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## Chemistry 12th Edition

(a) $CH_3COO^-$: base $CH_3COOH$: conjugate acid $HCN$: acid $CN^-$: conjugate base (b) $HCO_3^-$ : base $H_2CO_3$: conjugate acid $HCO_3^-$ : acid $CO_3^{2-}$ : conjugate base (c) $H_2PO_4^-$: acid $HPO_4^{2-}$: conjugate base $NH_3$: base $NH_4^+$: conjugate acid (d) $HClO$: acid $ClO^-$: conjugate base $CH_3NH_2$: base $CH_3NH_3^+$: conjugate acid (e) $CO_3^{2-}$: base $HCO_3^-$: conjugate acid $H_2O$: acid $OH^-$: conjugate acid
Analyzing the direct reaction (left to right), the acid is the compound that loses one $(H^+)$, and the base is the one that receives this $(H^+)$. They are on the right side. The conjugate base will be the compound in the right side, that is the acid that lost its ($H^+$). The conjugate acid will be the compound in the right side, that is the base that received the ($H^+$).


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# Thread: Game Show Possibilities Problem...

1. ## Game Show Possibilities Problem...

Hi Everyone,

Would the following problem be a permutation, or a combination?

"For a gameshow, there are 9 spots to select on a board. You get to choose 3 of them in order to try to find the grand prize. How many different ways are there to select your spots?"

I think it's a permutation, because order matters. In that case, can I still solve it using 9C3 (9 choose 3)? Or do I need to do something else to account for the position of the answers?

Also, this one:

"If a Congressional committee is to be made of six people selected from a group of eight Democrats and five Republicans, how many ways are there to select exactly one Republican?"

The part that throws me off here is the "exactly one" republican. It wouldn't be as simple as a 13C6 calculation, but would that be part of it?

Thank you!
-matt

2. ## Re: Game Show Possibilities Problem...

Originally Posted by BackstreetZAFU
"For a gameshow, there are 9 spots to select on a board. You get to choose 3 of them in order to try to find the grand prize. How many different ways are there to select your spots?"

I think it's a permutation, because order matters. In that case, can I still solve it using 9C3 (9 choose 3)? Or do I need to do something else to account for the position of the answers?
There is no way to answer this unless we know the exact way the game is played. What are the rules?

Originally Posted by BackstreetZAFU
"If a Congressional committee is to be made of six people selected from a group of eight Democrats and five Republicans, how many ways are there to select exactly one Republican?"
This is multiplicative: $\dbinom{5}{1}\cdot \dbinom{8}{5}$

3. ## Re: Game Show Possibilities Problem...

[QUOTE=Plato;946524]There is no way to answer this unless we know the exact way the game is played. What are the rules?

I'm assuming the problem means that you choose three spots (out of 9), and then reset. So, I could choose spots 1, 2, and 3 (in that order), and then reset. Then I could choose spots 1, 3, and 2 (in that order), and reset. And so on. Does that make more sense?

Regarding the second problem, is that the same as 5C1 x 8C5?

-matt

4. ## Re: Game Show Possibilities Problem...

Also, regarding the Congressional problem, if it read:

Using the situation from above, how many ways are there to select exactly three Republicans?

Would it be?

5 Choose 3 x 8 Choose 3
5C3 x 8C3 =
10 x 56 = 560

Thank you again!
-matt

5. ## Re: Game Show Possibilities Problem...

[QUOTE=BackstreetZAFU;946527]
Originally Posted by Plato
I'm assuming the problem means that you choose three spots (out of 9), and then reset. So, I could choose spots 1, 2, and 3 (in that order), and then reset. Then I could choose spots 1, 3, and 2 (in that order), and reset. And so on. Does that make more sense?
NO! Sorry that makes no sense at all.
Let this be a lesson to you for future posts. If you are assigned a question then post the exact, Wording of that question.
If you post your understanding of the question it is possible that any confusion is at least doubled.
So post the exact wording of a question. Post your understanding of that question. Post your efforts to solve that question.

6. ## Re: Game Show Possibilities Problem...

Originally Posted by BackstreetZAFU
Hi Everyone,

Would the following problem be a permutation, or a combination?
It is better not to worry about that at all!

