Upcoming SlideShare
×

# 03 finding roots

937 views

Published on

A lecture about Finding Root in Computational Physics course at Physics Department, Institut Teknologi Bandung, Indonesia

Published in: Education, Technology
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
937
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
19
0
Likes
0
Embeds 0
No embeds

No notes for slide

### 03 finding roots

1. 1. 0101011000000011101000110001 1 0 Finding Roots Sparisoma Viridi Nuclear Physics and Biophysics Research Division Institut Teknologi Bandung, Bandung 40132, Indonesia dudung@fi.itb.ac.id v2011.09.20 FI3102 Computational Physics 1
2. 2. 0101011000000011101000110001 Outline 1 • Function and roots 0 • A simple physical problem and its solution • Several methods in finding roots • Graphical method • Newton’s method, secant method • Bisection method • References v2011.09.20 FI3102 Computational Physics 2
3. 3. 0101011000000011101000110001 Function 1 • Function: 0 – Output: f(x) – Input: x (the argument) • The argument can be a real number or elements of a given set (matrix, vector, etc) • See reference [1] v2011.09.20 FI3102 Computational Physics 3
4. 4. 0101011000000011101000110001 Roots 1 • Argument that makes a function to 0 produce zero output [2, 3] • Argument that make two functions to produce same output [4] • Question 1. Is the second definition actually the same as the first one? v2011.09.20 FI3102 Computational Physics 4
5. 5. 0101011000000011101000110001 Roots (cont.) 1 • Example 0 of roots for f(x) = cos (x) in interval [-2π,2π], see [5] v2011.09.20 FI3102 Computational Physics 5
6. 6. 0101011000000011101000110001 Roots (cont.) 1 • Say that there are two functions: 0 – First function is f(x) – Second function is g(x) • The second definition of roots is argument that makes two functions produce same output • If x is a root than f(x) = g(x) v2011.09.20 FI3102 Computational Physics 6
7. 7. 0101011000000011101000110001 Roots (cont.) 1 • Or, we can introduce a new function 0 – The new function is h(x) = f(x) – g(x) • If x is a root, it makes f(x) = g(x) • If x is a root, it also makes h(x) = 0 • So, both definitions are actually the same (Answer 1) v2011.09.20 FI3102 Computational Physics 7
8. 8. 0101011000000011101000110001 A simple physical problem 1 • A ball is launched from bottom of a incline 0 plane as illustrated below, find where and when does it hit the plane! v0 H (xH, yH) α β O (0, 0) v2011.09.20 FI3102 Computational Physics 8
9. 9. 0101011000000011101000110001Solving the problem analytically 1 • The ball motions in x and y direction as 0 function of time t are x(t) = v0 [cos (α + β)] t y(t) = v0 [sin (α + β)] t – ½ gt2 then y(t) = y[t(x)] = y(x) will give y = f(x) = [tan (α + β)] x – [g / 2v02 cos2(α + β)] x2 • The equation of incline plane is y = g(x) = [tan β] x v2011.09.20 FI3102 Computational Physics 9
10. 10. 0101011000000011101000110001 The roots of the problem 1 • In this case we can use second definition 0 of roots, so f(x) = g(x) → x is a root • Or we construct a new function h(x) h(x) = f(x) – g(x) = [tan (α + β) – tan β] x – [g/2v02 cos2(α + β)] x2 • The view of both definition in finding roots will be shown in the next slide v2011.09.20 FI3102 Computational Physics 10
11. 11. 0101011000000011101000110001 Graphics of f(x), g(x), and h(x) 1 0 h(x) f(x) root g(x) v2011.09.20 FI3102 Computational Physics 11
12. 12. 0101011000000011101000110001 Graphics of .. (cont.) 1 • Even both graphics show different curve, 0 but they give the same value of x (the root we want to find) • Parameters are: α = π/6, β = π/6, v0 = 5, g = 10, x0 = 0, y0 = 0, ∆x = 0.025 • The root should be around x = 1.4 • And vertical position around y = 0.85 • v2011.09.20 FI3102 Computational Physics 12
