I tried function "at" with some function inside it and then give the output to some variable. Maxima successfully differentiated the expression, but then "at" fails and the output is "at( --some successfully done function--, z=l)=0". I need "at" to work properly, to give the result to a variable.
(%i34) a: 45*z^2*l-1; /*expression*/
eq1: at(diff(a, z, 1), z = l)=0; /*giving the meaning of the operations to eq1*/
at(diff(a, z, 1), z = l)=0; /*trying the same without giving the result to a variable*/
ev(eq1, eval); /*trying ev*/
(a) l*z^2* 45-1
(eq1) at(2*l*z* 45,z=l)=0
(%o34) 2*l^2* 45=0
(%o35) at(2*l*z* 45,z=l)=0
So when I don't give the result of at to other variables, it's fine but when I try to - it fails even with additional evaluation. How does that work? And also this was tried on Linux. On Windows I don't have the same problem.
I get the following output. Isn't %o3 what you are looking for?
(%i2) a: 45*z^2*l-1;
2
(%o2) 45 l z - 1
(%i3) eq1: at(diff(a, z, 1), z = l)=0;
2
(%o3) 90 l = 0
I am working w/ Maxima 5.42.2 on MacOS. What does build_info(); report on your Linux system? Some Linux distributions package an ancient version of Maxima; maybe you can get a newer version. It is actually pretty easy to build Maxima from a source tarball on a Linux system; I can help if you want to go down that road.
Related
I want to use Maxima to get the prime factorization of a random positive integer, e.g. 12=2^2*3^1.
What I have tried so far:
a:random(20);
aa:abs(a);
fa:ifactors(aa);
ka:length(fa);
ta:1;
pfza: for i:1 while i<=ka do ta:ta*(fa[i][1])^(fa[i][2]);
ta;
This will be implemented in STACK for Moodle as part of a online exercise for students, so the exact implementation will be a little bit different from this, but I broke it down to these 7 lines.
I generate a random number a, make sure that it is a positive integer by using aa=|a|+1 and want to use the ifactors command to get the prime factors of aa. ka tells me the number of pairwise distinct prime factors which I then use for the while loop in pfza. If I let this piece of code run, it returns everything fine, execpt for simplifying ta, that is I don't get ta as a product of primes with some exponents but rather just ta=aa.
I then tried to turn off the simplifier, manually simplifying everything else that I need:
simp:false$
a:random(20);
aa:ev(abs(a),simp);
fa:ifactors(aa);
ka:ev(length(fa),simp);
ta:1;
pfza: for i:1 while i<=ka do ta:ta*(fa[i][1])^(fa[i][2]);
ta;
This however does not compile; I assume the problem is somewhere in the line for pfza, but I don't know why.
Any input on how to fix this? Or another method of getting the factorizing in a non-simplified form?
(1) The for-loop fails because adding 1 to i requires 1 + 1 to be simplified to 2, but simplification is disabled. Here's a way to make the loop work without requiring arithmetic.
(%i10) for f in fa do ta:ta*(f[1]^f[2]);
(%o10) done
(%i11) ta;
2 2 1
(%o11) ((1 2 ) 2 ) 3
Hmm, that's strange, again because of the lack of simplification. How about this:
(%i12) apply ("*", map (lambda ([f], f[1]^f[2]), fa));
2 1
(%o12) 2 3
In general I think it's better to avoid explicit indexing anyway.
(2) But maybe you don't need that at all. factor returns an unsimplified expression of the kind you are trying to construct.
(%i13) simp:true;
(%o13) true
(%i14) factor(12);
2
(%o14) 2 3
I think it's conceptually inconsistent for factor to return an unsimplified, but anyway it seems to work here.
I have a fairly large expression that involves a lot of subexpressions of the form (100*A^3 + 200*A^2 + 100*A)*x or (-A^2 - A)*y or (100*A^2 + 100*A)*z
I know, but I don't know how to tell Maxima this, that it in this case is valid to make the approximation A+1 ~ A, thereby effectively removing anything but the highest power of A in each coefficient.
I'm now looking for functions, tools, or methods that I can use to guide Maxima in dropping various terms that aren't important.
