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 Borel summation and other continuation/summability methods for continuum sums mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/29/2009, 10:55 AM (This post was last modified: 12/29/2009, 10:55 AM by mike3.) Hi. Here's a new possibility for tetration. It's based on the use of Borel summation and other types of methods to attempt to extend continuum sums to a wider domain. Now as you may know, the continuum sum from Faulhaber's formula: $\sum_{n=0}^{z-1} f(n) = \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} \frac{a_{n-1}}{n} {n \choose k} B_{n-k}\right) z^k$ (here we use $B_0 = -\frac{1}{2}$) where $f(z) = \sum_{n=0}^{\infty} a_n z^n$ is a holomorphic function, only works for a very limited range of functions. Namely, it won't work for non-entire functions, or (it seems) entire functions not of exponential type or of exponential type greater than $2\pi$. Tetration fails both requirements: it is not entire ($\mathrm{tet}_b(-2 - n)$ for nonnegative integer $n$ is a singularity, a branch point in fact, and there may be others too), and except for $e^{-e} \le b \le e^{1/e}$, it is not of exponential type due to insanely rapid growth. These are well-known. Because of this, the inner sums will not converge. So the question arises: could one assign a meaning to the formula even when that is the case? Note that we already know some continuum sums for functions that fail this criterion, e.g. $\sum_{n=0}^{z-1} \frac{1}{n+1} = \gamma + \digamma(n + 1)$ $\sum_{n=0}^{z-1} b^n = \frac{b^z - 1}{b - 1},\ b \ne 1$ (even for $\log(b) \ge 2\pi$!) $\sum_{n=0}^{z-1} \log(1+n) = \log(\Gamma(z+1))$ etc. and for functions that can be represented via exponential series, i.e. $f(z) = \sum_{n=0}^{\infty} a_n e^{nuz}$, $\sum_{n=0}^{z-1} f(z) = a_0 z + \sum_{n=1}^{\infty} a_n \frac{e^{nuz} - 1}{e^{nu} - 1}$. So we would expect that any such divergent/continuation/summability method that's up to the task should preserve these, while allowing us to continuum-sum more things, and it should be "natural" in some way (what that means is, of course, the biggest question). One option I've thought of is Borel summation. It works like: $\sum_{n=0}^{\infty} u_n {\'\'=\'\'} \int_{0}^{\infty} e^{-t} g(t) dt$ where $g(t) = \sum_{n=0}^{\infty} \frac{u_n}{n!} t^n$ if this can be analytically continued to all $t > 0$ and grows at most exponentially. Thus we get the "regularized Faulhaber coefficients" $b_k = \int_{0}^{\infty} e^{-t} \left(\sum_{n=0}^{\infty} \left(\frac{a_{n-1}}{n n!} {n \choose k} B_{n-k}\right) t^n\right) dt$ with the inner sum analytically continued, so the continuum sum is $\sum_{n=0}^{z-1} f(n) = \sum_{k=1}^{\infty} b_k z^k$. Another method that might be useful is the one I mentioned here: http://math.eretrandre.org/tetrationforu...93#pid4293 Perhaps it could give a still wider range of functions. There are some functions for which these do not appear to work -- consider $f(z) = \sum_{n=0}^{\infty} \frac{z^{2n}}{B_{2n}}$. The first coefficient of the continuum sum $b_1$ by the Faulhaber's formula gives the divergent sum $1 + 1 + 1 + 1 + ...$ which does not look to be Borel summable. However it seems it could work for other functions, and the big question is, of course, could it work for tetration, and if so, do the obtained extensions of tetration agree with the ones already made, yet enable expansion to a much wider variety of bases? sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 12/30/2009, 09:19 PM (12/29/2009, 10:55 AM)mike3 Wrote: Hi. Here's a new possibility for tetration. It's based on the use of Borel summation and other types of methods to attempt to extend continuum sums to a wider domain. Now as you may know, the continuum sum from Faulhaber's formula: $\sum_{n=0}^{z-1} f(n) = \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} \frac{a_{n-1}}{n} {n \choose k} B_{n-k}\right) z^k$ (here we use $B_0 = -\frac{1}{2}$) where $f(z) = \sum_{n=0}^{\infty} a_n