"For a gameshow, there are 9 spots to select on a board. You get to choose 3 of them in order to try to find the grand prize. How many different ways are there to select your spots?"
You have 9 choices for the first spot. After you have made that choice there are 8 spots left so you have 8 choices for the second spot. Then there are 7 spots left for the third choice.. You can do this in 9*8*7= 504 ways. That is the same as $\displaystyle \frac{9*8*7*6*5*4*3*2*1}{6*5*4*3*2*1}= \frac{9!}{6!}= \frac{9!}{(9- 3)!}= P(9, 3)$ but it is better to think about a problem than to memorize formulas.

I think it's a permutation, because order matters. In that case, can I still solve it using 9C3 (9 choose 3)? Or do I need to do something else to account for the position of the answers?

Also, this one:

"If a Congressional committee is to be made of six people selected from a group of eight Democrats and five Republicans, how many ways are there to select exactly one Republican?"

The part that throws me off here is the "exactly one" republican. It wouldn't be as simple as a 13C6 calculation, but would that be part of it?

Thank you!
-matt
Start by choosing the republican first- there are 5 ways to do that. Now choose the first democrat. There are 8 ways to do that. That leaves 7 democrats so there are 7 ways to choose the second democrat. There are then 6, 5, and 4 ways to choose the third, fourth, and fifth democrats. The total number of ways to choose these 6 people, in this order, is 5*8*7*6*5*4= 5*8!. However, the order in which we choose the people are chosen is not relevant so we divide by 6! to remove the 6! different orders:
$\displaystyle \frac{5*8!}{6!}= 5*8*7= 280$.

7. ## Re: Game Show Possibilities Problem...

[QUOTE=Plato;946539]
Originally Posted by BackstreetZAFU
NO! Sorry that makes no sense at all.
Let this be a lesson to you for future posts. If you are assigned a question then post the exact, Wording of that question.
If you post your understanding of the question it is possible that any confusion is at least doubled.
So post the exact wording of a question. Post your understanding of that question. Post your efforts to solve that question.
That WAS the exact wording of the question. Sorry you couldn't figure it out.

8. ## Re: Game Show Possibilities Problem...

Originally Posted by BackstreetZAFU
Regarding the second problem, is that the same as 5C1 x 8C5?
Huh? Where's the first problem???

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# An Inequality

Algebra Level pending

Find the minimum value of the expression below

$$\dfrac{ab}{(a-b)^2} + \dfrac{bc}{(b-c)^2} + \dfrac{ca}{(c-a)^2}$$

where $$a$$, $$b$$ and $$c$$ are distinct real numbers.

×

Problem Loading...

Note Loading...

Set Loading...


text:
On the stability for Alexandrov’s soap bubble theorem.(English)Zbl 1472.53013

Summary: Alexandrov’s Soap Bubble Theorem dates back to 1958 and states that a compact embedded hypersurface in $$\mathbb{R}^N$$ with constant mean curvature must be a sphere. For its proof, A. D. Alexandrov invented his reflection principle. In [Indiana Univ. Math. J. 26, 459–472 (1977; Zbl 0391.53019)], R. C. Reilly gave an alternative proof, based on integral identities and inequalities, connected with the torsional rigidity of a bar. In this article we study the stability of the spherical symmetry: the question is how near is a hypersurface to a sphere, when its mean curvature is near to a constant in some norm?
We present a stability estimate that states that a compact hypersurface $$\Gamma \subset \mathbb{R}^N$$ can be contained in a spherical annulus whose interior and exterior radii, say $$\rho_i$$ and $$\rho_e$$, satisfy the inequality $\rho_e - \rho_i \leq C \| H - H_0\|_{L^1(\Gamma)}^{\tau_N},$ where $$\tau_N = 1/2$$ if $$N = 2, 3$$, and $$\tau_N = 1/(N + 2)$$ if $$N \geq 4$$. Here, $$H$$ is the mean curvature of $$\Gamma,\ H_0$$ is some reference constant, and $$C$$ is a constant that depends on some geometrical and spectral parameters associated with $$\Gamma$$. This estimate improves previous results in the literature under various aspects. We also present similar estimates for some related overdetermined problems.

