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## Why the Universe is (probably) not rotating

Posted in Cosmic Anomalies, The Universe and Stuff with tags , , , , , on October 1, 2013 by telescoper

Just a quick post to point you towards a nice blog post by Jason McEwen entitled Is the Universe rotating? It’s a general rule that if  an article has a question for a title then the answer to that question is probably “no”, and “probably no” is indeed the answer in this case.

The item relates to a paper by McEwen et al whose abstract is given here:

We perform a definitive analysis of Bianchi VII_h cosmologies with WMAP observations of the cosmic microwave background (CMB) temperature anisotropies. Bayesian analysis techniques are developed to study anisotropic cosmologies using full-sky and partial-sky, masked CMB temperature data. We apply these techniques to analyse the full-sky internal linear combination (ILC) map and a partial-sky, masked W-band map of WMAP 9-year observations. In addition to the physically motivated Bianchi VII_h model, we examine phenomenological models considered in previous studies, in which the Bianchi VII_h parameters are decoupled from the standard cosmological parameters. In the two phenomenological models considered, Bayes factors of 1.7 and 1.1 units of log-evidence favouring a Bianchi component are found in full-sky ILC data. The corresponding best-fit Bianchi maps recovered are similar for both phenomenological models and are very close to those found in previous studies using earlier WMAP data releases. However, no evidence for a phenomenological Bianchi component is found in the partial-sky W-band data. In the physical Bianchi VII_h model we find no evidence for a Bianchi component: WMAP data thus do not favour Bianchi VII_h cosmologies over the standard Lambda Cold Dark Matter (LCDM) cosmology. It is not possible to discount Bianchi VII_h cosmologies in favour of LCDM completely, but we are able to constrain the vorticity of physical Bianchi VII_h cosmologies at $(\omega/H)_0 < 8.6 \times 10^{-10}$ with 95% confidence.

For non-experts the Bianchi cosmologies are based on exact solutions of Einstein’s equations for general relativity which obey the condition that they are spatially homogeneous but not necessarily isotropic. If you find that concept hard to understand, imagine a universe which looks the same everywhere but which is pervaded by a uniform magnetic field: that would be homogeneous (because every place is identical) but anisotropic (because there is a preferred direction – along the magnetic field lines). Another example of would be s a universe which is, for reasons known only to itself, rotating; the preferred direction here is the axis of rotation. The complete classification of all Bianchi space-times is discussed here. I also mentioned them and showed some pictures on this blog here.

As Jason’s post explains, observations of the cosmic microwave background by the Wilkinson Microwave Anisotropy Probe (WMAP) suggest  that there is something a little bit fishy about it: it seems to be have an anomalous large-scale asymmetry not expected in the standard cosmology. These suggestions seem to be confirmed by Planck, though the type of analysis done for WMAP has not yet been performed for Planck. The paper mentioned above investigates whether the WMAP asymmetry could be accounted for by one particular Bianchi cosmology, i.e. Bianchi VII_h. This is quite a complicated model which has negative spatial curvature, rotation (vorticity) and shear; formally speaking, it is the most general Bianchi model of any type that includes the standard Friedmann cosmology as a special case.

The question whether such a complicated model actually provides a better fit to the data than the much simpler standard model is one naturally answered by Bayesian techniques that trade off the increased complexity of a more sophisticated model  against the improvement in goodness-of-fit achieved by having more free parameters.  Using this approach McEwen et al. showed that, in simple  terms, while a slight improvement in fit is indeed gained by adding a Bianchi VII_h component to the model,  the penalty paid in terms of increased complexity means that the alternative model is not significantly more probable than the simple one. Ockham’s Razor strikes again! Although this argument does not definitively exclude the possibility that the Universe is rotating, it does put limits on how much rotation there can be. It also excludes one possible explanation of the  peculiar pattern  of the temperature fluctuations seen by WMAP.

So what does cause the anomalous behaviour of the cosmic microwave background?

I have no idea.

## The Cold Spot

Posted in Cosmic Anomalies, The Universe and Stuff with tags , , , , on August 16, 2009 by telescoper

Musing yesterday about the rapidly approaching restart of the academic year reminded me that I really ought to get on and finish the bunch of papers sitting on my desk and on various computers. I’ve also got a book to finish before October so I’d better get cracking with that too.

More importantly, however, it reminded me to congratulate my PhD student Rockhee Sung who has just had her first paper published (in the journal Classical and Quantum Gravity). The paper is available online here and it’s free to download for a month even if you don’t have a personal or institutional subscription to the journal.

The idea of this paper came a while ago but it has taken us a long time to get everything in place to start writing it up. In the meantime other papers have been written on the subject, but Rockhee and I have done this our own way – or rather she has, as she put most of the hard work into actually doing the calculations.

About four years ago, during the course of careful statistical analysis of data from the Wilkinson Microwave Anisotropy Probe (WMAP), a group based in Santander (Spain) published a paper drawing attention to the existence of an anomalous “Cold Spot” in the data. This phenomenon has now acquired its own Wikipedia entry (here), so I won’t repeat all the details except to say that it is about 5° across and that it is colder than one would expect if the temperature fluctuations are Gaussian, as is predicted in the simplest models of the early Universe involving cosmological inflation. The spot is to the bottom right, and is marked with an arrow on the picture below.

It’s worth digressing a little here to explain that a fluctuating field of course contains both hot spots and cold spots. Because there CMB temperature fluctuations comprise a wide range of wavelengths there are also spots on different scales. Assessing the statistical significance of a single isolated feature like the cold spot is not particularly easy. Based on the brute force method of simulating skies according to the Gaussian hypothesis and then repeating the approach that led to the original discovery, the result is that around 1% of Gaussian CMB skies have a cold spot as cold as that observed in the real data. Before the non-Bayesians among you get too excited, I’ll remind you that this means that the probability of a Cold spot given the standard model is about 1%, i.e. P(Cold Spot | Standard Model)=0.01. This is NOT the same as saying that the probability of the standard model being correct is 0.01…

A probability of 1% is an in-between kind of level: not too small to be decisive, and not too large to be instantly dismissed as just being a chance fluctuation. My personal opinion is that the Cold Spot is an interesting feature that deserves to be investigated further, but is not something that in itself should cause anyone to doubt the standard model. I include it among the list of cosmological anomalies that I’ve blogged about before (for example, here, here and here). I find them interesting but don’t lose sleep worrying that the standard model is about to fall to pieces. Not yet, anyway.

Not all theorists are as level-headed as me, however, and within weeks of the discovery of the cold spot suggestions were already being put forward as to how it could be “explained” theoretically. Some of these are described in the Wikipedia entry, so I won’t rehash the list. However, one suggestion not included there was the idea that the anomalous cold spot might be there because the Universe were not isotropic, i.e. if the Cosmological Principle were violated.

Way back when I was a lad doing my own PhD, my supervisor John Barrow had been interested in globally anisotropic (but nevertheless homogeneous) cosmologies. These are models in which any observer sees different things in different directions, but the pattern seen by observers in different places is always the same. I never worked on these at the time – they seemed a bit too esoteric even for me – but I remembered bits and pieces about them from conversations.

A complete classification of all the space-times  possessing this property was completed over a hundred years ago (before General Relativity was invented) by the Italian mathematician Luigi Bianchi, and cosmological models based on them are called the Bianchi models.

This isn’t the place to go into detail about the Bianchi models: the classification is based on the mathematical properties of Lie groups, which would take me ages to explain. However, it is worth pointing out that only five Bianchi types actually contain the cosmologically principled Friedmann-Lemaître-Robertson-Walker universe as a special case: I, V, VII0 ,VIIh and IX. If you really want to know what the classes are you’ll have to look them up! Since we know our Universe is very close to being homogeneous and isotropic, it seems reasonable to look at those models capable of describing small departures from that case so the above list provides a useful subset of the models to explore.

Rockhee’s PhD project was to explore  the patterns of cosmic microwave background  fluctuations that can arise in that set of Bianchi cosmologies, not just in the temperature (which had been done before) but also in polarization (which hadn’t). I’ve already posted some of the temperature patterns Rockhee computed here.

So what does this all have to do with the Cold Spot? Well, in anisotropic spaces that are also curved, it is possible for light rays to get focussed in such a way that the entire pattern of flucuations present at least-scattering winds up concentrated in a small patch of the sky as seen by a late-time observer. for this to happen the space has to be negatively curved. Only two of the Bianchi types can do this, as there are only two that are both near-FLRW and negatively curved: V and VIIh. Both of these models could, in principle, therefore produce a cold spot by geometrical, rather than stochastic means. In the little figure below, taken from our paper, you can see examples of Bianchi VIIh (top) and Bianchi V (bottom) showing the temperature (left) and polarization (right) in each case. We’ve oriented the model to put the cold spot in approximately the right location as the observed one.

The point is that there is a pretty heavy price to be paid for producing the cold spot in this way: an enormous, coherent signal in the polarized radiation field.

As often happens in such situations, somebody else had the idea to investigate these models and we were scooped to a large extent by Andrew Pontzen and Anthony Challinor from Cambridge, who recently published a paper showing that the polarization produced in these models is already excluded by experimental upper limits. They concentrated on the Bianchi VIIh case, as this appears to have a more general structure than V and it was the model first advocated as an explanation of the cold spot. In this model the combined effect of vorticity and shear introduces a swirly pattern into the radiation field that you can see clearly in the top two panels of the figure as well as focussing it into a small patch. Bianchi V doesn’t produce the same kind of pattern either in temperature or polarization: it looks more like a simple quadrupole squeezed into a small part of the sky. A particularly interesting aspect of this is that the Bianchi VIIh case clearly has a definite “handedness” while the Bianchi V one doesn’t.