z^n$ is a holomorphic function, only works for a very limited range of functions. Namely, it won't work for non-entire functions,There is always base $\eta$, but of course, it can be solved by other means. The upper superexponential at base $\eta$ is entire with no singularities, and with super-exponential growth. Perhaps if you can get the Borel summation to work for $\eta$, you can figure out how to make it work for other bases. Quote:or (it seems) entire functions not of exponential type or of exponential type greater than $2\pi$. Tetration fails both requirements: it is not entire ($\mathrm{tet}_b(-2 - n)$ for nonnegative integer $n$ is a singularity, a branch point in fact, and there may be others too), and except for $e^{-e} \le b \le e^{1/e}$, it is not of exponential type due to insanely rapid growth. These are well-known. Because of this, the inner sums will not converge. So the question arises: could one assign a meaning to the formula even when that is the case? Note that we already know some continuum sums for functions that fail this criterion, e.g. $\sum_{n=0}^{z-1} \frac{1}{n+1} = \gamma + \digamma(n + 1)$ $\sum_{n=0}^{z-1} b^n = \frac{b^z - 1}{b - 1},\ b \ne 1$ (even for $\log(b) \ge 2\pi$!) $\sum_{n=0}^{z-1} \log(1+n) = \log(\Gamma(z+1))$ etc. and for functions that can be represented via exponential series, i.e. $f(z) = \sum_{n=0}^{\infty} a_n e^{nuz}$, $\sum_{n=0}^{z-1} f(z) = a_0 z + \sum_{n=1}^{\infty} a_n \frac{e^{nuz} - 1}{e^{nu} - 1}$. So we would expect that any such divergent/continuation/summability method that's up to the task should preserve these, while allowing us to continuum-sum more things, and it should be "natural" in some way (what that means is, of course, the biggest question). One option I've thought of is Borel summation. It works like: $\sum_{n=0}^{\infty} u_n {\'\'=\'\'} \int_{0}^{\infty} e^{-t} g(t) dt$ where $g(t) = \sum_{n=0}^{\infty} \frac{u_n}{n!} t^n$ if this can be analytically continued to all $t > 0$ and grows at most exponentially. Thus we get the "regularized Faulhaber coefficients" $b_k = \int_{0}^{\infty} e^{-t} \left(\sum_{n=0}^{\infty} \left(\frac{a_{n-1}}{n n!} {n \choose k} B_{n-k}\right) t^n\right) dt$ with the inner sum analytically continued, so the continuum sum is $\sum_{n=0}^{z-1} f(n) = \sum_{k=1}^{\infty} b_k z^k$. Another method that might be useful is the one I mentioned here: http://math.eretrandre.org/tetrationforu...93#pid4293 Perhaps it could give a still wider range of functions. There are some functions for which these do not appear to work -- consider $f(z) = \sum_{n=0}^{\infty} \frac{z^{2n}}{B_{2n}}$. The first coefficient of the continuum sum $b_1$ by the Faulhaber's formula gives the divergent sum $1 + 1 + 1 + 1 + ...$ which does not look to be Borel summable. However it seems it could work for other functions, and the big question is, of course, could it work for tetration, and if so, do the obtained extensions of tetration agree with the ones already made, yet enable expansion to a much wider variety of bases? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/30/2009, 09:51 PM (12/30/2009, 09:19 PM)sheldonison Wrote: There is always base $\eta$, but of course, it can be solved by other means. The upper superexponential at base $\eta$ is entire with no singularities, and with super-exponential growth. Perhaps if you can get the Borel summation to work for $\eta$, you can figure out how to make it work for other bases. And it's such superexponential growth that would be why it wouldn't work, at least without the Borel, etc. methods. And if we apply them, the presence of singularities should not be a problem, and so we should be able to solve for the lower superexponential as well. I'm not sure how having the upper would be helpful anyway, especially considering the failure of the "base change method" to produce an analytic function. « Next Oldest | Next Newest »

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