Factorial simplification error

asked 2011-06-23 16:17:42 +0100

garret
43 ●1 ●2 ●6

updated 2011-06-23 16:49:53 +0100

I've run into an issue when running simplify on an equation containing a factorial:

sage: var('n')
sage: z1 =  (-1)^(1/2*(n - 1))*factorial(n) == 0
sage: z2 = 0 == (-1)^(1/2*(n - 1))*factorial(n)
sage: z1.simplify()
n*(-1)^(1/2*n - 1/2) != 0
sage: z2.simplify()
0 == (-1)^(1/2*n - 1/2)*factorial(n)

My best guess is when sage sends the equation down to maxima for simplification it's interpreting

(-1)^(1/2*(n - 1))*n!  =  0

(-1)^(1/2*(n - 1))*n   !=   0

Any ideas on how I can work around this?

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answered 2011-06-23 19:49:53 +0100

Mike Hansen

4088 ●24 ●48 ●90

I made this ticket 11539. As a temporary workaround, you can run the following commands which should fix things.

sage.calculus.calculus.maxima.eval(r":lisp (defun msize-factorial (x l r) (setq l (msize (cadr x) (revappend '(#\f #\a #\c #\t #\o #\r #\i #\a #\l #\() l) nil 'mparen 'mparen) r (list 1 #\) ) ) (list (+ (car l) (car r)) l r)) ")
sage.calculus.calculus.maxima.eval(r":lisp (defprop mfactorial msize-factorial grind)")

This changes things in Maxima so that factorial(x) is printed as factorial(x) instead of x!.

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Wow, that's quite a workaround!

kcrisman ( 2011-06-23 21:17:47 +0100 )edit

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answered 2011-06-23 17:49:48 +0100

parzan
977 ●6 ●15 ●36

Here is one option (not the most elegant):

sage: z1.lhs().simplify() == z1.rhs().simplify()
(-1)^(1/2*n - 1/2)*factorial(n) == 0

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