Editing Symbolic package

The {{Forge|symbolic|symbolic package}} is part of the [[Octave Forge]] project.

[[Category:Octave Forge]]

=== 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.

<source lang="octave">

## 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

</source>



* '''Demo of how to graph symbolic functions (by converting SYMBOLIC functions into ANONYMOUS functions)'''

<source lang="octave">

##The following code will produce the same vector field plot as Figure 1.14 from Example 1.6 (pg. 39) from A Student's Guide to Maxwell's Equations by Dr. Daniel Fleisch.

##Make sure symbolic package is loaded and symbolic variables declared.
pkg load symbolic;
syms x y

##Write a Vector Field Equation in terms of symbolic variables 
vector = [sin(pi .* y ./2 ); -sin(pi .* x ./2 )];

##Vector components are converted from symoblic into anonymous form which allows them to be graphed
##The '1' & '2', in vector(1) & vector(2) are indeces for the vector.
##The " 'vars', [x y]" syntax tells function_handle that each component is a function of both 'x' & 'y'
iComponent = function_handle (vector(1), 'vars', [x y]); 
jComponent = function_handle (vector(2), 'vars', [x y]); 

##Setting the Grid up to graph all axes for all variables from -.5 to .5, with spacings .05 apart. 
[X,Y] = meshgrid([-.5:.05:.5]);

##Draw the Vector Field!
figure
quiver(X,Y,iComponent(X,Y),jComponent(X,Y))

</source>


* '''Demo of Anonymous function to symbolic function and back to anonymous function and then the use of the interval pkg.'''

<source lang="octave">
% 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]
</source>


* '''Demo of inputting a function at the input prompt and making an Anonymous function.'''

<source lang="octave">
# 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 '' ")
</source>


* '''Demo of ODE with a step input and initial conditions.'''

<source lang="octave">

 ## 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) 
 sqrt2=sym(1.41421);
 de =diff(y, 3 ) +sqrt2*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 

</source>