• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26

Numerical Method (Maths Investigation)

Extracts from this document...

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.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    The Gradient Function

    5 star(s)

    In this investigation, in the last section, for each value of n I have attempted to prove each gradient function is valid using general proof. Here, I shall try and do the same for when x = n. As a guideline, I shall use the previous proof of x� along the way to compare to it to x^n.

  2. MEI numerical Methods

    and F=(K+2)) C= F/K Thus C = 0.678176364 Assuming c = common factor, then ? at k =4 x C will equal ? when K = 5. 0.197900255 x 0.678176364 = 0.134211275 0.134211275 = 0.165395943, therefore this isn't a geometric progression.

  1. Sequences and series investigation

    nave obtained 'a' we can now substitute its value into the equation to find the other values. We can now find 'b' by substituting in 15 into the equation as follows: 5(_) + b = 4Y 5(_) = 6Y b = 4Y - 6Y b= -2 We have now found values a & b.

  2. In my coursework I will be using three equations to investigate their solutions using ...

    Staircase diagram method 1. choose a value of x 2. find the corresponding value of y 3. make the value of y into the new value of x 4.

  1. Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

    There are 8 columns to set up in order to use the method of bisection and they are as follows: - The number of bisection done. - Lower bound of the target root. - Output of the function when x=a - Upper bound of the target root.

  2. Newton Raphson Method for Solving 6x3+7x2-9x-7=0

    = 0.000011852. f(-0.634875) = -0.000041312 and f(-0.634885)= 0.000065017. The error bounds are ±0.000005 and the solution bounds are (-0.634875≤ x ≤-0.634885). I am going to find out the third root which is located between the intervals -1 and -2. The f(-2)

  1. C3 COURSEWORK - comparing methods of solving functions

    = 0.879385241 and f(0.879395) = 0.879385241 The root is trapped between 0.879385 and 0.879395 We can see that the root is 0.879385± 0.000005 or 0.87939 to 5 decimal places. Comparisons among all three methods Types of method Ranking of speed Number of calculations Change of sign 2nd 36 Newton Raphson 1st 5 x=g(x)

  2. Fractals. In order to create a fractal, you will need to be acquainted ...

    the pixels black if the value of c is bounded; else we color the pixel white if it goes to infinity.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work