i see, this was unexpected for me :P

But now that i know about it i can hopefully find ways around it :)

I have added a longer comment to the answer below.

Thx for your time and quick responce!

i see, this was unexpected for me :P

But now that i know about it i can hopefully find ways around it :)

I have added a longer comment to the answer below.

Thx for your time and quick responce!

ahh, yes i ofc mean x. And you are right about the not defined ( didnt think of it since in the original i use ^abs(..). ) . But that eval of the sum makes sence, i assume this can be done for r<0 and r>0 etc. Thx for your reply. Although i must say i find it odd that Maple automaticly does that. But i guess its ok aslong as im aware of it :)

Maybe for curiosity i can then extend the question to the reason i met this problem in the first place?

if i take the sum: 

sum(p1^abs(l)*p2^abs(l-2)*exp(-I*2*Pi*f*l*T), l = 1 .. infinity)

i get :    p1/(p2*(exp((2*I)*Pi*f*T)-p1*p2))

But here maple has not accounted for p2 being able to take the value of zero ( then the answer would be divide by zero ), but the sum is infact defined for p2=0  -> the answer is then simply p1^abs(2)*1*exp(-2*I*Pi*f*2*T) , because l=2 would be the only time it differs from 0.

So this is why i tried calculating the sum by inserting 0 for p2 ,and hence the earlier question. 

In maples definition of sum, it says that sum(..., from a..b) = sum(..., from a..n1) + sum(...,from (n1+1)..b) , so i figured lets see what happens if i divide it up like this. Then by taking the sum from 1:10 + 11:infinity and simplifying it, i get: 

 -p1*(-p2+p2^2*exp(-(2*I)*Pi*f*T)*p1-p1*exp(-(2*I)*Pi*f*T))/(exp((2*I)*Pi*f*T)-p1*p2)

which is reduced to p1^2 * exp(-4*I*Pi*f*T) when inserting p2=0. 

I assume its because for sums of a certain range, the individual terms are calculated, and for longer its some sort of algorithm or simplification.

So hopefully i have solved it, but im not sure that i trust the answer to be correct for other values of p1 and p2, any ideas how to check it? 

Again thx for all of your time!

ahh, yes i ofc mean x. And you are right about the not defined ( didnt think of it since in the original i use ^abs(..). ) . But that eval of the sum makes sence, i assume this can be done for r<0 and r>0 etc. Thx for your reply. Although i must say i find it odd that Maple automaticly does that. But i guess its ok aslong as im aware of it :)

Maybe for curiosity i can then extend the question to the reason i met this problem in the first place?

if i take the sum: 

sum(p1^abs(l)*p2^abs(l-2)*exp(-I*2*Pi*f*l*T), l = 1 .. infinity)

i get :    p1/(p2*(exp((2*I)*Pi*f*T)-p1*p2))

But here maple has not accounted for p2 being able to take the value of zero ( then the answer would be divide by zero ), but the sum is infact defined for p2=0  -> the answer is then simply p1^abs(2)*1*exp(-2*I*Pi*f*2*T) , because l=2 would be the only time it differs from 0.

So this is why i tried calculating the sum by inserting 0 for p2 ,and hence the earlier question. 

In maples definition of sum, it says that sum(..., from a..b) = sum(..., from a..n1) + sum(...,from (n1+1)..b) , so i figured lets see what happens if i divide it up like this. Then by taking the sum from 1:10 + 11:infinity and simplifying it, i get: 

 -p1*(-p2+p2^2*exp(-(2*I)*Pi*f*T)*p1-p1*exp(-(2*I)*Pi*f*T))/(exp((2*I)*Pi*f*T)-p1*p2)

which is reduced to p1^2 * exp(-4*I*Pi*f*T) when inserting p2=0. 

I assume its because for sums of a certain range, the individual terms are calculated, and for longer its some sort of algorithm or simplification.

So hopefully i have solved it, but im not sure that i trust the answer to be correct for other values of p1 and p2, any ideas how to check it? 

Again thx for all of your time!