"Fractional-index sums"
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From: mike3 on 9 Aug 2010 18:19 Hi. I was wondering about this. Consider the finite forward difference of the exponential: Delta e^(ux) = (e^u - 1) e^(ux) This suggests that the "indefinite sum" of the exponential is sum_x e^(ux) = (e^(ux))/(e^u - 1) Note that this isn't the only possible one: if we add any 1-periodic theta(x) to it, we get another. However, we may take this as more "natural" in a sense, for one, the Delta operator sends exponentials to exponentials, and the sum_x operator here, a sort of "inverse" does the same. This can be used to define "fractional index sums", like sum_{n=0...x-1} e^(un) = (e^(ux) - 1)/(e^u - 1). For example, we can sum the powers of 2 "indexed from" 0 to 1/2 as 2^(3/2) - 1 = 1.828427124... (just set u = log(2) and x = 3/2 in the above.) We can even set complex values, e.g. the sum "from" 0 to i is 2^(1+i) - 1 = 0.538477802... + 1.277922552...i. Now consider a periodic function given by a Fourier series, i.e. f(x) = sum_{n=-inf...inf} a_n e^(2pii/P nx) where P is the period. Applying our sum_x, we get sum_x f(x) = sum_{n=-inf...inf} a_n/(e^(2pii/P n) - 1) e^(2pii/P nx) with the caveat that the term at n = 0 is interpreted as a_0 x (we use an additional "natural" formula, namely that sum_x k = kx, k being a constant.). So my question, then, is, assuming the Fourier series for f(x) converges everywhere on the real line, when does that for sum_x f(x) converge? Obviously, it'd be whenever the Fourier series formed on the right hand side of that last equation is also convergent, but *what does that being so imply about the structure/behavior of the function f(x)*?
From: mike3 on 12 Aug 2010 14:19
On Aug 9, 4:19 pm, mike3 <mike4...(a)yahoo.com> wrote: > Hi. > > I was wondering about this. Consider the finite forward difference of > the exponential: > > Delta e^(ux) = (e^u - 1) e^(ux) > > This suggests that the "indefinite sum" of the exponential is > > sum_x e^(ux) = (e^(ux))/(e^u - 1) > > Note that this isn't the only possible one: if we add any 1-periodic > theta(x) to it, we get another. However, we may take this as more > "natural" in a sense, for one, the Delta operator sends exponentials > to exponentials, and the sum_x operator here, a sort of "inverse" does > the same. > > This can be used to define "fractionalindexsums", like > > sum_{n=0...x-1} e^(un) = (e^(ux) - 1)/(e^u - 1). > > For example, we can sum the powers of 2 "indexed from" 0 to 1/2 as > 2^(3/2) - 1 = 1.828427124... (just set u = log(2) and x = 3/2 in the > above.) We can even set complex values, e.g. the sum "from" 0 to i is > 2^(1+i) - 1 = 0.538477802... + 1.277922552...i. > > Now consider a periodic function given by a Fourier series, i.e. > > f(x) = sum_{n=-inf...inf} a_n e^(2pii/P nx) > > where P is the period. Applying our sum_x, we get > > sum_x f(x) = sum_{n=-inf...inf} a_n/(e^(2pii/P n) - 1) e^(2pii/P nx) > > with the caveat that the term at n = 0 is interpreted as a_0 x (we use > an additional "natural" formula, namely that sum_x k = kx, k being a > constant.). > > So my question, then, is, assuming the Fourier series for f(x) > converges everywhere on the real line, when does that for sum_x f(x) > converge? Obviously, it'd be whenever the Fourier series formed on the > right hand side of that last equation is also convergent, but *what > does that being so imply about the structure/behavior of the function > f(x)*? Any comments? |