User gottfried helms - MathOverflow

What is the (fractional) half-derivative of zeta at $s=0$ (and how to compute it)?

2013-05-04T07:49:02Z

(I asked this in MSE before but there was only a general reference which did not help for my specific question)

I think I understood the concept of fractional derivatives applied to monomials to consider some general cases and also applied to exponential terms as they occur in the formal expression of the Riemann zeta as Dirichlet series and their integer or fractionally indexed derivatives.

For integer indexed derivatives of the zeta at zero I can numerically evaluate the alternating zeta (Dirichlet's eta) instead using some common version of divergent summation, say Euler-summation ($\mathfrak E$) and then formally refer to $$ {d^m \over dx^m } \eta (0) =\eta^{(m)}(0) \underset{\mathfrak E}= \sum _{k=0}^\infty (-1)^k (-\log(1+k))^m$$
and then express the m'th derivatives of the zeta at zero recursively using a binomial formula (and writing $ \zeta_m =\zeta^{(m)}(0)$ and $ \eta_m =\eta^{(m)}(0) $ and $u=\log(1/2)$ for convenience) $$ \zeta_m = - \left(\eta_m + 2 \cdot \sum_{k=0}^{m-1} \left[ \binom{m}{k}u^{m-k} \zeta_k\right]\right) $$

However, I cannot generalize this to the according fractional derivatives, in this question the half-derivative, because the finite binomial sum would become an infinite series, likely also divergent and where I also would not know, in which direction I should re-interpret the binomial sum which is symmetric in its indexes for the case of integer exponents.

For the half-derivative of the $\eta$ I get by the Euler-summation and using the setting $\sqrt{-\log(1+k)} = i\sqrt{\log(1+k)} $ the approximation $$\eta_{1/2} \sim -0.347006596200 i $$

Q1: How could I express the half-derivative $\zeta_{0.5} = \zeta^{(0.5)}(0) $ formally
Q2: and what is a meaningful value?

Efficient (divergent) summation for sum of zetas at negative arguments?

2013-04-18T14:21:29Z

In a question in MSE (see bottom of my own answer) I'm considering the following series, depending on a parameter m: $$ L(m) = -\zeta(1m)/1 - \zeta(2m)/2 - \zeta(3m)/3 - \ldots $$ where I want to make sense of that sums at negative m.

For the numerical evaluation I use a customized version of the Noerlund-summation ($ \mathfrak N $) (with the software Pari/GP), but which converges only poorly and even with 128 terms I get not much more than some 15 correct digits. However, at least for $m=-1$ I get a convincing guess using mathematica (at wolfram alpha) such that $$L(-1) \underset{\mathfrak N}{=} 1- \log( \sqrt{2 \pi}) \qquad \small \text{ // guessed from 15 digits }$$ For fractional m between $0 \gt m \gt -1$ the summation still works in principle but with even less reliable digits precision. Here are some guesses for $B(m) = \exp(L(m))$ $$ \small \begin{array} {r|lr} m & B(m) (\text{ using } \mathfrak N ) \\ \hline -1 & 1.0844375514192275466 & \qquad (=\exp(1)/ \sqrt{2 \pi} = 1-A_{-1})\\ -1/2 & 1.2904007518681174634 \\ -1/3 & 1.48044921903 \\ -1/4 & 1.65184851943 \\ \end{array} $$

Unfortunately for $m \lt -1$ my procedures seem to be completely useless.
So I'd like to ask here: Q: Which (efficient) procedure can I use to get meaningful evaluations for $L(m)$ at negative m ?

[update]
To respond to @joro's computation: I did also a Borel-summation. The result for $L(-1)$ was $$L(-1) \sim 0.08106146679532725821967026359438236013860...$$

I proceeded this way. My function to be summed at -1 is $$ L(m) = - \sum_{k=1}^\infty \zeta(km)/k \qquad \text{ at } m=-1$$

The Borel-transform is $$ \mathfrak B L(-1) = - \sum_{k=0}^\infty \zeta(-1-k)/(1+k) \cdot x^k/k! $$ and we define $$ B_0(x) = - 1/x \sum_{k=1}^\infty \zeta(-k) \cdot x^k/k! $$ (where the index is also conveniently adapted)

Then the Borel-sum is computed by the integral $$ L(-1) \underset{\mathfrak B}{=}\int_0^\infty \exp(-t) B_0(t) dt $$

Now using the software Pari/GP and the sumalt-procedure for $B_0(x)$ we are still confined to small x, so the integral cannot be evaluated at high values of t . But the $B_0(x)$ can be expressed in a closed form using only the exponential, which I denote here as $B_1(x)$: $$ B_1(x)= \left(\frac 12- \frac 1x -{1 \over 1-\exp(x)}\right) \cdot \frac 1x $$ This integral can now be evaluated numerically by Pari/GP with a far better interval: $$ L(-1) \underset{\mathfrak B}{\sim}\int_{1e-20}^{1e6} \exp(-t) B_1(t) dt $$ and gives the above value to about 30 digits precision (I don't think it is a rational value).

Unfortunately, I cannot generalize that transformation to closed form for sequences of $\zeta(1m)/1,\zeta(2m)/2 , \zeta(3m)/3, ...$ where a negative $m$ is different from $-1$, say $m=-1/2$ or $m=-3$ ... [/update]

Efficient way for computation of derivatives of $f(x) = \zeta(1-x) + 1/x $ at integer x?

2013-03-08T17:18:31Z

[This question is copied from math.stackexchange, it didn't get answers so far]

For some exercises with (divergent) summation of the Stieltjes constants,see also MSE I'm trying a formula, which involves derivatives of the $\zeta()$ -function at negative integers; perhaps better formulated as $$ f(x) = \zeta(1-x) + 1/x \qquad x \in \mathbb N $$ and then the scaled derivatives $$ g(x,d) = x^d \cdot {f^{(d)}(x)\over d!} $$.

In Pari/GP there is a very nice procedure for the computation of $\zeta$ implemented, but I don't see anything for the derivatives except to implement numerical differentiation along the line ${f(x+\epsilon/2) - f(x-\epsilon/2)\over \epsilon}$ and so on, but which becomes inaccurate for higher derivatives.
For small x I can use the representation of the function $f(x)$ as a power series involving the Stieltjes-constants themselves, but a suitable computation is then possible only for a small range of x-values if I use only, say 100 or 200 terms of the power series.

Q: What would be a more efficient representation of $f(x)$ and/or $g(x)$ for the actual computation, where -say- x and d both can go up to 100 or 200 ? Note, that for high c (the index for the partial sums, see below) there is the significant effect , that either (x is small and d is high) or converse, such that d+x in the terms of one partial sum is constant.

More context: what I'm finally after is to be able to explore the partial sums and its single terms in
$$ S \underset{\mathfrak E}{=} \lim_{c\to \infty} S(c) = \sum_{k=0}^{c} (-1)^k g( c+1-k ,k) $$ which should be a representation for the (divergent) sum of the Stieltjes constants, taken by some "Eulerian" summation procedure $\mathfrak E$ (see an earlier post here on MO).

Answer by Gottfried Helms for Math for a cake

2012-12-29T13:59:42Z

One which I like much is $$ \exp \left(\begin{bmatrix} . & . & . & . & .\\ 1 & . & . & . & . \\ . & 2 & . & . & . \\ . & . & 3 & . & . \\ . & . & . & 4 & . \\ \end{bmatrix} \right)= \begin{bmatrix} 1 & . & . & . & . \\ 1 & 1 & . & . & . \\ 1 & 2 & 1 & . & . \\ 1 & 3 & 3 & 1 & . \\ 1 & 4 & 6 & 4 & 1 \\ \end{bmatrix}$$ It is practically easier and a bit more iconic if we reduce it a bit - although for me it is not so pleasing, because the immediate remembering of the Pascal-triangle comes with the 1-4-6-4-1-row: $$ \Large \exp \small \left(\begin{bmatrix} . & . & . & . \\ 1 & . & . & . \\ . & 2 & . & . \\ . & . & 3 & . \\ \end{bmatrix} \right)= \begin{bmatrix} 1 & . & . & . \\ 1 & 1 & . & . \\ 1 & 2 & 1 & . \\ 1 & 3 & 3 & 1 \\ \end{bmatrix}$$

With a bit explanation which might be useful for other guests http://go.helms-net.de/math/binomial/index-Dateien/image008.png

Answer by Gottfried Helms for Math for a cake

2012-12-29T14:05:26Z

A geometric one, where the zero can be made a cake (circle) itself $$ x^2 + y^2 -1 = \Huge \circ $$