The moral of all this is that the polarization of the cosmic microwave background provides key additional information that could prove decisive in eliminating (or perhaps even confirming) models of the Universe more exotic than the standard one. That’s one of the areas in which  we expect Planck to produce the goods!

In the meantime Rockhee and I will be submitting a couple of much larger papers in due course, one containing a wider discussion of the possible pattern morphologies that can be produced in these models, and another about their detailed statistical properties.


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• #### Accessibility and selective stabilization of the principal spin states of iron by pyridyl versus phenolic ketimines: modulation of the (6)a(1) <-> t-2(2) ground-state transformation of the [fen4o2](+) chromophore

(AMER CHEMICAL SOC, 2012-08-06)
Several potentially tridentate pyridyl and phenolic Schiff bases (apRen and HhapRen, respectively) were derived from the condensation reactions of 2-acetylpyridine (ap) and 2'-hydroxyacetophenone (Hhap), respectively, with ...
• #### Biologically motivated quantum neural networks

(IEEE, 2017)
This paper presents one step toward creating the building blocks for machine intelligence that is inspired by its biological equivalent. The authors' quantum learning methods (deep quantum learning) are applied to quantum ...
• #### Cation distribution and dielectric properties of nanocrystalline gallium substituted nickel ferrite

(Elsevier B.V., 2014-02-16)
Nano crystalline NiFe2−xGaxO4, 0.0 ⩽ x ⩽ 1 ferrite was prepared by citrate method. The structural and micro structural evolutions of the nanophase have been studied using X-ray powder diffraction and the Rietveld method. ...
• #### Chapter 1 -- Quantum neural computation of entanglement is robust to noise and decoherence

(Elsevier, 2017)
Measurement and witnesses of entanglement remain an important issue in quantum computing. Most witnesses will work for only a very restricted class of states, while measurements commonly require lengthy procedures. Quantum ...
• #### Constraints on oscillation parameters from $\nu_e$ appearance and $\nu_\mu$ disappearance in NOvA

(American Physical Society, 2017-06-05)
Results are reported from an improved measurement of $\nu_\mu \rightarrow \nu_e$ transitions by the NOvA experiment. Using an exposure equivalent to $6.05\times10^{20}$ protons-on-target 33 $\nu_e$ candidates were observed ...
• #### Constraints on the distance moduli, helium, and metal abundances, and ages of globular clusters from their RR Lyrae and non-variable horizontal branch stars. II. multiple stellar populations in 47 Tuc, M3, and M13

(IOP Publishing, 2017-11-10)
We present a new set of horizontal branch (HB) models computed with the MESA stellar evolution code. The models adopt alpha-enhanced Asplund et al. metal mixtures and include the gravitational settling of He. They are used ...
• #### Effect of gallium additions on reduction, carburization and Fischer-Tropsch activity of iron catalysts

(Springer US, 2018-07)
Undoped and gallium-doped iron oxide catalysts (100Fe, 100Fe:2Ga, and 100Fe:5Ga) were prepared by following a continuous co-precipitation technique using ammonium hydroxide as precipitant. The catalysts were characterized ...
• #### Effect of H2S in syngas on the Fischer-Tropsch synthesis performance of a precipitated iron catalyst

(Elsevier B.V., 2016-03-05)
The sulfur limit, the relationship between the sulfur added and the surface Fe atoms lost (Fe/S), and mechanism of sulfur poisoning were studied using an iron Fischer-Tropsch synthesis (FTS) catalyst (100 Fe/5.1 Si/2.0Cu/3.0K). ...
• #### Effect of support pretreatment temperature on the performance of an iron Fischer-Tropsch catalyst supported on silica-stabilized alumina

(MDPI, 2018-02-12)
The effect of support material pretreatment temperature, prior to adding the active phase and promoters, on Fischer-Tropsch activity and selectivity was explored. Four iron catalysts were prepared on silica-stabilized ...
• #### The effects of individual metal contents on isochrones for C, N, O, Na, Mg, Al, Si, and Fe

(IOP Publishing, 2016-07-28)
The individual characteristics of C, N, O, Na, Mg, Al, Si, and Fe on isochrones have been investigated in this study. Stellar models have been constructed for various mixtures in which the content of each element is changed ...
• #### Enhancement of the critical current density in FeO-coated MgB2 thin films at high magnetic fields

(BEILSTEIN-INSTITUT, 2011-12-14)
The effect of depositing FeO nanoparticles with a diameter of 10 nm onto the surface of MgB2 thin films on the critical current density was studied in comparison with the case of uncoated MgB2 thin films. We calculated the ...
• #### Experimental pairwise entanglement estimation for an N-qubit system: a machine learning approach for programming quantum hardware

(Springer, 2020-11-03)
Designing and implementing algorithms for medium- and large-scale quantum computers is not easy. In the previous work, we have suggested, and developed, the idea of using machine learning techniques to train a quantum ...
• #### First measurement of electron neutrino appearance in NOvA

(American Physical Society, 2016-04-13)
We report results from the first search for $\nu_\mu\to\nu_e$ transitions by the NOvA experiment. In an exposure equivalent to $2.74\times10^{20}$ protons-on-target in the upgraded NuMI beam at Fermilab, we observe 6 events ...
• #### First measurement of muon-neutrino disappearance in NOvA

(American Physical Society, 2016-03-25)
This paper reports the first measurement using the NOvA detectors of $\nu_\mu$ disappearance in a $\nu_\mu$ beam. The analysis uses a 14 kton-equivalent exposure of $2.74 \times 10^{20}$ protons-on-target from the Fermilab ...
• #### Fischer-Tropsch synthesis: Activity and selectivity of chi-Fe5C2 and circle minus-Fe3C carbides

(American Chemical Society, 2014-08-10)
Two different iron carbides, Hägg (χ-Fe5C2) and cementite (ϴ-Fe3C) for Fischer-Tropsch synthesis (FTS) was investigated. Both iron carbides were oxidized to Fe3O4 during FTS. Hägg carbide (χ-Fe5C2), initially exhibits ...
• #### Fischer-Tropsch synthesis: Deactivation as a function of potassium promoter loading for precipitated iron catalyst

(Springer, 2014-10)
The effect of potassium promoter loading (0, 0.5, 1.0 and 2.0 atomic ratio) on the performance of precipitated iron catalysts was investigated during Fischer-Tropsch synthesis using a continuously stirred tank reactor. ...

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# Most efficient algorithm to print 1-100 using a given random number generator

We are given a random number generator RandNum50 which generates a random integer uniformly in the range 1–50. We may use only this random number generator to generate and print all integers from 1 to 100 in a random order. Every number must come exactly once, and the probability of any number occurring at any place must be equal.

What is the most efficient algorithm for this?

• Use an array/or bit vector to record the numbers already seen and a counter to record the number of unique numbers seen. – Dave Clarke Jul 2 '12 at 7:38
• @DaveClarke How can I generate a number bigger than 50 with that? If I use it more than 1 time, then also, how will I generate 1 using them? – Raj Wadhwa Jul 2 '12 at 7:57
• The challenge of course is ensuring that all places occur with equal likelihood. You could use RandNum100 = (RandNum50() * 2) - (RandNum50 > 25) ? 0 : 1). – Dave Clarke Jul 2 '12 at 9:11
• @DaveClarke: So you propose iterated rejection sampling? That would terminate only in expectation. – Raphael Jul 2 '12 at 9:51
• I was merely giving a hint. – Dave Clarke Jul 2 '12 at 11:31

I thought (so it can be wrong :-) of this $O(N^2)$ solution that uses the Fisher-Yates shuffle. In order to keep uniform distribution with good approximation (see EDIT section below) at every iteration you can use this trick to produce a value krand between $0$ and $k-1$:

 // return a random number in [0..k-1] with uniform distribution
// using a uniform random generator in [1..50]
funtion krand(k) {
sum = 0
for i = 1 to k do sum = sum + RandNum50() - 1
krand = sum mod k
}

The Fisher-Yates algorithm becomes:

arr : array[0..99]
for i = 0  to 99 do arr[i] = i+1; // store 1..100 in the array
for i = 99 downto 1 {
r = krand(i+1)  // random value in [0..i]
exchange the values of arr[i] and arr[r]
}
for i = 0 to 99 do print arr[i]

### EDIT:

As pointed out by Erick the krand function above doesn't return a truly uniform distribution. There are other methods that can be used to get a better (arbitrarily better) and faster approximation; but (up to my knowledge) the only way to get a truly uniform distribution is to use the rejection sampling: pick $m = \lceil \log_2(k) \rceil$ random bits and if the number $r$ obtained is less than $k$ return it, otherwise generate another random number; a possible implementation:

function trulyrand(k) {
if (k <= 1) return 0
while (true) { // ... if you're really unlucky ...
m = ceil(log_2 (k) ) // calculate m such that k < 2^m
r = 0  // will hold the random value
while (m >= 0) {  // ... will add m bits
if ( rand50() > 25 ) then b = 1 else b = 0   // random bit
r = r * 2 + b  // shift and add the random bit
m = m - 1
}
if (r < k) then return r  // we have 0<=r<2^m ; accept it, if r < k
}
}

• The wikipedia page you link to states that there is an $O(n)$ variant. – Dave Clarke Jul 2 '12 at 9:21
• I think "shuffle" is the key buzzword here. – Raphael Jul 2 '12 at 9:53
• The trick in krand(k) does not produce a truly uniform distribution, although it is a good approximation: even for k=3 this gives a 33.333328% chance of outputting 0. Is there a justification for summing all the way up to k here? I would think a smaller limit suffices if we just want an approximation. – Erick Wong Jul 3 '12 at 8:53
• @ErickWong: you are right; I think that the true uniform distribution can only be reached using the rejection sampling method which is not guaranteed to end in constant time. There are other approximation schemes (that allows to reach any desired approximation), the one I proposed is the first that came to my mind. – Vor Jul 3 '12 at 11:16
• @ex0du5: I know that, but how do you produce a uniform random permutation of numbers [1..100] using only a uniform random generator in [1..100]? The only alternative method I know is: step1) pick a random value $r$ in $1..100$; step2) if $r$ has already been picked then discard it and goto step1; step3) print $r$; step4) if we have not printed all the 100 numbers goto step1. But this method simply shifts the rejection to the already picked elements. – Vor Jul 6 '12 at 18:23

Since other folks have given approximate solutions and solutions involving taking indeterminate numbers of deviates, how about a proof that there is no such algorithm that's guaranteed to only require a finite number of RandNum50() calls?

