On Feb 1, 11:53 pm, Simon wrote:
> Thank you both for your replies.
> It's interesting that in Andrej's examples, the e^x and exp(x) form do
> not yield the same result,
> since in maxima (which I know next to nothing about) we have
> (%i1) declare(m, integer);
> (%o1)
Thank you both for your replies.
It's interesting that in Andrej's examples, the e^x and exp(x) form do
not yield the same result,
since in maxima (which I know next to nothing about) we have
(%i1) declare(m, integer);
(%o1)done
(%i2) rectform(exp(2*m*%i*%pi));
(%o2
>
> In Maxima you would use rectform to convert the expression from polar
> to rect form:
>
> sage: int._maxima_().rectform()
> 0
> sage: e^(i*2*pi*m)._maxima_().rectform()
> e^(2*I*pi*m)
> sage: exp(i*2*pi*m)._maxima_().rectform()
> 1
Thanks, Andrej, that is very helpful. Am I correct in assumin
On Feb 1, 4:47 pm, Simon 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, 'integ
On Feb 1, 10:47 am, Simon 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, 'in
