10/07/2010, 04:32 AM

You ask a ton of questions but I'll try my best...

Yes. Or , where is the continuum sum of . And of course, the opposite, i.e. that .

The Fourier is only used to sum the approximations -- the limit is the continuum sum of the aperiodic function. But if you want to extend those approximations to the whole plane, or even the limiting continuum sum, why not use the difference equation? The Fourier converges in a strip, the Taylor only in a disk. And the method by which the continuation is accomplished does not really matter from the theoretical side, in where only the continuation itself is relevant.

You mean like taking a sequence of periodic approximations with real period (which is what that is equivalent to)? There are some problems there, namely that the continuum sum fails for a harmonic that is 1-periodic. This means all integer periods are out of the question. Though sequences of periodic approximations of increasing period which there are no subsequences of periods which approach an integer period might work. That last requirement can be formulated as "there must be some for which there is no such that ".

Yeah, when the function is not entire. Take a periodic approximation (or periodic function with singularities), expand it on one side of the singularity, then expand it on the other and compare the continuum sums.

For a given periodic function, such as a periodic approximation, a Fourier series can only be generated along the line of periodicity (that is, lines in the plane with slope equal to the slope of the complex period vector, parameterized by , where is the period and is an arbitrary complex constant.). If the strips of convergence intersect then that means there was no singularity between the lines of expansion, i.e. they were all on one or the other side of the singularity. In that case, both will give the same result. It's if they're on different sides that there's a problem. This means the continuum sum of the given periodic function is ambiguous in a way that is not easily resolved (continuum sum is ambiguous by nature, but the Fourier series provides a "natural" definition for it applied to periodic functions, at least ones that are entire or otherwise singularity-free).

What does that mean?

It shows the two equally "good" continuum sums that come from considering "summing to the right" of the singularities and "summing to the left".

(10/06/2010, 12:44 PM)tommy1729 Wrote: continuum-sum recurrence equations ?

you mean f(z+1) = f(z) + delta f(z) , where delta is the antisum ?

Yes. Or , where is the continuum sum of . And of course, the opposite, i.e. that .

(10/06/2010, 12:44 PM)tommy1729 Wrote: why not turn the fourier series into taylor and do ordinary analytic continuation. mittag leffler ?

or do they give different results ? i dont think so.

The Fourier is only used to sum the approximations -- the limit is the continuum sum of the aperiodic function. But if you want to extend those approximations to the whole plane, or even the limiting continuum sum, why not use the difference equation? The Fourier converges in a strip, the Taylor only in a disk. And the method by which the continuation is accomplished does not really matter from the theoretical side, in where only the continuation itself is relevant.

(10/06/2010, 12:44 PM)tommy1729 Wrote: what if we take the fourier series at I(z) = 0 ?

You mean like taking a sequence of periodic approximations with real period (which is what that is equivalent to)? There are some problems there, namely that the continuum sum fails for a harmonic that is 1-periodic. This means all integer periods are out of the question. Though sequences of periodic approximations of increasing period which there are no subsequences of periods which approach an integer period might work. That last requirement can be formulated as "there must be some for which there is no such that ".

(10/06/2010, 12:44 PM)tommy1729 Wrote: so the problems occur when we have 2 fourier series expanded on different lines that are not entire and require continuation ?

Yeah, when the function is not entire. Take a periodic approximation (or periodic function with singularities), expand it on one side of the singularity, then expand it on the other and compare the continuum sums.

(10/06/2010, 12:44 PM)tommy1729 Wrote: is it true that if the radiuses intersect , the problem cannot occur ?

is it false that when both are entire they have to agree ?

For a given periodic function, such as a periodic approximation, a Fourier series can only be generated along the line of periodicity (that is, lines in the plane with slope equal to the slope of the complex period vector, parameterized by , where is the period and is an arbitrary complex constant.). If the strips of convergence intersect then that means there was no singularity between the lines of expansion, i.e. they were all on one or the other side of the singularity. In that case, both will give the same result. It's if they're on different sides that there's a problem. This means the continuum sum of the given periodic function is ambiguous in a way that is not easily resolved (continuum sum is ambiguous by nature, but the Fourier series provides a "natural" definition for it applied to periodic functions, at least ones that are entire or otherwise singularity-free).

(10/06/2010, 12:44 PM)tommy1729 Wrote: when are there complex continu solutions that satisfy both fourier expansions ?

What does that mean?

(10/06/2010, 12:44 PM)tommy1729 Wrote: i dont get your digamma(-z) argument ...

sorry if i ask trivial questions.

It shows the two equally "good" continuum sums that come from considering "summing to the right" of the singularities and "summing to the left".