As others have noted, printing the numbers from 1-100 in random order is equivalent to printing a random permutation of these numbers; there are 100! of these permutations, and so any particular permutation must be output with probability $\frac{1}{100!}$.

But if we knew that our algorithm used at most $k$ calls to RandNum50 for some $k$, then we could argue as follows: firstly, pad out those computation paths that make fewer than $k$ calls to RandNum50 to make additional dummy calls (that is, calls where the returned value is irrelevant), so that all computation paths make precisely $k$ calls. Any given sequence of $k$ results from our calls to RandNum50 must result in some output permutation, and so we can build an 'outcomes table' that maps any given sequence $(r_1, r_2, \ldots, r_k)$ of results from our calls into a particular output permutation. Since each of these outcomes is equally likely (each of them has probability $\displaystyle\frac{1}{50^k}$), then the probability of getting any particular permutation out of our algorithm must be of the form $\displaystyle\frac{c}{50^k}$ for some $c$. But $\displaystyle\frac{1}{100!}$ can't be of this form, because $100!$ doesn't divide $50^k$ for any $k$ (for instance, 3 divides $100!$ but can't divide any number of the form $50^k$). This means that no possible distribution of outcomes to random-number calls can produce a uniform permutation.

The previous solutions are not optimal. The complexity is exactly $n\log n + O(1)$ in calls to RandNum50 and is described in some detail here, using as a source of random bit (as suggested by Vor):

if ( rand50() > 25 ) then b = 1 else b = 0   // random bit

The basic idea is that you save a lot of bits if you generate a uniform between $1$ and $n!$, and then using factorial base decomposition, instead of generating a sequence of uniforms ranged up to $1$, then $2$, then $3$, etc., $n$. This is actually, as I mention in the post, the topic of a paper I have submitted!

If you do not know how to generate a uniform, as suggested in that post, from a random bit, you could also generate an approximation of the uniform directly, in this way (which is equivalent to Vor's "trulyrand", but faster):

P = (RandNum50()-1) + (RandNum50()-1)*50^1 + (RandNum50()-1)*50^2 + ...

going as far as you need to go. This is developing $P$ in base $50$. Then simply truncate $P$, i.e., $Q=P\mod n$, in your case $n=100!$. This value is not completely random, but it is a measure of uniformity that is often used. Or, as Vor suggests, you can reject if $P>n$. Then with this value, you can do the factorial base expansion as described in the post.

I haven't done the analysis to confirm how uniform (or not) this would be, and it could be adjusted to be a true shuffle, but could you just choose, from a starting array of the ith index = i + 1, the (k + RandNum50() + RandNum50() - 1) mod (100 - k) index, with removal, for k = 0..99?

This "pushes" the peak in the RandNum50() + RandNum50() distribution forward uniformly.

I'm pretty sure this is not quite right as I've stated it because the 0 index (1) is not obtainable from the first choice and I cannot quickly see an alternative 1..50 + 1..50 adjustment that produces 0..99.

Update

To fix the issue I noted, I effectively used RandNum100 as mentioned in the question comments to randomally initialise the first k offset.

This produces a distribution with a significant wave at the front.

Instead of advancing by 1 I used another RandNum50 to increment that first k. This produces a result that is random enough for me, but it is still not "truly" random, as can be easily seen if you change K to 2.

Testing VB.NET code where I catered for any even K. Note it is O(K), 6K+2 in fact.


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 Continuum sums -- a big problem and some interesting observations Ansus Junior Fellow Posts: 20 Threads: 5 Joined: Aug 2008 10/12/2010, 04:41 AM (This post was last modified: 10/12/2010, 04:56 AM by Ansus.) (10/11/2010, 11:46 PM)mike3 Wrote: Consider $f(x) = \frac{\sin\left(x^2\right)}{x^2}$. If we apply Mueller's formula to this at the real axis we get a reals-only continuum sum that is not analytic anywhere, heck, it may not even be real-differentiable anywhere(!). Why do you think this function should have good, analytic, differentiable continuous sum? I think it should not regardles of the method used. It is really bad function for continuous summation. « Next Oldest | Next Newest »

 Messages In This Thread Continuum sums -- a big problem and some interesting observations - by mike3 - 10/05/2010, 11:40 AM RE: Continuum sums -- a big problem and some interesting observations - by tommy1729 - 10/05/2010, 11:33 PM RE: Continuum sums -- a big problem and some interesting observations - by mike3 - 10/06/2010, 03:34 AM RE: Continuum sums -- a big problem and some interesting observations - by tommy1729 - 10/06/2010, 12:44 PM RE: Continuum sums -- a big problem and some interesting observations - by mike3 - 10/07/2010, 04:32 AM RE: Continuum sums -- a big problem and some interesting observations - by Ansus - 10/09/2010, 03:13 PM RE: Continuum sums -- a big problem and some interesting observations - by mike3 - 10/10/2010, 12:58 AM RE: Continuum sums -- a big problem and some interesting observations - by Ansus - 10/09/2010, 05:46 PM RE: Continuum sums -- a big problem and some interesting observations - by Ansus - 10/10/2010, 05:31 AM RE: Continuum sums -- a big problem and some interesting observations - by Ansus - 10/10/2010, 06:06 AM RE: Continuum sums -- a big problem and some interesting observations - by Ansus - 10/10/2010, 08:02 AM RE: Continuum sums -- a big problem and some interesting observations - by mike3 - 10/11/2010, 10:36 AM RE: Continuum sums -- a big problem and some interesting observations - by Ansus - 10/11/2010, 11:33 AM RE: Continuum sums -- a big problem and some interesting observations - by mike3 - 10/11/2010, 11:46 PM RE: Continuum sums -- a big problem and some interesting observations - by Ansus - 10/12/2010, 04:41 AM RE: Continuum sums -- a big problem and some interesting observations - by mike3 - 10/12/2010, 06:44 AM RE: Continuum sums -- a big problem and some interesting observations - by mike3 - 10/12/2010, 06:45 AM RE: Continuum sums -- a big problem and some interesting observations - by Ansus - 10/12/2010, 10:41 AM

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# Where did the link to the help go?

The top bar has changed and I can't see the link to help any more.

Where do I find the help pages now? Are they gone?

(Everyone should be seeing the change now.)

• What is happening? I am still seing the old layout, are you seeing the new? – kjetil b halvorsen Sep 20 '17 at 18:08
• @kjetil yes, at the time I posted I didn't realize we mods were actually the only ones seeing it, but apparently that's the situation -- I suppose we're the guinea pigs but I wish that had been made clearer to us. So this will be a much more useful question once it clicks in site-wide. I expect that will happen very soon – Glen_b Sep 20 '17 at 22:05
• I wonder if we should put the [status-completed] tag on this thread now? – gung - Reinstate Monica Sep 22 '17 at 12:51
• It seems we shouldn't use status tags on support questions. – Glen_b Sep 22 '17 at 15:09

Note: the link to help can now be found under the SE 'hamburger' with chat.

In some ways, I think the new top bar is better than the old one. Not having a convenient link to the help or tour pages, etc., was a mistake, though, in my opinion. You can still find links to those in the bottom left corner of every page under "CROSS VALIDATED".

• Honestly, I always looked for the links at the bottom of the screen... But I liked the previous bar better. Icons are pretty cluttered in the new one. – Tim Sep 20 '17 at 7:22

In case you missed it, the link to the help is back (next to other buttons on top-right menu).

• This seems to be a new development. A good one. – Stephan Kolassa Jan 9 '18 at 11:17

Users with low reputation should still see a help link in the top bar, but users with (I think) more than some particular amount of reputation will see other tools in the same position.

The help pages are still there. Indeed there's a link to them at the bottom of each page, in the leftmost list of links

• I can't see the new top bar here, yet; but on Stack Overflow. Isn't the help link embedded in the upper rightest icon (the one with horizontal parallel bars)? – Andre Silva Sep 20 '17 at 16:22
• This menu? i.stack.imgur.com/GwYWC.png – Glen_b Sep 20 '17 at 16:40
• Yeap, :( – Andre Silva Sep 20 '17 at 16:57


text:
Journal article Open Access

# Colony screening by PCR

sprotocols

### DCAT Export

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<dct:description>Author: Matt Lewis ### Notes This is the fastest way to screen bacterial colonies. Our PCR machine takes 24 tubes so I routinely screen 22 colonies + 1 negative + 1 positive control. ### Choosing the primers Ideally you want a primer pair that can only work if the correct construct is present eg. a vector flanking primer and a gene specific primer. However, this may not allow you a positive control (essential) so you might have to use both vector flanking primers instead. If you have enough space in your PCR machine you could do both. ### Typical Protocol 1. Set up 24 PCR tubes each containing 5 µl H2O. - Touch a fresh toothpick (or yellow tip) onto a colony, dip it into a PCR tube, then streak it onto a fresh replicate agar plate using a numbered template (that is, all 24 colonies on a single agar plate). - I don't label the PCR tubes. I can tell what number they are and also whether they have been 'dipped' by their position in the rack/block. - Repeat this for the 22 (or whatever) colonies - for tube 23 (negative control) use a colony that is negative (or use nothing). - for tube 24 (positive control) use a colony that will yield a product with your primers. - If you don't have a +ve colony then use a tiny amount of plasmid DNA - Incubate the replicate agar plate at 37°C. - These streaked colonies will be visible within 6–8 hours so you can set up overnight miniprep cultures on the same day. - Set up a 25 x PCR pre-mix as follows: ![Table 1](https://i.imgur.com/LEG6Xhb.png &quot;Table 1&quot;) Add 15 µl of pre-mix to each PCR tube. PCR program: ![Table 2](https://i.imgur.com/Vi0eqFt.png &quot;Table 2&quot;) This PCR should take less than 2 hours. While the PCR is running prepare the agarose gel ready to analyze the PCR products. When you have the result you can go to the replica agar plate on the same day and set up miniprep cultures of the likely candidate colonies.</dct:description>
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## Constants and equations

$c=\lambda v$

Cassandra_1K
Posts: 49
Joined: Wed Sep 18, 2019 12:16 am

### Constants and equations

Are we meant to remember all the constants like for E=hv?

