Problem with plotting and simplifying a function

[new one]

I have a question that troubles me , and I'll appreciate your advices.
Given Sum of : X_N(t) =Sum [ Exp(j2*Pi*k*t) ] with lower limit -N and upper limit +N , -2<= t <= 2
I need to calculate x(t) ,so I used Euler :
X_N(t) = Sum e^(j2*pi*k*t) = Sum [ cos(2pi*k*t) + j*sin (2pi*k*t) ].

Now ,if I'm mistaken the sinus part becom equal to zero , because the cycle
of sinus is PI . But I have a problem with the Cosinus part , how can I
simplify it ? the "t" parameter interrupts with the simplifying ......
maybe separating it to conditions ? N=2R ------> even , N=2R+1 --> odd ?

thanks in advance

[Jens-Peer Kuska]

Hi,

what do you want to simplify with

1+2 Sum[Cos[2*Pi*k*t],{k,1,n}]

?

Regards
Jens

[Sjoerd C. de Vries]

I doubt whether this is about Mathematica. It looks more like a math
question.

Given that this is a Mathematica newsgroup a few remarks from this
point of view, just in case:

Your syntax is not even close to that of the Mathematica language.
-Underscores are not allowed in variable r function names (they are
part of the pattern match syntax).
-Use square brackets for the Exp function (and all other functions for
that matter)
-Lower case e does not equal the base of the natural log; that's
written as either E or esc-ee-esc.
-cosine and sine are written as Cos and Sin (starting with a capital
letter as does any function in Mathematica)
-You don't specify over which variable you're summing. Is it j2, k or
t?
-You use a lowercase j to indicate the square root of -1. In
Mathematica you either use capital I, or esc-ii-esc or esc-jj-esc. For
the Eulerian expression to be true j2 must be I as well, or does it
have a deeper meaning?

Now to the math:
-You state that the cycle (I prefer the term 'period') of the sine is
Pi. That's not true. It's 2Pi. Its zeros are spaced Pi apart, but
that's not the same.
-You state that sin (2pi*k*t) is zero. That's not generally true, only
for integer values of of 2 k t. You never say that they are integer.
-However, assuming 2 k t is integer (since you seem sure the sin
equals zero), cos(2pi*k*t) equals 1, and your summation is 2N+1.

Cheers -- Sjoerd

> maybe separating it to conditions ? N=2R ------> even , N=2R+1 --> od=
d ?
>
> thanks in advance

[dh]

Hi,

simply write:

Sum[Cos[2 Pi k t], {k, -n, n}]

this gives:

-1 + 2 Cos[n \[Pi] t] Csc[\[Pi] t] Sin[(1 + n) \[Pi] t]

Further, note that the Sin part is zero because Sin is an odd function.

Daniel