Given a function:
f(r, phi) := [r * cos(phi), r * sin(phi)];
I can find the derivative w.r.t. the first argument by providing arguments that turn it into an expression, like this:
f1(a, b) := diff(f(a, b), a);
But that's kind of wordy and round about. Is there a way to directly say "derivative of f w.r.t first argument?"
Related
Sometimes I need to use a declaration of a function inside another function. For example, I made the following in Mathematica:
Ie: There is a function f and when I compute f[Cos], it declares h as Cos[x]. Observe that I can't do the same by computing f[x] because it does not recognize $x$ as a function, although I can circumvent that by using Mathematica's notation for pure function: f[#&], now it works perfectly fine.
I noticed I can do the same with Maxima:
Although I don't know how to do the f[#&] in Maxima. Ie: Tell Maxima "x" is a function just as I did in Mathematica. Is there a way to do the same in Maxima? Also If I try to compile:
Without asking it to compute g(x), f won't work in the same way it worked before. It's not clear to me why this happens.
:= always defines a global function, even if it's within another function or block. As it stands, when you call f twice, the second definition of g clobbers the first one -- you can't have two different g functions.
I think what you want is an unnamed function, lambda([x], ...). The tricky part is that the body of lambda is not evaluated, so when you write lambda([x], y(x)), the value of y doesn't appear in the result. There are a few different ways to achieve that; I'll describe one way using subst.
subst('y = something, lambda([x], y(x))) constructs an unnamed function and then pastes something into it, replacing y. The result is lambda([x], something(x)) which I think is what you want.
To put this in the framework you outlined,
(%i3) f(x, y) := block ([g: subst ('y = y, lambda ([x], y(x)))], [g, g(x)]);
(%o3) f(x, y) := block([g : subst('y = y, lambda([x], y(x)))], [g, g(x)])
(%i4) f(1/2, cos);
1
(%o4) [lambda([x], cos(x)), cos(-)]
2
I have a formlua in DNF form, say:
abcx + abcy + abz
Is there any way to take out the common variables, to get the follwing formula:
ab (cx + cy + z)
A followup question, can it be done recursively, like
ab ( c(x+y) + z)
Sure.. Here's one way:
from z3 import *
a, b, c, x, y, z = Ints('a b c x y z')
print simplify(a*b*c*x + a*b*c*y + a*b*z, hoist_mul=True)
This prints:
a*b*(c*(x + y) + z)
which is exactly what you're looking for.
And for your next question, how did I find about hoist_cmul=True argument? Simply run:
help_simplify()
at your Python prompt, and it'll list you all the options simplify takes.
Note that you should in general not count on what the simplifier will give you. It's mostly heuristic driven, and in the presence of other terms what you get may not match what you expected. (It'll of course still be an equivalent expression.) There's no notion of "simplest" when it comes to arithmetic expressions, and what you consider simple and what z3 considers simple may not necessarily match.
Here is an example from the "Programming with Erlang" (2nd edition):
count_chars([], Result) ->
Result;
count_chars([C|Substring], #{C := N}=Result) ->
count_chars(Substring, Result#{C := N+1 });
count_chars([C|Substring], Result) ->
count_chars(Substring, Result#{C => 1}).
..which mercylessly yields the following error:
variable 'C' is unbound
So I am kind of stuck here; to my view, variable 'C' is bound, namely it must be a head of the string (just a linked list of chars, right?). Yet Erlang disagrees with me, breaking example from the (probably, outdated?) book I'am reading right now.
So what's wrong? What's the right way to pattern-match in this particular example?
P.S. A screenshot from the book. Pay attention at slightly different syntax, which also doesn't work for me:
P.P.S. I am using the latest version of Erlang I've managed to download from the official site.
C must be bound before the expression #{C := N}=Result is evaluated.
You consider that C is bound since the first parameter [C|Substring] was evaluated before: #{C := N}=Result. In fact it is not the case. There is no real assignment until a head evaluation succeed and the function enters the body.
Writing count_chars([C|Substring], #{C := N}=Result) -> is exactly the same as count_chars([C1|Substring], #{C2 := N}=Result) when C1 =:= C2 ->
During the head evaluation, each element is stored in a different element (a place in the heap) to check if all the parameters match the head definition. In your case the compiler want store the value C in an element, let's say x1 and the key C? in another element, let's say x2, and then verify that x1 and x2 are equals. the second operation is not possible without a deep modification of the compiler behavior.
I wrote a small example to show how it works, and compiled it with the option 'S' to see the result of the compilation:
test([K|_],K,M) -> % to see how the test of parameter equality is done
#{K := V} = M, % to verify that this pattern works when K is bound.
V.
the assembly result is :
{function, test, 3, 33}.
{label,32}.
{line,[{location,"mod.erl",64}]}.
{func_info,{atom,mod},{atom,test},3}.
{label,33}.
{test,is_nonempty_list,{f,32},[{x,0}]}. % the whole list is assigned to x0
{get_hd,{x,0},{x,3}}. % the list's head is assigned to x3
{test,is_eq_exact,{f,32},[{x,1},{x,3}]}. % the second parameter K is assigned to x1, verify if x1 =:= x3
{test,is_map,{f,34},[{x,2}]}. % the third parameter M is assigned to x2, check if it is a map if not go to label 34
{get_map_elements,{f,34},{x,2},{list,[{x,3},{x,0}]}}. % get the value associated to key x3 in the map x2 and store it into x0, if no suck key go to label 34
return.
{label,34}. % throw a badmatch error
{line,[{location,"mod.erl",65}]}.
{badmatch,{x,2}}.
Now, to code your function you can simply write:
count_chars([], Result) ->
Result;
count_chars([C|Substring], Result) ->
N = maps:get(C, Result, 0) +1,
count_chars(Substring, Result#{C => N }).
I want to calculate the seventh derivative of tan(x) at x=pi/4 in Maxima:
f(x) := diff(tan(x), x, 7);
f(%pi / 4);
Yet I cannot get the result. Ay ideas?
I would do it like this,
at(diff(tan(x),x,7),[x=%pi/4]);
The function diff returns a function as its result. You can evaluate this function at a point by using the at function.
Another way of doing is like so,
f: diff(tan(x), x, 7);
at(f, [x=%pi/4]);
Now f is just a variable that holds the output of diff and then at is used to evaluate it at a point.
I hope this helps.
When you define a function via :=, the function body is quoted (i.e., not evaluated). You can tell Maxima to evaluate an expression by using the quote-quote '' operator.
(%i1) display2d : false $
(%i2) f(x) := ''(diff (tan(x), x, 7));
(%o2) f(x):=64*sec(x)^2*tan(x)^6+1824*sec(x)^4*tan(x)^4+2880*sec(x)^6*tan(x)^2
+272*sec(x)^8
(%i3) f(%pi / 4);
(%o3) 34816
Note that '' has the possibly-surprising property that it is only applied once, when an expression is entered, not every time the expression is evaluated.
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.