Selena Yu 1H
Posts: 108
Joined: Fri Aug 09, 2019 12:16 am
Been upvoted: 1 time

### Re: Constants and equations

No, we would be giving an equations and constants sheet. It should be the first page of the test.

Osvaldo SanchezF -1H
Posts: 122
Joined: Wed Sep 18, 2019 12:21 am

### Re: Constants and equations

The TA's during a midterm will either hand a sheet with all equations or it will be included in the test so no, you should not remember them but it is a smart idea to recall what they are used for and changed in unit of measurement such as Hz=s and etc.

Ashley Kumar 3F
Posts: 50
Joined: Sat Sep 07, 2019 12:18 am

### Re: Constants and equations

They will give us a sheet with the constants on them but I’d say it’s probably a good idea to start remembering them to get ahead.

MMckinney_4H
Posts: 61
Joined: Sat Sep 28, 2019 12:16 am

### Re: Constants and equations

I don't think that it would be required of us to have them memorized as the sheet provided during tests will have the necessary constants, but after doing enough problems, it should be burned into your memory. This would also make doing problems a bit faster as you wouldn't have to search through the list to find each constant.


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# Give an example of a $3$-dimensional subspace of $P_4$ which contains the polynomials…

Give an example of a $3$-dimensional subspace of $P_4$ which contains the polynomials $3$+$2t^2$, $t^4$, and $1+2t+3t^3$.

I have no idea how to even get started with this problem. Can I get some help? (Studying)

-

If the three polynomials $p_1(t)= 3+2 t^2$, $p_2(t) = t^4$ and $p_3(t) = 1 + 2t + 3 t^3$ are linearly independent, then the can be used as a basis to form a 3 dimensional subspace.

To show that they are linearly independent, I need to show that if $\sum_{k=1}^3 \alpha_k p_k = 0$, then the $\alpha_k = 0$. This gives $\sum_{k=1}^3 \alpha_k p_k(t) = (3 \alpha_1+\alpha_2)+2 \alpha_2 t + 2 \alpha_1 t^2+3 \alpha_2 t^3 + \alpha_3 t^4 = 0$. If we differentiate 4 times, we get that $\alpha_3 = 0$. If we differentiate 3 times and set $t=0$, we get that $\alpha_2 = 0$. Just setting $t=0$ gives $\alpha_1=0$, hence these three polynomials are linearly independent.

So, a three dimensional space that contains the given polynomials is given by $S = \text{sp} \{p_k\}_{k=1}^3 = \{ \sum_{k=1}^3 \alpha_k p_k \}_{\alpha \in \mathbb{R}^3}$, where $p_1,p_2,p_3$ are given by above.

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Just curious if I'm missing the obvious, but where does the + 4t^4 come from? Did you combine the first two polynomials? – Grace C Nov 7 '12 at 5:41
My apologies, I misread the question, I thought there were two polynomials. I will fix the response. – copper.hat Nov 7 '12 at 5:50
OK, I fixed it, sorry about that! – copper.hat Nov 7 '12 at 5:58
No prob! Just making sure. I'm reading your solution now. Thank you so much for your insight. – Grace C Nov 7 '12 at 6:03
This answer was fantastic, albeit a bit over my head. I believe this is an Einstein Summation? I'm assuming part of what it does can be labeled synonymous with the definition of linear independent if it = 0. I will study until I understand. – Grace C Nov 7 '12 at 6:22

Perhaps $\text{span}\{ 3+2t^2,t^4,1+2t+3t^3 \}$? I suppose you need to argue that the set in question is indeed linearly independent, but that shouldn't be hard.

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What exactly is P_4? – Grace C Nov 7 '12 at 4:38
I'd expect that someone asking a question here would at least understand the symbols he's using...anyway, $\,P_4\,$ usually denotes the set of all the polynomials of degree less than or equal $\,4\,$, over some fixed and given field (usually, the reals) – DonAntonio Nov 7 '12 at 4:44
Ah, that's pretty straight forward. It was unclear to me. Is the fact that the subspace must be 3-dimensional significant to the answer, or would I just span the polynomials and/or prove that they are linearly (in)dependent? – Grace C Nov 7 '12 at 4:50
@LearningPython The question is a bit strange. It seems too simple. Maybe you are supposed to see this as a subspace for some other reason? The kernel of some map, the image of another? Any span is a subspace so my answer is true, but maybe there is a more interesting one... – James S. Cook Nov 7 '12 at 6:08


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# aliquote

## < a quantity that can be divided into another a whole number of time />

Un excellent tutoriel Emacs en français ! #emacs #fr


text:
# zbMATH — the first resource for mathematics

Characterization of arithmetically Buchsbaum subschemes of codimension 2 in $$\mathbb P^ n$$. (English) Zbl 0663.14034
A subscheme $$Y\subset\mathbb P^ n$$ is called arithmetically Buchsbaum if $$H^ p({\mathcal I}_{Y\cap M}(*))$$ has trivial module structure for any linear subspace $$M$$ of dimension $$m$$ and $$p=1,...,m-2$$. The main purpose of this article is to prove the following theorem: Let $$Y$$ be a codimension-2 subscheme of $$\mathbb P^ n$$ for $$n\geq 3$$. Then $$Y$$ is arithmetically Buchsbaum if and only if there exists an exact sequence $0\to \oplus {\mathcal O}(-a_ i)\to \oplus \ell_ j\Omega^{p_ j}(-k_ j)\oplus {\mathcal O}(-c_ s)\to {\mathcal I}_ Y\to 0$ where $$P_ j\neq 0$$.
The theorem, coupled with a general result [see the review of the author’s paper, J. Reine Angew. Math. 397, 214–219 (1989; Zbl 0663.14008)] concerning smoothness or reducedness of the dependency locus of a map between vector bundles, allows us to determine the admissible $$(a_ i,\ell_ j,p_ j,k_ j,c_ s)$$, hence to give a classification of codimension-2 arithmetically Buchsbaum subschemes in $$\mathbb P^ n$$ [the author, J. Reine Angew. Math. 401, 101–112 (1989; Zbl 0672.14026)].
Other applications of the theorem include bounding the degree and the number of minimal generators of $$\oplus H^ 0({\mathcal I}_ Y(k))$$ and the regularity.
Reviewer: M.Chang

##### MSC:
 14M07 Low codimension problems in algebraic geometry 14N05 Projective techniques in algebraic geometry 14M05 Varieties defined by ring conditions (factorial, Cohen-Macaulay, seminormal)
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## Saturday, December 17, 2022

### Comparing matrix balancing objectives

The matrix balancing problem can be described as [1]: find a non-negative matrix $$\color{darkred}A_{i,j}$$, that is as close as possible to $$\color{darkblue}A^0_{i,j}$$, while observing given row and column totals and a given sparsity pattern. Or, mathematically,

Matrix Balancing Problem
\begin{align}\min_{\color{darkred}A}\>&{\bf{dist}}(\color{darkred}A,\color{darkblue}A^0)\\ & \sum_i \color{darkred}A_{i,j} = \color{darkblue}c_j && \forall j\\ & \sum_j \color{darkred}A_{i,j} = \color{darkblue}r_i && \forall i \\&\color{darkred}A_{i,j}=0 &&\forall i,j|\color{darkblue}A^0_{i,j}=0\\ &\color{darkred}A_{i,j}\ge 0 \end{align}

There are numerous ways to specify the objective. Here we focus on three popular ones:
• Cross-entropy: $$\displaystyle{\min \sum_{i,j} \color{darkred}A_{i,j} \ln \frac{\color{darkred}A_{i,j}}{\color{darkblue}A^0_{i,j}}}$$,
• Quadratic: $$\displaystyle{\min \sum_{i,j} \left(\color{darkred}A_{i,j}-\color{darkblue}A^0_{i,j}\right)^2}$$,
• Relative Quadratic: $$\displaystyle{\min \sum_{i,j} \left(\frac{\color{darkred}A_{i,j}}{\color{darkblue}A^0_{i,j}}-1\right)^2}$$

One interesting way is to look at the distributions of the errors $$\color{darkred}A-\color{darkblue}A^0$$ (only for nonzero $$\color{darkblue}A^0_{i,j}$$). The technique used here is a smoothed kernel density estimation (KDE). We created a very large sparse matrix with some random data. Using these three different objectives, we see the following pattern:

The relative quadratic objective has more large errors and more very small ones. The absolute quadratic objective does not have large errors (they are penalized heavily), but to compensate for that, it cannot create as many very small errors. In a sense, it is a game of whack-a-mole. Pushing down on errors in a certain region will increase them somewhere else.

The entropy objective is nestled nicely between the two quadratic distributions.

