# Numerical Method (Maths Investigation)

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Introduction

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 12 Rearrangement Method and its Failure 17 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: X-value F(X) ...read more.

Middle

EQUATION USED:, the required root is at [1.5,2] DECIMAL SEARCH A Table has been imported from Microsoft Excel producing the result of the changing of sign method. X F(X) 1.5 -0.01831093 2 1.389056099 This is the 1st step of the decimal search. Then come a few more steps in another table. X F(X) 1.51 -0.003269206 1.52 0.012225195 1.53 0.028176822 1.54 0.044590271 1.55 0.061470183 1.56 0.078821245 1.57 0.096648194 1.58 0.114955811 1.59 0.133748928 x F(X) 1.511 -0.001740211 1.512 -0.000206685 1.513 0.001331377 1.514 0.00287398 1.515 0.004421127 1.516 0.005972823 1.517 0.007529074 1.518 0.009089883 1.519 0.010655255 X F(x) 1.5121 -5.30826E-05 1.5122 0.000100565 1.5123 0.000254258 1.5124 0.000407996 1.5125 0.000561779 1.5126 0.000715608 1.5127 0.000869482 1.5128 0.001023402 1.5129 0.001177367 x F(x) 1.51211 -3.77199E-05 1.51212 -2.23567E-05 1.51213 -6.99314E-06 1.51214 8.37092E-06 1.51215 2.37354E-05 1.51216 3.91004E-05 1.51217 5.44658E-05 1.51218 6.98317E-05 1.51219 8.5198E-05 Average of Upper and Lower Limit = Error Bound = 0.000005 Hence, the approximation root value is 1.512135 0.000005, which is equal to the example shown above in Newton-Raphson section. NEWTON-RAPHSON METHOD ***Shown in NEWTON-RAPHSON METHOD SECTION, Page 5 REARRANGEMENT METHOD The two possibilities of the equation are: CASE 1: CASE 2: Test each case whether rearrangement method can be apply here by substituting each x in CASE 1 and X in case 2 with the interval of [ 1.5,2] which is 1.75. Using a calculator, When x = 1.75, CASE 1 g/(X) = 1.9182 (> 1), therefore I can't use this case as it will lead me to the failure of basis iteration. When X = 1.75, CASE 2 g/(x) = 0.5714 (<1 and >-1). Since it match the condition and the rearrangement method can be used here. From Microsoft Excel: Iteration Xn Xn+1 Difference between Xn and Xn+1 1 1.50000 1.50408 0.004077397 2 1.50408 1.50679 0.002714577 3 1.50679 1.50860 0.001803185 4 1.50860 1.50979 0.001195989 5 1.50979 1.51058 0.000792469 6 1.51058 1.51111 0.000524749 7 1.51111 1.51146 0.000347321 8 1.51146 1.51169 0.000229819 9 1.51169 1.51184 0.00015204 10 1.51184 1.51194 0.000100571 11 1.51194 1.51200 6.65204E-05 12 1.51200 1.51205 4.39958E-05 13 1.51205 1.51208 2.90972E-05 14 ...read more.

Conclusion

For NEWTON-RAPHSON METHOD Table of Formulas that I have use imported from Microsoft Excel XP. A B C D Xn f(Xn) f'(Xn) Xn+1 1.5 = EXP(x1_) - 3*x1_ =EXP(x1_) - 3 =(x1_) - (y1_ / m) =D15 = EXP(x1_) - 3*x1_ =EXP(x1_) - 3 =(x1_) - (y1_ / m) =D16 = EXP(x1_) - 3*x1_ =EXP(x1_) - 3 =(x1_) - (y1_ / m) =D17 = EXP(x1_) - 3*x1_ =EXP(x1_) - 3 =(x1_) - (y1_ / m) This is more complicated than the Decimal Search as more formulas need to apply and names need to assign to. Since after the first time, X0, we assign our own value, but later, X1, X2, etc, we use the values we find in Xn+1, so I make the second row of Xn as "=D15" which is the first row of column Xn+1. I have to think of the differentiation of the f(Xn) my self and type in the formula under column f ' (Xn). In Xn+1, the m is the name I assigned for f '(Xn). I use the iteration formula to calculate my Xn+1. For REARRANGEMENT METHOD Table of formula below is what I used in my Coursework for examples and Comparison. A B C D Xn Xn+1 ***Difference between Xn and Xn+1 f(Xn) 1.5 =LN(3*x1_) =x2_-x1_ =EXP(x1_) - 3*x1_ =B43 =LN(3*x1_) =x2_-x1_ =EXP(x1_) - 3*x1_ =B44 =LN(3*x1_) =x2_-x1_ =EXP(x1_) - 3*x1_ =B45 =LN(3*x1_) =x2_-x1_ =EXP(x1_) - 3*x1_ =B46 =LN(3*x1_) =x2_-x1_ =EXP(x1_) - 3*x1_ =B47 =LN(3*x1_) =x2_-x1_ =EXP(x1_) - 3*x1_ =B48 =LN(3*x1_) =x2_-x1_ =EXP(x1_) - 3*x1_ Basis Iteration, Xn+1 = g(Xn) is what I use to find Xn+1. The g(X) of f(X) = ex-3x is ln (3x). Therefore you can see under Xn+1 column, "=LN(3*x1_)". Here, like Newton-Raphson Method, Xn always takes in Xn+1 value except the first one. I add in two extra items, the difference between Xn and Xn+1 and f(Xn). I can see that the difference between Xn and Xn+1 is getting smaller and smaller while f(Xn) is getting nearer and nearer to 0 which indicates the distance from the required root. ~~~END OF APPENDIX A~~~ P2 Coursework Cody Tang - 1 - ...read more.

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