Continuum sums -- a big problem and some interesting observations Ansus Junior Fellow Posts: 20 Threads: 5 Joined: Aug 2008 10/10/2010, 08:02 AM (This post was last modified: 10/10/2010, 08:11 AM by Ansus.) So my advice is to avoid confusion by using continous sum but not indefinite continous sum, but a continous sum with fixed lower limit which is defined unambiguiously. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 10/11/2010, 10:36 AM (This post was last modified: 10/11/2010, 10:38 AM by mike3.) (10/10/2010, 08:02 AM)Ansus Wrote: So my advice is to avoid confusion by using continous sum but not indefinite continous sum, but a continous sum with fixed lower limit which is defined unambiguiously. So, then we have $\sum_x f(x)$ as definitely ambiguous and not well defined if $f$ is not entire, even with Fourier sums, as powerful as they are, while $\sum_{n=n_0}^{x-1} f(n)$ is not. The direction of the singularities is toward the right of a singularity in $f$ if $\Re(n_0) < \Re(S)$, where $S$ is the singularity point, and to the left otherwise. And if $\Re(n_0) = \Re(S)$, it goes in both directions, i.e. $\frac{1}{2}\left(\psi(-x) + \psi(x+1)\right)$. Thus, the laws $\sum_{n=a + \alpha}^{b + \alpha} f(n) = \sum_{n=a}^{b} f(n + \alpha)$ and $F(b+1) - F(a) = \sum_{n=a}^{b} f(n)$, where $F(z) = \sum_{n=n_0}^{z-1} f(z)$ for some $n_0$, are only guaranteed to hold when all numbers $a$, $b$, and $\alpha$ are integers, or $f$ is entire. Also, I made an interesting observation about the function $F(x) = \sum_{n=0}^{x-1} e^{-n^2}$, a sort of sum-analogue of the error function. With the Fourier-based sum, it does not settle to a limit as $x \rightarrow \infty$, but instead goes to a small 1-cyclic wobble about the limit of the discrete sum. Namely, $F(1/2) = \frac{1}{2}$, and so $\lim_{n \rightarrow \infty} F\left(n + \frac{1}{2}\right) = \frac{1}{2} + \sum_{n=0}^{\infty} e^{-{\left(n + \frac{1}{2}\right)}^2} \ne \sum_{n=0}^{\infty} e^{-n^2}$. The value 1/2 for the function at 1/2 can also be obtained from Faulhaber's formula when it is organized as a sum over the Bernoulli polynomials. I think the half-integers are the only other places than the integers where that formula converges. The message here being, of course, that we can't necessarily expect continuum sums and discrete sums to behave in the same way, so the idea we may have to sacrifice some identities may not be a surprise. Ansus Junior Fellow Posts: 20 Threads: 5 Joined: Aug 2008 10/11/2010, 11:33 AM (This post was last modified: 10/11/2010, 11:52 AM by Ansus.) (10/11/2010, 10:36 AM)mike3 Wrote: a sort of sum-analogue of the error function. With the Fourier-based sum, it does not settle to a limit as $x \rightarrow \infty$, but instead goes to a small 1-cyclic wobble about the limit of the discrete sum. I believe in such cases the periodic component can be easily excluded (i.e. by requireing monotonous derivatives of higher orders). This is related to positive real axis. So to obtain the "natural solution" to the continous sum, just purify it of the periodic component on the positive real axis. Of course in any case the periodic component on the complex plane still will remain. You can also use checks by other methods such as Mueller's formula (this formula is well-applicable to the function you presented as an example). If $\lim_{x\to{+\infty}}f(x)=0$ then $\sum _x f(x)=\sum_{n=0}^\infty\left(f(n)-f(n+x)\right)+ C$ mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 10/11/2010, 11:46 PM (This post was last modified: 10/12/2010, 12:30 AM by mike3.) (10/11/2010, 11:33 AM)Ansus Wrote: I believe in such cases the periodic component can be easily excluded (i.e. by requireing monotonous derivatives of higher orders). This is related to positive real axis. So to obtain the "natural solution" to the continous sum, just purify it of the periodic component on the positive real axis. Of course in any case the periodic component on the complex plane still will remain. You can also use checks by other methods such as Mueller's formula (this formula is well-applicable to the function you presented as an example). If $\lim_{x\to{+\infty}}f(x)=0$ then $\sum _x f(x)=\sum_{n=0}^\infty\left(f(n)-f(n+x)\right)+ C$ The problem with the Mueller formula, and part of the reason for the introduction of the Fourier continuum sum, is that it does not preserve analyticity. Consider $f(x) = \frac{\sin\left(x^2\right)}{x^2}$. If we apply Mueller's formula to this at the real axis we get a reals-only continuum sum that is not analytic anywhere, heck, it may not even be real-differentiable anywhere(!). The Fourier method also preserves the result from Faulhaber's formula. Canceling the wobble would mean that we no longer have agreement with Faulhaber's formula. Ansus Junior Fellow Posts: 20 Threads: 5 Joined: Aug 2008 10/12/2010, 04:41 AM (This post was last modified: 10/12/2010, 04:56 AM by Ansus.) (10/11/2010, 11:46 PM)mike3 Wrote: Consider $f(x) = \frac{\sin\left(x^2\right)}{x^2}$. If we apply Mueller's formula to this at the real axis we get a reals-only continuum sum that is not analytic anywhere, heck, it may not even be real-differentiable anywhere(!). Why do you think this function should have good, analytic, differentiable continuous sum? I think it should not regardles of the method used. It is really bad function for continuous summation. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 10/12/2010, 06:44 AM (10/12/2010, 04:41 AM)Ansus Wrote: Why do you think this function should have good, analytic, differentiable continuous sum? I think it should not regardles of the method used. It is really bad function for continuous summation. Why should it not? What makes a function "too bad" to have a continuum sum, other than that some proposed definition of continuum sum does not yield a continuum sum for it? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 10/12/2010, 06:45 AM (This post was last modified: 10/12/2010, 08:07 AM by mike3.) (10/12/2010, 04:41 AM)Ansus Wrote: Why do you think this function should have good, analytic, differentiable continuous sum? I think it should not regardles of the method used. It is really bad function for continuous summation. Why should it not? What makes a function "too bad" to have a continuum sum, other than that some proposed definition can't sum it? Ansus Junior Fellow Posts: 20 Threads: 5 Joined: Aug 2008 10/12/2010, 10:41 AM By the way, why do you think Mueller's formula will give non-analytic result here? I have just took a closer look to the function and I see no problem with continuous summation here. « Next Oldest | Next Newest »

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