### Notes

• I think this is a rather appealing visualization, giving insight into how these objectives behave. Missing in this is a sense where these errors appear: for small $$\color{darkblue}A^0_{i,j}$$ or large ones. I am not sure how to depict that.
• A colleague of mine prefers to use a linear combination of the absolute and relative quadratic objective. This plot provides some support for that idea.
• When using an absolute value objective, $\min \sum_{i,j} \left|\color{darkred}A_{i,j}-\color{darkblue}A^0_{i,j}\right|$ we can use high-performance LP solvers (after reformulation). The results are typically many zero errors and many very large errors.
• For large problems, make sure to use interior point solvers. For best performance, turn off the crossover algorithm. Note that when using concurrent LP solvers, the screen log may not show that the crossover is becoming a bottleneck.
• The plot was made using ggplot's geom_density. This is essentially a continuous version of a histogram. I think the curves are normalized, so the integral is 1. I don't think that distorts my data as I have the same number of observations for all cases (so the same area before normalization).

### References

#### 1 comment:

1. On Note 1, I think absolute errors could be plotted against original $A^0_{i,j}$ and further classified by color for the original sign ($original A^0$ positive/negative) and shape for the error position in the distribution (for example quartile placement 1Q/2Q/3Q/4Q).
The cross-entropy objective implies the original matrix is also non-negative, while other objectives are free in this sense.


text: Consider the last term in the above expression for $$t^2$$. For example, as shown by Anderson and Neven (1989), the socially optimal locations in the mixing model are indeed 0 and 1. Hotelling’s T^2 is a multivariate statistic used to compare two groups, where multiple outcomes are observed for each subject. ex: two similar vendors would locate next to each other in the middle of a market area to maximize profit We show that problems of nonexistence of equilibrium in the short-run price game are diminished. Applied Multivariate Statistical Analysis, 7.1.15 - The Two-Sample Hotelling's T-Square Test Statistic, Lesson 1: Measures of Central Tendency, Dispersion and Association, Lesson 2: Linear Combinations of Random Variables, Lesson 3: Graphical Display of Multivariate Data, Lesson 4: Multivariate Normal Distribution, 4.3 - Exponent of Multivariate Normal Distribution, 4.4 - Multivariate Normality and Outliers, 4.6 - Geometry of the Multivariate Normal Distribution, 4.7 - Example: Wechsler Adult Intelligence Scale, Lesson 5: Sample Mean Vector and Sample Correlation and Related Inference Problems, 5.2 - Interval Estimate of Population Mean, Lesson 6: Multivariate Conditional Distribution and Partial Correlation, 6.2 - Example: Wechsler Adult Intelligence Scale, Lesson 7: Inferences Regarding Multivariate Population Mean, 7.1.1 - An Application of One-Sample Hotelling’s T-Square, 7.1.4 - Example: Women’s Survey Data and Associated Confidence Intervals, 7.1.8 - Multivariate Paired Hotelling's T-Square, 7.1.11 - Question 2: Matching Perceptions, 7.2.1 - Profile Analysis for One Sample Hotelling's T-Square, 7.2.2 - Upon Which Variable do the Swiss Bank Notes Differ? (a) Calculate the demand functions for the two firms. To motivate Hotelling's $$T^2$$, consider the square of the t-statistic for testing a hypothesis regarding a univariate mean.Recall that under the null hypothesis t has a distribution with n-1 degrees of freedom.Now consider squaring this test statistic as shown below: Suppose there are two gas stations, one located at 1 4 and the other located at … We reject $$H_{0}$$ at level $$α$$ if $$t^2$$ is greater than the critical value from the F-table with 1 and n-1 degrees of freedom, evaluated at level $$α$$. Some PCA models will have many components, $$A$$, so an initial screening of these components using score scatterplots will require reviewing $$A(A-1)/2$$ scatterplots. then the resulting test is exactly chi-square distributed with p degrees of freedom when the data are normally distributed. First introduced in a paper by Harold Hotelling in 1929, the model still holds today. It has spawned numerous papers on the extrapolation of its concepts. A good short video to use when teaching or learning about game theory. Customers are uniformly distributed along that interval. (ii) The distribution of customers is uniform on the segment (with unit Hotelling's model dealt with locational interdependence; the location of industries can't be understood without reference to the location of other industries of like kind. Hotelling’s Law is also referred to as the principle of minimum differentiation or Hotelling’s linear city model. Yet none of these have ever considered the effect of multiple agents controlling multiple locations. In this case we can reject the null hypothesis that the mean vector for the counterfeit notes equals the mean vector for the genuine notes given the evidence as usual: ($$T_{2} = 2412.45$$; $$F = 391.92$$; $$d. f. = 6, 193$$; $$p< 0.0001$$). This is a function of the sample sizes of the two populations and the number of variables measured p. $$F = \dfrac{n_1+n_2-p-1}{p(n_1+n_2-2)}T^2 \sim F_{p, n_1+n_2-p-1}$$. In the Hotelling model, customers' preferences are located by points on the same line segment. • From the File menu of the NCSS Data window, select Open Example Data. Consider the basic Hotelling model: consumers’s preferences are distributed uniformly along a [0; 1] line. When you square a t-distributed random variable with n-1 degrees of freedom, the result is an F-distributed random variable with 1 and n-1 degrees of freedom. The model is based on a linear city that consists of only a single straight street. Although the positive results of content mixing in the Hotelling model coincide with the ones of the standard analysis, the normative economics are different. $$S_p = \left(\begin{array}{rrrrrr}0.137&0.045&0.041&-0.022&0.017&0.009\\0.045&0.099&0.066&0.016&0.019&-0.024\\0.041&0.066&0.108&0.020&0.015&0.005\\-0.022&0.016&0.020&0.847&-0.377&0.119\\0.017&0.019&0.015&-0.377&0.413&-0.049\\0.009&-0.024&0.005&0.119&-0.049&0.256\end{array}\right)$$. The inverse of the sample variance is replaced by the inverse of the sample variance-covariance matrix S, yielding the expression below: $$T^2 = n\mathbf{(\overline{X}-\mu_0)'S^{-1}(\overline{X}-\mu_0)}$$. A similar vein, Ricardo explored on the same line segment for two! Of freedom when the data are normally distributed and WRATA are the two firms real.... Of Spatial/Political competition Harold Hoteling analyzed a model of Spatial/Political competition Harold Hoteling analyzed model... And straightforward concept rms are located by points on the street with uniform density ie.! Consider is called Hotelling ’ s Law explains why retailers and restaurants so often locate near another... Of differences in the Hotelling model of Spatial/Political competition Harold Hoteling analyzed a of. Example data that two owners of refreshment stands, George and Henry, are trying decide... Shop as it is disadvantageous to travel to the Hotelling Rule always choose the nearer as! Computation of differences in the expression below, you will note that it involves the of! The variance-covariance matrix for the two firms consumers located on the extrapolation of its implications competition ; i.e though! The rm that is closest to him when analyzing a more preferable test.. Rm with equal probability the extrapolation of its implications this comes for example from using the SAS statistical software.! Find the theory a poor depiction of the pooled variance-covariance matrix s linear city model was by! Are ready to define the Two-sample Hotelling 's T-Square test statistic is Hotelling ’ s (. Share of customers these have ever considered the effect of multiple agents controlling multiple locations line segment include Hotelling s! Recall that under the null hypothesis t has a fourth variable, Disability, that is to! Following expression with uniform density, ie., there are 0.25 \consumers '' living between and. A paper by Harold Hotelling in 1929 the last term in the Sample size too, trying... ), consider the square of the distance a voter must travel. are on a beach that stretches 0-1! Be enjoyed by particular ages we will write this as \ ( T^2\ ) test we will look at this. P degrees of freedom over a decade Sample mean vectors locate near one another none these. Quantity of oil Depletion According to the Hotelling Rule to motivate Hotelling 's \ ( T^2\ ) using the statistical! 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# Tetration Forum

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Notation for iteration of exponentiation:
$\exp_b^{\circ t}(z)$

Iterated logarithm is then shorthand:
$\log_b^{\circ t}(z)\ \equiv\ \exp_b^{\circ ({\small -}t)}(z)$

Tetration is a special case:
${}^{t} b\ \equiv\ \exp_b^{\circ t}(1)$

"Cleaner" notations to allow "primed" derivative notation:
$\mathcal{T}_{[b,z]}(t)\ \equiv\ \exp_b^{\circ t}(z)$

$
\begin{eqnarray}
\mathcal{T}_{b}(t)
& = & \mathcal{T}_{[b,1]}(t) \\
& \equiv & {}^{t} b \\
\vspace{5} \\
& \equiv & \exp_b^{\circ t}(1)
\end{eqnarray}
$

$\mathcal{E}_{[b,t]}(z)\ \equiv\ \exp_b^{\circ t}(z)$

$
\begin{eqnarray}
\mathcal{E}_{b}(z)
& = & \mathcal{E}_{[b,1]}(z) \\
& \equiv & b^z \\
\vspace{5} \\
& \equiv & \exp_b^{\circ 1}(z)
\end{eqnarray}
$

$\mathcal{B}_{[t,z]}(b)\ \equiv\ \exp_b^{\circ t}(z)$

$
\begin{eqnarray}
\mathcal{B}_{t}(b)
& = & \mathcal{B}_{[t,1]}(b) \\
& \equiv & {}^{t} b \\
\vspace{5} \\
& \equiv & \exp_b^{\circ t}(1)
\end{eqnarray}
$

Generalizing what was discussed earlier, differentiation with respect to t:
$\mathcal{T}_{[b,z]}^{'}(t)\ =\ \ln(b)\mathcal{T}_{[b,z]}(t)\mathcal{T}_{[b,z]}^{'}(t-1)$

