# Difference between revisions of "Symbolic package"

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The symbolic package is part of the octave-forge project.

### Demos and usage examples

• 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);

% the next 2 line take the Anonymous function into a symbolic formula

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)

% this uses the interval pkg. to find all the roots between -15 an 10

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
# sring 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
```