I have attempted with subst, but that requires me to specify each and every factor separately, because:
subst([A+1=B], (A+2)*(A+1)*2);
subst([A+1=B], (A+2)*(A*2+2));
(%o1) 2*(A+2)*B
(%o2) (A+2)*(2*A+2)
(that is, I need to add one expression for each slightly different variant)
I tried with ratsimp, but that's too eager to change every occurrence:
ratsubst(B, A+1, A*(A+1)*2);
ratsubst(B, A+1, A*(A*2+2));
(%o3) 2*B^2-2*B
(%o4) 2*B^2-2*B
which isn't actually simpler, as I would have preferred the answer to have been given as 2*B^2.
In another answer, (https://stackoverflow.com/a/22695050/5999883) the functions let and letsimp were suggested for the task of substituting values, but I fail to get them to really do anything:
x:(A+1)*A;
let ( A+1, B );
letsimp(x);
(x)A*(A+1)
(%o6) A+1 --\> B
(%o7) A^2+A
Again, I'd like to approximate this expression to A^2 (B^2, whatever it's called).
I understand that this is, in general, a hard problem (is e.g. A^2 + 10^8*A still okay to approximate as A^2?) but I think that what I'm looking for is a function or method of calculation that would be a little bit smarter than subst and can recognize that the same substitution could be done in the expression A^2+A as in the expression 100*A^2+100*A or -A^2-A instead of making me create a list of three (or twenty) individual substitutions when calling subst. The "nice" part of the full expression that I'm working on is that each of these A factors are of the form k*A^n*(A+1)^m for various small integers n, m, so I never actually end up with the degenerate case mentioned above.
(I was briefly thinking of re-expressing my expression as a polynomial in A, but this will not work as the only valid approximation of the expression (A^3+A^2+A)*x + y is A^3*x + y -- I know nothing about the relative sizes of x and y.
We are giving students some exercises, where their solutions are evaluated with maxima.
The answer involves the unit step function. The evaluation in maxima seems to go okay, except that there seems to be missing some algebraic rules on the unit_step functions.
For example is(unit_step(x)*unit_step(x) = unit_step(x)) evaluates to false. It is quite unlikely that the student gives the answer in such a form, but still we don't want to have the possibility that the student gives a good answer, that is evaluated as incorrect.
Below is a screenshot of an answer we try to evaluate with maxima involving the unit_step function (that we defined as u):
Maxima doesn't know much about unit_step at present (circa Maxima 5.41). This is just a shortcoming, there's no reason for it, except that nobody has gotten around to doing the work. That said, it's not too hard to make some progress.
The simplifier for multiplication merges identical terms into powers:
(%i3) unit_step(x)*unit_step(x);
2
(%o3) unit_step (x)
So let's define a simplifier rule which reduces positive powers of unit_step. (I was going to say positive integer powers, but a moment's thought shows that the same identity holds for noninteger positive powers as well.)
(%i4) matchdeclare (aa, lambda ([e], e > 0)) $
(%i5) matchdeclare (xx, all) $
(%i6) tellsimpafter (unit_step(xx)^aa, unit_step(xx));
(%o6) [^rule1, simpexpt]
Let's try it.
(%i7) unit_step(x)*unit_step(x);
(%o7) unit_step(x)
(%i8) is (unit_step(x)*unit_step(x) = unit_step(x));
(%o8) true
(%i9) unit_step(t - 5)^(1/4);
(%o9) unit_step(t - 5)
(%i10) assume (m > 0);
(%o10) [m > 0]
(%i11) unit_step(2*u + 1)^m;
(%o11) unit_step(2 u + 1)
So far, so good. Of course this is just one identity and there are others that could be useful. Since this rule is not built-in, one would have to load that definition in order to make use of it; that would be bothersome if you intend for others to use this.
For the record, the only simplification for unit_step which I found in the Maxima source code is in share/contrib/integration/abs_integrate.mac, which contains a function unit_step_mult_simp, which applies the identity unit_step(a)*unit_step(b) --> unit_step(min(a, b)).
I would like to make matrix calculator, but I struggle a little bit, how to make an input of the program. I have commands that user can use in calculator. Some takes 1 argument, 2 arguments or 3 arguments. I was inspired by program on this website http://www.ivank.net/blogspot/matrix_pascal/matrices.pas
But I don't really understand, how the input is made. Program from the website use parse, split procedures, but I don't know, how does it work. Does it exists some website, where it is good explained (Parse in Pascal)? I would like to really understand it.
