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 Generalized recursive operators andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 02/11/2008, 09:47 PM Ivars Wrote:=...=0,007297583=1/137,0316766 I'm not sure, it seems like no one has mentioned that this value is also the ratio between the action potentials of (electron/photon), better known as the "fine structure constant", also known as alpha. However, your sum is only accurate to 4 decimal places, which is still better than the previous approximation of 1/137, which is only accurate to 3 decimal places. Good job. Andrew Robbins GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 02/11/2008, 10:53 PM (This post was last modified: 02/11/2008, 11:03 PM by GFR.) Concerning the two asymptotes of y = eta[5]x, with eta = e^(1/e): Ivars Wrote:GFR, ... The double values returned are also interesting, the positive asymptote value should correspond to some very slow operation- may be inverse pentation, what ever it means-is there a definition?. How did You show (it can be shown...) that what You have shown ? Analytically? ....As I said, it can be shown that, for b = eta = e^(1/e) = 1.444667861.., we have two fixpoints in y = b[4]x and that they probably (...) are: x = {-Pi/2, +Pi/2} = {-1.570796327..., +1.570796327..}, corresponding to two symmetrical horizontal asymptotes of y = eta[5]x, with those values (positive, for x -> +oo, and negative, for x -> -oo). Unfortunately, I was not able (to date) to show it analytically, but only graphically. This explains the word "... probably". See the attachment. GFR Attached Files   Two tetra fixpoints.pdf (Size: 5.98 KB / Downloads: 343) Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/14/2008, 06:05 PM andydude Wrote:Ivars Wrote:=...=0,007297583=1/137,0316766 I'm not sure, it seems like no one has mentioned that this value is also the ratio between the action potentials of (electron/photon), better known as the "fine structure constant", also known as alpha. However, your sum is only accurate to 4 decimal places, which is still better than the previous approximation of 1/137, which is only accurate to 3 decimal places. Good job. Andrew Robbins Hi, Andy, Thanks. Of course I was aiming for fine structure constant, assumed everyone would notice- I live in assumptions, sometimes. I do not know how accurate are the assymptotic values You and JFox and GFR have derived. I they are accurate to enough decimal places, than this indeed is rather accurate SIMPLE symbolic approximation of fine structure constant (I mean You can always approximate any number using arbitrary values and constants,or infinite sums ) - I used only 5 terms, integer coefficients, integer powers , simple logic how they change, and the value of pentatition asymptote which must be a true mathematical constant. Perhaps this can be even written down as simple formula , just need a Symbol for the asymptote value. From my point of view, hyperoperations are exactly the place to look for alpha value. And having 5 terms in a 5-tation sum also seemed a nice fit. I would expect this formula is somehow extendable giving even better fit with either higher hyperoperations, or involving hyperoperations in formula itself, but for that there is a need to understand what does this pentation asymptote might mean, physically. Which is why I started to draw spirals, so far for me it remains unresolved issue. Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/07/2008, 06:58 PM andydude Wrote:I just found the asymptotes of pentation, hexation, heptation, octation, and beyond! And they're fascinating:  \begin{align} \lim_{b \rightarrow -2}(a \begin{tabular}{|c|}\hline 4 \\\hline\end{tabular} b) & = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) & = -2 \\ \lim_{b \rightarrow -4}(a \begin{tabular}{|c|}\hline 6 \\\hline\end{tabular} b) & = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) & = -4 \\ \lim_{b \rightarrow -6}(a \begin{tabular}{|c|}\hline 8 \\\hline\end{tabular} b) & = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 9 \\\hline\end{tabular} b) & = -6 \end{blign} ... I suppose you could see this from the integer versions of these operators, but I think the continuous (or