Answer by Gottfried Helms for Singular Value Decomposition of Noisy Matrices

2012-11-09T07:25:08Z

It seems to be a classical question (handled by methods) of "factor analysis". There the problem is attacked using the covariance-, or correlation matrix, in this case possibly simply $\small \widetilde C = \widetilde U \cdot \widetilde U^{\Tiny T} /n $ (without or with centering) and then some "pca","paf" or "maximum likelihood" factor extraction appended, based on your assumtion of characteristics of $E$ resp. $ \small E \cdot E^{\Tiny T} $, i.e. estimates about the variances of the error.
But there is less "classical" method, possibly even better suited for your case, called ICA - unfortunately I do not exactly know how this is computed and how it is approached at all. (The wikipedia links may give a first impression, there is much more available even in internet sources)

Answer by Gottfried Helms for Eigenvalues of infinite matrices

2012-10-15T00:48:40Z

One simple example with a special matrix, which has somehow "a continuum" as eigenvalue...
Consider some function $ f(x) = K + ax + bx^2 + cx^3 + ... $ having a nonzero radius of convergence. Then think of the infinite matrix of the form $$ \small \begin{bmatrix} K & . & . & . & \cdots \\ a & K & . & . & \cdots \\ b & a & K & . & \cdots \\ c & b & a & K & \cdots \\ \vdots & \vdots & \vdots& \vdots & \ddots \end{bmatrix} $$ From the properties of finite matrices we would expect, that K is an eigenvalue. But consider a type of an infinite vector

$$ V(x) = [1,x,x^2,x^3,x^4,\ldots ] $$ with a scalar parameter $x$ from the range of convergence, then $$ V(x) \cdot F = f(x) \cdot V(x) $$ This means also: any vector $V(x)$ is an eigenvector of the matrix F and corresponds to the eigenvalue $f(x)$. If now $f(x)$ is entire, for instance the exponential function $ f(x)=\exp(x)$, then any value from the complex plane (except $0$ because $\exp(x)$ is never $0$) "is an eigenvalue" of F contradicting the "naive" extrapolation from the finite truncation of the matrix ...

"Bell" or "Jabotinsky"-matrix - What's the canonical name (if any)?

2012-10-15T10:12:35Z

I'm just reading J. Cigler's script for his talks "Konkrete Analysis" where I find the term "Jabotinsky-matrix" for that matrix, which I've (informally) been taught to call "Bell-matrix" (see at least Wikipedia where it is subsumed under "Carleman-matrix", but can be found in various papers). Then I find the same in D. Knuth's 1992-article on "convolution polynomials" where he develops/refers to that same idea and attributes it to Eri Jabotinsky(1947) (gives as an example the formal power series for the "half-iterate" of $ \exp(x)-1$ )
(The carleman-matrix is a transpose and similarity-transform using the factorials).

Q: is one of the names "canonized"? And if, which?
Q2: And, if not: which should one prefer in writing?

An infinite set of identities using Stirling numbers 1st kind - are they all zero?

2012-08-09T17:12:46Z

I have the following set of series involving the Stirling numbers 1'st kind and binomials, which can be understood as a set of dot-products of row- and column-vectors of two infinite matrices (where R and C indicate rows and columns, beginning at zero):

$$ w_{R,C} =\sum_{k=\max(R,C)}^\infty (-1)^k {s_1(1+k,1+k-R)\over k!} \cdot (-1)^C (1+C)^k \cdot \binom {1+k}{1+C} $$

I've tested this heuristically for several R and C and always approximated zero; also wolfram-alpha can evaluate this explicitely to zero if feeded with

sum (-1)^k * StirlingS1(k+1,1+k-R)/k! * (1+C)^k * binomial(1+k,1+C), for k=max(C,R) to infty

where we replace $C$, $R$ and $\max(C,R)$ with actual values.

However, I've no option to let wolfram-alpha answer this in general.

I've proved this for $C=0$ and the first few R using exponential generating functions, but again, a general proof is out of reach for me (possibly I'm overlooking something trivial like telescoping...), so I ask for help here.

The convention for Stirling numbers first kind as in Math'ica, indexes beginning at zero:

$ \small \qquad \qquad \begin{array} {rrrrr} 1 & . & . & . & . & . \\ 0 & 1 & . & . & . & . \\ 0 & -1 & 1 & . & . & . \\ 0 & 2 & -3 & 1 & . & . \\ 0 & -6 & 11 & -6 & 1 & . \\ 0 & 24 & -50 & 35 & -10 & 1 \end{array} $

If some background is of interest: here are the questions on MSE
http://math.stackexchange.com/questions/16228 // question of some user which motivated me to look at an example
http://math.stackexchange.com/questions/89853 // my follow-up question dealing with the current problem
and a more worked out treatize on this in a pdf-file http://go.helms-net.de/math/divers/InverseNullmatrix.pdf

Answer by Gottfried Helms for Matrix Inversion Lemma for Infinite Matrices

2012-08-29T14:45:24Z

Hmm, if all dotproducts in $X = H^T D H$ are convergent, then I don't see any special problem: the $X$ matrix simply has finite rank, and you do not encounter any specific problems due to this configuration. However, the rank of $(A+X)$ might be reduced - but this is a general problem for any $(A+X)^{-1}$-problem...

How to check numerical precision of my computation of Stieltjes-constants?

2012-08-26T14:31:37Z

In a thread in MSE I proposed an older routine of mine for the efficient computation of coefficients; I use a very similar routine for the quick&dirty computation of the Stieltjes-constants.

This motivated me to try to improve my earlier toy-computations to calculate now the first 512 Stieltjes to 1000 dec digits precision. I'm unable to estimate the number of correct digits by analytical arguments; at least wolframalpha allowed me to display StieltjesGamma[511] to 400 digits, which met my own computations.

The only freely available table around seems to be that of S. Plouffe (linked via wikipedia) but they display only the first 78 numbers to 256 digits precision.

Update2: This is the effective formula to which the Pari/GP code reduces:

Let $ \qquad h_c = {1\over c!} \sum_{k=0}^\infty (-1)^k {\ln(1+k)^c\over1+k}$ This is done using the sumalt-procedure.

Next let $ \qquad r_c = - {\ln(2)^{c-1}\over c!} b_c$ where $b_c$ are the bernoulli numbers

Then $ \qquad \gamma_c = c! \sum_{d=0}^{c+1} h_d \cdot r_{c+1-d} $

So my question:

how could I possibly get an educated guess for the number of correct digits based on my Pari/GP-routine?*

Alternatively:

is there some table with comparable precision around such that I can at least check the match for the first m digits (where m should optimally go to 1000)?

(here is the table with my current computations of 512 coeffs by 1000 digits)

Update1:
Heuristically I find, that beginning with some precision, say $300$ dec digits at the first $\gamma_0$ , I simply lose one digit precision per step in the index, so in $\gamma_k$ are roughly $300-k$ digits correct, maybe a handful less.
For this I used differences when computed with precision $200,300,400,500,600,700$ from that with precision $800$, $\gamma_0$ had just nearly all leading digits constant, when precision was increased, so that was always correct to the full precision.
That would mean, that if I want $1000$ correct digits for $\gamma_{511}$ I need dec precision of (at least) $1550$ . Simple, if that is true...

Here is my routine. I reduced the precision-parameter so that this can just be copied & pasted to a Pari/GP-environment. For precision of 1000 dec digits and 512 coefficients this must be optimized due to exorbitant increase of stack and computation-time otherwise

Prepare computations with parameters for precision of computation

termsforseries = 32
digitstocompute = 200; digitstoshow = 12;
default(realprecision,digitstocompute)
default(format,Str("g0.",digitstoshow))
default(seriesprecision,termsforseries)

Compute the coefficients of the Laurent-expansion of the zeta by conversion from the same series-type of the eta-function (the alternating zeta)

\\ ========= Zeta Laurent-expansion providing Stieltjes-coefficients ====
ps_eta = sumalt(k=0,taylor((-1)^k/(1+k)^(1-x),x))

    tmp = Vec(1-2*2^(-(1-x)));
    tmp[1]=0; \\ make the first zero exact. this step is needed for
              \\ allowing the reciprocal of the powerseries
ps_etatozeta=1/Ser(tmp)

ps_zeta = ps_eta * ps_etatozeta \\ contains now the Stieltjes-coefficients

    tmp=Vec(ps_zeta);tmp=vector(#tmp-1,c,tmp[1+c]) \\ remove the first coefficient (at 1/x)
sti = vector(#tmp,r,tmp[r]*(r-1)!)  \\ extract Stieltjes-constants by mult with factorials

Answer by Gottfried Helms for Do complex iterates of functions have any meaning?

2011-07-27T22:37:58Z

I'm discussing this from the view of iterated exponentiation (although the technical process should be the same with other functions as well).

