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 sexp by continuum product ? tommy1729 Ultimate Fellow Posts: 1,436 Threads: 348 Joined: Feb 2009 06/12/2011, 11:50 PM (This post was last modified: 06/13/2011, 08:39 AM by bo198214.) We all know $integral$ $dx/x = ln(x)$ $integral$ $dx/(x ln(x)) = ln(ln(x))$ $integral$ $dx/( x ln(x) ln(ln(x)) ) = ln(ln(ln(x)))$ So now the idea occurs : let $ctp_k (f(x,k))$ denote continuum product with respect to k , $integral$ $dx / ctp_k (sexp(slog(x) - k)) = sexp(slog(x) - k)$ And the continuum product is defined as e^(continuum sum ( log (a_k )) We use the q-method for the continuum sum. This method is explained by mike 3: (04/20/2010, 02:48 AM)mike3 Wrote: We have $f(z) = \sum_{n=0}^{\infty} a_n b^{nz}$ or, even better $f(z) = \sum_{n=-\infty}^{\infty} a_n b^{nz}$. Then, $\sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} a_n \frac{b^{nz} - 1}{b^n - 1}$ with the term at $n = 0$ interpreted as $a_0 z$, so, $\sum_{n=0}^{z-1} f(n) = \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1}\right) + a_0 z + \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1} b^{nz}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1} b^{nz}\right)$. However, when viewed in the complex plane, we see that exp-series are just Fourier series $f(z) = \sum_{n=-\infty}^{\infty} a_n e^{i \frac{2\pi}{P} n z}$. which represent a periodic function with period $P$. Thus it would seem that only periodic functions can be continuum-summed this way. Tetration is not periodic, so how could this help? Well, we could consider the possibility of continuum-summing an aperiodic function by taking a limit of a sequence of periodic functions that converge to it. The hypothesis I have is that if $f_0, f_1, f_2, ...$ is a sequence of periodic analytic functions converging to a given $f$, then their continuum sums, if they converge to anything, converge to the same thing, regardless of the sequence of functions. An example. Let $f(z) = z$, the identity function. We can't continuum-sum it with the exp-series directly. But now let $f_u(z) = u \sinh\left(\frac{z}{u}\right)$, so that $\lim_{u \rightarrow \infty} f_u(z) = f(z)$, a sequence of periodic (with imaginary period $2 \pi i u$) entire functions converging to $f(z)$. The continuum sum is, by using the exponential expansion of sinh giving $f_u(z) = -\frac{u}{2} e^{-\frac{z}{u}} + \frac{u}{2} e^{\frac{z}{u}}$, $\sum_{n=0}^{z-1} f_u(n) = \frac{u}{2} \left(\frac{e^{\frac{z}{u}} - 1}{e^{\frac{1}{u}} - 1} - \frac{e^{-\frac{z}{u}} - 1}{e^{-\frac{1}{u}} - 1}\right)$. Though I didn't bother to try to work it out by hand, instead using a computer math package, the limit is $\frac{x(x-1)}{2}$ as $u \rightarrow \infty$, agreeing with the result from Faulhaber's formula. Another example is the function $f(z) = \frac{1}{z}$, or better, $f(z) = \frac{1}{z+1}$. We can construct periodic approximations like $f_u(z) = \frac{1}{u \sinh\left(\frac{z}{u}\right) + 1}$, and take the limit at infinity. The terms in the Fourier series are even worse. If we use a numerical approximation of the series with period , I get the continuum sum from 0 to -1/2 as ~0.6137 (rounded, act. more like something over 0.61369) suggesting the process is recovering the digamma function, as can be seen by setting 1/2 in the canonical formula $-\gamma + \Psi(x + 1) = \sum_{n=0}^{x-1} \frac{1}{x+1}$ yielding 0.61370563888011... Trying it with $log(1 + z)$, so as to attempt to evaluate $z! = \exp\left(\sum_{n=0}^{z-1} \log(1+z)\right)$ yields values that agree with the gamma function, providing more evidence that gamma is the natural extension of the factorial function to the complex plane. Thus it seems this continuum sum is recovering all the expected sums and extensions. So we ( try to ... ) use this idea to compute a (probably unique and analytic ) sexp for bases > sqrt(e). regards tommy1729 Moderator's note: If you quote things, please mark it accordingly best with link to its origin. Otherwise I would consider it as wilful deceit. I