Mittag-Leffler series for generating continuum sum? - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Mittag-Leffler series for generating continuum sum? (/showthread.php?tid=387) Pages: 1 2 3 RE: Mittag-Leffler series for generating continuum sum? - mike3 - 12/11/2009 Geez, the number of terms in that puppy grows up insanely fast! $g_4$ already has over a trillion terms... Is there any way to actually "use" this formula? Especially if we put in the Bernoulli polynomials for the continuum sum, then the number of terms gets even bigger and it gets even hairier(!). So I'd be curious if this is the same as the other formula (the one supposedly mentioned in the 1905 book) or not. EDIT: got that wrong, that's a trillion operations not terms. The maximum degree for $g_4$ is just $4^{2*4} + 4^{2*3} + 4^{2*2} + 4^{2*1} = 69904$ But still, it's a lot of operations! RE: Mittag-Leffler series for generating continuum sum? - bo198214 - 12/12/2009 (12/11/2009, 11:16 AM)bo198214 Wrote: $G_0(x)=g_0(x)=\phi(0)$ $G_n(x)=g_n(x)-g_{n-1}(x)$ for $n>0$, $\phi(x)=\sum_{n=0}^\infty G_n(x)$ Well this G is of no use because $\phi(x)=\lim_{n\to\infty} g_n(x)=\sum_{\lambda_1=0}^{n^{2n}}\lim_{n\to\infty}\sum_{\lambda_2=0}^{n^{2n-2}} \dots\sum_{\lambda_n=0}^{n^2} \frac{\phi^{(\lambda_1+\dots+\lambda_n)}(0)}{\lambda_1!\dots\lambda_n!} \left(\frac{x}{n}\right)^{\lambda_1+\dots+\lambda_n}$ If we sort the terms for $m=\lambda_1+\dots+\lambda_n$ we can see that this is a sequence of polynomials $g_n(x)=\quad\sum_{m=0}^{n^2+\dots+n^{2n}}\quad c^{(n)}_m \frac{\phi^{(m)}(0)}{m!} x^m$, where the coefficients $c^{(n)}_m$ are independent on $\phi$ and on the region. I guess these huge number of terms is the drawback of the formula and the reason why you nearly can find it anywhere in the net (if it was of big practical use one would get hundrets of hits for the formula). Other coefficients I guess there are also other coefficients possible, the only thing that is needed in the proof is a sequence of polynomials $P_n(z)=\sum_{m=0}^{M(n)}c^{(n)}_m z^m$ that converge compactly to $\frac{1}{1-z}$ on the region $\mathbb{C}\setminus \{z: z\ge 1\}$. The basic idea is that $f(z)=\frac{1}{2\pi i} \int_{\gamma} \frac{f(\zeta)}{\zeta-z}d\zeta$, where you replace $\frac{1}{\zeta-z}=\frac{1}{(1-\frac{z}{\zeta})\zeta}$ with the $P_n(\frac{z}{\zeta})\frac{1}{\zeta}=\sum_{m=0}^{M(n)} c^{(n)}_m z^m \frac{1}{\zeta^{m+1}}$ Then $f(z)=\lim_{n\to\infty} \sum_{m=0}^{M(n)} c^{(n)}_m z^m \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta^{m+1}}d \zeta = \lim_{n\to\infty} \sum_{m=0}^{M(n)} c^{(n)}_m \frac{\phi^{(m)}(0)}{m!} z^m$ So give me any such polynomial sequence and I give you a Mittag-Leffler expansion. Computing the original expansion To get an impression I computed the original expansion for f(z)=log(z+1). And this is the result: MLE $g_2$ has degree $2^2+2^4=20$: [attachment=663] MLE $g_3$ has degree $3^2+3^4+3^6=819$: [attachment=664] Normal powerseries of log(z+1) with as many terms as $g_3$: [attachment=665] As comparison: $g_2(x)=-\frac{969}{4194304} x^{20} + \frac{255}{524288} x^{19} - \frac{1343}{1572864} x^{18} + \frac{189}{131072} x^{17} - \frac{2517}{1048576} x^{16} + \frac{647}{163840} x^{15} - \frac{1471}{229376} x^{14} +\\ \frac{1093}{106496} x^{13} - \frac{397}{24576} x^{12} + \frac{281}{11264} x^{11} - \frac{193}{5120} x^{10} + \frac{1}{18} x^{9} - \frac{163}{2048} x^{8} + \frac{99}{896} x^{7} - \frac{19}{128} x^{6} + \frac{31}{160} x^{5} - \frac{1}{4} x^{4} + \frac{1}{3} x^{3} - \frac{1}{2} x^{2} + x$ one sees that up to $x^4$ the development matches the development of log(x+1). Here the direct coefficients each sequence starting with index 0: $c^{(1)}=(1,1)$ $c^{(2)}=(1, 1, 1, 1, 1, \frac{31}{32}, \frac{57}{64}, \frac{99}{128}, \frac{163}{256}, \frac{1}{2}, \frac{193}{512}, \frac{281}{1024}, \frac{397}{2048}, \frac{1093}{8192}, \frac{1471}{16384}, \frac{1941}{32768}, \frac{2517}{65536}, \frac{3213}{131072}, \frac{4029}{262144}, \frac{4845}{524288}, \frac{4845}{1048576})$ $c^{(3)}=(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \frac{59048}{59049}, \frac{177124}{177147}, \frac{531152}{531441}, \frac{1591696}{1594323}, \frac{4763648}{4782969}, \frac{14226880}{14348907}, \frac{42360320}{43046721}, \frac{125616640}{129140163}, \frac{370626560}{387420489}, \frac{1086986240}{1162261467}, \frac{3166363648}{3486784401}, \frac{9155108864}{10460353203}, \frac{26261389312}{31381059609},\dots)$ RE: Mittag-Leffler series for generating continuum sum? - mike3 - 12/12/2009 *sighs* I guess its back the drawing board, then. This continuum sum thing seems really difficult to generalize to arbitrary analytic functions, as is needed for tetration. RE: Mittag-Leffler series for generating continuum sum? - bo198214 - 12/12/2009 (12/12/2009, 08:18 PM)mike3 Wrote: *sighs* I guess its back the drawing board, then. This continuum sum thing seems really difficult to generalize to arbitrary analytic functions, as is needed for tetration. Well, even if there was a simple Mittag-Leffler expansion, the