If you can use the Schroeder-function for the continuous iteration, then the iteration-height-parameter (say "h") goes into the exponent of some basis (the log of the fixpoint, often denoted as $ \small \lambda$ ). Imaginary heights h then switch the value of the schröder-function to the negative; this allows then to extend the iteration beyond "infinite height".

For instance, use base $ \small b = \sqrt 2 $ for iterated exponentiation, $ \small z_0=x, z_1=b^x , z_2=b^{b^x}, \ldots $. Then if you begin at, say, $ \small z_0=x=1$ you can iterate to infinite height to approach the limit at $ \small z_\infty = 2$ . If you start at $ \small z_0=x=3$ you can approach $ \small z_\infty = 2$ or even $ \small z_{-\infty}=4$ . But you cannot iterate from a value $ \small z_m<2 $ to a value $ \small 2 < z_w < 4 $ using real heights, even when infinite.

But if you use the imaginary unit height you iterate directly from $ \small z_m=1$ to something like $ \small z_{m+i}=2.4 $.

Assume again $ \small z_0=1$. Then the value of the schröder-function (which is assumed to be normed to have the powerseries $ \small \sigma(x)= 1x+\sum_{k>1} a_k x^k $ ) is about $ \small s=-0.316049330525 $. Then $ \small \sigma^{o-1}( \lambda^1 s)\cdot 2 +2=b=\sqrt 2$ because that is the iteration of height 1 (in the exponent of $ \small \lambda$ ).
If we replace that exponent by $ \small h_w = i \cdot {\pi \over \ln \lambda } $ then we get $ \small \sigma^{o-1}( \lambda^{h_w} s) \cdot 2 +2=2.46791405022...$ which is, in some sense "beyond infinity" with respect to the iteration height.

Answer by Gottfried Helms for How to interpret this class of numbers?

2012-07-28T09:57:50Z

In addition to the answer of Harun Siljak one should perhaps mention that for the slightly different expression $$ \operatorname{dxp}\small([topexponent],[base],[iterationheight])=\operatorname{dxp}(x,t,h) $$ where $$ \operatorname{dxp}(x,t,0)= x
\\ \operatorname{dxp}(x,t,1)= t^x - 1
\\ \operatorname{dxp}(x,t,2)= t^{t^x - 1 } - 1
\\ \cdots $$ there is a solution for real $h$ based on the power series for the exponential function minus the constant term. Usually this is discussed for the function $$ \operatorname{dxp}(x,e,1)= \exp(x) - 1 $$ and fractional or even irrational heights $h$ and a parametrization for the coefficients for $$ \operatorname{dxp}(x,t,1)= \sum_{k=1}^\infty u^k{x^k \over k!}
\\ \operatorname{dxp}(x,t,h)= \sum_{k=1}^\infty \mathcal{P}(u,h,k){x^k \over k!}
\\ \text{where I wrote }u \text{ for } \ln(t) $$ where $\mathcal{P}$ denotes a polynomial in iteration-height, $\ln(t)$ and the series-index $k$.

For series like this and its iterations it is accepted, that the indicated family of iteration heights form a semigroup, where the height-parameter $h$ can be non-integer and can even be complex. This can already be found in L.Comtet's "advanced combinatorics" but also elsewhere.

Unfortunately, although the iterations of dxp() and exp() can be converted into each other (simply by a change of base) for integer heights, this is not uniquely determined for the fractional heights (the reason is, that for the same base in $b^x$ we have multiple bases $t$ in $t^x-1$ and the various $t$ give different results for the same $x$ and height $h$ if $h$ is fractional). Which then leads to the comment in the other answer, that there is not (yet) a commonly accepted interpretation for the noninteger heights in your original problem.

Can the relative count of the primefactors in $\small \lim_{w\to\infty}\prod_{k=1}^w (p_k-1) $ be determined analytically?

2011-12-19T22:13:57Z

(I've posted this question earlier to MSE but did not receive answers, so I'll try it here. I also condensed the wording, hopefully not too much)

Let
$\displaystyle \small \qquad f_w = (2-1)(3-1)(5-1)\ldots(p_w-1) \qquad = \prod_{k=1}^w (prime(k)-1) $
or in general with a natural number for the exponent n
$\displaystyle \small (1) \qquad f_w(n) = (2^n-1)(3^n-1)(5^n-1)\ldots(p_w^n-1) \qquad = \prod_{k=1}^w (prime(k)^n-1) $
with w going to infinity.

Then let's denote the canonical primefactorization of that product
$\displaystyle \small (2) \qquad f_w(n) = 2^{a_{n,1}} \cdot 3^{a_{n,2}} \cdot 5^{a_{n,3}} \cdot \ldots \cdot q_k^{a_{n,k}} \cdot \ldots $
using q for the primefactors here to avoid confusion between the two representations.

I am interested, whether there is an analytical expression for the relative frequencies
$\small (3) \qquad r_w(n,k) = a_{n,k} / w $
in the limit in the latter expression.

Empirically (using the first 600000 primes in formula (1)) I found approximations to rational values for the relative frequencies of the first few primefactors q in formula (2) giving a somehow meaningful table, where, after scaling near to integers, for small primes q the error was in the near of 1/1000 . However, I cannot determine, whether the deviations from my estimated analytical formula are random and are vanishing in the limit or whether they keep a bias. Especially the primefactor q=2 in the formula (2) seems to have a nonrandom bias which might survive in the limit.

Here is the table. The entries $\small e_{n,q}$ give the rounded empirical frequencies $\small e_{n,q} \approx a_{n,k}/w \cdot (q-1)^2 $

$\small \qquad \begin{array} {r|rrrrrrrrrrrr} n&2&3&5&7&11&13&17&19&23& (\ldots \text{ primefactor }q)\\ \hline \\ 1&2&3&5&7&11&13&17&19&23 \\ 2&4&6&10&14&22&26&34&38&46 \\ 3&2&5&5&21&11&39&17&57&23 \\ 4&5&6&20&14&22&52&68&38&46 \\ 5&2&3&9&7&55&13&17&19&23 \\ 6&4&10&10&42&22&78&34&114&46 \\ 7&2&3&5&13&11&13&17&19&23 \\ 8&6&6&20&14&22&52&136&38&46 \\ 9&2&7&5&21&11&39&17&171&23 \\ 10&4&6&18&14&110&26&34&38&46 \\ 11&2&3&5&7&21&13&17&19&253 \\ 12&5&10&20&42&22&156&68&114&46 \\ 13&2&3&5&7&11&25&17&19&23 \\ 14&4&6&10&26&22&26&34&38&46 \\ 15&2&5&9&21&55&39&17&57&23 \\ 16&7&6&20&14&22&52&272&38&46 \\ 17&2&3&5&7&11&13&33&19&23 \\ 18&4&14&10&42&22&78&34&342&46 \end{array} $

The heuristical formula that I extrapolated (letting w increase towards infinity) has two forms:

if q=2 and n is even (gcd(n,q)=2):
$\small \qquad e_{n,2} = (3 + \operatorname{val}( n,2 ) ) $
where the function val(n,q) means: the exponent, to which primefactor q occurs in n

For all other cases
$\small \qquad e_{n,q} = \gcd(n,q-1) \cdot (q + (q-1)\cdot \operatorname{val}(n,q) ) $

Then
$\small \qquad \displaystyle a_{n,q} = { e_{n,q} \cdot w \over (q-1)^2 } $

Can the guessed formula be confirmed by an analytical argument?

Answer by Gottfried Helms for Eigencircles of n x n matrices?

2012-04-27T03:45:07Z

As I understand the problem it can be generalized. The $\small \lambda,\mu $-matrix can be seen as scaled rotation, and using $ \small r^2 = \lambda^2 + \mu^2 $ we can, because the $\small \lambda,\mu$ - matrix is invertible and is the left factor on the rhs (thus it is not really an eigenvalue analogon in my opinion) we can differently rewrite your original equation as

$\qquad \small \frac1{r^2} \begin{bmatrix} \lambda & -\mu \\ \mu & \lambda \end{bmatrix} \cdot \begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} x \\ y \end{bmatrix} \cdot 1 $

or $\qquad \small \begin{bmatrix} \cos(\varphi) & -\sin(\varphi) \\ \sin(\varphi) & \cos(\varphi) \end{bmatrix} \cdot \begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} x \\ y \end{bmatrix} \cdot r $

or
$\qquad \small (T \cdot A) \cdot X = X \cdot r$

where then the eigen-problem is
$\qquad \small \left\vert T \cdot A - r\cdot I \right\vert = 0 $

The left factor T is simply a rotation-matrix and r the eigenvalue of the rotated matrix $\small T\cdot A$.