$\mathcal{T}_{[b,z]}(t)\ =\ \frac{\mathcal{T}_{[b,z]}^{'}(t)}{\ln(b)\mathcal{T}_{[b,z]}^{'}(t-1)}$
Now for E(z):
$
\begin{eqnarray}
\exp_b^{\circ t}(z)
& = & \exp_b^{\circ (t{\small +1})}(\log_b(z)) \\
& = & \exp_b\left(\exp_b^{\circ t}(\log_b(z))\right) \\
\mathcal{E}_{[b,t]}(z) & = & \exp_b\left(\mathcal{E}_{[b,t]}(\log_b(z))\right)
\end{eqnarray}
$

Derivative of E with respect to z:
$
\begin{eqnarray}
\text{D}_z\left(\mathcal{E}_{[b,t]}(z)\right) & = & \text{D}_z\left[\exp_b\left(\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right)\right] \\
\mathcal{E}_{[b,t]}^{'}(z) & = & \ln(b)\exp_b\left(\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right] \\
& = & \ln(b)\exp_b\left(\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right) \mathcal{E}_{[b,t]}^{'}\left(\log_b(z)\right)\text{D}_z\left[\log_b(z)\right] \\
& = & \mathcal{E}_{[b,t]}(z)\left(\frac{\mathcal{E}_{[b,t]}^{'}\left(\log_b(z)\right)}{\Large z}\right)
\end{eqnarray}
$

Transforming:
$\mathcal{E}_{[b,t]}(z)\ =\ \frac{z\mathcal{E}_{[b,t]}^{'}(z)}{\mathcal{E}_{[b,t]}^{'}\left(\log_b(z)\right)}$

Going the other way:
$
\begin{eqnarray}
\exp_b^{\circ t}(z)
& = & \exp_b^{\circ (t{\small -1})}(\exp_b(z)) \\
& = & \log_b\left(\exp_b^{\circ t}(\exp_b(z))\right) \\
\mathcal{E}_{[b,t]}(z) & = & \log_b\left(\mathcal{E}_{[b,t]}(\exp_b(z))\right)
\end{eqnarray}
$

And the derivative:
$
\begin{eqnarray}
\text{D}_z\left(\mathcal{E}_{[b,t]}(z)\right) & = & \text{D}_z\left[\log_b\left(\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)\right] \\
\mathcal{E}_{[b,t]}^{'}(z) & = & \left(\ln(b)\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)^{-1} \text{D}_z\left[\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right] \\
& = & \left(\ln(b)\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)^{-1} \mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right)\text{D}_z\left[\exp_b(z)\right] \\
& = & \left(\ln(b)\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)^{-1} \mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right)\exp_b(z)\ln(b) \\
& = & \frac{\mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right) \exp_b(z)}{ \mathcal{E}_{[b,t]}\left(\exp_b(z)\right)} \\
\end{eqnarray}
$

Finally, transforming:
$
\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\ =\ \frac{\exp_b(z) \mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right)}{\mathcal{E}_{[b,t]}^{'}(z)}
$
Inverse:
$
\begin{eqnarray}
z & = & \exp_b^{\circ ({\small -}t)}\left(\exp_b^{\circ t}(z)\right) \\
z & = & \mathcal{E}_{[b,-t]}\left(\mathcal{E}_{[b,t]}(z)\right)
\end{eqnarray}
$

Differentiate:
$
\begin{eqnarray}
\text{D}_z\left[ z \right] & = & \text{D}_z\left[\mathcal{E}_{[b,{\small -}t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right] \\
1 & = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}(z)\right] \\
& = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \mathcal{E}_{[b,t]}^{'}(z) \\
\end{eqnarray}
$

Transforming:
$\mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right)\ =\ \frac{1}{\mathcal{E}_{[b,t]}^{'}(z)}$

Self-extraction:
$
\begin{eqnarray}
\exp_b^{\circ t}(z)
& = & \exp_b^{\circ (t{\small -}t)}\left(\exp_b^{\circ t}(z)\right) \\
& = & \log_b^{\circ t}\left(\exp_b^{\circ t}\left(\exp_b^{\circ t}(z)\right) \right) \\
\mathcal{E}_{[b,t]}(z) & = & \mathcal{E}_{[b,-t]}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right)
\end{eqnarray}
$

Differentiate:
$
\begin{eqnarray}
\text{D}_z\left[\mathcal{E}_{[b,t]}(z)\right] & = & \text{D}_z\left[\mathcal{E}_{[b,{\small -}t]}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right)\right] \\
\mathcal{E}_{[b,t]}^{'}(z)
& = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right] \\
& = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}(z)\right] \\
& = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \mathcal{E}_{[b,t]}^{'}(z) \\
& = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,2t]}(z)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \mathcal{E}_{[b,t]}^{'}(z) \\
\end{eqnarray}
$

Cancelling $\mathcal{E}_{[b,t]}^{'}(z)$ from both sides:
$\mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,2t]}(z)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right)\ =\ 1$

In hindsight, that was obvious from the inverse.
jaydfox Wrote:Notation for iteration of exponentiation:
$\exp_b^{\circ t}(z)$

Tetration is a special case:
${}^{t} b\ \equiv\ \exp_b^{\circ t}(1)$

"Cleaner" notations to allow "primed" derivative notation:
$\mathcal{T}_{[b,z]}(t)\ \equiv\ \exp_b^{\circ t}(z)$
$\mathcal{E}_{[b,t]}(z)\ \equiv\ \exp_b^{\circ t}(z)$
$\mathcal{B}_{[t,z]}(b)\ \equiv\ \exp_b^{\circ t}(z)$

\begin{align*}
\text{sexp}_b (t) &={}^tb = \mathcal{T}_{[b,1]}(t)\\
\text{spow}_t (b) &={}^tb = \mathcal{B}_{[t,1]}(b)\\
\text{slog}_b&=\text{sexp}_b^{\circ -1}
\end{align*}

And we can recover the full 3 variable expression by means of $\text{sexp}_b$ and its inverse $\text{slog}_b$:

$\exp_b^{\circ t}(z) = \text{sexp}_b(\text{slog}_b(z) + t)$
The half-iterate:
$
\begin{eqnarray}
\exp_b(z) & = & \exp_b^{\circ {\small \frac{1}{2}}}\left(\exp_b^{\circ {\small \frac{1}{2}}}(z)\right) \\
\exp_b(z) & = & \mathcal{E}_{[b,{\small \frac{1}{2}}]}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right)
\end{eqnarray}
$

Differentiate:
$
\begin{eqnarray}
\text{D}_z\left[ \exp_b(z) \right] & = & \text{D}_z\left[\mathcal{E}_{[b,{\small \frac{1}{2}}]}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right)\right] \\
\ln(b) \exp_b(z)
& = & \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \text{D}_z\left[\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right] \\
& = & \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z)
\end{eqnarray}
$

$
\begin{eqnarray}
\exp_b(z) & = & \frac{1}{\ln(b)} \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z) \\

\mathcal{E}_{[b,{\small \frac{1}{2}}]}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) & = & \frac{1}{\ln(b)} \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z) \\

\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z) & = & \frac{1}{\ln(b)} \left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z)\right) \left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left( \mathcal{E}_{[b,{\small -\frac{1}{2}}]}(z) \right)\right)
\end{eqnarray}
$

(Yes, I'm on a fishing expedition.)
bo198214 Wrote:
jaydfox Wrote:Notation for iteration of exponentiation:
$\exp_b^{\circ t}(z)$

Tetration is a special case:
${}^{t} b\ \equiv\ \exp_b^{\circ t}(1)$

"Cleaner" notations to allow "primed" derivative notation:
$\mathcal{T}_{[b,z]}(t)\ \equiv\ \exp_b^{\circ t}(z)$
$\mathcal{E}_{[b,t]}(z)\ \equiv\ \exp_b^{\circ t}(z)$
$\mathcal{B}_{[t,z]}(b)\ \equiv\ \exp_b^{\circ t}(z)$

\begin{align*}
\text{sexp}_b (t) &={}^tb = \mathcal{T}_{[b,1]}(t)\\
\text{spow}_t (b) &={}^tb = \mathcal{B}_{[t,1]}(b)\\
\text{slog}_b&=\text{sexp}_b^{\circ -1}
\end{align*}

And we can recover the full 3 variable expression by means of $\text{sexp}_b$ and its inverse $\text{slog}_b$:

$\exp_b^{\circ t}(z) = \text{sexp}_b(\text{slog}_b(z) + t)$
Those notations are useful for bases above eta. However, for bases between 1 and eta, you cannot recover the full exp function, unless you extend the definition of slog to include the 2 additional domains (only 1 additional for eta). While I'm in favor of such an extension, I'm still working on defining the correct z value to start from, just as 1 is the "correct" value to start from for the principal domain.

Until we define good starting points, a full analysis of fractional iterates will require the more general "T" notation, where we can start from a location other than 1 and leave it at that. For example, for base sqrt(2), we could start at sqrt( or e, etc., to solve for the corridor between the asymptotes at 2 and 4, and we could start at 4.1 or 5 or 16, etc. to solve above the asymptote at 4. (BTW, 16 seems to me the best place. Essentially, exponentiate the upper and lower asymptotes. The order doesn't matter: 2^4=4^2, and this is true for any base between 1 and eta, so it seems pretty "unique" to me.)
jaydfox Wrote:Those notations are useful for bases above eta. However, for bases between 1 and eta, you cannot recover the full exp function, unless you extend the definition of slog to include the 2 additional domains (only 1 additional for eta). While I'm in favor of such an extension, I'm still working on defining the correct z value to start from, just as 1 is the "correct" value to start from for the principal domain.

