I cannot find anything about this, sorry.
If I have expressions with the symbolic function f(x) and now I want to replace in these expression f(x) by its explicit form how to do it?
For example:
I have
f(x):= x^2+sin(x)
and in the differentiation
diff (%e**sqrt(f(x)*a), x,2);
I want to replace now f(x) by the expression above?
Thanks
Karl
(%i1) i: integrate(f(x)*f(4*x), x, 0, 1) $
(%i2) f(x):= x^2+sin(x) $
(%i3) ev(i, f);
1
/
[ 2 2
(%o3) I (sin(x) + x ) (sin(4 x) + 16 x ) dx
]
/
0
-- Function: ev (<expr>, <arg_1>, ..., <arg_n>)
Evaluates the expression <expr> in the environment specified by the
arguments <arg_1>, ..., <arg_n>. The arguments are switches
(Boolean flags), assignments, equations, and functions. 'ev'
returns the result (another expression) of the evaluation.
Related
I need to make tons of simple computations and present each step in my report with predefined manner:
(for ex i got B=2, C=3):
A=B+12-6/C^2; A=2+12-6/3^2=13.333;
I can get 1st block and answer like this:
B:2$ C:3$
A:'(B+12-6/C^2)$
print("A=",A,"; ","A= ??? =",ev(A, numer) );
and get:
6
A= (- --) + B + 12 ; A= ??? = 13.33333333333333
2
C
What i need instead of '???' to get desired output?
Maxima distinguishes two parts of figuring out a result: evaluation and simplification. Evaluation = substituting one thing (the value) for another thing (a variable or a function). Simplification = applying mathematical identities to get a "simpler", equivalent result.
In your problem, it appears you want to postpone simplification. You can say simp: false to do that. Here's one possible approach. I'll disable simplification, substitute values into the expression, print the substituted expression, and then re-enable simplification to get the final result.
(%i2) expr: A=B+12-6/C^2;
6
(%o2) A = (- --) + B + 12
2
C
(%i3) simp: false $
(%i4) subst ([B = 2, C = 3], expr);
- 2
(%o4) A = 12 + 2 + (- 6) 3
(%i5) simp: true $
(%i6) %o4;
40
(%o6) A = --
3
Note that many operations in Maxima happen by simplification (e.g. adding numbers together), so in general, Maxima will act noticeably different when simp is false. But in this case that's what you want.
EDIT: OP points out that the result after substitution is displayed in a somewhat different from. The reason for this has to do with some obscure implementation details of Maxima. Be that as it may, it's possible to work around that behavior by using the Lisp substitution function SUBST (referenced in Maxima as ?subst) instead of Maxima subst. SUBST is a little different than Maxima subst; the syntax is ?subst(new_thing, old_thing, some_expression). After substituting via SUBST, it's necessary to resimplify explicitly; one way to do that is to say expand(..., 0, 0) (which doesn't expand anything, the only effect is to resimplify).
(%i2) expr: A=B+12-6/C^2;
6
(%o2) A = (- --) + B + 12
2
C
(%i3) simp: false $
(%i4) ?subst (3, C, ?subst (2, B, expr));
6
(%o4) A = (- --) + 2 + 12
2
3
(%i5) simp: true $
(%i6) expand (%o4, 0, 0);
40
(%o6) A = --
3
Since SUBST is has a different effect on the internal representation, it is possible you could create an invalid expression, for some choices of new_thing, old_thing, and some_expression. I won't try to sort that out here.
I'm struggling to find a way to switch from a symbolic declaration of the differential operator to its implementation
I give you an example.
F: (10-'diff(x(t),t)^2 -2*x(t)*'diff(x(t),t) -5*x(t)^2)*%e^(-t);
E: ratsimp(diff(F, x(t)) - diff(diff(F, 'diff(x(t),t)), t));
sol: ode2(E, x(t), t);
sol: ev(sol, [%k1 = C1, %k2=C2]);
trans_cond: diff(F, 'diff(x(t), t));
trans_cond: ev(trans_cond, sol);
trans_cond: at(trans_cond, [t=1]);
The corresponding output maintains the symbolic notation whereas I would like to evaluate the diff() obtained after the last substitution.
