bio | website | go.helms-net.de |
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location | Germany | |
age | 61 | |
visits | member for | 4 years, 2 months |
seen | 1 hour ago | |
stats | profile views | 1,150 |
Mathematics is only a hobby - although I have done undergrad courses in the 70ties. 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-term 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.
Sep 15 |
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Prove that the Dirichlet eta function is monotonic
Your focused variation in the behave of the function $g(p)$ might be overcome if we apply Eulersum to the terms of the series $\eta(p)$ for $ p \gt 0$. Possibly this is not too difficult to prove by the construction of the transformation-terms of the Eulersum-method, but I have not yet an idea how to do this... |
Sep 12 |
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Inequality of arithmetic, geometric and harmonic means
An even more general solution for the value m for the tuple of $d-1$ is: let $\rho_d$ be the root of $g_d(x)= x \cdot ({x^{-d}+d-1 \over d}+{d \over x^d+d-1})-2$ then simply $m_d=\rho_d^d$. It gives for $m_3 \approx 6.638$ and $m_4 \approx 22.59$ and so on. |
Sep 12 |
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Inequality of arithmetic, geometric and harmonic means
I've found another representation for the value of around $m \approx 6.64$: let $w$ be the cubic root of complex unity $w=1/2 - 0.866... î$ then $m = (w^{1/3}+w^{-1/3})^3$ (in my answer I had just $x=\log(m)$ ) |
Sep 12 |
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Inequality of arithmetic, geometric and harmonic means
Hmm, @Aaron, I don't get it at the moment, I'll think about it. Anyway I find it not surprising that the visual image of a shape loses (or modifies) its visual symmetries under an exponential or logarithmic map. I'll try to get it more precise later. |
Sep 12 |
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Inequality of arithmetic, geometric and harmonic means
@Aaron: yes, I think that in the d-dimensional case it is just like this. I've tried to confirm the intuition for the 3-d case, drawing the pictures with an additional variable $z$ with layers defined by constant values for $z$. It comes out to be as expected, only the shape of the "leaves" changes: they become longer and more slim. I'd like to find an estimate for the extension on the negative axes. |
Sep 11 |
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Inequality of arithmetic, geometric and harmonic means
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Sep 11 |
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Inequality of arithmetic, geometric and harmonic means
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Sep 11 |
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Inequality of arithmetic, geometric and harmonic means
@joro: true. Everything is continuous and a sign-change occurs, so a zero must occur. (I just was in a hurry and didn't want to write something wrong...) |
Sep 11 |
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Inequality of arithmetic, geometric and harmonic means
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Sep 11 |
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Inequality of arithmetic, geometric and harmonic means
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Sep 11 |
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Inequality of arithmetic, geometric and harmonic means
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Sep 11 |
answered | Inequality of arithmetic, geometric and harmonic means |
Aug 31 |
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Smoothness in Ecalle's method for fractional iterates
Everything is fine. Thank you very much! |
Aug 31 |
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Smoothness in Ecalle's method for fractional iterates
Please let me see your jpegs. I think my email occurs in my profile-page. |
Aug 25 |
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A family Mersenne composite numbers?
:-) Nice approach to sort things out... |
Aug 24 |
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A family Mersenne composite numbers?
@Joro: true; for some small $t$ the factorizing looked a bit systematic, but in a second view for larger $t$ that simple pattern did not hold. |
Aug 24 |
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A family Mersenne composite numbers?
Hmm, so you mean $M=2^m-1$ where $m=2^{2t+1}-1+2t=4\cdot 2^w+w $. If $m$ is composite, then $M$ must also be composite. So it might be easier to prove that $m$ is composite? |
Aug 19 |
accepted | An infinite set of identities using Stirling numbers 1st kind - are they all zero? |
Aug 19 |
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An infinite set of identities using Stirling numbers 1st kind - are they all zero?
Very nice, ideed. I suspected some separation into two sums in the manner as you did it with the difference of pairs of $s_1()$ but couldn't get the key entry. I'll have to go through your solution step by step and check the relevant identities with the $s_1$, although I think that this answers my question and it's ok to accept it now. Thank you very much, this has now been an open problem for 3 years... |
Aug 19 |
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An infinite set of identities using Stirling numbers 1st kind - are they all zero?
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