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 Is this entire expression for tetration legal JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 12/10/2012, 03:43 PM (This post was last modified: 12/11/2012, 02:09 PM by JmsNxn.) I'm going to prove to you an entire expression for tetration in one post for all b greater than eta. Is this valid? We start by defining: $\mathcal{E} f(s) = \frac{1}{\Gamma(-s)} \int_0^\infty f(-u) u^{-s-1}\, \partial u$ We call this the exponential derivative because: $\mathcal{E} f(s) = \frac{d^s f(t)}{dt^s} |_{t=0}$ Where this complex derivative is evaluated according to the Riemann-Liouville differ-integral where exponentiation is the fix point. By this definition we find: $\mathcal{E}(b^s) = \ln(b)^s$ Now we define a new multiplication across FUNCTIONS; i.e. you cannot plug in numbers into this multiplication. We refer to $\mathbb{E}$ as the complex linear span of $e^{\lambda s}\,\,;\,\,\lambda \in \mathbb{C}$. We now define: $\forall \alpha \in \mathbb{C}\,\,\,\,;\,\,\,\,\forall f,g,h \in \mathbb{E}$ $f \times g \in \mathbb{E}$ $\alpha \times f = \alpha f$ $f \times g = g \times f$ $(f \times g) \times h = f \times (g \times h)$ $f \times (g + h) = (f \times g) + (f \times h)$ With this we define $\forall \omega, \nu \in \mathbb{C}$: $(\Gamma(\omega + 1) (^\omega e)^s) \times(\Gamma(\nu + 1) (^\nu e)^s) = \Gamma(\omega + \nu + 1) (^{\omega+ \nu } e)^s$ This result gives us a beautiful isomorphism $\psi: \mathbb{E} \to \mathbb{C}[s]$: $\mathbb{U} = \{ 1, e^s, (^2 e)^s, (^3 e)^s, ..., (^N e)^s,...\}$ $\psi(\alpha f + \beta g) = \alpha \psi f + \beta \psi g$ $\psi(f \times g) = (\psi f) \cdot (\psi g)$ $\psi(\mathcal{E} f) = \frac{d \psi(f)}{ds}$ $\psi((^N e)^s) = \frac{s^N}{N!}$ Now thanks to Newton we have: $s^\omega = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} (s - 1)^N$ $s^{\omega} = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} \sum_{k=0}^N \frac{N!}{(N-k)! k!}(-1)^{N-k}s^k$ Now we apply the isomorphism to get the result $\exists \phi \in \mathbb{E}$: $\Gamma(\omega + 1) \phi_\omega(s) = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} \sum_{k=0}^N \frac{N!}{(N-k)! k!}(-1)^{N-k}k!(^k e)^s$ We can reduce this equation to: $\phi_\omega(s) = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!}(^k e)^s}{\Gamma(\omega - N +1)}$ We know these functions satisfy the equations: $\phi_\omega \times \phi_\nu = \frac{\Gamma(\omega + \nu + 1)}{\Gamma(\omega + 1)\Gamma(\nu + 1)} \phi_{\omega + \nu}(s)$ $\mathcal{E} \phi_\omega = \phi_{\omega - 1}$ There is only one function which satisfies this: $\phi_{\omega}(s) = (^\omega e)^s$ We can do the same procedure to arrive at the more general expression: $(^\omega b)^s = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!}(^k b)^s}{\Gamma(\omega - N +1)}$ This is probably the fastest way to derive tetration. It's analytic, entire for $b > \eta$ when we let $\Re(s) < 0$which is easy to show by the ratio test since tetration grows ridiculously faster than the Gamma function. Anyone see any mistakes I'm making? Or did I just solve tetration in fifteen minutes of work. LOL Being quick I'll write the glorious formula; $\forall b \in \mathbb{R}\,\,;\,\, b > e^{\frac{1}{e}}$: $\frac{1}{^\omega b} = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!