Well, here's my problem:

I'm finding the first few terms for a Fourier transform for a function. I
found a_n and b_n. a_n turns out to always be zero (except a_0), but b_n is
defined as:

(1/pi)-(cos(pi*n)/pi)

When I put this in

sum(b_n*sin(n*x),n,1,5)

To find the sum terms of the function I get:

-1/15*(-5*I*e^(2*I*x) - 15*I*e^(4*I*x) + 15*I*e^(6*I*x) + 5*I*e^(8*I*x)
+ 3*I*e^(10*I*x) - 3*I)*e^(-5*I*x)/pi

I know that that is the representation of the complex form and it is
equivalent, but I need it in cosine form only.

Thanks

On Thu, May 27, 2010 at 8:15 AM, kcrisman <[email protected]> wrote:

> Dear Jaasiel,
>
> Thanks for your request.  Can you be a little more specific about your
> question?  It sounds very vague.  Do you mean something like this?
>
> - kcrisman
>
> ----------------------------------------------------------------------
> | Sage Version 4.4.2, Release Date: 2010-05-19                       |
> | Type notebook() for the GUI, and license() for information.        |
> ----------------------------------------------------------------------
> sage: var('n k')
> (n, k)
> sage: sum(binomial(n,k),k,0,n)
> 2^n
>
>
> On May 27, 9:32 am, Jaasiel Ornelas <[email protected]> wrote:
> > Hi!
> >
> > Is there a way to have sage give a certain number of terms in a sum?
> > And, is there a way to define a function in terms of that sum?
> >
> > Thanks Very Mucho.
>
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