if not continuous, mostly real-valued) versions make it easier to see. First, Andrew, these are really fascinating findings. andydude Wrote:$\lim_{N\rightarrow\infty} (a \begin{tabular}{|c|}\hline N \\\hline\end{tabular} b) = b+1$ for all $a>1, b<0$ meaning, in the limit, all hyper-operators return to the successor operation, like the circle of life... I dare a proof by induction which only needs the integer operations. Proposition: If we have a sequence of operations [n] on the natural numbers (>0) that satisfy b[n+1]1=b, b[n+1](x+1)=b[n](b[n+1]x) for n$\ge$ 1, then we can extend the domain of the right operand of [n] to integer k with k$\ge$ 3-n and the only way to do so still satisfying the above conditions and injectivity of the functions f(x)=b[n]x is by b[n](-k)=-k+1 for 0$\le$ k$\le$ n-3. Proof: We prove by induction over k that b[n](-k)=-k+1 for all n$\ge$ k+3. Induction Start k=0: b[n]1=b=b[n+1]1=b[n](b[n+1]0), for n$\ge$2, by injectivity follows 1=b[n+1]0 for n+1$\ge$3=0+3 Induction Step k=k+1: by induction assumption for n$\ge$k+3 : b[n](-k)=-k+1=b[n+1](-k)=b[n](b[n+1]-(k+1)) by injectivity: -k = b[n+1]-(k+1) which is the induction assertion: -(k+1)+1=b[n+1]-(k+1) for n+1$\ge$k+1+3 GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 03/10/2008, 11:56 AM Now, Henryk will say: "But, Gianfranco, you are always late and .... approximated !" This will happen concerning: bo198214 Wrote:andydude Wrote:I just found the asymptotes of pentation, hexation, heptation, octation, and beyond! And they're fascinating:  \begin{align} \lim_{b \rightarrow -2}(a \begin{tabular}{|c|}\hline 4 \\\hline\end{tabular} b) & = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) & = -2 \\ \lim_{b \rightarrow -4}(a \begin{tabular}{|c|}\hline 6 \\\hline\end{tabular} b) & = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) & = -4 \\ \lim_{b \rightarrow -6}(a \begin{tabular}{|c|}\hline 8 \\\hline\end{tabular} b) & = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 9 \\\hline\end{tabular} b) & = -6 \end{blign} ...... these are really fascinating findings.Indeed, they are! But, I think, they are not completely correct (Please, Andydude, correct me, if I'm wrong!! Please also remember my last msg to you "Sure thing"!). In fact, take the first two lines: $ \lim_{b \rightarrow -2}(a \begin{tabular}{|c|}\hline 4 \\\hline\end{tabular} b) = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -2 \ $ I agree that the first line is correct, for any $a > 1$. In case of $a < 1$, the limit is +oo. For $a =1$ and $a <0$, we are in trouble. Concerning the second line, I should like to recall the following relations, the first of which valid for tetration: a[3]b = a ==> h(a) = a[4]oo, depending on base a; which also should imply the following, for pentation: a[4]b = a ==> h(a) = a[5]oo, also depending on a. I mean that the fixpoints (of exponentiation) determine the limit heights of tetration (we know that). Similarly, the fixpoints (of tetration) should determine the limit heights of pentation. We had (I still have ...) a lot of problems in linking real and complex fixpoints with the tetration limit heights h and, I think, we are (I ... am) not yet completely prepared to study the fixpoints of tetration, for any base a. Nevertheless, there is an area of tetration where the "plots" are more or less symmetrical, for various bases "a", in the negative domains of the slog and of the sexp right (or second) operands, i.e.: [/a]sexp(x) and [/a]slog(x), for x < 0. In fact, let us consider: y = a[4]x and y = [/a]slog(x) The two plots, always increasing "functions" for x < 0, have their two "tails" that cross themselves in one fixpoint, the coordinate(s) of which we may call "Sigma", with definitely: -2 < Sigma < 0. (see: http://forum.wolframscience.com/attachme...ostid=4192 ). The values of Sigma are depending on base "a", i.e.: [b --> -oo] lim a[5]b = Sigma(a). We already found some approximated values of Sigma, in the following cases (always for operamd b < 0): Sigma(e) = - 1.841..., obtained by the KAR/GFR linear approximation (Andydude can do better) Sigma(Eta) = Sigma(e^(1/e)) = - Pi/2 = 1.570796327.. , graphically obtained by GFR, for b < 0 .... Sigma(oo) = -2 (is