# Difference between revisions of "Symbolic package"

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m (→Demos and usage examples: proofread a new example: fix typos and style) |
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## | ## | ||

syms y(x) | syms y(x) | ||

− | de =diff(y, 3 ) + | + | 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) | f = dsolve(de, y(0) == 0, diff(y,1)(0) == 0 , diff(y,2)(0) == 1) | ||

ff=function_handle(rhs(f)) | ff=function_handle(rhs(f)) |

## Revision as of 12:03, 6 April 2018

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

```
## 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
vectorfield = [sin(pi*y/2); -sin(pi*x/2)];
## Vector components are converted from symbolic into "anonymous functions" which allows them to be graphed.
## The "'vars', [x y]" syntax ensures each component is a function of both 'x' & 'y'
iComponent = function_handle (vectorfield(1), 'vars', [x y]);
jComponent = function_handle (vectorfield(2), 'vars', [x y]);
## Setup a 2D grid
[X,Y] = meshgrid ([-0.5:0.05:0.5]);
figure
quiver (X, Y, iComponent (X, Y), jComponent (X,Y))
```

**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)
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
```