If this characterization of the $\small \lambda, \mu$-matrix is the interesting one (which I assume) it can then be generalized to more dimensions in the obvious way and the $\small \lambda ,\mu $'s have to be determined as scalings of the cos/sin-parameters by the eigenvalues of the multidimensionally rotated matrix.

[Update] Here is a picture of the construction according to my interpretation of the problem. I take some example-matrix A $\small \begin{bmatrix} 12&24\\ 13&83 \end{bmatrix} $ then leftmultiply by the rotationmatrices of angles between 0 and $\small 2 \pi $. Each has then two eigenvalues $\small w_0 , w_1 $. The $\small \lambda,\mu$ -matrix is then $\small T \cdot w_0 $. Note, that the eigenvalues and thus the $\small \lambda,\mu$-matrix may become complex for some rotations. Then I plot the log of the absolute value of the two eigenvalues in different colors. Here is the picture:

Answer by Gottfried Helms for How to compare two similarity matrices?

2012-03-12T22:41:55Z

As far as you use the cosine as similarity measure, the matrix is a correlation matrix. For this situation in statistics there is the concept of "canonical correlation", and this might be then the most appropriate for your case: it gives an index how much "variance of one set of variables is explained by the other". The two set of variables are the two sets of vectors $\small v_i $ here.

Another option could be to compute the cholesky factors ("factor loadings matrices") L1 and L2 of each of the correlation matrices R1 and R2 and do a target-rotation of L1 to L2. Then, for instance, the squared distances of the vector-tips of each related vector in rotated(L1) and L2 could be summed and this could be understood as similarity measure of the matrices(!) - but this is no standard method as far as I know...

Answer by Gottfried Helms for Interesting result on the Euler-Maschroni constant - what is the background?

2012-03-12T22:25:45Z

Adding an example to @quid's answer:

Using Pari/GP the harmonic numbers minus Euler-gamma can be obtained using the psi-function. If we subtract 1 from the $\small a_i $-values the composition of the result by i-digits long blocks of decimal expansion of the bernoulli-numbers becomes then immediately visible. With $\small i \gt 20 $ or so it becomes even more impressive:

fmt(200,60)   \\ user function: internal prec 200, display prec 60 digits
i=6
psi(10^i)-1 - i*log(10)
 %681 = -1.00000050000008333333333332500000000000396825396824980158730
(psi(10^i)-1 - i*log(10))*6
 %683 = -6.00000300000049999999999995000000000002380952380949880952381
(psi(10^i)-1 - i*log(10))*42
 %684 = -42.0000210000034999999999996500000000001666666666664916666667
(psi(10^i)-1 - i*log(10))*42*30
 %685 = -1260.00063000010499999999998950000000000499999999999475000000

etc...

For an approach to the Hadamard-matrix-problem: is there a proof, that the iterative plane-wise orthogonal rotations (Quartimax/Varimax) converge to global maximum?

2012-03-08T09:09:51Z

I've asked this question at stat-exchange and at the "Semnet"-mailing list of professionals in statistics. The reference to some articles in Psychometrica (for instance ten Berge 1995, Jennrich 2001) were insofar helpful, that I saw, that reasons for nonsuccessful rotation were handled - but the explicite statement, whether the iteration with plane-wise rotations using the Varimax- or Quartimax-criterion is always converging /can always be made converging was not made/referred to. (This is different for instance for the case of using the principal components-criterion, for which I think I've come across an explicite proof ~20 years ago).

The reason to ask this is an approach to the Hadamard-matrix-problem: the Hadamard-matrix is defined to consist of entries +1 and -1 only and being orthogonal.
Being orthogonal means to have the form of a rotation-matrix, simply scaled by one scalar factor such that all entries have the desired value -1 or +1. That means, the variance of the squared entries is zero. So if I begin with any initial rotation-matrix, and rotate to minimal (instead of maximal) variance using the inverse of the "varimax" or of the "quartimax"-criterion, say "varimin" or "quartimin", I should arrive at a proper Hadamard-matrix, scalar scaled by the reciprocal of the square of the number of rows/columns.

Indeed this works for small sizes (up to n=20, so nxn=20x20) and "sometimes" for n=24, but then the Hadamard-form is found extremely seldom.
Ten Berge (1995) gave insight in the rotation procedure, noting, that sometimes convergence is not reached because of systematic missing of optimization of some planes in the iterative process, but which might be overcome by some workaround. The global convergence question however was not explicitely mentioned as solved.

Question: is a proof for the convergence to the global optimum by the iterative, plane-wise rotation using the quartimax/varimax-rotation known? Or is it known, that it does not always converge?) (If it is simple enough it might be done here, or else a reference were good. I've it -surprisingly- not seen in the monographies on factor analysis of for instance S.Mulaik(74) or K.Überla(68) .

(This "varimin"-rotation-approach to the Hadamard-matrix problem has its further charme, because it is also applicable to matrices which have other dimensions than n=4*m, and is thus some more general formulation/notion for that problem: "the lower bound for the variance of the squared entries of orthogonal matrices of size nxn where n=4m is zero" )

Examples

//commands in "Matmate" to reproduce the approach
n=8
A = gettrans(randomu(n,n),"drei")  // creates a random rotation-matrix in A
A = sqrt(n) * A                    // rescales it with the scalar  sqrt(n)
T = gettrans(A,"-varimax")         // gets the rotationmatrix T which rotates A
                                   //     using the optimization criterion "Varimin"
H = A * T                          // gets the Hadamardmatrix of size 8x8 
                                   //     by column-rotation

$\qquad \small \text{ A =} \begin{array} {rrrrrrrr} 0.95& 0.24& 1.32& 0.34& -0.87& -1.08& 1.54& -0.94\\ 1.45& 0.06& -1.03& -0.57& 1.71& -0.41& 0.05& -1.19\\ 1.29& -0.91& -0.50& 1.41& -0.98& 1.29& -0.54& -0.59\\ 1.02& 1.14& -1.36& 0.36& -0.67& -0.75& 0.35& 1.60\\ 0.53& -0.28& 1.21& 1.36& 1.14& -0.97& -1.14& 0.86\\ 0.20& 1.76& 0.72& 0.57& 0.87& 1.69& 0.62& 0.13\\ 0.64& 1.26& 0.62& -1.03& -0.92& -0.11& -1.84& -0.57\\ 1.26& -1.03& 0.85& -1.48& 0.08& 0.78& 0.36& 1.31 \end{array} $

and the Hadamardmatrix H

$\qquad \small \text{ H =} \begin{array} {rrrrrrrr} -1.00& -1.00& 1.00& 1.00& -1.00& -1.00& 1.00& -1.00\\ 1.00& -1.00& -1.00& -1.00& 1.00& -1.00& 1.00& -1.00\\ 1.00& -1.00& -1.00& 1.00& -1.00& 1.00& -1.00& -1.00\\ 1.00& 1.00& -1.00& 1.00& -1.00& -1.00& 1.00& 1.00\\ 1.00& -1.00& 1.00& 1.00& 1.00& -1.00& -1.00& 1.00\\ 1.00& 1.00& 1.00& 1.00& 1.00& 1.00& 1.00& -1.00\\ 1.00& 1.00& 1.00& -1.00& -1.00& -1.00& -1.00& -1.00\\ 1.00& -1.00& 1.00& -1.00& -1.00& 1.00& 1.00& 1.00 \end{array} $

For a dimension not a multiple of 4 we do not get the variance of the squares of the matrix-entries to equal zero but I think, we get the solution of minimal variance, and the squares of the entries are not 1.
Example n=6:

$\qquad \small \text{ A =} \begin{array} {rrrrrr} 1.00& 1.29& 1.32& -1.10& 0.33& 0.52\\ 0.48& 0.23& 1.14& 1.56& -1.39& -0.22\\ 1.51& -1.51& -0.08& 0.09& 0.12& 1.18\\ 1.38& 0.62& -1.39& -0.40& -0.85& -0.94\\ 0.74& -0.32& 0.49& 0.63& 1.55& -1.52\\ 0.21& 1.22& -0.88& 1.34& 0.89& 1.04 \end{array} $
and
$\qquad \small \text{ H =} \begin{array} {rrrrrr} 0.00& 1.10& 1.10& -1.10& 1.10& 1.10\\ 1.10& 1.10& 1.10& 1.10& -1.10& 0.00\\ 1.10& -1.10& -0.00& -1.10& -1.10& 1.10\\ 1.10& 1.10& -1.10& -1.10& 0.00& -1.10\\ 1.10& -1.10& 1.10& 0.00& 1.10& -1.10\\ 1.10& -0.00& -1.10& 1.10& 1.10& 1.10 \end{array} $

where the nonzero entries are exactly $\small \pm \sqrt{6 / 5} $

[update] For the n=6-case I've got one alternate solution, which is not optimal but which is stationary under the varimin/quartimin-criterion. This is
$\qquad \small \text{ H =} \begin{array} {rrrrrr} 0.75580438&0.75580439&1.32408152&1.12540822&-1.12540822&-0.75580438\\ 1.32408152&-1.12540822&-0.75580438&0.75580439&-0.75580438&1.12540822\\ 1.12540822&-1.32408152&0.75580439&-0.75580438&0.75580438&-1.12540822\\ 1.12540822&1.12540822&0.75580438&-0.75580438&0.75580439&1.32408152\\ 0.75580439&0.75580438&-1.12540822&-1.32408152&-1.12540822&-0.75580438\\ 0.75580438&0.75580438&-1.12540822&1.12540822&1.32408152&-0.75580439 \end{array} $
This is just plain vanilla-"varimin", I did not yet include the ten Berge-workaround here.

