Theme 1: Solving Nonlinear Equations
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1 Theme 1: Solving Nonlinear Equations Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 1 / 22
2 Sources of Errors Finite decimal representation (Rounding): Finite decimal representation will be used to represent numbers (computer memory is finite). Irrational numbers do not have exact decimal representation, anyway. Need to decide on number of decimals to represent numbers! Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 2 / 22
3 Sources of Errors I Examples: 2 = using 16 digits π = = such numbers do not have an exact representation, truncated! Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 3 / 22
4 Rounding Solving Nonlinear Equations For example, approximate 2, π using 5 decimal places (digits): 2 = becomes 2 = (rounding and chopping) π = becomes π = (chopping), π = (rounding) Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 4 / 22
5 Rounding Solving Nonlinear Equations 2 3 using 5 decimal places (digits): 2 3 = becomes = (rounding) = (chopping), Note: if the 6th decimal digit 5 add 1 to the 5th digit - rounding! Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 5 / 22
6 Rounding Let x be such that 0 x < 1: x = 0.d 1 d 2... = lim n (d d d n 10 n ) d i {0, 1, 2,..., 9} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, i = 1, 2, 3,..., n Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 6 / 22
7 Rounding Example: 2 3 x = = lim n ( n ) Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 7 / 22
8 Rounding Approximate x by y with only k decimal digits: where y = 0.d 1 d 2... d k d i = d i for i k 1; d k = d k if d k+1 4 and d k = d k + 1 if d k+1 5. Note: x y 5 10 k 1 < 10 k Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 8 / 22
9 Rounding Solving Nonlinear Equations Example: 2 3 with k = 5 and x = i < k 1: i.e. i 4, di = 6: y 1 = ; d5 = d = 7 since d 6 5; y = y 2 = x y = < 10 5, as expected! Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 9 / 22
10 Significant Digits Write a number r with finite decimal representation in the form: where 0.1 x < 1 r = x 10 n correct decimals in x are significant digits of r; rounding r to k significant digits amounts to rounding x to k decimals. Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 10 / 22
11 Significant Digits 2 3 = , rounded to how many signifiant digits? π = , and π = rounded to how many significant digits? r = and r = , rounded to how many significant digits? Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 11 / 22
12 Significant Digits If y = x correct to k decimals, we say correct to k significant digits. r = y 10 n = while = correct to 3 significant digit = correct to 4 significant digits. 129 = 130 correct to 2 significant digits. Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 12 / 22
13 Finite Digit Arithmetic Arithmetic Operations with rounded numbers errors! Round the real numbers after each arithmetic operation. Single precision: 16 significant digits. Double precision: 32 significant digits. Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 13 / 22
14 Finite Digit Arithmetic - Example Calculate with 4-digit arithmetic. Solution: = = = = Note: the digits in red have to be removed by rounding! Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 14 / 22
15 Sources of Errors II Solving Nonlinear Equations Inaccurate data: Measurements are never exact, see Example 3 on Page Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 15 / 22
16 Bisection Method Bisection Method The bisection method is a interval division (bracketing) method. Find a solution of f (x) = 0, when it is known that in [a, b] the function f (x) is continuous and equation has a root i.e. f (x) will have opposite signs at endpoints of interval. Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 16 / 22
17 Bisection Method Solving Nonlinear Equations Bisection Method If a function changes sign over an interval, the function value at the midpoint is evaluated. Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 17 / 22
18 Bisection Method Solving Nonlinear Equations Bisection Method The location of the root is then determined as lying within the subinterval where the sign change occurs (x NS is numerical solution). Figure: A.Gilat, V. Subramaniam, Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 18 / 22
19 Bisection Method The absolute error is reduced by a factor of 2 for each iteration. Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 19 / 22
20 Bisection Method Figure: Source: A.Gilat, V. Subramaniam, Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 20 / 22
21 Bisection Method Bisection Method, contd/... Algorithm: (1) Choose interval [a, b] in which root lies i.e. f (a)f (b) < 0. (2) Calculate the first estimate of numerical solution x NS1 using: x NS1 = a + b 2. Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 21 / 22
22 Bisection Method Bisection Method, contd/... Algorithm: (3) Determine on which side of x NS1 the root lies: if f (a)f (x NS1 ) < 0, the true solution is between a and x NS1. if f (a)f (x NS1 ) > 0, the true solution is between x NS1 and b. (4) Select the interval that contains root as the new interval [a, b], and go back to Step 2. Prof. Mapundi K. Banda (Tuks) WTW263 Semester II 22 / 22
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