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 general sums tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 06/22/2010, 10:29 PM i was thinking about " general sums ". that means continuum iterations of continuum sums. i believe q-analogues and fourrier series play an important role in this. its still vague , but i wanted to throw it on the table. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/23/2010, 01:16 AM (06/22/2010, 10:29 PM)tommy1729 Wrote: i was thinking about " general sums ". that means continuum iterations of continuum sums. i believe q-analogues and fourrier series play an important role in this. its still vague , but i wanted to throw it on the table. What do you mean? You mean continuous iteration of the sum operator? Hmmmmm... Well, for $e^{ux}$, the basis of Fourier series, we have $\Delta e^{ux} = \Sigma^{-1} e^{ux} = (e^u - 1) e^{ux}$. $\Delta^2 e^{ux} = \Delta \Delta e^{ux} = \Delta (e^u - 1) e^{ux} = (e^u - 1) \Delta e^{ux} = (e^u - 1)(e^u - 1) e^{ux} = (e^u - 1)^2 e^{ux}$. Induction shows that $\Delta^n e^{ux} = (e^u - 1)^n e^{ux}$. Thus we have the (indefinite!) continuum sum $\frac{e^{ux}}{e^u - 1}$ by setting $n = -1$, and we can "formally" continuously iterate the summation and difference operator by setting fractional, real, and complex values for $n$. Iteration of the difference operator seems to have been studied before -- look up "fractional finite differences". The generalization above may remind one of generalizing the derivative to non-integer order. For a Fourier/exp-series, $f(x) = \sum_{n=-\infty}^{\infty} a_n e^{nux}$, the fractional forward difference is $\Delta^t f(x) = \sum_{n=-\infty}^{\infty} a_n (e^{nu} - 1)^t e^{nux}$ which only gives iterations of the formal continuum sum if $a_0 = 0$. If $a_0 \ne 0$, then we get $0^t$ which is undefined for negative t and even t = 0, meaning we can't even apply the operator 0 times. I'm not sure how to extend it in those cases. tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 06/23/2010, 04:26 PM an intresting paper related to continuum sum ( but not continuum iterations of it ) is this : http://www.math.tu-berlin.de/~mueller/HowToAdd.pdf especially " 3. Basic Algebraic Identities " where the geometric part is what mike3 uses (together with fourrier expansion) to get his continuum sum. the idea of ' removing the period ' is also known and the origin of this 'geometric part equation' is as old as " q-math " ( q-series and q-analogues and fourrier series ) i knew id seen it before ... in fact i used it myself even way before that paper was written , although probably similar papers have been written much earlier. not to mention eulers example given in the paper. intresting is the continuum product product x ; sin(x) + 5/4. or equivalent the continuum sum sum x ; ln(sin(x) + 5/4). and the question if these sums resp products are periodic themselves. and the question if these sums resp products are divergent ( lim x -> oo does not equal +/-oo or 0) ( it is known that integral 0,2pi log(sin(x) + 5/4) = 0 ) regards tommy1729 kobi_78 Junior Fellow Posts: 11 Threads: 3 Joined: Dec 2009 06/24/2010, 06:53 PM There exists a formula for iterated sums. If $Sum_n({ a_n }) = S_n$ $S_0 = 0$ $S_n = S_{n - 1} + a_n$ Then $Sum_n^k({ a_n }) = \sum_{j=1}^{n} { {k + (n - j) - 1}\choose{k - 1} } \cdot a_j$ tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 06/24/2010, 07:45 PM (06/24/2010, 06:53 PM)kobi_78 Wrote: There exists a formula for iterated sums. If $Sum_n({ a_n }) = S_n$ $S_0 = 0$ $S_n = S_{n - 1} + a_n$ Then $Sum_n^k({ a_n }) = \sum_{j=1}^{n} { {k + (n - j) - 1}\choose{k - 1} } \cdot a_j$ thats great ! now all we need to do is take the continuum sum of that. where did you get that formula btw ? newton ? gauss ? thanks regards tommy1729 kobi_78 Junior Fellow Posts: 11 Threads: 3 Joined: Dec 2009 06/24/2010, 07:56 PM Hi, I discovered it myself To derive it, use the rule that the multiplication of two power series is the discrete convolution of their coefficients and that $(1 - x)^{-1} = \sum_{n=0}^{\infty}{x^n}$. Now try to multiply a formal power series with $(1 - x)^{-1} = \sum_{n=0}^{\infty}{x^n}$. Repeat it k times and use the formula of $(1 - x)^{-k}$ to derive the formula. It however reminds Cauchy's formula for repeated integration (with all umbral calculus variations). « Next Oldest | Next Newest »

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