Alternating sums of powers of the lngamma ($\small f_p(x) = \sum \log(k!)^p x^k $ at $\small x=-1$)

2012-01-31T07:23:43Z

I'm still fiddling with this recent question and come to a detail, whether I can find closed forms for the sums of the $\small lngamma() $ function. Precisely
$$ \small f_p(x) = \sum_{k=0}^{\infty} \log(k!)^p \cdot x^k $$ and my first question is to find closed forms for $\small f_p(-1) $ for consecutive p.

They are all non-converging series, but which can nicely be summed by, for instance, Euler-summation, but I wish to find closed forms (or simple forms of other series with more known analytical properties). For p=0 this is $\small \eta(0) $ (the "Dirichlet's eta", or "alternating zeta", function), for p=1 this is $\small \eta(0)' $ and I expected, that $\small f_2(-1) $ would be something composed by the square of $\small f_1(-1)$ or the second derivative of the $\small \eta $ at zero, but didn't succeed so far.

The numerical values for the first few p seem to be (using Pari/GP sumalt-procedure)
$\small \qquad \begin{array} {rl} p & f_p(-1) \\ \hline \\ 0 & 0.500000000000 \\ 1 & 0.112895676322 \\ 2 & -0.0380319653072 \\ 3 & 0.0135052530749 \\ 4 & 0.0183298626301 \\ 5 & -0.107164190642 \\ 6 & 0.331363715855 \\ 7 & -0.482387386451 \\ 8 & -2.91602127867 \\ 9 & 32.5904726686 \\ 10 & -154.360744999 \\ 11 & -162.033212532\\ \end{array} $

mertens-function in the light of divergent summation - what summation method were best adapted

2010-11-26T23:15:54Z

Just reading about the Mertens-function in the other thread Mertens function I remember an earlier attempt to apply divergent summation to the series which is constructed of the Moebius-function at consecutive arguments, or in other words of which the Mertens-function-values represent the partial sums.

Eulersummation, although relatively poorly adapted, suggested that the (divergent) sum should be meaningfully evaluated to -2. But that sequence of partial sums (although seldom exceeding only the squareroot of its current index) seems to be a specific difficult case for such common summation methods - the approximation is relatively poor even for 128 terms. I tried Nörlund-means/Cesaro-sum, Euler-sums of different orders and also a selfmade matrix summation-method using the eulerian numbers (with which I could -on the other hand- well handle the even strongly diverging $ 0!-1!+2!-3!+...$ -series), but I tried not yet for instance Abel and Borel.

Q1: What method would be most appropriate to sum the series $ S = \sum _{k=1}^{\inf} moebius(k) $

Evaluation of 128 terms (Euler,Cesaro) suggested the result $S = -2$

Q2: And how could it be determined whether the Cesaro- and/or Euler-summation are at all capable to evaluate that series to a final value?

Here is a plot of the summation.

Answer by Gottfried Helms for existence of polynomial equation system solution

2012-01-26T21:20:16Z

[update] By an example of 4x4-matrices the ansatz below could not be used to solve the problem. The matrix $\small Q_K $ cannot in general be made lower triangular by choices of the $\small k_i $. I'll delete this answer soon if I cannot improve the ansatz.

I do not yet have a full answer but possibly a first step into one. I think that this can be answered unsing two facts:

a) There is a similarity transformation with a rotation T such that $\small P = T^{-1}\cdot A \cdot T = T' \cdot A \cdot T $ where P is triangular and has the eigenvalues of A on its diagonal.

b) Your sum-expression of vector outer products, let it be called matrix $\small E_K = \sum_{i=1}^n B_i k_i C_i = \sum_{i=1}^n k_i (B_i C_i)= \sum_{i=1}^n k_i E_i $ is a weighted (by the $\small k_i $ weights) sum of rank-1-matrices $\small E_i $

From the "similarity rotated version" of all matrices

$\qquad \small Q_K = T' E_K T = \sum_{i=1}^n k_i (T' E_i T) = \sum_{i=1}^n k_i Q_i $ (which should be made triangular by choices of $\small k_i $ ) and
$\qquad \small R = T' D T $ which is then also triangular

we get your final equation in its form with triangular matrices

$\qquad \small R = P + Q_K $

We'll have a solution if the weights $\small k_i $ for the non-triangular, generic but rank-1-matrices $\small Q_i $ can be chosen such that their sum $\small Q_K$ becomes triangular and its diagonal equals the negative diagonal in $\small P $.

I've a vague speculation that the equation with this triangular matrices can easier be shown to be "almost always" possible, but have not yet a further concrete approach. At least this construction exhibits that the rank-1-matrices $\small E_i $ (and thus $\small Q_i$ ) cannot be simple scalar multiples of each other if A has full rank and thus the restriction to "for almost all" is unavoidable and possibly mainly consists of this property.

Answer by Gottfried Helms for Euler's divergent series sum n!*(-1)^n: what is known about the resulting constant?

2012-01-14T19:18:03Z

A small - surely not authoritative- bit of information: http://www.groupsrv.com/science/about477640.html

Answer by Gottfried Helms for Is `$\sum\limits_{n=0}^\infty x^n / \sqrt{n!}$` positive?

2012-01-06T05:15:32Z

Another "not-yet-answer"...

I've tried another idea. Assume the function f(x) is expressed by the following composition: $$\small x' = \exp(x)-1 $$ $$\small f(x) = g(x') = g(exp(x)-1) $$ The idea is, that the unavoidable big "hump" in the partial sums, after which the sequence of partial sums begins to decrease, may be absorbed by the function $\small g(x)$ - because $\small \exp(x) $ is really small for large negative x and x' is then very little above -1. I did not yet arrive at a conclusive result; but the power series for $\small g(x) $ begins with the smooth looking form (and gives the partial sums for $\small x'=\exp(-100)-1 $):
$\qquad \small \begin{array} {r|r} \text{powerseries} & \text{partial sums for x' } \\ \hline \\ 1.00000000000 & 1.00000000000 \\ +1.00000000000x & 3.72007597602E-44 \\ +0.207106781187x^{2} & 0.207106781187 \\ +0.0344748426106x^{3} & 0.172631938576 \\ -0.0100670743762x^{4} & 0.162564864200 \\ +0.00821765977664x^{5} & 0.154347204423 \\ -0.00654357122833x^{6} & 0.147803633195 \\ +0.00537330847179x^{7} & 0.142430324723 \\ -0.00451702185603x^{8} & 0.137913302867 \\ +0.00386915976824x^{9} & 0.134044143099 \\ -0.00336528035075x^{10} & 0.130678862748 \\ +0.00296428202807x^{11} & 0.127714580720 \\ -0.00263893325448x^{12} & 0.125075647465 \\ +0.00237058888853x^{13} & 0.122705058577 \\ -0.00214611388717x^{14} & 0.120558944690 \\ +0.00195602261228x^{15} & 0.118602922077 \\ -0.00179331457091x^{16} & 0.116809607506 \\ +0.00165272361723x^{17} & 0.115156883889 \\ -0.00153022060566x^{18} & 0.113626663284 \\ +0.00142267593977x^{19} & 0.112203987344 \\ -0.00132762563657x^{20} & 0.110876361707 \\ +0.00124310598493x^{21} & 0.109633255722 \\ -0.00116753462507x^{22} & 0.108465721097 \\ +0.00109962364925x^{23} & 0.107366097448 \\ \end{array} $

The the question is, for some large negative x, say $\small x=-100 \qquad x'=exp(-100)-1 = -1+ \epsilon $ the series $\ g(x') $ converges to zero. Unfortunately - although we've translated the original problem to one with nice small numbers I don't see, how to really come nearer a solution, because the convergence of $\small g(-1+\epsilon) $ is really slow - if it converges at all to a positive value... So this is not yet a solution, but perhaps a suggestion for a path to try...

