Items tagged with indets

indets(AA(MM(AA(X1,X2),B2),MM(AA(B2,X3),X5)));
this can get a list of X1,X2,B2,X3,X5

propositionsentence := MM(AA("it", "run"), "maple"):
indets(propositionsentence);

but this can not get "it", "run", "maple"

how to do this in case this in maple 12?
 

A wealth of knowledge is on display in MaplePrimes as our contributors share their expertise and step up to answer others’ queries. This post picks out one such response and further elucidates the answers to the posted question. I hope these explanations appeal to those of our readers who might not be familiar with the techniques embedded in the original responses.

The Question: Variable Identification

Don Carota wanted to know the best approach for finding variables that are not assigned to a value within an equation. He wrote:

I got a set of equations to solve, like this one:

eq[1]:=W[1,0]*(a*HRa[1,0]+b*ga[1,0]+c)=d*SR[1,1]/grid

a,b,c,d are numbers, like 2.0458 and so on.

When I want to solve the set, I need to tell Maple the command solve:

solve( {seq(eq[i],i=1..N)},{variables});  (N is an integer of course)

To set the variables, one must check each equation to write: {W[1,0],HRa[1,0],ga[1,0]...} and so on.

I know that I can use the command is(variable,assignable) to check if a variable has not a value assigned already and, according to true/false I can construct the set {variables} and solve the set of equations.

That´s an easy solution if I need to check variables under a certain pattern, like: X[1], X[2], X[3] since I can create a loop and check them one after the other. But my problem is that I have different names for variables or that variables change to assignable from assigned according to other conditions, so I can never be sure if W[1,0] is going to be a variable to compute in all steps instead of SR[1,1].

for example:

if a>3 then
SR[1,1]:=1/2;
else
W[1,0]:=300:
end if;

So, when I need to type solve, the {variables} part is different according to each case. Is there any command that allows me to insert an expression and Maple can return me the variables or parameters in the expression that are not numeric already?

(note that the link added to the is command above was added by me, not the original author)

dharr and Carl Love provided solutions that use the indets command.

The code provided by dharr is as follow:

  1. indets(eq[1],name);

Result: gives the indeterminates: {a, b, c, d, HRa[1, 0], SR[1, 1], W[1, 0], ga[1, 0]}

The code provided by Carl Love is as follows:

1.       indets(eq[1], assignable(name));

or, doing all equations at once,

2.       indets({entries(eq, nolist)}, assignable(name));

 

Further Explaining the indets and type commands.

Both dharr and Carl Love provided an answer that used the indets command. In essence the indets command used in this example contains two arguments: indets(expr, typename). Expr is a rational expression that only uses the operations such as addition, subtraction, division, and multiplication. Typename is a parameter used when the desired return is a set containing all subexpressions in expr that are of type typename.

Carl Love used the assignable(name) argument  for the typename parameter in order to return all the variables that can be assigned a value, excluding constants such as Pi that are also considered names. Indeed, assignable is a type and can be used without an additional argument. For example, the command indets(x+f(x)+y=1, assignable) returns {x,y,f(x)} because all three symbols can be assigned values. However, indets(x+f(x)+y=1, assignable(name)) returns just {x,y} because f(x) is of type function, not of type name. Similarly, indets(x+y=Pi, assignable) would return just {x,y} because Pi is not considered to be something that can be assigned to.

Carl’s second command used ({entries(eq, nolist)} as the expr parameter. In this version, eq is the table whose members are the individual equations. Remember, the syntax x[1] creates a table whose name is x, and whose entry is the object assigned to x[1]. The entries(t) function returns a sequence of the table members, each member being placed in list brackets. By including the option nolist, the return is then a sequence of table members without list brackets. 

Finally, note that different programmers will use different approaches to finding “indeterminants” in expressions. Dr. Lopez pointed out that some years ago he asked three different programmers about extracting the “assignable names” in an expression such as q:=x+Pi+cos(a). The naive indets(q) returns {a,x,cos(a)}, whereas indets(q,name) returns {Pi,a,x}. However, select(type,indets(q),name) returns {a,x}, as does indets(q,And(symbol,Not(constant))).

Don Carota’s question is able to showcase some of the different types that are within Maple’s platform. Therefore, it is important to go over what the type-checking function is and what it does. In many contexts, it is not necessary to know the exact value of an expression; instead it is enough to know if the value belongs to a group of expressions that have similarities. Such groups are knows as types.

Maple’s engine uses the type function in every single procedure to direct and maintain the flow of control in algorithms and to decide if the user’s input is valid. There are about 240 different types that Maple recognizes including: prime, string, symbol, list, etc.  Let’s see some examples of how the function works using elements from this question. 

