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Why Sage cannot pass a value of variable from one function to another nested function?

asked 2013-10-31 07:09:56 -0600

terces907 gravatar image

updated 2013-10-31 07:50:01 -0600

The first i ran this:

sage: f(x)=(2/n)*(sin(n*x)*(-1)^(n+1))
sage: sum(f, n, 1, 2) #using summation function
-sin(2*x) + 2*sin(x)

So, In this case the result was evaluated correctly.

But if i tried to combine the first line and the second line together:

sage: f(x,k) = sum((2/n)*(sin(n*x)*(-1)^(n+1)), n, 1, k)
#where n = 1,2,3 ... k
sage: f(x,2)
-2*sum((-1)^n*sin(n*x)/n, n, 1, 2)

The result wasn't finished!

Why sage cannot evaluate mathematical expression in this case?

Another tried to prove that Sage can pass its variable from left function to right function even though the right function was a nested function:

sage: f(x) = sin(arcsin(x)) 
sage: f(0.5)
0.500000000000000

Edit: (See the same question on SO.)

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answered 2013-10-31 07:24:50 -0600

kcrisman gravatar image

updated 2013-11-01 05:53:24 -0600

sage: n, k = var('n, k')
sage: f(x,k) = sum((2/n)*(sin(n*x)*(-1)^(n+1)), n, 1, k)
#where n = 1,2,3 ... k
sage: f
(x, k) |--> -2*sum((-1)^n*sin(n*x)/n, n, 1, k)

I'm not sure what you think is wrong here. The 2 and a factor of -1 were both factored out, that's all.

However, I do agree that this doesn't expand. What is happening is that we are sending the sum to Maxima

if algorithm == 'maxima':
    return maxima.sr_sum(expression,v,a,b)

and then ordinarily when it returns, it is still a Maxima object (which may be a bug?). But when we put it in the function, it becomes a Sage object - but we don't have a Sage "sum" object. So I think that is what would have to be fixed.


That this is possible is shown by the following Maxima example (which I put on the ticket):

(%i1) f: -2*'sum((-1)^n*sin(n*x)/n,n,1,2);
                                2
                               ====       n
                               \     (- 1)  sin(n x)
(%o1)                      - 2  >    ---------------
                               /            n
                               ====
                               n = 1

(%i8) f, nouns;
                                 sin(2 x)
(%o8)                       - 2 (-------- - sin(x))
                                    2
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That mean i cannot define the function like that right? It will be very nice if i can define it similar to the second example. it will be consistent with the real equation on my paper.

terces907 gravatar imageterces907 ( 2013-10-31 07:58:54 -0600 )edit

I agree with kcrisman that at first sight the result seems to be ok and only be presented in a different form. But as the example sage: g(x)=(2/n)*(sin(n*x)*(-1)^(n+1)) sage: f(x,k) = sum((2/n)*(sin(n*x)*(-1)^(n+1)), n, 1, k) sage: (f(x,2)-sum(g, n, 1, 2)).full_simplify() doesn't return zero I would call this a bug.

twch gravatar imagetwch ( 2013-11-01 03:26:21 -0600 )edit

Well, I don't know about a bug, since deciding for an arbitrary expression whether it happens to be zero is very, very hard... but it is decidedly suboptimal, so I'll open a ticket for this issue.

kcrisman gravatar imagekcrisman ( 2013-11-01 05:31:32 -0600 )edit

This is now [Trac #15346](http://trac.sagemath.org/ticket/15346).

kcrisman gravatar imagekcrisman ( 2013-11-01 05:52:47 -0600 )edit
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answered 2013-11-01 03:36:13 -0600

twch gravatar image

You can avoid this problem if you don't use a symbolic function but define a python function:

sage: var('n')
sage: def g(x,k):
sage:    return sum((2/n)*(sin(n*x)*(-1)^(n+1)), n, 1, k)
sage: print g(x,2)
-sin(2*x) + 2*sin(x)
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I like this! I tried this using lambdas but it didn't work correctly, but this is a good workaround.

kcrisman gravatar imagekcrisman ( 2013-11-01 05:31:00 -0600 )edit

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Asked: 2013-10-31 07:09:56 -0600

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Last updated: Nov 01 '13