This is, how it should looks like:
command: sum X Y
command: multiply X
command: transpose X
In the sample which inspired you, all the calculation is realized by the 'procedure parse(command:String);'.
The first step consists to extract the command and all parameters by:
com := Split(command, ' ');
In your case, you will obtain for 'command: sum X Y':
Length(com) = 3
com[0] = 'sum'; com[1] = 'X'; com[2] = 'Y';
But, be carefull, the 'X' and 'Y' parameters shall not have characters between numbers.
I'm trying to teach myself Prolog. Below, I've written some code that I think should return all paths between nodes in an undirected graph... but it doesn't. I'm trying to understand why this particular code doesn't work (which I think differentiates this question from similar Prolog pathfinding posts). I'm running this in SWI-Prolog. Any clues?
% Define a directed graph (nodes may or may not be "room"s; edges are encoded by "leads_to" predicates).
room(kitchen).
room(living_room).
room(den).
room(stairs).
room(hall).
room(bathroom).
room(bedroom1).
room(bedroom2).
room(bedroom3).
room(studio).
leads_to(kitchen, living_room).
leads_to(living_room, stairs).
leads_to(living_room, den).
leads_to(stairs, hall).
leads_to(hall, bedroom1).
leads_to(hall, bedroom2).
leads_to(hall, bedroom3).
leads_to(hall, studio).
leads_to(living_room, outside). % Note "outside" is the only node that is not a "room"
leads_to(kitchen, outside).
% Define the indirection of the graph. This is what we'll work with.
neighbor(A,B) :- leads_to(A, B).
neighbor(A,B) :- leads_to(B, A).
Iff A --> B --> C --> D is a loop-free path, then
path(A, D, [B, C])
should be true. I.e., the third argument contains the intermediate nodes.
% Base Rule (R0)
path(X,Y,[]) :- neighbor(X,Y).
% Inductive Rule (R1)
path(X,Y,[Z|P]) :- not(X == Y), neighbor(X,Z), not(member(Z, P)), path(Z,Y,P).
Yet,
?- path(bedroom1, stairs, P).
is false. Why? Shouldn't we get a match to R1 with
X = bedroom1
Y = stairs
Z = hall
P = []
since,
?- neighbor(bedroom1, hall).
true.
?- not(member(hall, [])).
true.
?- path(hall, stairs, []).
true .
?
In fact, if I evaluate
?- path(A, B, P).
I get only the length-1 solutions.
Welcome to Prolog! The problem, essentially, is that when you get to not(member(Z, P)) in R1, P is still a pure variable, because the evaluation hasn't gotten to path(Z, Y, P) to define it yet. One of the surprising yet inspiring things about Prolog is that member(Ground, Var) will generate lists that contain Ground and unify them with Var:
?- member(a, X).
X = [a|_G890] ;
X = [_G889, a|_G893] ;
X = [_G889, _G892, a|_G896] .
This has the confusing side-effect that checking for a value in an uninstantiated list will always succeed, which is why not(member(Z, P)) will always fail, causing R1 to always fail. The fact that you get all the R0 solutions and none of the R1 solutions is a clue that something in R1 is causing it to always fail. After all, we know R0 works.
If you swap these two goals, you'll get the first result you want:
path(X,Y,[Z|P]) :- not(X == Y), neighbor(X,Z), path(Z,Y,P), not(member(Z, P)).
?- path(bedroom1, stairs, P).
P = [hall]
If you ask for another solution, you'll get a stack overflow. This is because after the change we're happily generating solutions with cycles as quickly as possible with path(Z,Y,P), only to discard them post-facto with not(member(Z, P)). (Incidentally, for a slight efficiency gain we can switch to memberchk/2 instead of member/2. Of course doing the wrong thing faster isn't much help. :)
I'd be inclined to convert this to a breadth-first search, which in Prolog would imply adding an "open set" argument to contain solutions you haven't tried yet, and at each node first trying something in the open set and then adding that node's possibilities to the end of the open set. When the open set is extinguished, you've tried every node you could get to. For some path finding problems it's a better solution than depth first search anyway. Another thing you could try is separating the path into a visited and future component, and only checking the visited component. As long as you aren't generating a cycle in the current step, you can be assured you aren't generating one at all, there's no need to worry about future steps.
The way you worded the question leads me to believe you don't want a complete solution, just a hint, so I think this is all you need. Let me know if that's not right.