Handout 2 - Root Finding using MATLAB

Transcription

1 Handout 2 - Root Finding using MATLAB Middle East Technical University MATLAB has couple of built-in root finding functions. In this handout we ll have a look at fzero, roots and solve functions. It is called as follows. fzero fzero(function, initial guess, options) It locates only one root. Function is given as a function of x. Inıtial guess can be a single value or an interval. The last options parameter is optional. To solve the Mars robot wheel example, we can first define the function as anonymous and then call fzero. f 0.4-x+sin(x); fzero(f, 0) The result is , which is exact. Or we can provide the function directly in fzero as a string in single quotes. fzero('0.4-x+sin(x)', 0) Warning: To see all digits of the result as given above, you need to first enter format long command on the command window. To see less details use format short. Warning: When you copy-paste things from Word document or a PDF file into MATLAB, MATLAB may complain about certain characters, such as the single quotes given above. You need to correct them in the way MATLAB wants to see them. Let s try to locate the first positive root of the vibrating cantilever beam problem fzero('cos(x)*cosh(x)+1', 0) The result is , which is the first negative root. To locate the first positive root, we either need to provide an initial guess that is closer to it such as 1, or we can provide an interval fzero('cos(x)*cosh(x)+1', [0, 5]) The result is , which is the exact first positive root. With fzero, it is NOT possible to locate multiple roots at once. We need to search for them one by one. fzero('cos(x)*cosh(x)+1', [3, 7]) fzero('cos(x)*cosh(x)+1', [7, 10]) This will find This will find

2 Warning: So, you need to know what you re doing. You cannot use fzero blindly. Middle East Technical University If you enter an interval with no root in it, you ll get an error message about the sign of the function being the same at the end points of the interval. fzero('cos(x)*cosh(x)+1', [0, 1]) What happens if you enter an interval that contains two (even number of) roots? fzero('cos(x)*cosh(x)+1', [0, 6]) Actually there are two roots inside [0, 6], but fzero cannot locate them. What happens if you enter an interval that contains three (odd number of) roots? fzero('cos(x)*cosh(x)+1', [0, 9]) Actually there are three roots inside [0, 9], and fzero finds the first one. It is unaware of the other too. Warning: So, you need to know what you re doing. Again. Let s try to find the root of the bungee jumper example fzero('sqrt(9.81*x/0.25)*tanh(sqrt(9.81*0.25/x)*4)-35', 0) The result is Not meaningful. Mass cannot be negative. Let s try a different initial guess. fzero('sqrt(9.81*x/0.25)*tanh(sqrt(9.81*0.25/x)*4)-35', 10) The result is , another negative number. Not physical. Warning: MATLAB knows nothing about bungee jumping, but you know. Let s try a different initial guess, which is closer to the root fzero('sqrt(9.81*x/0.25)*tanh(sqrt(9.81*0.25/x)*4)-35', 80) The result is This is what we are looking for. But finding it took many trials. Question: What s going on here? How does fzero work? 2

3 Let s try to set an option to see iteration details. We do this using the optimset command of MATLAB. We can set many different options with it. Below we set display option to iter, meaning that we want to see iteration details options = optimset('display', 'iter'); fzero(f,80, options) This will show the following details Search for an interval around 80 containing a sign change: Func-count a f(a) b f(b) Procedure initial interval search search search search search search search search Search for a zero in the interval [54.4, 105.6]: Func-count x f(x) Procedure initial interpolation e-06 interpolation e-09 interpolation interpolation Zero found in the interval [54.4, 105.6] ans = e+02 fzero uses a hybrid root finding technique known as Brent s method. You can read about it in your textbook but you are not responsible for it. It is a combination of several bracketing and open methods. When you provide fzero only a single initial guess, not an interval, it first tries to find an interval where the function changes sign. To find the interval, it goes to the right and to the left of the initial guess and check for sign changes. This is like incremental search, but more clever, i.e. the increment is not constant, adjusted automatically by looking at the properties of the function. Starting from the given initial guess of x = 80, it peforms many iterations and at iteration 17, it locates an interval [54.4, 105.6] where the sign changes. Than it uses a proper root finding method such as the bisection, the quadratic interpolation (see your textbook for this one, but you are not responsible for it) or the secant method. It decides on what to use by its own. Sometimes it uses a combination of these together. In the above example the convergence is achieved at iteration 21. The default tolerance value fzero uses for stopping is a low value around 10 16, which can be changed using the optimset command. 3

4 If instead of a single initial guess, we provide an interval, initial search part is skipped, as seen below fzero(f,[0,200], options) Func-count x f(x) Procedure initial interpolation bisection interpolation interpolation interpolation e-07 interpolation e-11 interpolation e-14 interpolation e-15 interpolation interpolation Here the root is found at 12 iterations by using bisection and quadratic interpolation methods. Exercise: Using the options argument as explained above, see what fzero actually did for the earlier cases where things went wrong. roots This function finds the roots of polynomials. Fzero can also be used to find roots of polynomials, but there are methods specially designed to find the roots of polynomials, such as the Müller s method or the Bairstow s method. See your textbook for details (you are not responsible for them). To find the roots of the polynomial 3x 2 2x 4 we use roots([3-2 -4]) The results is As seen the coefficients of the polynomial are provided as an array, starting from that of the highest order term. If one of the coefficients is zero do not skip it, instead give zero to it. For example to find the roots of 3x 2 4 we need to write roots([3 0-4]) If we do not write the middle zero, MATLAB will understand the polynomial to be 3x 4 4

5 solve This function solves equations and equation sets symbolically using the Symbolic Mathematics Toolbox of MATLAB. Therefore, it provides analytical (exact) answers To find the roots of the polynomial x 2 3x 4 we can use it as follows eqn = x^2-3*x - 4; The results are given as follows, which are exact and obtained without any numerical root finding procedure. 1/3-13^(1/2)/3 13^(1/2)/3 + 1/3 To find the root of sin(x) cos(x) = 0 function use it as follows eqn = sin(x) - cos(x); which gives the result as pi/4. This is the exact first negative root. Actually there are infinitely many roots. But the use of the solve function is limited. To find the root of the Mars robot wheel example, we can try to use it as follows, but it gives a warning and instead uses a numerical technique (probably the fzero function) to find the root. eqn = 0.4-x+sin(x); Warning: Unable to solve symbolically. Returning a numeric approximation instead. Same thing happens for the vibrating cantilever beam and bungee jumper examples, too. 5