The summation $\sum_{n=-N}^{N}|e^{J\frac{\pi}{4}n}|^2$ contains exactly 2N + 1 terms. This conclusion arises from counting the terms in the specified range of n, which includes N negative integers (from -N to -1), the zero term, and N positive integers (from 1 to N). The magnitude of the complex exponential function squared is equal to one, confirming that each term contributes equally to the sum.
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Drain Brain said:I just want to know how do I determine the number of terms will be in this summation. The answer to this 2N+1 terms. I can only arrive at preliminary steps of solving this. can you tell why 2N+1 is the number of terms? I know that the magnitude of complex exponential function squared would result to one.
$\sum_{n=-N}^{N}|e^{J\frac{\pi}{4}n}|^2$
kaliprasad said:as n is from -N to N the total number of terms is N ( -N to -1) + 1 ( zero) + N ( 1 to N) = 2N + 1
Within parenthesis I have mentioned the range
Drain Brain said:Hi kaliprasad! How did you choose the range?
kaliprasad said:In the sum that is (sigma) n is from -N to N and hence the range
Drain Brain said:why it is only -N to -1, 0, and 1 to N? I'm thinking of other ranges like -N to -2 etc.. I'm confused. Please help.