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 Continuum sum formula "rescued"? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 10/22/2009, 07:26 AM (This post was last modified: 10/23/2009, 07:31 AM by mike3.) Hi. I was wondering about a way by which one could theoretically "rescue" the continuum sum formula of Ansus. The idea is this. Let $f(x)$ be a function that decays to 0 as $x \rightarrow \infty$. Then, there exists a "natural" notion of continuum sum, namely, $\sum_{n=1}^{x} f(n) = \sum_{n=1}^{\infty} f(n) - f(n + x)$. provided the sum converges. We can modify this for a function that decays to a fixed point as well. If $f(x)$ decays to a fixed point $L$, then $f(x) - L$ will decay to 0, and so we have $\sum_{n=1}^{x} f(n) - L = \sum_{n=1}^{\infty} f(n) - f(n + x)$ By linearity, $\sum_{n=1}^{x} f(n) - L = \sum_{n=1}^{x} f(n) - \sum_{n=1}^{x} L$ so $\sum_{n=1}^{x} f(n) = \left(\sum_{n=1}^{x} f(n) - L\right) + \sum_{n=1}^{x} L$ and the sum of a constant, of course, has a natural definition, and so we get $\sum_{n=1}^{x} f(n) = Lx + \sum_{n=1}^{\infty} f(n) - f(n + x)$ as the Ls cancel out when we plug $f(n) - L$ into our sum formula. Now take Ansus' continuum sum formula, here for base e, $\frac{\mathrm{tet}'(z + z_0)}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=0}^{z-1} \mathrm{tet}(n + z_0)\right)$ and we reverse $z$ to its negative, yielding $\frac{\mathrm{tet}'(-z + z_0)}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=0}^{-z-1} \mathrm{tet}(n + z_0)\right)$ Now, we can apply the identity $\sum_{n=0}^{-x-1} f(n) = \sum_{n=1}^{x} -f(-n)$. Then we get $\frac{\mathrm{tet}'(-z + z_0)}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=1}^{z} -\mathrm{tet}(-n + z_0)\right)$ and if the tetrational decays exponentially to a fixed point in the left half of the complex plane, the continuum sum is well defined. The only problem I see here is that $z_0$ cannot be 0, otherwise there will be singularities at the integers and then the sum formula won't work. If we use a displacement along the imaginary axis then it might work but how then do you evaluate that derivative at $z_0$, or integrate to -1 to obtain the iterating formula? Hmm... What do you think about this? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/10/2009, 12:42 AM (This post was last modified: 11/10/2009, 05:02 AM by mike3.) Hmm. It looks like this may not work, unfortunately: apparently though the continuum sum has the property $\sum_{n=0}^{-x-1} f(n) = \sum_{n=1}^{x} -f(-n)$ for integer values of $x$, it seems to not necessarily hold for non-integer ones. Thus a different way will be needed to generalize $\sum_{n=1}^{x} f(n) = \sum_{n=1}^{\infty} f(n) - f(n + x)$ to ideally, general analytic functions at least. However, we can apply the formula to those bases who converge to a fixed point at positive infinity, like how the regular formula can. Probably not, however, for the endpoints $b = e^{-e}$ and $b = e^{1/e}$ as their convergence rate is too slow (hyperbolic at best) for the above infinite sum to exist. Base $\sqrt{2}$ Consider, for example, $b = \sqrt{2}$. We can already apply the continuum sum formula above in the "forward" direction in this case due to the convergence toward the fixed point 2 as the tower grows. We use the summing formula $\sum_{n=1}^{x} f(n) = xL + \sum_{n=1}^{\infty} f(n) - f(n + x)$. and since for tetration/hyper4, $f(0) = 1$, we get $\sum_{n=0}^{x-1} f(x) = 1 + (x-1)L + \sum_{n=1}^{\infty} f(n) - f(n + x - 1)$. We can then approximate a solution using the iterating form of Ansus' formula: $f(x) = K \int_{-1}^{x} (\log(b))^t \exp_b\left(\sum_{k=0}^{t-1} f(k)\right) dt$. by sampling an interval of length 1 on the real axis (here, I use [-1, 0]) with nodes for an approximation (this test was done with quadratic spline), and then use the tetration recurrence formulas to fill up the real axis and iterate over the nodes with the continuum sums and the integrals. The constant K is determined at each step by the reciprocal of the value of the function at 0, or the last node in the array. Using 129 nodes and 8 iterations, starting with a linear initial guess and normalizing the formula at each iteration using the last node's value, and capping the infinite sum at $n = 100$ I get $^{1/2} \sqrt{2} \approx 1.243621$ which agrees well with the value obtained via the regular iteration. More nodes could be used to get more accuracy. A graph constructed this way is shown here:     (x-axis = tower, y-axis = value of tetration) Base $\frac{1}{4}$ We do this the same way as before. The formula still converges, only this time to complex values instead of real ones. The fixed point is $L = \frac{1}{2}$. For example, we have $^{\frac{1}{2}} \left(\frac{1}{4}\right) \approx 0.460022 + 0.311787i$ which, again agrees with the value from the regular formula. Graphs of this function for real height are given here: Real part:     (x-axis = tower, y-axis = real part of value of tetration) Imag part:     (x-axis = tower, y-axis = imag part of value of tetration) mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/10/2009, 11:17 AM (This post was last modified: 11/10/2009, 11:22 AM by mike3.) This also makes me think about the nature of the continuum sum operator as a subject in itself. Examining the formula for the continuum sum of a function that converges quickly to a limit at infinity, it is important to note its extremely "global" nature: unlike say, the derivative, which depends on the value of the function only very near ("infinitely" near) the point at which it's taken, or even the integral, which depends on the values over some interval, but which may be finite and bounded, the continuum sum at a point not only depends on the value there, but on that at the point plus 1, at the point plus 2, at the point plus 3, and so on, all the way to infinity, and on the value at the point 1, at the point 2, at the point 3, and so on, to infinity as well! (if we start our summation at index 0) In other words, almost any change of any substantial nature, even very far away along the number line or right complex half-plane, that causes little or no change in the function near the origin (or whatever the lower bound on the summation is) -- doesn't matter if its 10 numbers away or 10 billion or a googolplex or a giggolplex (i.e. $^{^{100} 10} 10$) or a moser or even Graham's Number of places away -- can still cause significant change in the continuum sum there. This might be part of the reason "why" Faulhaber's formula fails to converge (beyond the obvious one that its terms don't settle: why don't they, on a deeper level?) for series with finite convergence radii: although locally they look much like the function they are approximating, globally they don't at all: the truncated series take on huge values over most of the real line or the complex plane. If the general "canonical" continuum sum operator (which, I believe, is what we need to construct the general tetration of arbitrary complex bases to arbitrary complex heights or towers: Ansus' formula bridges the tetration problem to the continuous summation problem -- if we have a "natural" definition for continuum sum, then it only makes sense tetration should be consistent with it through the Ansus formula) is also "global" like this, and the Faulhaber formula is a special case of that general operator, which makes sense (it can actually be obtained fairly naturally from the convergent continuum sum special case I mentioned, I can show you how it's done if you want and don't know already), then it should not be surprising that it will not work. « Next Oldest | Next Newest »

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