On Feb 1, 11:53 pm, Simon simonjty...@gmail.com 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)
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,
On Feb 1, 4:47 pm, 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,
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 assuming that
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));