MSC:

 53A07 Higher-dimensional and -codimensional surfaces in Euclidean and related $$n$$-spaces 53A10 Minimal surfaces in differential geometry, surfaces with prescribed mean curvature 49Q10 Optimization of shapes other than minimal surfaces

Zbl 0391.53019
Full Text:

References:

 [1] Adams, R. A., Sobolev Spaces (1975), New York: Academic Press, New York [2] Alessandrini, G.; Morassi, A.; Rosset, E., The linear constraints in Poincaré and Korn type inequalities, Forum Math., 20, 557-569 (2008) · Zbl 1151.26319 [3] Aleksandrov, A. D., Uniqueness theorems for surfaces in the large. V, Vestnik Leningrad Univ., 13, 19, 5-8 (1958) [4] Alexandrov, A. D., A characteristic property of spheres, Ann. Math. Pura Appl., 58, 303-315 (1962) · Zbl 0107.15603 [5] Brandolini, B.; Chiacchio, F.; Trombetti, C., Optimal lower bounds for eigenvalues of linear and nonlinear Neumann problems, Proc. Roy. Soc. Edinburgh Sect. A, 145, 31-45 (2015) · Zbl 1321.35117 [6] Brasco, L.; Magnanini, R.; Salani, P., The location of the hot spot in a grounded convex conductor, Indiana Univ. Math. Jour., 60, 633-660 (2011) · Zbl 1328.35063 [7] Ciraolo, G.; Maggi, F., On the shape of compact hypersurfaces with almost constant mean curvature, Comm. Pure Appl. Math., 70, 665-716 (2017) · Zbl 1368.53004 [8] Ciraolo, G.; Magnanini, R.; Vespri, V., Hölder stability for Serrin’s overdetermined problem, Ann. Mat. Pura Appl., 195, 1333-1345 (2016) · Zbl 1348.35146 [9] Ciraolo, G.; Vezzoni, L., A sharp quantitative version of Alexandrov’s theorem via the method of moving planes, J. Eur. Math. Soc., 20, 261-299 (2018) · Zbl 1397.53050 [10] Esposito, L.; Nitsch, C.; Trombetti, C., Best constants in Poincaré inequalities for convex domains, J. Convex Anal., 20, 253-264 (2013) [11] Giusti, E., Direct Methods in the Calculus of Variations (2003), River Edge, NJ: World Scientific, River Edge, NJ [12] Heintze, E.; Karcher, H., A general comparison theorem with applications to volume estimates for submanifolds, Ann. Sci. École Norm. Sup., 11, 451-470 (1978) · Zbl 0416.53027 [13] Korevaar, N., Convex solutions to nonlinear elliptic and parabolic boundary value problems, Indiana Univ. Math. J., 32, 603-614 (1983) · Zbl 0481.35024 [14] Meyers, N. G., Integral inequalities of Poincaré and Wirtinger type, Arch. Ration. Mech. Anal., 68, 113-120 (1978) · Zbl 0416.46024 [15] Payne, L.; Schaefer, P. W., Duality theorems in some overdetermined boundary value problems, Math. Meth. Appl. Sciences, 11, 805-819 (1989) · Zbl 0698.35051 [16] Payne, L. E.; Philippin, G. A., Some applications of the maximum principle in the problem of torsional creep, SIAM J. Appl. Math., 33, 446-455 (1977) · Zbl 0378.73028 [17] Payne, L. E.; Weinberger, H. F., An optimal Poincaré inequality for convex domains, Arch. Ration. Mech. Anal., 5, 286-292 (1960) · Zbl 0099.08402 [18] Reilly, R. C., Applications of the Hessian operator in a Riemannian manifold, Indiana Univ. Math. J., 26, 459-472 (1977) · Zbl 0391.53019 [19] Reilly, R. C., Mean curvature, the Laplacian, and soap bubbles, Amer. Math. Monthly, 89, 180-188 (1982) [20] Ros, A., Compact hypersurfaces with constant higher order mean curvatures, Rev. Mat. Iberoamericana, 3, 447-453 (1987) · Zbl 0673.53003 [21] Serrin, J., A symmetry problem in potential theory, Arch. Ration. Mech. Anal., 43, 304-318 (1971) · Zbl 0222.31007 [22] Weinberger, H. F., Remark on the preceding paper of Serrin, Arch. Ration. Mech. Anal., 43, 319-320 (1971) · Zbl 0222.31008
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Does the negative apply before or after an exponent

I have been having an argument with a friend, and they claim $-8^0$ is $1$, and I claim that $-8^0$ is really $-(8^0)$ and is therefore $-1$. Who is right?

