# Changes

,  07:41, 10 June 2019
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Remove redundant Category:Packages.
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The {{Forge|symbolic|symbolic package}} is part of the octave-forge project.
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The {{Forge|symbolic|symbolic package}} is part of the [[Octave Forge]] project.
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[[Category:Octave-Forge]]
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[[Category:Octave Forge]]
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=== Demos and usage examples ===
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* '''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?)".'''
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In general, you should be very careful when converting floating point ("doubles") to symbolic variables, that's why the warning is bothering you.
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<source lang="octave">
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## Demo of how to use a number (which was calculated in an octave
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## variable) in a symbolic calculation, without getting a warning.
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## use octave to calculate some number:
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a = pi/2
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## now do some work with the symbolic pkg
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syms x
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f = x * cos (x)
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df = diff (f)
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## Now we want to evaluate df at a:
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# subs (df, x, a)    # this gives the "rats" warning (and gives a symbolic answer)
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dfh = function_handle (df)
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dfh (a)
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ans = -1.5708
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## And you can evaluate dfh at an array of "double" values:
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dfh ([1.23 12.3 pi/2])
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ans =
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-0.82502  4.20248  -1.57080
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</source>
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* '''Demo of how to graph symbolic functions (by converting SYMBOLIC functions into ANONYMOUS functions)'''
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<source lang="octave">
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## 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.
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## Make sure symbolic package is loaded and symbolic variables declared.
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syms x y
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## Write a Vector Field Equation in terms of symbolic variables
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vectorfield = [sin(pi*y/2); -sin(pi*x/2)];
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## Vector components are converted from symbolic into "anonymous functions" which allows them to be graphed.
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## The "'vars', [x y]" syntax ensures each component is a function of both 'x' & 'y'
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iComponent = function_handle (vectorfield(1), 'vars', [x y]);
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jComponent = function_handle (vectorfield(2), 'vars', [x y]);
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## Setup a 2D grid
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[X,Y] = meshgrid ([-0.5:0.05:0.5]);
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figure
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quiver (X, Y, iComponent (X, Y), jComponent (X,Y))
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</source>
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* '''Demo of Anonymous function to symbolic function and back to anonymous function and then the use of the interval pkg.'''
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<source lang="octave">
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% have fun and change it if you want to.
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f = @(x) x.^2 + 3*x - 1 + 5*x.*sin(x);
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% these next lines take the Anonymous function into a symbolic formula
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syms x;
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ff = f(x);
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% now calculate the derivative of the function
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ffd = diff(ff, x);
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% and convert it back to an Anonymous function
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df = function_handle(ffd)
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% this uses the interval pkg to find all the roots between -15 an 10
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fzero (f, infsup (-15, 10), df)
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ans ⊂ 4×1 interval vector
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[-5.743488743719015, -5.743488743719013]
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[-3.0962279604822407, -3.09622796048224]
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[-0.777688831121563, -0.7776888311215626]
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[0.22911205809043574, 0.2291120580904359]
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</source>
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* '''Demo of inputting a function at the input prompt and making an Anonymous function.'''
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<source lang="octave">
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# This prog. shows how to take a
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# string input and make it into an anonymous function
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# this uses the symbolic pkg.
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disp("Example input")
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disp("x^2 + 3*x - 1 + 5*x*sin(x)")
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fucn_sym=sym(str_fucn)
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f=function_handle(fucn_sym)
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# now back to symbolic
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syms x;
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ff=formula(f(x));
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% now calculate the derivative of the function
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ffd=diff(ff);
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% and convert it back to an Anonymous function
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df=function_handle(ffd)
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% now lets do the second derivative
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ffdd=diff(ffd);
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ddf=function_handle(ffdd)
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% and now plot them all
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x1=-2:.0001:2;
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plot(x1,f(x1),x1,df(x1),x1,ddf(x1))
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grid minor on
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legend("f","f '", "f '' ")
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</source>
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* '''Demo of ODE with a step input and initial conditions.'''
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<source lang="octave">
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## This is a demo of a second order transfer function and a unit step input.
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## in laplace we would have        1                      1
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##                              _______________        *  _____
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##                            s^2 + sqrt(2)*s +1          s
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##
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## So the denominator is s^3 + sqrt(2) * s^2 + s
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# and for laplace initial conditions area
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##            t(0)=0 t'(0) =0  and the step has initial condition of  1
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## so we set  t''(0)=1
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## In the code we use diff(y,1)(0) == 0 to do t'(0)=0
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##
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## I know that all this can be done using the control pkg
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## But I used this to verify that this solution is the
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##  same as if I used the control pkg.
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## With this damping ratio we should have a 4.321% overshoot.
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##
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syms y(x)
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sqrt2=sym(1.41421);
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de =diff(y, 3 ) +sqrt2*diff(y,2) + diff(y) == 0;
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f = dsolve(de, y(0) == 0, diff(y,1)(0) == 0 , diff(y,2)(0) == 1)
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ff=function_handle(rhs(f))
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x1=0:.01:10;
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y=ff(x1);
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plot(x1,y)
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grid minor on
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</source>
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