On Feb 1, 10:47 am, Simon <simonjty...@gmail.com> wrote:
> Hi, this is hopefully an easy question:
>
> As a simple exercise, I'm trying to show that \int_0^{2\pi} e^{i (m-n)
> x}dx = 2\pi\delta_{mn} for integer m, n.
> Here's how I did it:
>
> sage: var('m,n'); w = SR.wild(0);
> sage: assume(n, 'integer');assume(m, 'integer')
> sage: int = integrate(e^(i*(m-n)*x),x,0,2*pi)
> sage: print int.limit(m=n)
> sage: print int.subs({e^(w):cosh(w)+sinh(w)}).simplify_trig()
> 2*pi
> 0
>
> The bit I don't like is using the substitution... it should not be
> nesc. The problem, I think, lies at
>
> sage: sin(2*pi*n).simplify_trig() # this works
> sage: e^(i*2*pi*m).simplify_full() # this doesn't work
> 0
> e^(2*I*pi*m)
>
> Any suggestions?
Our assumption framework is directly from Maxima. Perhaps Maxima does
not simplify exponentials in these cases? It would be worth checking
whether there is another simplification routine there we have missed,
though perhaps Maxima intentionally does not have this simplification.
Unfortunately, progress on this in Sage has stopped until people can
agree on a way to reorganize the simplification wrappers from Maxima.
- kcrisman
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