modified your post accordingly. JmsNxn Long Time Fellow Posts: 456 Threads: 84 Joined: Dec 2010 06/13/2011, 02:23 AM That's interesting, especially about the values converging to the gamma function. That pretty much sets me in that the method works. The obvious open question though is how to exactly apply it to tetration. I see your argument that: $\frac{d}{dx} \ln^{\circ 0.5}(x) = \frac{1}{\prod_{k=0}^{-0.5} \ln^{\circ k}(x)}$ or generally: $\frac{d}{dx} \ln^{\circ r}(x) = \frac{1}{\prod_{k=0}^{r-1} \ln^{\circ k}(x)}$ Hmm, just doing some algebra: $\frac{d}{dx}\ln^{\circ r}(\ln^{\circ -r}(x)) = \frac{\frac{d}{dx}(x) \ln^{\circ -r}(x)}{\prod_{k=0}^{r-1} \ln^{\circ k-r}(x)} = 1$ which means: $\frac{d}{dx} \ln^{\circ -r}(x) = \prod_{k=0}^{r-1} \ln^{\circ k-r}(x)$ which means: $\exp^{\circ r}(x) = \int e^{\sum_{k=0}^{r-1} \ln^{\circ k-r+1}(x)} dx$ this contradicts $\exp^{\circ -r}(x) = \int \frac{dx}{e^{\sum_{k=0}^{r-1} \ln^{\circ k+1}(x)}$ so i guess we have restrictions on r, unless, we have the equivalency: $e^{\sum_{k=0}^{-r-1} \ln^{\circ k+r+1}(x)} = \frac{1}{e^{\sum_{k=0}^{r-1} \ln^{\circ k+1}(x)}$ $\sum_{k=0}^{-r-1} \ln^{\circ k+r+1}(x) = -1 \cdot (\sum_{k=0}^{r-1} \ln^{\circ k+1}(x))$ which does not hold for r = -1; and probably not for all r < 0 I wonder for what values it does hold, if any, if it does for r > 0, it already produces some interesting, though probably false identities: r = 1 $\sum_{k=0}^{-2} \ln^{\circ k+2}(x) = -1 \cdot (\sum_{k=0}^{0} \ln^{\circ k+1}(x)) = -\ln(x)$ and r = 0 $\sum_{k=0}^{-1} \ln^{\circ k+1}(x) = -1 \cdot (\sum_{k=0}^{-1} \ln^{\circ k+1}(x)) = 0$ (0 since only zero times negative one is equal itself.) I'm a little iffy on the restrictions to applying this continuum sum methods. I may just be doing something horribly wrong . Still though, this method looks very interesting. tommy1729 Ultimate Fellow Posts: 1,436 Threads: 348 Joined: Feb 2009 06/13/2011, 04:42 AM hi james i think you did something horribly wrong :p already your 3rd formula seems wrong. actually quite trivially wrong , it already fails for both positive and negative integer values of r. note that the continuum product is suppose to be taken before substitution. tommy1729 bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 06/13/2011, 08:45 AM (06/12/2011, 11:50 PM)tommy1729 Wrote: So we ( try to ... ) use this idea to compute a (probably unique and analytic ) sexp for bases > sqrt(e). So whats the progress to Mike's method? tommy1729 Ultimate Fellow Posts: 1,436 Threads: 348 Joined: Feb 2009 06/18/2011, 05:33 AM (This post was last modified: 06/18/2011, 05:33 AM by tommy1729.) i think i can show uniqueness for bases > sqrt(e). however 1) this might have already been done (here) 2) no idea if its analytic 3) uniqueness , not existance ... tommy1729 tommy1729 Ultimate Fellow Posts: 1,436 Threads: 348 Joined: Feb 2009 06/30/2011, 12:27 PM to test this it seems the (q-)continuum sum defined by mike ( rediscovered ) is not practical in a numerical sense. however there are other continuum sum methods that give the same result as long as the values are real and twice differentiable. those others give the same numerical value ( proof ref desired ) as for instance " Riemann's continuum sum " : sum a,b f(x) = integral a,b f(x) dx + integral f ' (x) ( x - floor(x) -1/2) dx + (f(a) + f(b))/2 if 'a' is an integer and f ' (x) is continu. together with log^[3/2] = around log log log log 2sinh^(1/2) ( exp exp x ) , it seems the OP conjecture can be numericly tested ! sorry for no tex use , im in a hurry. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,436 Threads: 348 Joined: Feb 2009 06/30/2011, 10:07 PM or the euler-maclaurin formula for the continuum sum , which is equivalent to " Riemann's continuum sum " ( and the q-continuum sum )but computed differently. btw many formula's come from the amazing darboux theorem such as this continuum sum and taylor series expansion. « Next Oldest | Next Newest »

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