13. 13. 0101011000000011101000110001 Analytical solution 1 • Using quadratic formula we can find the 0 roots • But this case is simpler since other root are 0 • Then it can found that x = [tan (α + β) – tan β] [2v02 cos2(α + β)] / g = 1.443376 f(x) = f(1.443376) = 0.833333 v2011.09.20 FI3102 Computational Physics 13
14. 14. 0101011000000011101000110001 Analytical solution (cont.) 1 • Question 2. If we already know the 0 analytical solution, then for what is it the numerical solution? v2011.09.20 FI3102 Computational Physics 14
15. 15. 0101011000000011101000110001 Analytical solution (cont.) 1 • Answer 2. It is for testing whether the 0 numerical solution has the same value as the analytical one • You can not (or should not) use your method or algorithm or code to some problem, before you test it • Test it for the problem that you have already known its solution v2011.09.20 FI3102 Computational Physics 15
16. 16. 01010110000000 Several methods in finding roots11101000110001 1 • Graphical method 0 • Newton method • Secant method • Einschlussverfahren method [4]: bisection, regula falsi, pegasus, Anderson-Björck, Zeroin, Illinois, King, Anderson-Björck- King, etc) v2011.09.20 FI3102 Computational Physics 16
17. 17. 0101011000000011101000110001 Graphical method 1 • It uses graphics to find the root 0 • It can accommodate the first and second definition of the root • It is rather not accurate since it depends on our judgment to determine whether f(x) = g(x) or h(x) = 0 • Some software, i.e. gnuplot, has feature to return the position of mouse pointer v2011.09.20 FI3102 Computational Physics 17
18. 18. 0101011000000011101000110001 Graphical method (cont.) 1 • We (actually) already used this method 0 root analytical solution x = 1.443376 graphical y = 0.833333 solution x = 1.4 y = 0.85 v2011.09.20 FI3102 Computational Physics 18
19. 19. 0101011000000011101000110001 Newton’s method 1 • Taylor’s expansion around x = a will give 0 f (x) ≈ f (a) + (x – a)f (a) with assumption that (x – a) small enough to terminate next term in the series • By setting x = 0 (the target) we can find that x = a – f (a) / f (a) • Or, the iterative expression is xi+1 = xi – f (xi) / f (xi) v2011.09.20 FI3102 Computational Physics 19
20. 20. 0101011000000011101000110001 Newton method (cont.) 1 • Question 3. Write the iterative expression 0 for previous problem using Newton’s method [use h(x) instead of f(x) and g(x)] and show also the result • From previous problem: h(x) = [tan (α + β) – tan β] x – [g/2v02 cos2(α + β)] x2 h(x) = ? xi+1 = xi + ? v2011.09.20 FI3102 Computational Physics 20
21. 21. 0101011000000011101000110001 Newton method (cont.) 1 • Answer 3. It can be found that 0   g  2  [tan (α + β ) − tan β ]x i −  xi   2 v o cos (α + β )   2 2  xi +1 = xi −    [tan (α + β ) − tan β ] −  g  2    2 xi   v o cos (α + β )  2   v2011.09.20 FI3102 Computational Physics 21
22. 22. 0101011000000011101000110001 Newton method (cont.) 1 • And the result is 0 • Only need about 6 iterations • x = 1.443376 (analytical solu- tion is 1.443376) • In a good agree- ment wtih a.s. v2011.09.20 FI3102 Computational Physics 22
23. 23. 0101011000000011101000110001 Secant method 1 • In Newton’s method we must know or 0 calculate the derivative of f(x) • In secant method it is calculated using f (xi) ≈ [f(xi) – f(xi-1)] / (xi – xi-1) • And this gives the iterative expression xi+1 = xi – (xi – xi-1) f (xi) / [f(xi) – f(xi-1)] with xi and xi-1 as starting values v2011.09.20 FI3102 Computational Physics 23
24. 24. 0101011000000011101000110001 Secant method (cont.) 1 • Secant method (right) is more efficient 0 than Newton’s method (left) as illustrated in [3] v2011.09.20 FI3102 Computational Physics 24
25. 25. 0101011000000011101000110001 Secant method (cont.) 1 • The result is 0 • Only (also) need ~ 6 iterations • x = 1.443376 (analytical solu- tion is 1.443376) • In a good agree- ment wtih a.s. v2011.09.20 FI3102 Computational Physics 25