Answer by Gottfried Helms for Is `$\sum\limits_{n=0}^\infty x^n / \sqrt{n!}$` positive?

2012-01-05T22:29:13Z

Additional data for Liviu's plots. I used Pari/GP with 1200 digits dec prec, documenting also the required number of terms after which the absolute values of the summands of the series decrease below 1e-100. There seems to be no local minimum...

$\small \begin{array} {rl|r} & & \text{# of terms}\\ x & f(x) & \text{ required} \\ \hline \\ -1 & 0.438599896749 & 201 \\ -2 & 0.247539616819 & 201 \\ -3 & 0.162554775870 & 211 \\ -4 & 0.117399404501 & 257 \\ -5 & 0.0903120618145 & 304 \\ -6 & 0.0726061182760 & 354 \\ -7 & 0.0602796213492 & 407 \\ -8 & 0.0512783927864 & 464 \\ -9 & 0.0444561508357 & 525 \\ -10 & 0.0391295513879 & 589 \\ -11 & 0.0348689168813 & 658 \\ -12 & 0.0313919770798 & 730 \\ -13 & 0.0285063993737 & 808 \\ -14 & 0.0260770215882 & 889 \\ -15 & 0.0240063146159 & 976 \\ -16 & 0.0222222780410 & 1067 \\ -17 & 0.0206706877888 & 1162 \\ -18 & 0.0193099849974 & 1263 \\ -19 & 0.0181078191003 & 1369 \\ -20 & 0.0170386561852 & 1479 \\ -21 & 0.0160820905671 & 1595 \\ -22 & 0.0152216309789 & 1715 \\ -23 & 0.0144438135509 & 1841 \\ -24 & 0.0137375438980 & 1972 \\ -25 & 0.0130936024884 & 2108 \\ -26 & 0.0125042681404 & 2250 \\ -27 & 0.0119630281606 & 2396 \\ -28 & 0.0114643528377 & 2548 \\ -29 & 0.0110035182996 & 2705 \\ -30 & 0.0105764661081 & 2867 \\ -31 & 0.0101796910429 & 3035 \\ -32 & 0.00981015071575 & 3208 \\ -33 & 0.00946519223932 & 3386 \\ -34 & 0.00914249232841 & 3569 \\ -35 & 0.00884000806032 & 3758 \\ -36 & 0.00855593615550 & 3953 \\ -37 & 0.00828867911422 & 4152 \\ -38 & 0.00803681690505 & 4357 \\ -39 & 0.00779908317617 & 4567 \\ -40 & 0.00757434517200 & 4783 \\ -41 & 0.00736158670179 & 5004 \\ -42 & 0.00715989363457 & 5231 \\ -43 & 0.00696844149585 & 5462 \\ -44 & 0.00678648482039 & 5700 \\ -45 & 0.00661334797911 & 5942 \\ -46 & 0.00644841724806 & 6190 \\ -47 & 0.00629113392871 & 6444 \\ -48 & 0.00614098836080 & 6703 \\ -49 & 0.00599751469633 & 6967 \\ -50 & 0.00586028632445 & 7236 \\ -51 & 0.00572891185489 & 7511 \\ -52 & 0.00560303158255 & 7792 \\ -53 & 0.00548231436720 & 8078 \\ -54 & 0.00536645487311 & 8369 \\ -55 & 0.00525517112099 & 8666 \\ -56 & 0.00514820231209 & 8968 \\ -57 & 0.00504530688991 & 9275 \\ -58 & 0.00494626080983 & 9588 \\ -59 & 0.00485085599129 & 9907 \\ -60 & 0.00475889893049 & 10230 \\ -61 & 0.00467020945455 & 10560 \\ -62 & 0.00458461960073 & 10894 \\ -63 & 0.00450197260623 & 11234 \\ -64 & 0.00442212199624 & 11580 \\ -65 & 0.00434493075923 & 11931 \\ -66 & 0.00427027059992 & 12287 \\ -67 & 0.00419802126157 & 12649 \\ -68 & 0.00412806991028 & 13016 \\ -69 & 0.00406031057475 & 13388 \\ -70 & 0.00399464363573 & 13766 \end{array} $

Answer by Gottfried Helms for Implication for m-cycles in Collatz-type problems.

2011-10-15T19:01:58Z

[update 2] Rereading the question after my own answer I note, that the doubly used symbol m for length of a series of collatz-transformation and the m for the minimal element (M for the maximal) irritated me and made me misunderstanding the question.
Anyway - the already given answer might be interesting for the background of my argument below as well as for further analysis of the problem at hand, so I'll leave it here for the cursory reader.

To answer the real question : "is there some requirement, that in a cyclic Collatz-trajectory the maximal element is a perfect-power-of-2-multiple of the minimal element (M=2^k m) ?" : there is no such requirement to all my knowledge.
Also this would only define a single special case according to my notation for a cycle, which by convention might begin with the smallest element m : $\small m = T(m; A_1,A_2,\ldots ,A_{n-1},A_n) $.
Here your requirement simply means, that the maximal element M must occur after the transformation $\small A_{n-1} $ (whose exponent must thus equal 1) frome where then the $\small A_n$ is the k in M=2^k m in your question. I didn't come across such a restriction in my own study or in any seen literature, and cannot see any reason for it (see derivations below). [end update 2]

If you write a collatz-transformation this way $ \small a_{k+1} = (3 a_k + 1)/2^A $ where A is the number of all following even (divide-by-2) steps, then also all $\small a_k$ must be odd.

With this write for one transformation $\small a_{k+1} = T(a_k;A) $ and for more $\small a_{k+m} = T(a_k;A_1,A_2,\ldots,A_m) $.
For $\small A=1 $ these steps are increasing and if $\small A>1$ the steps are decreasing.

A cycle occurs, if $\small a_m=T(a_0;A_0,A_1,\ldots,A_m) = a_0 $ thus $\small a_m = a_0 $.

Now let's look at two different types of cycles - the "general" one, where the exponents $\small A_k $ have no a priori restriction, and the "primitive" one, where all but the last $ \small A_k = 1 $, thus $\small a_0 = T(a_0;1,1,1,1,\ldots,1,A) $ with say $\small N-1 $ times 1 and only one $\small A_{N} \gt 1$

For the primitive cycle of the length N the smallest element $\small a_0 $ must have the form $\small a_0 = 2^{N-1} \cdot 2w - 1$ where w is some positive odd integer and that primitive cycle increases then up to $\small a_{N-1} = 3^{N-1}\cdot 2w-1 $ from where it must decrease by a consecutive set of A "even" transformations.

The proof of Ray Steiner (and the subsequent proofs of J.Simons and B.de Weger) use that requirement of the "primitive cycle" (in their nomenclature "1-cycle", see wikipedia) to show that such a cycle cannot exist except $\small a_0 = T(a_0;2) \qquad a_0=1 $ (where no exponents of value 1 occur - the degenerate case (also called "circuit").

For the general cycle this is much more complicated and does definitely not have the properties which you sketch in your op.

[added] Answering to the comment. A characterization of the general cycle of m collatz-steps (m=N + S in my notation, where I refer -writing "steps"- to the number N of abbreviated steps $\small T(a;A)$

Assume again one step as on the form $\small a_{k+1} = (3 a_k +1)/2^{ A_k} = T(a_k;A_k) $. Then to have a cycle this means (ignore the index k here) $\small a=(3a +1)/2^A $ and also $\small 2^A=(3a +1)/a = (3 + 1/a) $ This can only be solved if a=1 and A=2, so there is only one general-cycle of length N=1, S=2 and m=N+S=3 and we see a characteristic of the exponent A, which is much descriptive: the number of even steps of the original notation of the collatz-transformation is 2.
Now we assume a 2-step cycle $\small a = T(a;A,B) $ and dissolve this in two steps: $\small b=T(a;A) \qquad a=T(b;B) $ thus $\small b = (3 a +1)/2^A \qquad a=(3b+1)/2^B $. I'm used to write S for the sum A+B meaning the whole number of even steps, and N for the number of "odd steps", both wrt the original Collatz-notation, and in my notation N is the number of steps (and the power of 3 involved). If we write the (trivial) product of the two involved elements a and b in their direct notation and in their transformed expression we get $\small a\cdot b = (3b+1)/2^A \cdot (3a+1)/2^B $ and this can be rewritten as $\small 2^{A+B} = (3+1/a)(3+1/b) $ This is an interesting form and easily generalized for the analysis with bigger N (and respectively m) . Here we see, that a 2-step general cycle can only exist if $\small (3+1/a)(3+1/b) $ is a perfect power of 2; now considering the whole set of odd positive integers for a and b we see, that that product can vary only between $\small 9 \ldots 16 $ and thus must be S=A+B=4 and this requires a = b = 1 and thus A=B=2 (which is then only a concatenation of the trivial cycle $\small 1=T(1;2,2) $ . Here m=N+S=2+4=6 .

