12,360 reputation
11554
bio website go.helms-net.de
location Kassel, Germany
age 61
visits member for 3 years, 10 months
seen 3 hours ago

Mathematics is only a hobby - although I have done undergrad courses in the 70s. But part of my job was doing statistics and this kept me near my favourite subject "linear algebra" (programmed factor-analysis and a matrix-orientated calculator MatMate). Around 2002 I came in contact with the internet community in math newsgroups and could improve my Collatz discussion. Next subject was the Bernoulli numbers, then integer matrices and since 2006 the problem of iterated exponentiation aka tetration. Serving half-time jobs in teaching here at the university I found time to fiddle with that subjects in depth - and found love with the exploratory approach and impulse of the 18'th century numbertheory, namely L. Euler, "the master of us all"... Due to lack of formal education I've to do my "research" widely on my own - but that's what I just like: to find structure, pattern, laws from the ground.


14h
revised LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?
added link to my text where this occurs.
Jul
5
comment Clarification on tetration
You might be interested in my earlier answer which describes a somehow "quick&dirty" method to compute fractional "iteration-heights"; the method is of course inaccurate but better than a linear or quadratic ansatz and even seems to approximate naturally the Knesersolution if the size of the involved Carleman-matrix is simply increased. See math.stackexchange.com/questions/238970/…
Jul
5
revised Clarification on tetration
typo "Kneser"
Jul
4
accepted A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)
Jul
4
comment A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)
@EuYu:Ah, well, very nice! If you'd convert it to an answer I'd just accept it and could "close the case" ;-)
Jul
4
comment A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)
@EuYu: hmm, perhaps I'm on some wrong or trivial track without seeing this. The question came up when I was considering the convergence-behave of the series $s$ and I simply thought (4) could be improved just by some well known additional coefficient - I didn't need to make a big research out of this. Perhaps I should retract the whole question: sometimes if 500 formulae dance around you, you begin to dance yourself... And possibly this is what happens just now to me. I simply should be satisfied having found formula (4) already...
Jul
4
comment A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)
@Leonbloy: please see my comment to EuYu
Jul
4
comment A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)
@EuYu: hmm, in the Stirling-formula the $n$ is also under the square-root; in (4) it has moved mystically into the gamma-expression, so this is a variant of the Stirling-formula. Maybe I've not well expressed what I wanted to say. My problem is that I have the lhs in (4) and at integer n. Then I found that heuristically I can express that by Gamma( f(n)) where f(n) is some function of n (in this case f(n)=n+1/2) . After this ansatz, I can't see how to get an even better approximation by a variation of the expression on the rhs. Hope I could make it clearer?
Jul
4
asked A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)
Jul
4
revised LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?
deleted 1034 characters in body
Jul
4
revised LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?
added 31 characters in body
Jul
4
revised LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?
added 31 characters in body
Jul
4
comment LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?
Ioannis, I've just "accepted" that answer; could you please take also a short look at my further understanding now and the picture (which I've done into a full answering box because it's too long for a comment)?
Jul
4
accepted LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?
Jul
4
answered LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jul
1
comment LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?
Hmm, another thought: doesn't this look, as if the function which leads to the numbers $q$ could be understood as a continuation of the LambertW for the range $0 \lt x \lt 1$, where LambertW is constant $1$? The $q(x)$ which I compute is a limit-expression dependent on $x$ but also on the value of $n$ when this goes to infinity and for which there is also a power series whose $n$ leading terms are fixed and the power series is related to that for the LambertW in some form which I'm still investigating. So from that powerseries perhaps a continuation for the $W$ might be possible?
Jun
30
comment LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?
Thanks Ioannis, good explanation; and so, also the $h(x)$ doesn't give hope. Hmmm. Perhaps it's best to write about that branching at $x=1$ and insert a "case-clause" in the text. I'll see...
Jun
30
revised LambertW: $ x=W(x\cdot e^{x}) $ for $ x \ge -1$ but not for $x \lt-1$. How do I express my formula/my text?
added description