Sin [Pi. (x-1)] / x-1; for x =1,2,3, .N

[pjunky]

Sin [Pi. (x-1)] / x-1; for x =1,2,3,...N

Sin [ Pi. (x-1) ] / (x-1) = ?

for x=1,2,3,4,...N

I think for Sin (x) / x =1

but, what about that.

 

[CompuChip]

For x > 1 it is rather obvious, you can just plug in the value.
For x = 1 you are dividing by 0, so it is undefined. You can however take the limit, as you are indicating.

 

[pjunky]

For x > 1 it is rather obvious, you can just plug in the value.
For x = 1 you are dividing by 0, so it is undefined. You can however take the limit, as you are indicating.

for all X values will I get '0'

 

[CompuChip]

I am assuming you mean: for all X values which are integer and at least 2. Then you are right. sin(n pi) = 0 for all integers n, and since you are dividing by something non-zero for n non-zero, you have sin(n pi)/n = 0/something = 0.

For n = 1, you get sin((n - 1) pi) / (n - 1) which looks like 0/0 if you plug in the numbers.