Giving the result:
% 4*C1-C2^(-2)
Another solution. The nouns option for ev causes the evaluation of symbolic derivatives, and also any other noun expressions such as symbolic integrals, symbolic summations, etc.
(%i2) 'diff(4*x^2, x);
d 2
(%o2) -- (4 x )
dx
(%i3) ev (%o2, nouns);
(%o3) 8 x
A shorter form of ev(..., nouns) is recognized by the interactive console. You can input ..., nouns instead.
(%i5) %o2, nouns;
(%o5) 8 x
Here is ev(..., nouns) applied to symbolic integral:
(%i6) 'integrate (x^2, x);
/
[ 2
(%o6) I x dx
]
/
(%i7) %, nouns;
3
x
(%o7) --
3
and here, to a symbolic summation:
(%i8) 'sum (f(k), k, 1, 3);
3
====
\
(%o8) > f(k)
/
====
k = 1
(%i9) %, nouns;
(%o9) f(3) + f(2) + f(1)
Found the answer, ev() carries the option diff which solves all the symbolic differentiation in the expression.
In Wolfram Mathematica we can define an operator acting on a function (i.e. function that returns function) e.g. as for the operator of multiplication by the first argument acting on a functions of two arguments in the example below
X[f_] = Function[{x, y}, x*f[x, y]]
Then we can apply this operators to any function with any 2 arguments
In[2]:= X[g][z, t]
Out[2]= z g[z, t]
There is a similar construction in Maple
X:=proc(f) option operator; local x,y;
unapply(x*f(x,y),x,y)
end;
with the similar mode of application
> X(g)(z,t);
z g(z, t)
In Maxima I tried this
X(f):=lambda([x,y],x*f(x,y));
but when trying to apply it I get
(%i5) X(g)(z,t)
(%o5) z*f(z,t)
so, it looks like f is not recognized as an argument of function X when I used lambda.
Is there any way to solve this problem?
In case of Maple and Mathematica this type of operators helps a lot to manipulate with linear differential operators
Maxima lambda doesn't evaluate any of the expressions inside the body of it, so f is not evaluated (to g). That's the behavior you are seeing. The motivation for it that I can see is that the lambda body might contain expressions which don't have the expected effect until some variables have values, e.g. length, for ..., print, etc.
You can get the expected behavior by substituting into the body. Here are two ways to do that. The first uses the function subst which I think is maybe the most obvious.
(%i1) X(f):= subst ('f = f, lambda([x,y],x*f(x,y)));
(%o1) X(f) := subst('f = f, lambda([x, y], x f(x, y)))
(%i2) X(g);
(%o2) lambda([x, y], x g(x, y))
(%i3) X(g)(z, t);
(%o3) z g(z, t)
The second uses the function buildq which is effectively a substitution function which quotes (does not evaluate) the expression into which something is substituted.
(%i4) X(f) := buildq ([f], lambda ([x, y], x*f(x, y)));
(%o4) X(f) := buildq([f], lambda([x, y], x f(x, y)))
(%i5) X(g);
(%o5) lambda([x, y], x g(x, y))
(%i6) X(g)(z, t);
(%o6) z g(z, t)
Finally, if you are interested in creating lambda expressions using evaluated expressions more often, you could create your own kind of lambda for that. I'll call it evlambda here.
(%i11) evlambda (a, b) := apply (lambda, [a, b]);
(%o11) evlambda(a, b) := apply(lambda, [a, b])
(%i12) X(f) := evlambda ([x, y], x*f(x, y));
(%o12) X(f) := evlambda([x, y], x f(x, y))
(%i13) X(g);
(%o13) lambda([x, y], x g(x, y))
(%i14) X(g)(z, t);
(%o14) z g(z, t)
The key here is that evlambda is defined as an ordinary function, so its arguments are evaluated. So by the time lambda is applied, b has been evaluated so it contains g.
Note that this evlambda won't do anything useful with length, for ..., and print, which is to be expected.