(^k b)}}{ \Gamma(\omega - N +1)}$ I'm currently finishing a paper on $\mathcal{E}$ and this result just happened to fall in my lap. I'm wondering if it's valid so that I can keep it in the paper. sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 12/13/2012, 02:30 AM (This post was last modified: 12/13/2012, 02:35 AM by sheldonison.) (12/10/2012, 03:43 PM)JmsNxn Wrote: ....Being quick I'll write the glorious formula; $\forall b \in \mathbb{R}\,\,;\,\, b > e^{\frac{1}{e}}$: $\frac{1}{^\omega b} = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!(^k b)}}{ \Gamma(\omega - N +1)}$Hey James, very interesting stuff, though its a little over my head. But I did try your formula for base e, for sexp(0.5), $^{0.5}e$ $\frac{1}{^z e} = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!(^k e)}}{ \Gamma(z - N +1)}$ I generated the first hundred terms, and took the reciprocal and got the result for sexp(0.5), $^{0.5}e$ = -0.0889470119254744974, which is a strange result. Also, the summation also doesn't appear to be converging. Here are the first 100 terms for N=0 to N=99 in the outer loop summation. As you can see, the sum is slowly decreasing (larger negative number), which means the reciprocal is slowly getting smaller, and perhaps approaching zero. By the way, the inner loop converges very nicely with only k=0 to k=3 (four terms) required. For integer values of z, there is also the problem that gamma has a singularity for zero and negative integers, but if you somehow ignore the terms with the singularity, it seems to work. - Sheldon Code:N=0 1.12837916709551257390 N=1 -0.356635834837458935278 N=2 -0.0558854027631410430498 N=3 -0.0206132530877876970179 N=4 -0.0141193897106172441830 N=5 -0.0148207070343604561859 N=6 -0.0178722393015305743419 N=7 -0.0217507480727162365390 N=8 -0.0258849705867867567488 N=9 -0.0300391250051902282142 N=10 -0.0341121984624033791621 N=11 -0.0380621797995234682754 N=12 -0.0418743225766593487278 N=13 -0.0455468128421149043589 N=14 -0.0490839363416516118949 N=15 -0.0524926981203164306902 N=16 -0.0557811147157649591746 N=17 -0.0589573488130329367398 N=18 -0.0620292803860181190169 N=19 -0.0650043080717605755730 N=20 -0.0678892727812415897521 N=21 -0.0706904456867871814295 N=22 -0.0734135490929702590636 N=23 -0.0760637929100568892563 N=24 -0.0786459172599043479163 N=25 -0.0811642360986323262444 N=26 -0.0836226791891616201340 N=27 -0.0860248311369194801384 N=28 -0.0883739669751719546777 N=29 -0.0906730842116677024391 N=30 -0.0929249314732365839175 N=31 -0.0951320339948591280177 N=32 -0.0972967162440782634464 N=33 -0.0994211219794611853738 N=34 -0.101507232030120356963 N=35 -0.103556880061972629931 N=36 -0.105571766571162804793 N=37 -0.107553471319117637288 N=38 -0.109503464398763159097 N=39 -0.111423116098393868509 N=40 -0.113313705708880295616 N=41 -0.115176429401406493529 N=42 -0.117012407286649168777 N=43 -0.118822689752075097934 N=44 -0.120608263161640512378 N=45 -0.122370054991416584391 N=46 -0.124108938465338699673 N=47 -0.125825736747199128971 N=48 -0.127521226738006875002 N=49 -0.129196142521777878991 N=50 -0.130851178497564809986 N=51 -0.132486992230976648134 N=52 -0.134104207054477737117 N=53 -0.135703414442310877300 N=54 -0.137285176182888040280 N=55 -0.138850026368874240602 N=56 -0.140398473222902578280 N=57 -0.141931000774856544457 N=58 -0.143448070404900858692 N=59 -0.144950122264901360038 N=60 -0.146437576589519465340 N=61 -0.147910834907073090058 N=62 -0.149370281159202759044 N=63 -0.150816282737450884553 