Exponents take precedence over multiplication. Your friend is wrong.

• So by process of elimination, user144807 is correct. – SimonT Apr 22 '14 at 3:03
• It should be noted that this is simply a convention -- there's no mathematical reason why it would have to be so. Nonetheless, this convention is pretty much universally accepted, so that any mathematician would understand $-a^b$ as $-(a^b)$, not as $(-a)^b$. It does have the convenient advantage that it lets us write polynomials without having to stuff them full of parentheses whenever a coefficient happens to be negative. On the other hand, plenty of formulas involving alternating sums like $\sum_k (-1)^k f(k)$ would get simpler if it were the other way, so... – Ilmari Karonen Apr 22 '14 at 3:32
• @SimonT We would have to eliminate all other possibilities to use the process of elimination, and $-8^0$ could be anything. – Potatoswatter Apr 22 '14 at 4:32
• I've actually had someone once tell me their calculator was broken, showing me $-4^2 = -16$ on the screen and saying that it was wrong because "the sign doesn't matter when squaring" etc etc... I added the brackets and gave the calculator back, and he understood. – Thomas Apr 22 '14 at 7:45
• "precedence over multiplication" So you consider negation a kind of multiplication? One could have chosen a convention where $-a^b$ meant $(-a)^b$ and still have multiplication lower in precedence, such that $-ca^b$ still meant $-(c(a^b))$ or equivalently $(-c)(a^b)$, not $(-(ca))^b$ or $((-c)a)^b$. – Jeppe Stig Nielsen Apr 22 '14 at 21:05

Neither of you is right - or at least, you are both half right.. $$-8^0=-8^{1/2-1/2}=\frac{-8^{1/2}}{-8^{1/2}}$$ Now we compromise so that everyone gets an equal chance: $$\frac{-2\sqrt 2}{2\sqrt 2 i}={-1\over i}=i$$ So the answer is $i$. The imaginary unit. The enigma of the conundrums. The unspeakable number.

I have divulged this information at great risk to our agents in the field. Please use this remarkable knowledge with discretion, and always remember: use more parentheses.

• I just want to say i love this answer, for its simultaneous insanity and reason – Asimov Apr 22 '14 at 1:44
• Um... please explain how you are not assuming the friend's approach from the second equality (in the first line) of this answer? – apnorton Apr 22 '14 at 2:21
• @anorton: You're right, he really should've used the compromise approach there too, and declared that $-8^{1/2-1/2}=-\frac{-8^{1/2}}{-8^{1/2}}$, which finally leads to the conclusion that $-8^0 = -i$. But that's fine, since it's really just another branch of the same multivalued function. ;-) – Ilmari Karonen Apr 22 '14 at 3:20
• Whoooooooooosh. sound of point going over heads – blue Apr 22 '14 at 21:31

In general, operations are done in this order:

1. Brackets
2. Exponents
3. Division and Multiplication

For example, $10 + 0 \times 5$ isn't read as "$10 + 0 = 10$, $10 \times 5 = 50$", it's read as "$0 \times 5 = 0$, $10 + 0 = 10$".

Your example is slightly trickier as you have to recognize what the minus sign really means. "$-n$" is essentially shorthand for $0-n$, and realizing that makes the answer clear.

Your equation is "really" $0-8^0$. Looking at the list from earlier, we see that exponents are done before subtraction, so that step comes next. $0-8^0$ becomes $0-1$, which we write as $-1$. You are correct.

Really, though, the question is less than ideal. Writing it as $-(8^0)$, as you say, would clear up a lot of problems. It's very easy to misread or misunderstand a problem and steps clearly were never taken to avoid that here.

• One might also say that $-n$ is shorthand for $(-1)n$, which of course does not change the answer. – jpmc26 Apr 22 '14 at 21:18
• @jpm I can't think of any situations where $0-n \neq (-1)n$ (for $n \in \mathbb{R}$, at least) so having different ways to think about it is always good. :) – undergroundmonorail Apr 22 '14 at 22:10

You are correct: without parenthesis around the (−8), the formula −80 is equivalent to (−1) * (80).

You trying to determine whether the equation, −80 = x, like −23 = x, will result in x being positive or negative. 1 − 23 = −7 can be written as 1 + −23 = −7 or 1 + (−1) * (23) = 7. Clearly the negative addition of terms is separate from the result of the exponent.