(%i15) foo : evlambda ([l], 1 + length(l));
length: argument cannot be a symbol; found l
-- an error. To debug this try: debugmode(true);
(%i16) bar : evlambda ([n], for i thru n do print (i));
Unable to evaluate predicate 1 > n
-- an error. To debug this try: debugmode(true);
(%i17) baz : evlambda ([x], print (x));
x
(%o17) lambda([x], x)
(%i18) baz(5);
(%o18) 5
The last one, with print, evaluates print when baz is defined (so x is the output), but then not again when baz(5) is evaluated -- this behavior is to be expected since evlambda evaluates its arguments.
I'm working on creating maxima functions to simplify the del operator on vectors. How can I pass a list/vector to a function in maxima? This works:
(%i7) dot(a,b) := a[1]*b[1]+a[2]*b[2]+a[3]*b[3];
(%o7) dot(a, b) := a b + a b + a b
1 1 2 2 3 3
(%i8) dot(a,b);
2
(%o8) 3 x y - 4 x
but this doesn't:
(%i13) grad(a) := diff(a[1],x) + diff(a[2],y) + diff(a[3],z);
define: argument cannot be an atom or a subscripted memoizing function; found:
a
-- an error. To debug this try: debugmode(true);
Maxima has extremely confusing rules about scope and subscripts. First of all, I'll apologize for that.
My guess is that you already have an array named a by the time you define grad. Try a different name for the argument of grad -- try something which you haven't used yet. Does it work that way?
Anyway, shouldn't the definition be:
grad(a) := [diff(a, x), diff(a, y), diff(a, z)];
??
With Maxima, it is possible to replace an unknown by a value using at() statement.
But this use a list, for the substitution, and the solve() statement don't return a list.
Code:
(%i1) g(x):=x^2+a;
2
(%o1) g(x) := x + a
(%i2) g(x),solve(x=3),a=2;
(%o2) 11
I managed to compute a result using commas, but I can't create a function to do so:
(%i3) f(y) := g(x),solve(x=3),a=y;
(%o3) f(y) := g(x)
(%i4) f(2);
2
(%o4) x + a
Is there a statement for which the commas acts like it acts directly in the line?
Edit:
Actually, it is possible to use at() with solve() to create the function f(), as solve() just return a list of lists. So the code would be:
(%i5) f(y) := at(at(g(x), solve(x=3)[1]), a=y);
(%o5) f(y) := at(at(g(x), solve(x = 3) ), a = y)
(%i6) f(2);
(%o6) 11
Notice the [1] after solve(x=3) in the (%i5). It select the the first item (solution) of list.
I'm not sure what you are trying to accomplish -- probably it would be best if you would back up a couple of steps and describe the larger problem you are trying to solve here.
My best guess as to what you want is that you are trying to use the result of 'solve' to find a value to substitute into some expression. If so you can achieve it like this: f(eq, u) := map (lambda ([e], subst (e, g(u))), solve (eq, x)); where eq is an equation to solve for x and then substitute into g(u). Note that 'solve' can return multiple solutions so that's why I use 'map' to apply something to each solution. Here is an example output:
(%i7) f(eq) := map (lambda ([e], subst (e, g(x))), solve (eq, x));
(%o7) f(eq) := map(lambda([e], subst(e, g(x))), solve(eq, x))
(%i8) solve (x^2 + 2*x + 2);
(%o8) [x = - %i - 1, x = %i - 1]
(%i9) f (x^2 + 2*x + 2);
(%o9) [g(- %i - 1), g(%i - 1)]
Of course you can define 'g' in whatever way is appropriate.
The answer to your specific question (which I believe is not actually very much relevant, but anyway) is to use 'block' to group together expressions to be evaluated. E.g. f(x) := block (...);
Perhaps I'm answering the wrong question. Maybe what you want is ev(foo, bar, baz) -- ev is the function that is actually called when you write foo, bar, baz at the console input prompt. So the function would be written f(y) := ev (g(x), solve(x=3), a=y).
However, bear in mind that there are several different kinds of functionality built into ev, so it is hard to understand (see the documentation for ev). Instead, consider using subst which is much simpler.