N=64 -0.152249191444038307074 N=65 -0.153669344383391754809 N=66 -0.155077064790327269347 N=67 -0.156472662800217813649 N=68 -0.157856436165959503295 N=69 -0.159228670926092608054 N=70 -0.160589642028024068590 N=71 -0.161939613909932020797 N=72 -0.163278841044604707844 N=73 -0.164607568448171808999 N=74 -0.165926032156421792429 N=75 -0.167234459671161047302 N=76 -0.168533070378856319747 N=77 -0.169822075943608775267 N=78 -0.171101680676333554542 N=79 -0.172372081881860968129 N=80 -0.173633470185532711615 N=81 -0.174886029840737105782 N=82 -0.176129939018709985571 N=83 -0.177365370081821244524 N=84 -0.178592489841470093563 N=85 -0.179811459801623841895 N=86 -0.181022436388954587949 N=87 -0.182225571170454848279 N=88 -0.183421011059346164175 N=89 -0.184608898510033494726 N=90 -0.185789371702802181506 N=91 -0.186962564718902961836 N=92 -0.188128607706623473485 N=93 -0.189287627038901537828 N=94 -0.190439745462995875203 N=95 -0.191585082242693475689 N=96 -0.192723753293499332602 N=97 -0.193855871311223385361 N=98 -0.194981545894351078562 N=99 -0.196100883660557713855 SUM = -11.2426486101395293427 andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/13/2012, 05:03 AM (12/10/2012, 03:43 PM)JmsNxn Wrote: $(^\omega b)^s = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!}(^k b)^s}{\Gamma(\omega - N +1)}$Mindblowing. I'll have to look at it a bit more, but this method reminds me a lot of Woon's approach, which many on this forum have lumped in with Newton. Code:JmsTetrate[b_, w_, s_, m_] := Sum[Sum[(-1)^(n-k)*Tetrate[b, k]^s/(n-k)!, {k, 0, n}]/Gamma[w - n + 1], {n, 0, m}]; I tested your method with the above snippet, and it works well for small m, but it doesn't take long to blow up and overflow/underflow. I don't remember if I read it on this forum or not, but I believe someone has researched which bases the Newton methods work well for, and perhaps I was using the wrong bases. Anyway, this is certainly an interesting approach, even if it is a spin on something known. Andrew Robbins Refs: S.C.Woon "Analytic Continuation of Operators" http://arxiv.org/abs/hep-th/9707206 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 12/15/2012, 10:02 PM (This post was last modified: 12/15/2012, 10:03 PM by tommy1729.) (12/10/2012, 03:43 PM)JmsNxn Wrote: Being quick I'll write the glorious formula; $\forall b \in \mathbb{R}\,\,;\,\, b > e^{\frac{1}{e}}$: $\frac{1}{^\omega b} = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!(^k b)}}{ \Gamma(\omega - N +1)}$ considering the equation above ; if we do $exp(-1/x)$ on both sides that should be equivalent to adding 1 to $\omega$. If that is not the case then the fundamental functional equation of tetration is not satisfied. So can you confirm that doing $exp(-1/x)$ on both sides does what it is suppose to do ? For those confused notice $exp(-1/x) = 1/exp(1/x)$ and $1/exp^{[n]}(1/x) = (1/exp(1/x))^{[n]}.$ regards tommy1729 cosurgi Junior Fellow Posts: 3 Threads: 0 Joined: Jan 2013 01/20/2013, 07:32 PM (12/10/2012, 03:43 PM)JmsNxn Wrote: With this we define $\forall \omega, \nu \in \mathbb{C}$: $(\Gamma(\omega + 1) (^\omega e)^s) \times(\Gamma(\nu + 1) (^\nu e)^s) = \Gamma(\omega + \nu + 1) (^{\omega+ \nu } e)^s$ I'm lost at this point. Why do you define tetetration to satisfy this equality? How do you know it's true? « Next Oldest | Next Newest »

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