

However, the Change of Sign method takes the least initial work; for the Newton-Raphson method, f'(x) needs to be calculated. For many possible equations, differentiating them is very tedious and for some advanced equations beyond my capabilities and  for each step, more calculations are required. The x=g(x) method requires rearranging the formula to start with, and in many cases, this results in a non-continuous function (e.g. involving odd roots). For continuous functions with more than one root, the G(x) method will usually fail to find at least one of them due to the differences in gradient steepness.

I believe that the Change of Sign method is the best for finding all the roots of a function, considering that I use specialized graph producing software (Autograph) to do all the calculations in these methods. The time taken to perform calculations is negligible, especially when placed in context with the time to calculate the f'(x) or g(x). Also, this method is the most reliable with most equations that I have encountered.

...read more.

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