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+--- Thread: general sums (/showthread.php?tid=459)
general sums - tommy1729 - 06/22/2010
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.
RE: general sums - mike3 - 06/23/2010
(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.
RE: general sums - tommy1729 - 06/23/2010
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
RE: general sums - kobi_78 - 06/24/2010
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
\( Sum_n^k({ a_n }) = \sum_{j=1}^{n} { {k + (n - j) - 1}\choose{k - 1} } \cdot a_j \)
RE: general sums - tommy1729 - 06/24/2010
(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
RE: general sums - kobi_78 - 06/24/2010
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).