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# What carries the information for the Pauli exclusion principle to occur?

I have been reading about the exclusion principle a little bit, but I have some questions about it.

How does the information about the state of electrons get "passed around" so that other electrons in similar state can not have that same state? Is there some kind've information carrier?

Is there some unique force created by a set of quantum numbers for a fermion so that, the force then prevents another fermion with the same set of quantum numbers from being permitted? How do electrons know the states of other electrons to determine if their state is allowed?

Sometimes people say that the Pauli exclusion principle says that "two electrons can't be in the same state." This is not correct. It's not as though each particle has its own "state" that it keeps to itself. It's actually much deeper: the electron field itself has one state it's in (that's the whole point of quantum field theory) and crucially there are simply no states of the electron field corresponding to two electrons with the same spin, position, etc. There is nothing for it to "know," it's just what the quantum electron field is.

Fermionic fields are different from bosonic fields. A boson field also has a single state it's in, but the difference between a bosonic field and a fermionic field is simply that the boson field does have states that correspond to boson particles with the same position, spin, etc. Bosonic fields have way more states than fermionic fields.

(Actually, going a bit further, EVERY field together is really in a single universal state, but that's not really so important for the question at hand.)

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind May 13 at 9:57
• Nitpick: That bosons have "way more" states than fermions is certainly "intuitively true", but it's not really true, is it? Both the fermionic and the bosonic Fock spaces should have the same cardinality - that of separable Hilbert spaces. – ACuriousMind May 13 at 10:02
• Is there a known or relatively easy to verify bijective map between fermionic and bosonic states? – lurscher May 14 at 21:05

Looking for this kind of intuitive understanding of quantum mechanics is one of the things that makes QM hard to learn. You are thinking about how classical particles and waves behave, and are comparing that to quantum behavior. You look for similarities and they make sense because classical physics makes sense. But then the differences do not make sense.

It may be better to think of how quantum mechanical entities behave, and get used to the fact that they are not like classical entities. That is, rather than think about how an electron or photon is kind of like a particle and kind of like a wave, think about the new and different thing electron or photon is. What are its properties?

There are still profound conceptual difficulties. You have to accept that the universe behaves not just in ways that are different from what you expect, but contradict what you expect.

Suppose you had only ever heard of particles. You understand they have a mass, position, and velocity. They have a trajectory. Now someone introduces you to waves. They are different from particles. They don't have a point like position. They extend over a region, perhaps with fuzzy indefinite boundaries. They move, and yet don't have a trajectory. And yet they are something like particles. They carry energy and momentum. And yet they are different. They pass right through each other without bouncing off. It would be very confusing to think of a wave as some sort of particle. It makes more sense when you get used to what a wave is.

Here is a post that explains a photon is neither a particle nor a wave. How can a red light photon be different from a blue light photon? It shows how your intuition can be somewhat helpful and somewhat can get in the way. It turns out that electrons behave like this too, though classically we are more inclined to think of light as waves and electrons as particles.

Here is a link that begins to get at your question. Does the collapse of the wave function happen immediately everywhere? It gets more into how an electron is not like a particle nor a wave. A spread out electron can pass through two slits and interfere with itself on the other side. But then it can hit a single atom. It is very reasonable to wonder how information gets passed around. And yet this is not the right question to ask. There is no answer.

Historically, quantum mechanics addressed this with two sets of rules. One set of rules uses the Schrödinger equation to tell you how the electron's wave function changes with time. This tells you how the electron moves like a wave. These rules apply so long as the electron is not disturbed.

Then the electron gets disturbed, or "measured". Say it hits an atom. The wave function "collapses" to a new state. We don't see what happens during the collapse. We only see that we get a new state and we can describe how it changes with the Schrödinger equation. We can't predict what the new state will be from the old state. There may be a number of possible new states. We can predict probabilities of arriving at each one. Quantum mechanics has no mechanism that shows how information is passed around and no definite answers ahead of time to where it goes. This is called the Copenhagen interpretation of quantum mechanics.

This is something of a messy theory. It was accepted because it fit experiment very well. But it has problems. It isn't very clear exactly what a "measurement" is. It requires two sets of rules to describe what goes on, where one set would be more reasonable. It isn't deterministic.

