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

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errordude
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suppose, [tex]s_{n}(f;t) = \sum_{k=-n}^{n}\widehat{f}(k)e^{ikt}[/tex]
and
[tex]\sigma_{N}(f;t)= \frac{1}{N+1}\sum_{n=0}^{N}s_{n}(f;t)[/tex].

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

to this


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

obviously one starts with:

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

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?