12,407 reputation
11554
bio website go.helms-net.de
location Kassel, Germany
age 61
visits member for 3 years, 10 months
seen 10 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.


Jul
16
comment Calculating a Factorial Base Representation
very cool idea. It needed two times reading for me to understand that and why this is so easy. Wow.
Jul
16
comment Example of two convergent series whose product is not convergent.
Hmm, the two series are only conditionally convergent. Are there also examples where they are absolutely convergent?
Jul
13
comment What is the range of convergence of $\sum_{k=0}^\infty (k \cdot x \exp(-x))^k\cdot {1 \over k!} \cdot {1\over k+1}$?
Ah, I needed two times to read and understand this completely. This explains even more for me than the other answer - if I hadn't "accepted"that before I'd accept this now. Thank you very much!
Jul
13
comment What is the range of convergence of $\sum_{k=0}^\infty (k \cdot x \exp(-x))^k\cdot {1 \over k!} \cdot {1\over k+1}$?
Ahh, yes. I got it; better for me were the form to write $e^x \ge 1+x$ - and it is simply visible because the following terms of the exponential-series $1+x+x^2/2!+x^3/3!+...$ are missing. So this is "obvious" for me. Thank you!
Jul
11
comment Formula for Sum of Logarithms $\ln(n)^m$
@RobJohn: (continued) A simple argument for "exactness" . The coefficients in column $0$ should provide the power series for the evaluation $\sum_{k=a}^{b-1} \log(k)^0 = \log(a)^0+\log(a+1)^0+... + \log(b-1)^0 = 1+1+1+...+1 = b-a$ which is the number of terms and obviously the expected answer. The column provides an "exact" solution because it is simply the exponential-series (with a modification of the constant term) and the arguments $\log(a)$ and $\log(b-1)$ can be evaluated "exactly" to the expected integer values.
Jul
11
comment Formula for Sum of Logarithms $\ln(n)^m$
@Robjohn: I think I should regard it as an "exact" answer; it provides a power series for which I have the hypothese, that it is an exact representation for the problem, like the mercator-series is assumed as an exact solution for the problem of the logarithm. Only the matrix-inversion of the infinite matrix is not yet analytically justified; Peter Walker in his article remarks the lack of such rigorous justification but mentions the apparent convergence to the coefficients of the required series. I think it would be good to do analytical research for this detailproblem - to nail it once down.
Jul
11
comment Formula for Sum of Logarithms $\ln(n)^m$
@robjohn: Just a "ping" : the question sleeps since when you wrote your answer which was regarded as not helpful by the OP. Because you're much more expert than me (amateur): would you mind to look at my proposal in my new answer in which I tried to apply an Bernoulli-/Faulhaber-analogy to the question of sums of like powers of logarithms ?
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
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
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
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
24
comment How to prove $ \lim_{n \to \infty} e^n \cdot \left( \sum_{k=0}^{n-1} ({k-n \over e})^k/k! \right)- 2 \cdot n = \frac 23$?
I've just put some background-information from my work with this at the MO-site's question, which you've linked to. Perhaps this is interesting,too
Jun
23
comment How to prove $ \lim_{n \to \infty} e^n \cdot \left( \sum_{k=0}^{n-1} ({k-n \over e})^k/k! \right)- 2 \cdot n = \frac 23$?
Wow, what an incidence (the MO-question)! The formula occurs by the matrix-summation-method based on the "Eulerian numbers" and of which I've asked&discussed several aspects (also here and in MO). Now it would be interesting whether also the generalizations might have similar relatives like that in the MO-question...
Jun
23
comment How to prove $ \lim_{n \to \infty} e^n \cdot \left( \sum_{k=0}^{n-1} ({k-n \over e})^k/k! \right)- 2 \cdot n = \frac 23$?
Perhaps it is useful to know, that I got that $f(n)$ as partial sums (always up to $n$) of a transform of the geometric series (with $q=1$), which must somehow be related to the Borel-summation method. Also in the analysis of the Borel-summation I've seen the consideration of an integral, however I don't know whether it is related. But perhaps this loose relation contains another clue?
Jun
23
comment How to prove $ \lim_{n \to \infty} e^n \cdot \left( \sum_{k=0}^{n-1} ({k-n \over e})^k/k! \right)- 2 \cdot n = \frac 23$?
Very neat -thank you! Only I'm illiterate with that type of integral so I do not know how to proceed here (and more: I doubt I can use/adapt this idea by myself to the other cases in my set of limit-epressions). I know I can look in wikipedia (and have done this from time to time earlier) for the Cauchy integral or circular/path integral but I've never got the key idea of it and how to operate with it actually...
Jun
23
comment How to prove $ \lim_{n \to \infty} e^n \cdot \left( \sum_{k=0}^{n-1} ({k-n \over e})^k/k! \right)- 2 \cdot n = \frac 23$?
@Alex: I got it, thanks! Just edited the title.