26. 26. 0101011000000011101000110001 Newton’s vs secant method 1 Iteration Newton’s Secant method Difference method 0 1 1.796544 1.675235 0.121309 2 1.501396 1.383598 0.117798 3 1.445534 1.434796 0.010738 4 1.443379 1.443749 -0.00037 5 1.443376 1.443373 2.23E-06 6 1.443376 1.443376 5.76E-10 7 1.443376 1.443376 -8.9E-16 8 1.443376 1.443376 -2.2E-16 9 1.443376 1.443376 4.44E-16 10 1.443376 1.443376 0 v2011.09.20 FI3102 Computational Physics 26
27. 27. 0101011000000011101000110001 Newton’s vs secant ... (cont.) 1 • For Newton’s method x0 = 1 0 • For secant method x0 = 1 and x1 = 1.1 • Both methods gives similar result at about 10 iterations (difference less than 10–16) v2011.09.20 FI3102 Computational Physics 27
28. 28. 0101011000000011101000110001 Bisection method 1 • Suppose that we have a function as in [6] 0 v2011.09.20 FI3102 Computational Physics 28
29. 29. 0101011000000011101000110001 Bisection method (cont.) 1 • The algorithm: 0 v2011.09.20 FI3102 Computational Physics 29
30. 30. 0101011000000011101000110001 Example: x2 – 4x + 3 1 0 v2011.09.20 FI3102 Computational Physics 30
31. 31. 0101011000000011101000110001 Example: x2 – 4x + 3 (cont.) 1 0 v2011.09.20 FI3102 Computational Physics 31
32. 32. 0101011000000011101000110001 Example: x2 – 4x + 3 (cont.) 1 • The roots have been found 0 x1 ≈ 2.999 (analytical solution is 3) x2 ≈ 1.001 (analytical solution is 1) • Further iteration (more than 10) can be performed to get nearer results • What about muliple roots? Any idea? v2011.09.20 FI3102 Computational Physics 32
33. 33. 0101011000000011101000110001 Multiple roots 1 • After finding one root the original equation 0 must be modified • Question 4. Do you know how to modify the original equation in order to obtain other root? v2011.09.20 FI3102 Computational Physics 33
34. 34. 0101011000000011101000110001 Multiple roots (cont.) 1 • Answer 4. Divide the original equation 0 with all previous found root(s) in order to avoid the previous root(s) to be found again • Could you write the algorithm together with the finding roots algorithm? v2011.09.20 FI3102 Computational Physics 34
35. 35. 0101011000000011101000110001 Multiple roots (cont.) 1 • Multiple roots algorithm (1st root) 0 – Finding roots algorithm • Multiple roots algorithm (2nd root) – Finding roots algorithm • Multiple roots algorithm (3rd root) – Finding roots algorithm • .. for the nth root .. v2011.09.20 FI3102 Computational Physics 35
36. 36. 0101011000000011101000110001 Summary 1 • Three different methods for finding roots 0 have been presented • A simple physical problem is used to illustrate the first two method • Other problems which need to find roots can also be solved using given methods • Algorithm to solve multiple roots problem has not yet been discussed (only the idea) v2011.09.20 FI3102 Computational Physics 36
37. 37. 0101011000000011101000110001 References 1 1. Wikipedia contributors, “Function (mathe- 0 matics)”, Wikipedia, The Free Encyclopedia, 18 September 2011, 10:26 UTC, oldid:451121927> [2011.09.19 07.31+07] 2. Paul L. DeVries, “A First Course in Computational Physics”, John Wiley & Sons, New York, First Edition, 1994, pp. 41-85 v2011.09.20 FI3102 Computational Physics 37
38. 38. 0101011000000011101000110001 References (cont.) 1 3. Steven E. Koonin and Dawn C. Meredith, 0 “Computational Physics Fortran Version”, Westview, Canada, First Edition, 1990, pp. 11-14 4. Gisela Engeln-Müllges, Klaus Niederdrenk, and Reinhard Wodicka, “Numerik-Algorithmen”, Springer, Berlin, 10. Auflage, 2011, pp.25-90 v2011.09.20 FI3102 Computational Physics 38
39. 39. 0101011000000011101000110001 References (cont.) 1 5. Wikipedia contributors, “Zero of a 0 function”, Wikipedia, The Free Encyclopedia, 12 September 2011, 02:23 UTC, oldid:449926806 [2011.09.19 07.48+07] 6. Sparisoma Viridi, “Computational Physics”, Version 3, 2011, pp. 49-53 v2011.09.20 FI3102 Computational Physics 39
40. 40. 0101011000000011101000110001 1 0 Thank you (for your patience) v2011.09.20 FI3102 Computational Physics 40