The symbolic package is part of the octave-forge project.
Demos and usage examples
- I'm trying to substitute a double value into an expression. How can I avoid getting "warning: Using rat() heuristics for double-precision input (is this what you wanted?)".
In general, you should be very careful when converting floating point ("doubles") to symbolic variables, that's why the warning is bothering you.
## Demo of how to use a number (which was calculated in an octave ## variable) in a symbolic calculation, without getting a warning. ## use octave to calculate some number: a = pi/2 ## now do some work with the symbolic pkg syms x f = x * cos (x) df = diff (f) ## Now we want to evaluate df at a: # subs (df, x, a) # this gives the "rats" warning (and gives a symbolic answer) ## So instead, try dfh = function_handle (df) dfh (a) ans = -1.5708 ## And you can evaluate dfh at an array of "double" values: dfh ([1.23 12.3 pi/2]) ans = -0.82502 4.20248 -1.57080
- Demo of Anonymous function to symbolic function and back to anonymous function and then the use of the interval pkg.
% this is just a formula to start with, % have fun and change it if you want to. f = @(x) x.^2 + 3*x - 1 + 5*x.*sin(x); % these next lines take the Anonymous function into a symbolic formula pkg load symbolic syms x; ff = f(x); % now calculate the derivative of the function ffd = diff(ff, x); % and convert it back to an Anonymous function df = function_handle(ffd) % this uses the interval pkg to find all the roots between -15 an 10 pkg load interval fzero (f, infsup (-15, 10), df) ans ⊂ 4×1 interval vector [-5.743488743719015, -5.743488743719013] [-3.0962279604822407, -3.09622796048224] [-0.777688831121563, -0.7776888311215626] [0.22911205809043574, 0.2291120580904359]
- Demo of inputting a function at the input prompt and making an Anonymous function.
# This prog. shows how to take a # string input and make it into an anonymous function # this uses the symbolic pkg. disp("Example input") disp("x^2 + 3*x - 1 + 5*x*sin(x)") str_fucn=input("please enter your function ","s") fucn_sym=sym(str_fucn) f=function_handle(fucn_sym) # now back to symbolic syms x; ff=formula(f(x)); % now calculate the derivative of the function ffd=diff(ff); % and convert it back to an Anonymous function df=function_handle(ffd) % now lets do the second derivative ffdd=diff(ffd); ddf=function_handle(ffdd) % and now plot them all x1=-2:.0001:2; plot(x1,f(x1),x1,df(x1),x1,ddf(x1)) grid minor on legend("f","f '", "f '' ")
- Demo of ODE with a step input and initial conditions.
## This is a demo of a second order transfer function and a unit step input. ## in laplace we would have 1 1 ## _______________ * _____ ## s^2 + sqrt(2)*s +1 s ## ## So the denominator is s^3 + sqrt(2) * s^2 + s # and for laplace initial conditions area ## t(0)=0 t'(0) =0 and the step has initial condition of 1 ## so we set t''(0)=1 ## In the code we use diff(y,1)(0) == 0 to do t'(0)=0 ## ## I know that all this can be done using the control pkg ## But I used this to verify that this solution is the ## same as if I used the control pkg. ## With this damping ratio we should have a 4.321% overshoot. ## syms y(x) de =diff(y, 3 ) +sqrt(2)*diff(y,2) + diff(y) == 0; f = dsolve(de, y(0) == 0, diff(y,1)(0) == 0 , diff(y,2)(0) == 1) ff=function_handle(rhs(f)) x1=0:.01:10; y=ff(x1); plot(x1,y) grid minor on