On Thu, Mar 4, 2010 at 10:21 AM, Ondrej Certik <ond...@certik.cz> wrote:
> On Thu, Mar 4, 2010 at 10:10 AM, Chad File <archeryguru2...@gmail.com> wrote:
>> Ondrej, thanks for the reply.
>>
>> What I mean is exactly that, i%2. If 'i' is even, i%2 = 0, if 'i' is odd,
>> i%2 = 1. How else would one write that?
>>
>> I have a very large program that actually uses this 'remainder' operator
>> many times. However, only in the summation do I have any problems with it.
>> I'm currently using the below line in place of what I want.
>>
>> In [1]: mpmath.calculus.nsum(lambda i: (i-i%2)/2, [1,10])
>> Out[1]: 25.0
>>
>> I haven't timed it to check which is faster (cannot time what I cannot
>> compute), but I would prefer to do away with the mpmath dependence on the
>> summation, if possible.
>
>
> Got it. The % operator isn't implemented for symbols, as you can see here:
>
> In [1]: var("i")
> Out[1]: i
>
> In [2]: i%2
> ---------------------------------------------------------------------------
> TypeError Traceback (most recent call last)
>
> /home/ondrej/repos/sympy/<ipython console> in <module>()
>
> TypeError: unsupported operand type(s) for %: 'Symbol' and 'int'
>
>
> So it'd had to be implemented and then the Sum() should be improved to
> take this into account. A workaround is to sum over odd and even "i"s
> separately and use a substitution i = 2*k and i=2*k+1, e.g.:
>
> sum_i = sum_i /even/ + sum_i /odd/ = Sum(2*k/2, (k, 1, 5)) +
> Sum((2*k+1-1)/2, (k, 0, 4))
>
> and when you doit:
>
> In [6]: (Sum(2*k/2, (k, 1, 5)) + Sum((2*k+1-1)/2, (k, 0, 4))).doit()
> Out[6]: 25
>
>
> Maybe you can improve your program to do the substitution for you.
Better thing is to use a substitution i=2*k-1 for the odd "i"s,
because then you can make these equivalent operations:
In [7]: (Sum(2*k/2, (k, 1, 5)) + Sum((2*k-1-1)/2, (k, 1, 5))).doit()
Out[7]: 25
In [8]: Sum(2*k/2+(2*k-1-1)/2, (k, 1, 5)).doit()
Out[8]: 25
And in general, the procedure is:
Sum(f(i), (i, 1, 2*n)) = Sum(f(2*k)+f(2*k-1), (k, 1, n))
where f(i) = (i-i%2)/2 and n=5 in your case.
Ondrej
--
You received this message because you are subscribed to the Google Groups
"sympy" group.
To post to this group, send email to sy...@googlegroups.com.
To unsubscribe from this group, send email to
sympy+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/sympy?hl=en.
