Numerical Method of Algebra.

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NUMERICAL METHOD

INTRODUCTION

It is very useful to use Numerical Method to find the roots of an equation that cannot be solved ALGEBRAICALLY. Quadratic equations in the format of can be solved by Quadratic Formula: . However, for polynomial equations, that have highest power more than 2, has to be solved through Trial and Error, which is very hard and tedious to determine their roots. Moreover, some roots of polynomial equations may be not Integer or Large numbers, which make things harder.

Therefore NUMERICAL METHOD is developed to help to solve polynomial equations. In P2, we learned that there are two types of Numerical Method:

* Change of Sign Method

*

o Decimal Search

o Interval Bisection

o Linear Interpolation

* Fixed Point Iteration

*

o Newton-Raphson Method

o Rearrangement Method

About Coursework

Things have to do for Coursework:

* One kind of Change of Sign Method;

* Newton-Raphson Method;

* Rearrangement Method;

* Failures of each methods;

* Error Bound of each methods;

* Comparison made with the three methods above; and

* Ease of use and Availability of Hardware and Software to do coursework.

CONTENTS:

PAGE No.

Description

2

Change of Sign Method and its Failure

8

Newton-Raphson Method and its Failure

2

Rearrangement Method and its Failure

7

Comparison made onto one roots of an equation with the 3 Methods above

22

Ease of use and Availability of Hardware and Software used, e.g.:

* T1 Calculator

* Graphmatica or Autograph

* Microsoft Excel

* Microsoft Word

24

Appendix A

CHANGE OF SIGN METHOD

I choose to use Decimal Search, as it is the easiest of all methods through the use of Microsoft Excel. It is to find the roots of an equation that crosses the x-axis. The positive value and negative values of the y-value has to be observed.

EQUATION USED: = 0 has only 1 roots in [0,-1]

From Graphmatica:

Results obtained from Microsoft Excel:

To show Microsoft Excel is correct:

When x = -1

X-value

F(X)

-1

-1.5

Tbl DS-01: Step 1

0

0.5

Observe the change of y-value from negative value to positive value in the table. That shows that the root of the equation where the y-value is 0, is between x = 0 and x = -1. Now this is step 1, we're moving on to step 2.

To show Microsoft Excel is correct:

When x = -1

X-value

F(X)

-0.9

-0.7882969

-0.8

-0.3497152

-0.7

-0.0723543

-0.6

Tbl DS-02: Step 2

0.1120064

Now the search is narrowing in and it comes to one decimal place in between x = -1 and x = 0. Look up for the change of sign here and it is between x = -0.6 and x = -0.7. Hence, the roots is now between x = -0.6 and x = -0.7.

Gph DS-02

Now I move on to step 3, step 4 and so on until the fifth decimal place. The more decimal place the more accurate the answer of the root is.

X

Y

-0.69

-0.050563533

-0.68

-0.029629888

-0.67

-0.009507116

-0.66

0.009848393

-0.65

0.028477721

-0.64

0.046419535

-0.63

0.063710194

-0.62

0.080383854

-0.61

0.096472572

Tbl DS-03: Step 3

X

Y

-0.669

-0.007537736

-0.668

-0.005575983

-0.667

-0.003621818

-0.666

-0.001675197

-0.665

0.000263921

-0.664

0.002195576

-0.663

0.004119811

-0.662

0.006036665

-0.661

0.007946179

Tbl DS-04: Step 4

X

Y

-0.6659

-0.001480948

-0.6658

-0.001286774

-0.6657

-0.001092676

-0.6656

-0.000898652

-0.6655

-0.000704703

-0.6654

-0.000510829

-0.6653

-0.000317029

-0.6652

-0.000123304

-0.6651

7.03455E-05

7.03455E-05 represents

Tbl DS-05: Step 5

X

Y

-0.66519

-0.000103936

-0.66518

-8.45685E-05

-0.66517

-6.52016E-05

-0.66516

-4.58355E-05

-0.66515

-2.64701E-05

-0.66514

-7.1055E-06

-0.66513

.22584E-05

-0.66512

3.16215E-05

-0.66511

5.09839E-05

Tbl DS-06: Step 6

I have stopped at Step 6, as it will get more and more numbers behind the decimal point if I continue on. However, I haven't got to an answer that I satisfy. I only got its range, -0.66514 < root of the equation used < -0.66513. So, I decide to get their error bound by:

Average of the upper and lower error bound =

=

Upper Error Bound =

Lower Error Bound =

The Error Bound of the root of this equation is

Approximate Root Value is

FAILURE OF CHANGE OF SIGN METHOD

Gph DS-05

However, this Change of Sign Method comes to a failure when solving a graph that has an equal root, i.e. having the x-axis as its tangent to the curve. For example the equation=0.

When x = 2
Join now!


When x = 2.6

By referring to the table below, this proves that Microsoft Excel is correct.

A Set of values from Microsoft Excel:

X

F(x)

0

Tbl DS-07: From 0 to 0.9

-1.6641

0.1

-1.4161

0.2

-1.1881

0.3

-0.9801

0.4

-0.7921

0.5

-0.6241

0.6

-0.4761

0.7

-0.3481

0.8

-0.2401

0.9

-0.1521

x

f(x)

-0.0841

.1

Tbl DS-08: From 1 to 1.9

-0.0361

.2

-0.0081
...

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