this true? I think it is ...). We should then write: $ \lim_{b \rightarrow -2}(a \begin{tabular}{|c|}\hline 4 \\\hline\end{tabular} b) = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = Sigma(a)$ And ... the rest should follow, accordingly. For base a > 1, I also agree on the limit value of a[s]n, for s -> oo and n -> -oo, which (by calling w the first infinite ordinal, or something like it ...) are indeed given by: a[w]n = n + 1 We should therefore write, for a > 1, n < 0: [s --> +oo] lim a[s]n = a[w]n = n + 1 Quickfur would probably say: "Much ado about nothing!" . We proposed to call "Omegation" such funny titanical operation. By the way, Henryk, well done, for your induction proof. Please check what I said and see if I didn't confuse issues, variables or signs. My concern is what would probably happen in case of base a < 1. I wonder if the entire sequence, mentioned by Andydude collapses, without even reaching n + 1. In fact, I have even doubts on the possibility of correctly defining the slog operation, for base a < 1. But ... , Henryk, this is another story. If you allow me, I shall put an ad-hoc thread about that. GFR andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 03/10/2008, 09:38 PM (This post was last modified: 03/10/2008, 09:44 PM by andydude.) GFR Wrote:Indeed, they are! But, I think, they are not completely correct (Please, Andydude, correct me, if I'm wrong!! Please also remember my last msg to you "Sure thing"!). In fact, take the first two lines: You're right, I'm wrong. $\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -1.85$ not -2 as I had first thought. Jay has found the more precise values in this thread which you should look to instead of this post, as I'm only accurate to 1 decimal place. Andrew Robbins GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 03/10/2008, 09:53 PM Agreed! Thank you Andydude! My "yellow zone" formula, as you know, is wrong too. I found another one, but I cannot "inverse" it. . I promised to discuss it with KAR. Then, I shall say something about it. I don't dare to do it now, because I might be wrong ... again. Please think also of the possible collapsation of the pentation "suite" in case of bases 0 < a < 1. What a strange mathematical landscape! GFR bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/11/2008, 07:11 AM andydude Wrote:You're right, I'm wrong. $\lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -1.85$ not -2 as I had first thought. Jay has found the more precise values in this thread which you should look to instead of this post, as I'm only accurate to 1 decimal place. May I ask if anyone has an idea how much this result depends on the actually chosen tetration extension to real numbers? I mean the b[n]-k=-k+1 for 0$\le$k$\le$n-3 does not depend on the extension. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 03/11/2008, 10:24 AM Mmmm! I think that we should have: [n -> + oo] lim (b[n](-k)) = -k + 1. In all other situations, b[n](-k) would depend on both n and k. Am I right ? But, perhaps I misunderstood the question. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/11/2008, 10:53 AM (This post was last modified: 03/11/2008, 10:58 AM by bo198214.) GFR Wrote:I think that we should have: [n -> + oo] lim (b[n](-k)) = -k + 1. Ya thats guarantied already for the (unique) integer version. We have b[n](-k)=-k+1 for 0$\le$k$\le$n-3. If n goes to $\infty$ the k-range for b[n](-k)=-k+1 increases and in the limit reaches all positive integer k, i.e. $\lim_{n\to\infty}$b[n](-k) = -k + 1 for all k$\ge$0. But my question was about whether the limits $\lim_{x\rightarrow-\infty}$b [2n+1] x depend on the concrete extension of the hyper operation to the reals (for example by Andrew's method, by regular iteration at some fixed point, by matrix operator method, by Jay's method, by linear/polynomial/exponential base function, etc.). For example b[5](-k)=$\text{slog}_b^{\circ k}(1)$ depends surely on the choice of $\text{slog}$, which we per default regard to be Andrew's. However perhaps it can be that *in the limit* the specific choice of $\text{slog}$ does not matter. « Next Oldest | Next Newest »

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