Is there a way to force Infix.ParseOrThrow and related functions from the Math.NET Symbolics library to allow equations such as "4x + 3" instead of requiring the input to be "4*x + 3"?
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I'm new to Maple and I'm looking for a simple way to automate some tasks. In particular, I'm looking for a way to define custom "action" that perform some steps automatically.
As as an example I would like to define a quick way to compute the determinant of the Hessian of a polynomial. Currently the way I do this is opening Maple, create a new worksheet than performing the following commands:
p := (x, y) -> x^2*y + 3*x^3 + y^3
with(VectorCalculus):
h := Hessian(p(x, y), [x, y])
Determinant(h)
What I would like to do is to compute the hessian determinant directly with something like
HessDet(p)
where HessDet would be a custom command that performs the operations above. How does one achieve something like this in Maple?
First things first: The value assigned to your p is a procedure which can return a polynomial expression, but not itself a polynomial. It's important not to muddle expressions and procedures. Doing so is a common cause of problems for new users.
Being able to throw around p(x,y) may be visually pleasing to your eye, but it serves little programmatic purpose here. The fact that the formal parameters of procedure p happen to be called x and y, along with the fact that you called procedure p with arguments x and y, is actually just another common source of confusion. Don't create procedures merely to call them in this way.
Also, your call p(x,y) makes it look magic that your code snippet "knows" how many arguments would be required by procedure p. So it's already a muddle to have your candidate HessDet accept p as a procedure.
So instead let's keep it straightforward, by writing HessDet to accept a polynomial rather than a procedure. We can programmatically ascertain the names in which this expression of of type polynom.
restart;
HessDet:=proc(p::algebraic)
local H,vars;
vars:=indets(p,
And(name,Non(constant),
satisfies(u->type(p,polynom(anything,u)))));
H:=VectorCalculus:-Hessian(p,[vars[]]);
LinearAlgebra:-Determinant(H);
end proc:
Now some examples of using it,
P := x^2*y + 3*x^3 + y^3;
HessDet(P);
p := (x, y) -> x^2*y + 3*x^3 + y^3;
HessDet(p(x,y));
HessDet(x^3-x^2+4*x);
HessDet(s^2*t + 3*s^3 + t^3);
HessDet(s[r]^2*t[r] + 3*s[r]^3 + t[r]^3);
You might also wonder how you could re-use this custom procedure across sessions, without having to type it in each time. Two reasonable ways are:
Put the (above) defining plaintext definition of HessDet inside a personal initialization file.
Create a (.mla) Maple Library Archive file, then Save your HessDet to that, and then augment the Library search path in your initialization file.
It might look like 2) is more effort, but only the Save step is needed for repeats, and you can store many custom procedures to the same archive. Your choice...
[edit] The OP has asked for clarification of the first part of the above procedure HessDet, which I suspect means the call to indets.
If P is assigned an expression then then the call indets(P,name) will return a set of all the names present in that expression. Basically, it returns the set of all indeterminate subexpressions of the expression which are of type name in Maple's technical sense.
For example,
P := x*y + sin(a*Pi)*x;
x y + sin(a Pi) x
indets( P,
name );
{Pi, a, x, y}
Perhaps the name of the constant Pi is not wanted here. Ie,
indets( P,
And( name,
Non(constant) ) );
{a, x, y}
Perhaps we want only the non-constant names in which the expression is a polynomial? Ie,
indets( P,
And( name,
Non(constant),
satisfies(u->type(p,polynom(anything,u))) ) );
{x, y}
That last result is an advanced way of using the following tests:
type(P, polynom(anything, x));
true
type(P, polynom(anything, y));
true
type(P, polynom(anything, a));
false
A central issue here is that the OP made no mention of what kind of polynomials are to be handled by the custom procedure. So I guessed with some defensive coding, in hope of less surprises later on. The original Question states that the input could be a "polynomial", but we weren't told what kind of coefficients there might be.
Perhaps the coefficients will always be real and exact or numeric. Perhaps the custon procedure should throw an error when not supplied such. These details weren't mentioned in the Question.
Sorry for the complicated headline.
my basics idea is:
I have a function, lets say z(t) = x(phi(t)) + y(phi(t))
The twist where I couldn't find anything is:
suppose I want to calculate the symbolic derivation dz(t)/dt BUT without knowing neither x(t) nor y(t) specifically.
Can Matlab deliver something aloong the line
dz/dt=dx/d phi * dphi/dt + dy/dphi * dphi/dt
And if so, how do I have to program that?
How can I define unknown functions that have certain algebraic properties?
Example: I want to define the conjugation operator with the properties:
bar(a+b) = bar(a) + bar(b) and bar(a*b) = bar(a)*bar(b).
The goal is that maple should then use these properties to simplify a polynomial expression involving bars.
The command define might help with this somewhat. This creates evaluation rules rather than simplification rules for your operator, however.
So, I'm trying to establish some new stability criteria for my simulations, and this involves a lot of convoluted inequalities. I've worked through the math a few times by hand, and it's very laborous; so, I wanted to figure out a way to automate the process (as I'm trying to find the best integration scheme from a stability perspective). Is there anyway to solve inequalities symbolically in Matlab? Here's what I'm trying to solve. In the following expression, x refers to the gradient of a force function with respect to x, and t is the time step. In general, x < 0 and t > 0:
-(t*x + (2*t^3*x + t^2*x^2 - 2*t*x + 4*t + 1)^(1/2) + 1)/(x*t^2 - 2) < 1
Based on what I've looked at online, this seems to be possible in MuPAD, but using the following code does not give me any valid results:
solve(-(t*x + (2*t^3*x + t^2*x^2 - 2*t*x + 4*t + 1)^(1/2) + 1)/(x*t^2 - 2) < 1, t)
Any idea what I can do to make this work and automate the process?