chance to have luck with a convergent Faulhaber application, was anyway quite low, wasnt it? If not even entire the continuum sum of exp(exp(x)) was convergent. Hm, so summarizing: exp(x) has a convergent Faulhaber sum. non-entire functions have no convergent Faulhaber sum. functions with over-exponential growth (like exp(exp(x))) have probably no convergent Faulhaber sum. However if there was a way to rearrange the terms, there might convergence be possible (as you greatly showed with exp(exp(x)) and 1/(x+1)). But then my question would be whether the value depends on how the terms are rearranged (I mean not only giving different branches, but a continuum of solutions). RE: Mittag-Leffler series for generating continuum sum? - mike3 - 12/13/2009 (12/12/2009, 09:53 PM)bo198214 Wrote: But then my question would be whether the value depends on how the terms are rearranged (I mean not only giving different branches, but a continuum of solutions). I.e. given two transseries representations for an analytic function, if both converge and so do their Faulhaber continuum sums, do those sums agree whenever both converge? It would be interesting to determine this, but I'm not sure what the proof would be like. Another thing I've wondered about is, could there be some way to assign some sort of value to the divergent sums of Faulhaber coefficients given an arbitrary power series of nonzero convergence radius as input? Failing the Faulhaber method, could there be some other way to define continuum sum that agrees with it but covers a lot more ground? Mittag-Leffler product ? Ramanujan's master theorem - tommy1729 - 12/13/2009 this might be slightly off topic , but i think ( wont explain ) that a product analogue of mittag-leffler is more usefull for tetration. product expansions are not so commenly seen but imho intresting in general. for instance ( in the spirit of " q " (analogue) ) we have exp(x) = (1 + a_1 x)(1 + a_2 x^2)...(1 + a_i x^i) where a_i = 1 , 1/2 , -1/3 , 3/8 , ... = A137852 * (i !) ( and note denominator a_p*q = p^q * q^p for primes p and q ) ( see also witt vectors ) of course this product expansion of exp(x) is not valid everywhere ( exp HAS NO ZEROS AND ' hint for the radius ' : a_i ^ (1/i) ) and a general product expansion might be complicated considering that functions have zero's ( or even dont ! ) but there might be a way around that problem perhaps. im aware that certain product expansions might not be unique for some or all functions. and even if we only have convergeance in a certain domain , it might still be sufficient to help at tetration. i dont think there are known ways to find real iterations of a product into a product without converting to integrals and sums (and back again) and other ' non-product forms ' , but i think some specificly chosen aid-functions might work and help in solving the tetration issues. ----------------- as for the continuum sum idea and other types of continuation/interpolation i think Ramanujan's master theorem might be usefull. since a continuum sum = sum of a continuum up to a constant imho , and clever substitions ( to avoid poles of gamma for instance ) together with the master theorem provide ( when converging ) a kind of continuation/interpolation/summation method ( which may or may not be equivalent to some other ). convergeance will probably be the biggest obstacle. you will probably benefit from adding an extra parameter , do substitutions , use master theorem , substitute again and fix parameter to arrive at tetration. this master theorem idea is not ( at least in my mind ) related to the product idea posted above. -------------- i had those 2 ideas for a very long time , its about time i posted them i wish i had more time to investigate , but im to busy with number theory at the moment. ( and i could use better hardware and software :p for christmas ? :p) regards tommy1729 RE: Mittag-Leffler product ? Ramanujan's master theorem - mike3 - 12/13/2009 Hmm. P.S. continuum sum means interpolation of sum operator to non-integer bounds, e.g. summing a function from, say, 0 to 1/2 or something (so, a sum whose bounds can range in the continuum). RE: Mittag-Leffler product ? Ramanujan's master theorem - tommy1729 - 12/13/2009 (12/13/2009, 08:19 PM)mike3 Wrote: Hmm. P.S. continuum sum means interpolation of sum operator to non-integer bounds, e.g. summing a function from, say, 0 to 1/2 or something (so, a sum whose bounds can range in the continuum). yes i know. so ?? RE: Mittag-Leffler product ? Ramanujan's master theorem - mike3 - 12/14/2009 (12/13/2009, 11:15 PM)tommy1729 Wrote: (12/13/2009, 08:19 PM)mike3 Wrote: Hmm. P.S. continuum sum means interpolation of sum operator to non-integer bounds, e.g. summing a function from, say, 0 to 1/2 or something (so, a sum whose bounds can range in the continuum). yes i know. so ?? Because in the other post you said it was "sum 'of a continuum' up to a constant", and I wasn't sure what that meant.
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