Quantum mechanics is about a century old. Sometimes it takes a century or two to work out the kinks of a theory, and this is certainly true of quantum mechanics. About 50 years ago, the Everett or "Many Worlds" interpretation was proposed. It is now beginning to get acceptance. Here are a couple links that explain it. Parallel worlds probably exist. Here's why. And What is the Many Worlds interpretation? The jury is still out on this, but it is getting serious attention now.

The Many Worlds interpretation makes exactly the same experimentally testable predictions as the Copenhagen interpretation. But it eliminates the two sets of rules and questions of what a measurement is. It says the universe is deterministic. Where the Copenhagen interpretation says various outcomes might happen, the Many Worlds says all of the possible outcomes do happen every time. When they do, the universe splits into different worlds that can't talk to each other. The wave function doesn't collapse. The randomness comes in because we are split into many different versions of ourselves, and each version is only aware of one world.

This still doesn't provide a satisfactory, intuitive answer to your question. The wave function is all the information about a system that exists. Information gets passed around as described by the Schrödinger equation. Even though there may be two electrons, there is only one wave function that describes everything in all the many worlds. Two electrons never evolve so they have the same state because the Schrödinger equation doesn't permit it. This link explains why not. What causes Pauli's Exclusion Principle?

• I don't see the benefit of discussing Copenhagen vs Many Worlds here. IMHO, that's rather tangential to the question about Pauli exclusion. – PM 2Ring May 12 at 13:55
• @PM2Ring - The OP is asking how information is passed around. In Copenhagen, there is a collapse. There is no way to say how information moves around. In MW, information in the wave function continues to flow according to the Scrodinger equation. – mmesser314 May 12 at 19:20
• @DescheleSchilder Hidden variable theories have always been preferred by people, and it took considerably more effort to disprove them than would be warranted if it just didn't sound like such a great way to get rid of the "quantum weirdness". Hidden variables sound far more intuitive to humans, whose brains are created to model the world of throwing rocks; but do the experiments, and you get the wrong answers - or the number of hidden variables explodes. The fact that we don't have a good hidden variable theory doesn't come from people not trying. – Luaan May 13 at 11:49
• @DescheleSchilder No, hidden variables (at least local ones - bound to single 'particles') contradict experimental results. There exist third interpretation besides Copenhagen and MW - pilot wave; all three give identical results and are mostly philosophical thing :) – Arvo May 13 at 14:24
• @DescheleSchilder "Nonlocal" sounds tame; it is not. Local means in this context, among other things, "bounded by the speed of light", and if you break that, you have acausal madness. But somehow blocked from allowing conventional information to be acausal. – Yakk May 14 at 19:41

The Pauli exclusion principle is the quantum mechanical principle which states that two or more identical fermions (particles with half-integer spin) cannot occupy the same quantum state within a quantum system simultaneously

Let us talk simple, rock bottom quantum mechanics, not the high language of quantum field theory, which in any case is founded on rock bottom quantum mechanics.

By rock bottom I mean the postulates and the axiomatic postulated elementary particles, as in the standard model.

An important postulate is the wave function, a solution of a differential quantum mechanical equation that describes specific quantum numbers for each energy level of the solution in a potential. Lets take the simple hydrogen wavefunctions. There are three quantum numbers for each energy level, $$n, l, m$$. The Pauli exclusion fits the experimental observation that there are only single electrons at a single set of $$n, l, m$$. This makes sense in multi-electron atoms, as if there were no such limit, all electrons of the atom would hover in the lowest energy state, making chemistry impossible. (Even if a different law held, limiting the number in the same energy level chemistry would be different than the one we have.)

In a complex case of more than one atom the potential will not be that simple as to get analytic wavefunctions, but the principle of energy levels as the solution and quantum numbers assigned uniquely to each energy level again holds, thus the simple illustration with the single atoms holds.

If the energy levels are almost a continuum, as they are in the band theory of solids, there is no difficulty in fulfilling the Pauli principle and a large number of electrons.

How does the information about the state of electrons get "passed around" so that other electrons in similar state can not have that same state? Is there some kind've information carrier?

The information is built in the wavefunction quantum numbers, no matter how complex the wavefunction is.

Information does not get "passed around" because it is not even a material thing, it only exists for us and in our imagination which makes us believe that some relative positions of some objects in space have some meaning. But for the nature itself information is just not a thing.