First, since Wolfram Alpha gives you an answer (that I presume you've checked for correctness), I assume you want to use Matlab to solve other similar problems. However, this is a very non-trivial inequality due to the roots of the polynomials. I haven't been able to get Matlab/MuPAD to do anything with it. As I stated, regular Matlab can solve inequalities and systems of equalities in many cases, e.g., in R2013b
syms x real;
solve(x^3-1>1,x)
Even Mathematica 9 has trouble (the Reduce function can be used instead of it's Solve, but the output is not easy to use).
You can, however, solve for the real roots where x < 0 and t > 0 via
syms x t real;
assume(x<0);
assume(t>0);
f = -(t*x+sqrt(2*t^3*x+t^2*x^2-2*t*x+4*t+1)+1)/(x*t^2-2);
s = solve(f==1,t)
which returns:
s =
-(x + (x*(x - 2))^(1/2))/x
This simplifies to sqrt((x-2)/x)-1. Thus t > sqrt((x-2)/x)-1, one of the bounds provided by Wolfram Alpha. (The other more complicated bound is always less than zero and actually is the condition that ensures that t is real.)
But, do you even need to be solving this problem symbolically? Do you need explicit expressions for the various intervals in terms of all x? If not, this type of problem is probably better suited numeric approaches – either via root solving (e.g., fzero) or minimization (e.g., fmincon).
Just starting with Modelica and having trouble understanding how it works.
In the below 'method' of the model, qInflow and qOutflow are used in the second line to evaluate der(h), but they have not received a value yet! (they were not defined in the 'data' of the method)? In what order is the code executed.
equation
assert(minV >= 0, "minV must be greater or equal to zero");
der(h)=(qInflow - qOutflow)/area;
qInflow=if time > 150 then 3*flowLevel else flowLevel;
qOutflow=Functions.LimitValue(minV, maxV, -flowGain*outCtr);
error=ref - h;
der(x)=error/T;
outCtr=K*(error + x);
end FlatTank;
From http://www.mathcore.com/resources/documents/ie_tank_system.pdf
This is an understandable point of confusion when coming from languages and systems that utilize imperative semantics. But Modelica doesn't work like that.
When working with Modelica it is important to understand that an equation section contains equations, not assignments. Consider this, if I gave you the following equations:
x + y = 3;
x + 2*y = 5;
If you understand that this is a mathematical context, you can then determine that x must have a value of 1 and y must have a value of 2. In other words, you have to solve a system of simultaneous equations. You'll note that the left hand side of these equations are not variables (in general), they are expressions. An equation is simply a relationship that equates one expression, on the left hand side, with another expression, on the right hand side. Furthermore, this relationship is always true and so order is irrelevant.
This is quite different from imperative programming languages with imperative semantics. But it is also very powerful because you can state these relationships (linear systems of equations, non-linear systems of equations, implicit equations, etc) and the compiler will work out the most efficient way to solve them.
Getting back to your example, when you look at the code in your question you are interpreting those equations as assignment statements. This notion is reinforced because they just happen to have variables on the left hand sides. But they are really equations. In an equation based system, you do not worry about whether a given variable has been assigned to previously. Instead, the requirement is simply that for every variable there exists (somewhere) an equation and that there are no extra equations. In other words, you should have the same number of variables as unknowns and that the system of equations has a unique solution. That is all that Modelica requires.
Now, Modelica supports the kind of imperative semantics you are used to. But they are only to be used in special cases because they constrain the interpretation of the mathematical behavior in such a way that it interferes with the symbolic manipulation that allows Modelica compilers to generate really fast code. So it is more than a question of style. You should use equations if at all possible and algorithms in Modelica should only be used as a last resort.
One last note. Some people may be wondering "Are you telling me that these equations will be put into some giant system of equations and solved by matrix inversion or Newton-Raphson or something? Why make it so complicated when it could obviously be solved in a much easier way!" But it will not be solved as a giant system of equations. If it can be solved as a simple set of assignments it will. That is one (among many) of the different symbolic manipulation techniques that will be applied. In fact, this is a key point about Modelica...you don't need to worry about optimizing the solution method, the tool will take care of that. And more importantly, if you connect components in such a way that a simultaneous system does arise, you don't need to worry about that either. Modelica tools can handle such "algebraic loops" for you, they will optimize it to find the most computationally efficient formulation and won't depend on you reformulating your model for those cases.
Does that help?
You cannot know the execution order of the equations in a Modelica model until you run a Modelica tool on it (you can re-order any equation in the source model and get the same result). And then the order is only true for this tool with the settings you used.
This was the order chosen by the OpenModelica compiler (omc +s +simCodeTarget=Dump model.mo):
error = ref - h;
outCtr = K * (error + x);
der(x) = DIVISION(error, T, #SHARED_LITERAL_2(String#);
qOutflow = LimitValue(minV, maxV, (-flowGain) * outCtr);
qInflow = if time > 150.0 then 3.0 * flowLevel else flowLevel;
der(h) = DIVISION(qInflow - qOutflow, area, #SHARED_LITERAL_3(String#);
This example was a little boring because the left and right sides of no equation changed place (h = error - ref would be viable if h was not chosen as a state variable, etc).