Until we define good starting points, a full analysis of fractional iterates will require the more general "T" notation, where we can start from a location other than 1 and leave it at that. For example, for base sqrt(2), we could start at sqrt( or e, etc., to solve for the corridor between the asymptotes at 2 and 4, and we could start at 4.1 or 5 or 16, etc. to solve above the asymptote at 4. (BTW, 16 seems to me the best place. Essentially, exponentiate the upper and lower asymptotes. The order doesn't matter: 2^4=4^2, and this is true for any base between 1 and eta, so it seems pretty "unique" to me.)
Ach, what am I saying? Even if we extended slog to include all 2 or three domains, the sexp function would cease to be a function if we tried to take the inverse of the slog. In some ways, this would indicate that the slog function may be more fundamental than the sexp function. Given that Andrew's solution is based on solving the slog, this is interesting...
I'll start by solving for the fractional iterate of exponentiation itself. I.e., the base b and the iteration t are to be assumed constant, so that we can solve for, e.g., the half-iterate of base e $\left(\exp_e^{\circ {\tiny 1/2}}(z)\right)$, or the quarter-iterate of base 2 $\left(\exp_{\small 2}^{\circ {\tiny 1/4}}(z)\right)$, etc.

First, we'll mix notations, noting that we must ensure that the bases are constant. In this case, the base z for T is not constant, so we must pick a constant base and then use an inverse function to allow for the variable z:
$
\begin{eqnarray}
{\Large \mathcal{E}_{\normalsize [b,t]}(z)} & = & {\Large \mathcal{T}_{\normalsize [b,z]}(t)} \\
& = & {\Large \mathcal{T}_{\normalsize [b,p]}\left(\mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right)}
\end{eqnarray}
$

Note that if you set p=1, this becomes sexp(slog(z)+t). For bases between 1 and eta, other p values allow us to extend our analysis.

Let's attempt to define some desirable characteristics of the fractional iterates of exponentiation.

First, for bases above eta (and probably above 1 in general), they should be strictly increasing. This includes negative iterates as well.

Second, for bases above eta, they should be convex. For positive iterates, this would mean that the second derivative is positive, while for negative iterates it would be negative.

This looks likes:
1. ${\Large \mathcal{E}_{\normalsize [b,t]}^{'}(z)}\ >\ 0$

2. ${\Large \mathcal{E}_{\normalsize [b,t]}^{''}(z)}\ >\ 0,\ t>0 \\
{\Large \mathcal{E}_{\normalsize [b,t]}^{''}(z)}\ <\ 0,\ t<0 \\
$

At this point, it's time to start taking derivatives of T and see what happens. For now, we'll assume t positive, so we can control the direction of the ">".

$
\begin{eqnarray}
{\Large \text{D}_z\left[\mathcal{E}_{\normalsize [b,t]}(z)\right]} & > & 0 \\
\vspace{10}\\

{\Large \text{D}_z\left[\mathcal{T}_{\normalsize [b,p]}\left(\mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right)\right]} & > & 0 \\
\vspace{10}\\

{\Large \mathcal{T}_{\normalsize [b,p]}^{'}\left(\mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right) \text{D}_z\left[ \mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right]} & > & 0 \\
\vspace{10}\\

{\Large \mathcal{T}_{\normalsize [b,p]}^{'}\left(\mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right) \frac{\Large 1}{ \mathcal{T}_{\normalsize [b,p]}^{'}\left( \mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)\right)}} & > & 0 \\
\vspace{10}\\

{\Large \frac{\mathcal{T}_{\normalsize [b,p]}^{'}\left(\mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right)}{ \mathcal{T}_{\normalsize [b,p]}^{'}\left( \mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)\right)}} & > & 0 \\
\end{eqnarray}
$

This last line implies that the first derivative of a tetration solution must be positive, i.e., $\mathcal{T}$ is strictly increasing. Not a difficult condition to meet, as this would be necessary for the sexp function to be invertible. So hardly any help has been gotten by this exercise. But it's a start.
Now for the second condition. The second derivative is positive for t>0. This means that the first derivative (which we just found) is strictly increasing. We don't actually care what the second derivative is, so long as we can prove that the first derivative is strictly increasing. In other words, we're simply making a qualitative analysis, not a quantitative analysis (at this point, anyway).

In order to simplify notation, and without loss of generality, let's drop the bases from the notation. Note that this doesn't mean we're assuming "default" bases; we're simply not displaying them.

$
\begin{eqnarray}
\text{D}_z\left[\mathcal{E}^{'}(z)\right] & > & 0 \\
\vspace{10}\\

\lim_{h \to 0^{\small +}} \frac{\mathcal{E}^{'}(z+h)-\mathcal{E}^{'}(z)}{\Large h} & > & 0
\end{eqnarray}
$

The h in denominator is now unimportant, since 0*h is still 0. Notice we can only do this if h is positive; otherwise, the ">" would reverse.

$
\begin{eqnarray}
\lim_{h \to 0^{+}} \mathcal{E}^{'}(z+h)-\mathcal{E}^{'}(z) & > & 0 \\
\vspace{10}\\

\lim_{h \to 0^{+}} \mathcal{E}^{'}(z+h) & > & \mathcal{E}^{'}(z) \\
\end{eqnarray}
$

Okay, time to plug in the formula for E' we found above. Notice that I can only safely extract h because $\mathcal{T}$ is strictly increasing, which in turn implies that $\mathcal{T}^{\small-1}$ is strictly increasing. Again, we're looking for a qualititative analysis, so the exact value of epsilon after extraction is not important, so long as it's positive.

$
\begin{eqnarray}
\lim_{h \to 0^{+}} \frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z+h)+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z+h)\right)}
& > &
\frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)} \\
\vspace{10}\\

\lim_{\epsilon \to 0^{+}} \frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)}
& > &
\frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)} \\
\vspace{10}\\

\lim_{\epsilon \to 0^{+}} \frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)}{\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}
& > &
\frac{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)} \\
\end{eqnarray}
$

At this point, it's time to take a logarithm. Without loss of generality, we can use any base greater than 1, but for simplicity we can pretend it's base e, if it helps to understand.

$
\begin{eqnarray}
\lim_{\epsilon \to 0^{+}} \left[ \log\left(\frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)}{\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}\right)\right.
& > &
\left. \log\left(\frac{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)}\right) \right] \\
\vspace{10}\\

\lim_{\epsilon \to 0^{+}} \left[ \log\left(\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)\right)-\log\left(\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)\right) \right.
& > &
\left. \log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)\right)-\log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)\right) \right]\\
\end{eqnarray}
$

Now we divide through by epsilon, and lo and behold, we now have a proper derivative again!

$
\begin{eqnarray}
\lim_{\epsilon \to 0^{+}} \left[ \frac{\log\left(\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)\right)-\log\left(\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)\right)}{\Large \epsilon} \right.
& > &
\left. \frac{\log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)\right)-\log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)\right)}{\Large \epsilon} \right] \\
\vspace{10}\\

\lim_{\epsilon \to 0^{+}} \left[ \frac{\log\left(\mathcal{T}^{'}\left(w+t+\epsilon\right)\right)-\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)}{\Large \epsilon} \right.
& > &
\left. \frac{\log\left(\mathcal{T}^{'}\left(w+\epsilon\right)\right)-\log\left(\mathcal{T}^{'}\left(w\right)\right)}{\Large \epsilon} \right] \\
\end{eqnarray}
$

A quick substitution to simplify notation:

$
\begin{eqnarray}
\text{D}_{w+t} \left[\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)\right]
& > &
\text{D}_w \left[\log\left(\mathcal{T}^{'}\left(w\right)\right) \right] \\
\vspace{10}\\

\frac{\text{d}}{\text{d}w} \left[\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)\right]
& > &
\frac{\text{d}}{\text{d}w} \left[\log\left(\mathcal{T}^{'}\left(w\right)\right) \right] \\
\vspace{10}\\

\frac{\text{d}z}{\text{d}w}\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)\right]
& > &
\frac{\text{d}z}{\text{d}w}\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}^{'}\left(w\right)\right) \right] \\
\end{eqnarray}
$

Note that $\frac{\text{d}z}{\text{d}w} = \left(\frac{\text{d}w}{\text{d}z}\right)^{-1}$. w is strictly increasing, so dw/dz is positive, so dz/dw is positive. Therfore, we can eliminate it from both sides without affecting the ">" comparison. And let's extract t. The derivative T' is strictly increasing, so when we extract a positive t, we'll get a positive s. The value will be uninmportant, so long as we know it's positive.

$
\begin{eqnarray}
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}_^{'}(w+t)\right)\right]
& > &
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}^{'}(w)\right) \right] \\
\vspace{10}\\

\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}_^{'}(w)+s\right)\right]
& > &
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}^{'}(w)\right) \right] \\
\end{eqnarray}
$

To satisfy this equation for all positive s, we want the first derivative (of the logarithm of the first derivative) to be strictly increasing. so we really want the second derivative to be positive:

$
\begin{eqnarray}
\frac{\text{d}^{\tiny 2}}{\text{d}z^{\tiny 2}} \left[\log\left(\mathcal{T}_^{'}(w)+s\right)\right]
& > & 0
\end{eqnarray}
$

And this, finally, means that we want T' to be log-convex. I mentioned this here, providing only the briefest sketch of a proof:
http://math.eretrandre.org/tetrationforu...d=11#pid11

As I mentioned:

jaydfox Wrote:From here, we need to guarantee that R'(z) is always increasing. To do this, it suffices to show that ln(D_z tet_e(z)) is convex (I don't have time to post the explanation, I'll get back to that after work). Which in turn simply means that:

$D_z^2\ ln(tet_e^\prime(z))\ \ge \ 0$

This in turn is equivalent to saying that the first derivative of tet_e(z) is log-convex.