This way you may proceed studying longer assumed cycles. For instance it shows that the whole distance between two numbers $\small a_k - a_j = 2^S$ where S is the number of even transformations can never occur because there are always "odd Collatz steps" interspersed; even less can the distance between the minimal and maximal member of a loop be $\small 2^{N+S} = 2^m $, which were the whole length of the cycle in the counting of original Collatz-steps.

This formula describes pretty well important properties of the exponents of 2 in relation to the length of a "general cycle" , so it might be useful for the answering of your question. However - I can't relate anything in that formula to the asked property of a connection between the distance minimal...maximal member and 2^m where m is the number of all Collatz steps and actually is $\small m=S+N$ nd so I think there is none.

Addendum : It might be interesting to look at the existing cycles in the domain of negative odd numbers; they can also be identified using the procedere exercised above. The 2-step cycle was $\small 2^{A+B} = (3+1/a)(3+1/b) $ and allowing negative a and b gives the example: $\small 2^{A+B} = (3-1/5)(3-1/7) = {14 \over 5} \cdot {20 \over 7} = {8 \over 1} = 2^3 $ which is a perfect power of 2 as required. Then $\small A+B=3$ and one of them must equal 1. In fact we have the 2-step-cycle $\small -5=T(-5;1,2) $

Finally: a short treatize using the notation here is in that article of mine

1 Steiner,R.P.; A theorem on the syracuse problem, Proceedings of the 7th Manitoba Conference on Numerical Mathematics,pages 553...559, 1977

Answer by Gottfried Helms for hyperplane least square through points

2011-10-08T10:50:11Z

Hmm, I'm not sure, whether I got your problem right, but well, I'll give it a try.

I assume the N datvectors as rowvectors consisting of n datapoints (where n << N). Then the usual PCA assumes the columns as axes in an n-dimensional coordinate-system having N vectors beginning at the origin, say the data-matrix $\small Z $. The usual principal components can then be found by rotations of the columns, in matrix-notation $\small Z \cdot T$ , where $\small T $ is a rotation-matrix.

Now I understand your question such that you want to express your N data-vectors as linear combinations of a smaller set of (pairwise orthogonal) component(-vectors).

This seems simply the question of rotating the rows of $\small Z$ to their PC-position, thus $\small T \cdot Z $ and you get (at most) n component-rowvectors which can be composed to represent $\small Z $ by the inverse row-rotation.
And the best representation of $\small Z$ by m components only, where $\small m \lt n$ would likely be done to use only the first m components ; so to say $ \small PC_n = T \cdot Z $ giving at most n non-zero component-vectors in $\small PC_n $ . Then to have the best representation by $\small m \lt n $ rowvectors set all vectors in $\small PC_n $ of indexes k where $\small m < k \le n$ to zero to get $\small PC_m $ and apply $\small Z_m=T^\tau \cdot PC_m $ where $\small Z_m $ might then be the best rank- m -approximation to $\small Z$

Example: (sorry, this is in my proprietary MatMate-code (don't have Maple/Matlab/Math'ca) but should illustrate the pseudocode sufficiently) see protocol of worked example :

//******  MatMate Version 0.1108 Beta *****************************

// MO-Problem: 
//    Express N rowvectors optimally by m component-vectors (least squares)
//    Proposed solution: use PCA on rows of datamatrix
// -------------------------------------------------------------

// 1) generate random-data
n=6    
v=20   // this is big-N in the problem description
m=3
         set randomstart=41
Z = randomn(v,n)        // generate normally distributed randomdata
Z = zvaluezl(abwzl(Z))  // center and standardize Z rowwise
//-------------------------------------------------------

// 2) find principal components rowwise;
//      note: due to centering of rowdata there are 
//            maximally only n-1 independent components!

  // center data columnwise
  ME = meansp(Z)  // get a rowvector containing the columnwise means
   C = Z - ME     // ME will implicitely be expanded to fit the dimension of Z
                  // C contains then the columnwise recentered data

  // get the required rotation-matrix T first
  // because in MatMate rotations are done on columns, we have
  // to transpose C as well as the result (using ' as transpose-operator)
  T = gettrans(C',"pca")'

PC_n = T * C    // the first n-1 rows contain the principal components

// 3) now set all rowvectors with index k>m to zero into a new matrix PC_m
PC_m = { PC_n[1..m,*], Null(v-m,n) }

// 4) reverse the rotation where only the m-components are used
C_m = T' * PC_m    
Z_m = C_m + ME     // the rowvector ME is automatically expanded to fit the C_m matrix

// 5) check quality of approximation
chk = (Z-Z_m) ^# 2   // Check differences, ^# 2 means: apply power of 2 elementwise
err = sqrt(sum(chk)) // check overall-error

Answer by Gottfried Helms for Fermat-like equation $c^n=a^{2n}+a^n b^n + b^{2n}$

2011-09-19T08:33:12Z

I'd like to add another line of proving this (and similar problems); the proof is not yet formally complete but the method as such seems helpful for many such problems, so I show my reasoning here:

[update] After rereading my own answer I think now, that that line of attack might be more incomplete for this problem than initially thought; that the holes cannot be simply filled with one or two little more thoughts; so possibly I'll retract the whole approach later. Thanks for the upvotings anyway and apologies for distractions ... [/update]
The original answer went as below:

Let us define the following notations

a) {n,p} the exponent to which a prime p occurs in the primefactorization of some number n
b) the iverson-bracket [n:p] which evaluates to 1 if p divides n and to 0 if not

c) $ \small f(a,b,n)=a^n - b^n $ , the short form for expressions where a,b are thought to be constant and n is seen as varying
d) $ \small \lambda_p $ the smallest k>0 such that p divides $ \small f(a,b,k) $ ,
e) $ \small w_p $ the exponent, to which p occurs in $ \small f(a,b,\lambda_p) $ , formally $ \small w_p=\{ f(a,b,\lambda_p),p \} $

We use that definitions to rewrite the analysis of the problem in terms of the Euler-totient-theorem and the concept of the order of cyclic subgroups modulo some primes p.
Here the idea is to compare the odd primefactors in the canonical primefactorizations of the lhs and rhs in the conveniently rewritten problem
$$ \small c^n-(ab)^n =^? a^{2n}+b^{2n} = {a^{4n}- b^{4n} \over a^{2n}- b^{2n} } . $$ It will be sufficient to compare the odd primefactors $ \small p, q \in \text{odd primes} $ only; so we refer to possible exponents of 2 with some anonymous exponents s and t only. Then the lhs is with $ \small q \in \text{odd primes} $
$$ \small f(c,ab,n)=2^s \prod q^{ [n:\lambda_q] (w_q+\{n,q\})} $$ and the rhs is with some exponent t at the primefactor 2 : $$ \small a^{2n}+b^{2n}={ a^{4n}-b^{4n} \over a^{2n}-b^{2n} }={f(a,b,4n)\over f(a,b,2n)} $$ and $$ \small {f(a,b,4n)\over f(a,b,2n)} = 2^t { \prod p^{ [4n:\lambda_p] (w_p+\{4n,p\})} \over
\prod p^{ [2n:\lambda_p] (w_p+\{ 2n , p \} ) } } $$ Here for all odd primes p we have $ \small \{ 4n , p \} = \{ 2n, p \} = \{ n,p \} $, $$ \small { f(a,b,4n)\over f(a,b,2n)} = 2^t \prod p^{ ([4n:\lambda_p]- [2n:\lambda_p]) (w_p+\{ n, p \} ) } $$

Conclusion: (updated)
In the rhs we get only that primefactors p, whose order divide 4n but not 2n and - having $ \small n = 2^m \cdot o, o \text{ odd } $ thus must be exactly $ \small 4 \cdot 2^m r $ where r is any odd divisor of n, while on the lhs we get all primes in the factorization whose order equal any divisor of n. But the sets of primes must be equal to allow a solution for the original problem.

$f(x) \ne g(x)$ but $f(f(x))=g(g(x))$ - is there a name/some discussion of this property?

2011-08-27T05:35:48Z

In the context of iteration of functions I look at the eigenvalues of the associated matrixoperator/Carleman-matrix .

