Computing $\sigma_N(f;t)$ from $s_n(f;t)$

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The discussion focuses on deriving the expression for $\sigma_{N}(f;t)$ from the given formula involving $s_{n}(f;t)$. The initial formula is $\sigma_{N}(f;t)= \frac{1}{N+1}\sum_{n=0}^{N}s_{n}(f;t)$, which can be expanded to $\sigma_{N}(f;t)=\frac{1}{N+1}\sum_{n=0}^{N}\sum_{k=-n}^{n}\widehat{f}(k)e^{ikt}$. The conversation also touches on the implications of reversing the order of summation, suggesting that this could lead to a different representation of the series. Participants express frustration over the slow pace of responses in the forum, but one notes the importance of patience in discussions. Overall, the thread highlights the mathematical manipulation of series and the dynamics of forum interactions.
errordude
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suppose, s_{n}(f;t) = \sum_{k=-n}^{n}\widehat{f}(k)e^{ikt}
and
\sigma_{N}(f;t)= \frac{1}{N+1}\sum_{n=0}^{N}s_{n}(f;t).

how do i get from this \sigma_{N}(f;t)= \frac{1}{N+1}\sum_{n=0}^{N}s_{n}(f;t).

to this


\sigma_{N}(f;t)= \sum_{n=-N}^{N}(1-\frac{|n|}{N+1})\widehat{f}(n)e^{int}

obviously one starts with:

\sigma_{N}(f;t)=\frac{1}{N+1}\sum_{n=0}^{N}\sum_{k=-n}^{n}\widehat{f}(k)e^{ikt}

thanks!
 
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And what happens when you reverse the order of summation ... the sum on k outside, the sum on n inside?
 
g_edgar said:
and what happens when you reverse the order of summation ... The sum on k outside, the sum on n inside?

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wow this must be slowest forum on the face of the planet
 
errordude said:
wow this must be slowest forum on the face of the planet

Perhaps, but remember we're not all free to check forums 25 hours a day, 8 days a week. Two hours 40 for what looks like a hint seems pretty good to me. Have you tried it?
 
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