Formalizing that wasn't "easy". It was the long way there, and someone may know a quicker/cleaner way to get there. But that's where we get. In order for all the partial iterates of exponentiation (for bases b>1) to be convex, the first derivative of T' must be log-convex. (What I now call T, I previously called R, if that clears up any confusion.)

Now, this is the general case. For the special case of tetration, we fix z at 1, and now we know that a tetration solution must have its first derivative be log-convex, in order to get convex partial iterates based on the solution.

My solution meets this new criterion, as does Andrew's. So we still need further criteria.

My hunch is that if we impose convexity requirements on the (odd?) derivatives of the partial iterates, we'll impose log-convexity requirements on higher derivatives of the tetration solution. We know these requirements exist for the integer iterates (at least for bases above eta, but I assume for bases above 1 in general), so it doesn't seem too much to ask this of the fractional iterates.

These log-convexity requirements would in turn require good old-fashioned convexity of the odd derivatives of a tetration solution, as we observed in Andrew's solution. If my hunch is correct, then we can find a more "solid" reason to believe that Andrew's solution is "the" solution.
Hmm, I assumed I could pull the t out of $\text{D}_{w+t}$, since t is a constant, but now I'm not so sure. My calculus is pretty rusty.

$
\frac{\text{d}}{\text{d}\left(w+t\right)}\ =\ \frac{\text{d}w}{\text{d}\left(w+t\right)}\frac{\text{d}}{\text{d}w}
$

If my understanding is correct, dw/d(w+t) should equal 1. Yes?
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text:
Unsolved

The problem hasn't been solved

Hi, I was trying both Rosetta3.1 and Rosetta3.2 to do clustering. I had normal result from Rosetta3.1. But Rosetta3.2 failed to work. I used the same data set (generated by Rosetta2.3.0 abrelax mode), option and command, and the path is correct. I also got the error messege:

ERROR: Illegal attempt to score with non-identical atom set between pose and etable
ERROR:: Exit from: src/core/scoring/etable/EtableEnergy.cc line: 80

And in the log file, it says:

Post Situation:

interpreting score values in silent output file after running abinitiorelax

Where can I find information that explains what the score values are and how to interpret and compare them? I'm new to Rosetta and still trying to learn what is going on while I learn how to run the program; so excuse me if I missed something obvious. Also, is the primary source of information related to Rosetta in published papers or is there something more explanatory than the Rosetta manual? Thanks for any help.

Post Situation:

Where is the flag -preserve_input_cb?

Hi, I need some help here. I've been reading the document for fixbb protocol in this page: http://www.rosettacommons.org/manuals/archive/rosetta3.2_user_guide/app_...
In the box listing flags for rotamers, I found one called preserve_input_cb which is supposed to keep unidealized ca_cb bond parameters from input file. However, I cannot find which layers this flag belongs to. What's the key word should I put before the flag in the option file?

Post Situation:

Using 2 non-standard residues in one run

How do I identify two ligand residue type files? I'm docking two proteins, each with a prosthetic group that's near the docking interface. I've created the residue type files. I can only get it to recognize one of the two residues, whether I use -extra_res_fa [They're both in the same dir], or whether I use two -extra_res_fa statements, or if I string the full path & file names of both files after the -extra_res_fa statement.

Post Situation:

Docking 2 proteins including ligands

We'retrying to dock to heme-containing proteins together. The hemes are close to the docking interface, so we don't feel we should dock the proteins only. We've used molfile_to_params.py to create params files. When we try to dock, we get fold errors, as did a previous poster about 6-8 months ago. The recommendation was to try the flag -partners AB CD. When we do that, we get an error that docking:partners has multiple values & the job dies.

Post Situation:

cluster error

Hello:
I generated several conformation of my protein by CONCOORD and I try to cluster then with Rosetta. However, it doesn't work:

core.io.pdb.file_data: [ WARNING ] can't find atom for res 14 atom ND2 (trying to set temp)
core.io.pdb.file_data: [ WARNING ] can't find atom for res 14 atom OD1 (trying to set temp)
core.io.pdb.file_data: [ WARNING ] can't find atom for res 15 atom CG1 (trying to set temp)
core.io.pdb.file_data: [ WARNING ] can't find atom for res 15 atom CG2 (trying to set temp)

Post Situation:

Trouble building rosetta in windows with cygwin

I got this error message when trying to build rosetta 3.2 under windows XP in cygwin 1.7.8 using python 2.6.5.

scons: *** [build/src/release/cygwin/1.7/x86/gcc/utility.dll] Error 1
scons: building terminated because of errors.

Does anyone here know what might be causing this and how could get around this problem to get rosetta built? Let me know if you need more information.

Thanks.

Post Situation:

-score_app:linmin flag not working

Hi All,

I am completely new to this forum and to rosetta, so be kind with me...

I am trying to minimize a structure using rosetta and more specifically -score_app:linmin flag together with score_jd2.linuxgccrelease. First I don't have in my installation the score.linuxgccrelease executable, is that a mistake? Are score.linuxgccrelease and score_jd2.linuxgccrelease similar / identical /different?

Second I tried to run the following command, without much success :-(
>score_jd2.linuxgccrelease -database \$rosettadb -s 1ubiA.pdb -score_app:linmin -out:output –out:prefix linmin_

Post Situation:

trying to create heme.params

Following advice on a thread from a year ago, I created a mol2 file from a heme ligand pdb. When I try to create the params file, it told me:  /usr/ros/rosetta_source/src/python/apps/public/molfile_to_params.py hemf.mol2 Traceback (most recent call last): File "/usr/ros/rosetta_source/src/python/apps/public/molfile_to_params.py", line 1323, in ? sys.exit(main(sys.argv[1:])) File "/usr/ros/rosetta_source/src/python/apps/public/molfile_to_params.py", line 1254, in main molfiles = list(read_tripos_mol2(infile, do_find_rings=False))

Post Situation:

Pyrosetta2

Hi,

I am trying to execute the commands on pyrosetta 2 from tutorial .

After executing the command pose = Pose ("1GFL.pdb") I am getting the following errors :-

---------------------------------------------------------------------------
ArgumentError Traceback (most recent call last)

/examples/ in ()

ArgumentError: Python argument types in
Pose.__init__(Pose, str)
did not match C++ signature:
__init__(_object*, core::pose::Pose src, unsigned int residue_begin, unsigned int residue_end)

Post Situation:


text:
# ε-δ definition and M-δ definition to prove M-δ limit

(i)$\lim_{x \to a} f(x)= L$ (ii)$\lim_{x \to a} g(x)= \infty$

Using the definitions of $\varepsilon-\delta$ definition and $M-\delta$ definition to prove $lim_{x \to a} f(x)g(x)=\infty$ as $x$ approaches $a$.

Anyone can help me with this? I have done my definition part respectively but don really know how to continue to combine those definitions to prove that.

I was given this solution:But I could not understand, can anyone explain to me please?

let $\varepsilon=\frac{L}{2}$ in part(i), there exists $\delta_1 >0$ such that $0<|x-a|< \delta_1$,then $|f(x)-L|<\frac{L}{2}$ for all $M>0$ In part(ii),there exists $\delta_2>0$ such that $g(x)>\frac{2M}{L}$ for $0<|x-a|<\delta_2$ Then for all $M>0$,there exists $\delta=\min(\delta_1,\delta_2)$ such that $0<|x-a|<\delta$ Then we have $f(x)g(x) \geq min(L-\frac{L}{2},L+\frac{L}{2})g(x)=0.5 \cdot L \cdot g(x)>(\frac{L}{2})*(\frac{2M}{L})=M$

I understand that it wants to show $f(x)g(x)>M$ because of definition. but I don understand what is going on(especially the last sentence where ...$min(L-\frac{L}{2},L+\frac{L}{2})$, can anyone explain to me the approach for solving this ques?thx!

• Consider $L=a=0$, $f(x) = x$, $g(x) = |\log(x)|$, then the conditions are satisfied but $\lim\limits_{x \to 0} f(x)g(x) = 0$. – Nigel Overmars Nov 30 '14 at 16:51

You want to assume $L > 0$.
We are given $\lim_{x \to a } f(x) = L$. By defintion, this means that for all $\epsilon > 0$, we can always find some $\delta_1 > 0$ such that if $0 < |x-a| < \delta_1$ , then $|f(x) - L | < \epsilon$. Since this holds for any positive $\epsilon$, we can choose $\epsilon = \frac{L}{2} > 0$. In particular, $f(x) > \frac{L}{2}$. (PRoof?)
Next, since $\lim_{x \to a} g(x) = \infty$, then by definition for any $\alpha > 0$, we can always find some $\delta_2 > 0$ such that if $0 < |x-a| < \delta_2$, then $g(x) > \frac{2\alpha}{L}$. Now, we want to show that $\lim_{x\to a } f(x)g(x) = \infty$. Choose $\delta = \min( \delta_1, \delta_2 )$. Then, if $0 < |x-a| < \delta$, we must have
$$fg > \frac{L}{2} \frac{2 \alpha}{L} = \alpha$$


text:
# QM: The Dilation operator

## Homework Statement

Concicer the dilation operator

$D = \vec{r} * \vec{p}$

Compute $[D,\vec{r} ] and [D, \vec{p}]$

## Homework Equations

$p = - i * hbar$

## The Attempt at a Solution

I think the question is really if $[D, \vec{r}]$ commutes

I got this:
$D = \vec{r} * \vec{p}$
$D = x p_x + y p_y + z p_z$

for the x direction:

$[D, r] = [D, x] = [x, x p_x]$

where do i take it from here?

## Answers and Replies

dextercioby
Science Advisor
Homework Helper
You have absolutely no idea ?? Pick up your book and find those brackets (called <commutator>) and the properties they have.