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Re: Problem with plotting and simplifying a function


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.


new one wrote:

> 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


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