If a function $\small f(x)$ has a negative eigenvalue in its associated carleman-matrix, then the definition of a half-iterate must handle the squareroot of that negative number (and fractional iteration in general). If I create the function $\small g(x)$ by taking the absolute value of that eigenvalue, then $\small f(f(x)) = g(g(x))$ , and moreover $\small f(g(x)) = g(f(x)) $ . (I also assume, that the commutativity makes the solution unique). Then I can do fractional iteration on $ \small g(x) $ and I'm interested in the general relation of that two functions.

Is there a name for that functional relation (for instance "g is the dual of f" or something else so I have a keyword for search) and/or some study online available?

[update 2] There was some discussion related to this, where I also was involved answering: "do complex iterates (..) have any meaning?" Here the change-of-sign of the schröder-value, which means an imaginary iteration "height", resembles in some way the change-of-sign of the eigenvalue in the way I looked at in the current question. However, I thought there might be some wider discussion (and possibly a common keyword) to that specific case and the indicated function with "complementary"(?) eigenvalue here.

[update 1] After the remark of Gerald Edgar it might be instructive to show at least one example. Motivated by some other question regarding the inversion of $ \small \zeta$ I constructed a carleman-matrix for the zeta-function. Recentering at one fixpoint ( $ \small \sim -0.29590...$ this matrix could be made triangular (call it ZT) and the basic eigenvalue $\small \lambda_1 \sim -0.51273 $ and the others the sequence $\small \lambda_1^0,\lambda_1, \lambda_1^2,\lambda_1^3, \ldots $ Then using diagonalization I replaced the set of eigenvalues by their absolute values, which can also be understood as taking the matrix-squareroot of the square of the matrix ZT, call it ZTA. The entries in ZTA give the coefficients for the function $\small g(x)$ ; the radius of convergence is about $ \small -1.3 < x <0.7$ In this range we find that $\zeta(\zeta(x))=g(g(x))$ at least with visible accuracy..
Here is a picture .

The range nearer to $\small 1-$ must be determined by other means; the twofold iteration of the zeta in this area becomes chaotic and can be seen by the wolfram-alpha call

Answer by Gottfried Helms for Negative values of Riemann zeta function on the critical line.

2011-08-18T01:14:13Z

@ Q1: After the counterexample of Noam Elkies I used Pari/GP to draw that parametric plot to get more visual impression;
[update] The visual impression in the 1:1000 zoomed picture had artifacts; I deleted the picture and provide a more precise one and corrected in my original answer [/update]

Plot 1 shows the known curve in the complex plane, when t increases from 0 to 100:

 \\ Pari/GP:
   ri_zeta(t)=local(tmp);tmp=zeta(1/2+I*t);return([real(tmp),imag(tmp)])
   ploth(x=0,100,ri_zeta(x),1)

From the drawing one cannot discern, whether there is some crossing of the negative real axis. Here is a rescaling; the values of the zeta-function are scaled by the tanh-function:

 \\ Pari/GP:
ri_zeta(t)=local(tmp);tmp=10*zeta(1/2+I*t);return([tanh(real(tmp)),tanh(imag(tmp))])
ploth(x=0,100,ri_zeta(x),1)

and then a strong scaling factor of 1:1000 applied. [update] To remove artifacts, there is an option "recursive" in the plot-routine to scatter the coordinates more regularly; the strong zoom separated the dots of the plot too much so that artifacts are likely to occur. With an improvement of the sampling no crossings of the negative real axis can be seen [/update]

 \\ Pari/GP:
ri_zeta(t)=local(tmp);tmp=1000*zeta(1/2+I*t);return([tanh(real(tmp)),tanh(imag(tmp))])

I used internal precision of 200 dec digits, [update] so I think the computation of the single points do not introduce artefacts, but the connection by lines may do due to the strong scaling required. This type of plotting seems to require much resources; I'll see whether it can verify the crossing in the near of t=282 visually; I'll update then this answer again.

Answer by Gottfried Helms for Stirling Number of first kind : Implementation

2011-08-14T05:36:49Z

In Pari/GP; one could simplify for either readability, speed or memory organisation for big matrices:

{ makemat_St1(dim=n) = local(f, M); 
   M=matid(dim);
   f=1;
   for(r=2,dim,   \\ comp diagonal and first column
         M[r,1]=f;f*=(r)
      );
   for(c=2,dim,   \\ compute core entries
       for(r=c+1,dim,
           M[r,c]=M[r-1,c-1]+(r-1)*M[r-1,c]
          )
       );
    f1=1;     \\ apply signs 
    for(r=2,dim,
          f1*=-1;f2=-f1;
          for(c=1,r-1,
                f2*=-1;M[r,c]*=f2
                 )
         );
  return(M) }

A shorter form is this

{makemat_st1(dim=6) = local(m);  \\ give it a default dimension of 6
 m=matrix(dim,dim);
 m[1,1]=1;
 for(r = 2,dim, 
    m[r,1]= 0 - (r-1)*m[r-1,1] ; \\ first column has no up-left neighbour
    for(c = 2,r, 
       m[r,c]= m[r-1,c-1]  -  (r-1)*m[r-1,c]
       );
    );
 return(m);}

Comment by Gottfried Helms

2013-04-20T18:19:53Z

I see; I've made a sign-error, sorry.

Comment by Gottfried Helms

2013-04-20T11:37:03Z

hmm, if I feed the last mentioned function into Pari/GP I get for $s=\epsilon$ the result of $\epsilon^{−1}$ for epsilon near zero

Comment by Gottfried Helms

2013-04-19T22:23:50Z

@i707107: no, I've no proof for the guessed algebaric identity; so far it is only the numerical allusion - however I've got it to 40 digits instead of 15 now...

Comment by Gottfried Helms

2013-04-19T08:44:32Z

@joro: are you sure that are methods for divergent summation? And if, are they strong enough? For instance, using "sumalt" in Pari/GP is not strong enough although it is in general a very useful procedure for divergent summation.

Comment by Gottfried Helms

2013-03-23T18:22:03Z

@Carlo - thank you very much; I'll take a deeper look at it later...

Comment by Gottfried Helms

2013-03-09T06:20:46Z

Ah, so I see, there was possibly a misunderstanding. I have the first 500 Stieltjes constants to 1000 digits by a nice, small procedure. What I was trying to do, was to give a sense to the sum-of-all-Stieltjes numbers, by some summation-procedure. That procedure requires to use the derivatives of the zetas in a certain manner in finite sums as written above. That finite sums, having more and more terms according to the increasing parameter c, approximate something like 0.4990749...xyz , but convergence is slow and needs many of the zeta's derivatives...

Comment by Gottfried Helms

2013-03-08T18:10:28Z

Hi Fredrik, did you mean to apply the Euler-Maclaurin formula directly at the Stieltjes-numbers?

Comment by Gottfried Helms

2012-11-21T01:07:37Z

The first primes, which have no leading sequence of composites is $3,7,13,31,37,61,73,79,97,109,127,157,193,199,223,229,241,271,277,283,$

Comment by Gottfried Helms

2012-11-21T00:43:23Z

very nice... I'd like to see a formulation for the inverse operation (which must have a "choice"-option). I did not yet get it myself correctly myself ...

Comment by Gottfried Helms

2012-11-11T14:11:12Z

@Sankara: Sorry, I had no time for this yesterday. Well, I on't have information about the amount of distortion by the error, which goes to the principal axes, after data is rotated into such a position. Only it is somehow obvious to me, that the relative amount of coordinate-distortion is smaller in the direction of the principal axis, where the squared values of coordinates by U are maximized. (But this is surely not that what your after, sorry)

Comment by Gottfried Helms

2012-11-02T09:29:01Z

Didn't you at least consider to look at wikipedia (en.wikipedia.org/wiki/…)?

Comment by Gottfried Helms

2012-10-16T14:37:45Z

Also, "carets" assume something behind them, so insert something at the position, where I put an X in the last formula: $P=\frac{P+P^}{2}+\frac{P-P^}{2}$ becomes then $P=\frac{P+P^X}{2}+\frac{P-P^X}{2}$

Comment by Gottfried Helms

2012-09-28T10:19:37Z

The remarks on the periods are especially triggering for me, very nice!

Comment by Gottfried Helms

2012-09-24T07:17:10Z

That reversion was incorrect; look at the wikipedia on the subject. This i also somehow obvious: the "greater than 1"-version fixes a very low upper bound on d,e,f. But the catalan-fermat-conjecture is surely concerned with a lower bound on that variables...

Comment by Gottfried Helms

2012-08-26T22:59:10Z

@Fredrik: I've re-computed the 1000-digit version with prec 1600 and uploaded to the given link; the last entry matches now exactly your reference where the last digit is rounded upwards (you have 3 digits more). So I seem to have at least a rough estimate for the lower bound of computational requirements...