Type has two parameters: an expression e, and a valid type expression t. To check that the output of the entries(eq,nolist) is indeed not a list, the type command can be used as follows:

As expected, the last command returns false! If you want to learn more about the type and indets commands you can visit their corresponding help pages: ?type, ?indets.

 

This blog was written by Maplesoft’s intern Pia under the supervision of Dr. Robert Lopez. We both hope that you find this useful. If there is a particular question on MaplePrimes that you would like further explained, please let us know. 

Hello,

I would like to extract from a equation only the list of variables depending of time.

eq:= x0(t)+sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))*xb[1]-sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))*zb[1]-sin(alpha0(t))*cos(beta0(t))*yb[1]+cos(alpha0(t))*sin(gamma0(t))*zb[1]+cos(alpha0(t))*cos(gamma0(t))*xb[1]-l2[1]*(sin(psi[1](t))*sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))-cos(psi[1](t))*sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))+sin(psi[1](t))*cos(alpha0(t))*cos(gamma0(t))+cos(psi[1](t))*cos(alpha0(t))*sin(gamma0(t)))-l3[1]*(sin(theta[1](t))*sin(psi[1](t))*sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))+sin(theta[1](t))*cos(psi[1](t))*sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))+cos(theta[1](t))*sin(psi[1](t))*sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))-cos(theta[1](t))*cos(psi[1](t))*sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))-sin(theta[1](t))*sin(psi[1](t))*cos(alpha0(t))*sin(gamma0(t))+sin(theta[1](t))*cos(psi[1](t))*cos(alpha0(t))*cos(gamma0(t))+cos(theta[1](t))*sin(psi[1](t))*cos(alpha0(t))*cos(gamma0(t))+cos(theta[1](t))*cos(psi[1](t))*cos(alpha0(t))*sin(gamma0(t)))-xp[1](t) = 0;

I think the indets function should help be.

indets(lhs(eq)-rhs(eq), function(t))

The last argument of the indets is wrong but should enable to explain my need.

Thanks a lot for your help

Can anyone tell me what is going on in the following worksheet? 


restart:

 

The following type  matches any second-or-higher-order derivative specified in D form with independent variables. You don't need to understand how this type works---which is, admittedly, convoluted---in order to understand the rest of this post.


TypeTools:-AddType(
     HODD,
     {typefunc(
           name,
           typefunc(
                name,
                {'`@@`'(identical(D), posint),
                 And(specindex(posint, D), satisfies(D-> nops([op(D)]) > 1))
                }
           )
      )
     }
);

 

Here's an expression which is simply a sum of various types of derivatives. Note that the first and last terms only differ in the dependent variable name.


expr:=
     D[1,2](u)(x,t) + diff(u(x,t), x, t) + diff(f(x),x$2) +
     diff(f(t),t) + (D@@3)(f)(x) + D(g)(x) + D[1,2](v)(x,t)
;

(D[1, 2](u))(x, t)+diff(diff(u(x, t), t), x)+diff(diff(f(x), x), x)+diff(f(t), t)+((D@@3)(f))(x)+(D(g))(x)+(D[1, 2](v))(x, t)

I1:= indets(expr, HODD);

{(D[1, 2](u))(x, t), (D[1, 2](v))(x, t), ((D@@3)(f))(x)}

The above result is as expected: Ds of order greater than 1 are selected; diffs are not. Now I try to extract the diffs also.

 

indets(expr, HODD &under (convert, D));

{diff(diff(f(x), x), x), diff(diff(u(x, t), t), x), (D[1, 2](v))(x, t), ((D@@3)(f))(x)}

The above result is missing D[1,2](u)(x,t) even though it contains the syntactically identical term D[1,2](v)(x,t)! Running trace  on `type/&under` shows that it never gets called for that missing term! So, the problem is not in &under. A further test shows that if the missing term is put elsewhere in expr (after restart, of course) then the indets works okay.

 

The following result is even weirder. One would expect that for any types T1 and T2 and any expression expr, that indets(expr, {T1, T2}) would equal indets(expr, T1) union indets(expr, T2), right? One would expect that even if there were a bug with one of the individual indets calls, right? So, note that I1 above contains the first term of expr, yet ...

indets(expr, {HODD, HODD &under (convert, D)});

{diff(diff(f(x), x), x), diff(diff(u(x, t), t), x), (D[1, 2](v))(x, t), ((D@@3)(f))(x)}

I get the same results if put the type explicitly in the indets calls rather than using AddType. I get the same results in Maple 16 and 18. So, what's going on?

 


Download indets.mw

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