Various particles of matter just keep on moving and interacting with each other until they happen to exchange their energy and more or less temporarily get into one of the locally stable states. How exactly they do the interactions deep down inside - I don't know, and I guess it is not even possible to know for sure, but today we have a way to estimate the shape of the stable regions of space and probabilities of going into ones or the others under various conditions.

The fact that the two electrons in a given state have one spin up and the other down suggests that the exclusion of others is a result of magnetic dipole interactions.

• All fermions obey Pauli exclusion. That includes neutrinos, which are oblivious to electromagnetism. – PM 2Ring May 12 at 14:24
• Ferromagnetism is a direct counterexample to this. Unpaired electrons on adjacent atoms (adjacent in the directions perpendicular to the spin axis) would prefer to have antiparallel spins if the dipole interaction were the dominant factor. However, antiparallel spins lead to symmetric spatial wavefunctions, which end up having a larger electrostatic interaction, which in turn is enough to overpower the dipole interaction. The net result is that parallel spins are energetically favorable (at least over short distances), in direct contrast with the "preference" of the dipole interaction. – J. Murray May 12 at 14:43
• +1 The magnetic dipole is primary and the spin is a manifestation of the interaction with an external magnetic field. physics.stackexchange.com/questions/631584/… – HolgerFiedler May 13 at 5:27


text:
# source:trunk/sources/HeuristicLab.Problems.Instances.Regression/3.4/Plugin.cs.frame@7826

Last change on this file since 7826 was 7826, checked in by sforsten, 7 years ago

#1784: changed the class name in the Plugin.cs.frame files.

File size: 1.5 KB
Line
2/* HeuristicLab
3 * Copyright (C) 2002-2012 Heuristic and Evolutionary Algorithms Laboratory (HEAL)
4 *
5 * This file is part of HeuristicLab.
6 *
7 * HeuristicLab is free software: you can redistribute it and/or modify
9 * the Free Software Foundation, either version 3 of the License, or
10 * (at your option) any later version.
11 *
12 * HeuristicLab is distributed in the hope that it will be useful,
13 * but WITHOUT ANY WARRANTY; without even the implied warranty of
14 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
15 * GNU General Public License for more details.
16 *
17 * You should have received a copy of the GNU General Public License
18 * along with HeuristicLab. If not, see <http://www.gnu.org/licenses/>.
19 */
20#endregion
21
22using HeuristicLab.PluginInfrastructure;
23
24namespace HeuristicLab.Problems.Instances.Regression {
25  [Plugin("HeuristicLab.Problems.Instances.Regression", "3.4.0.$WCREV$")]
26  [PluginFile("HeuristicLab.Problems.Instances.Regression-3.4.dll", PluginFileType.Assembly)]
27  [PluginDependency("HeuristicLab.Common", "3.3")]
28  [PluginDependency("HeuristicLab.Core", "3.3")]
29  [PluginDependency("HeuristicLab.Data", "3.3")]
30  [PluginDependency("HeuristicLab.Problems.DataAnalysis", "3.4")]
31  [PluginDependency("HeuristicLab.Problems.Instances", "3.3")]
32  [PluginDependency("HeuristicLab.Random", "3.3")]
33  public class HeuristicLabProblemsInstancesRegressionPlugin : PluginBase {
34  }
35}
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text:
# ODE concepts

Okay! Just making sure I have the concept of orders in differential equations right. So the order refers to the highest order of the derivatives, not the actual functions right?

So a function like y + yy' = ? would be first order, y + y'' would be second order, but something like y^4 +3yy' would still be first order right?

## Answers and Replies

jamesrc
Gold Member
Yep, you've got it.

dextercioby
Homework Helper
The "order of the functions" (the power that "y" or its derivatives have in the ODE) gives the nonlinear character...

Daniel.

quasar987
Homework Helper
Gold Member
I guess it's preferable to talk about the "degree of the ODE" instead of the "order of the function or its derivative". If degree > 1, the ODE is nonlinear.

HallsofIvy


text:
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01-12-2014 | Case study | Issue 1/2014 Open Access

# Airline new customer tier level forecasting for real-time resource allocation of a miles program

Journal:
Journal of Big Data > Issue 1/2014
Authors:
Jose Berengueres, Dmitry Efimov
Important notes

## Electronic supplementary material

The online version of this article (doi:10.1186/2196-1115-1-3) contains supplementary material, which is available to authorized users.

## Competing interests

We do not have any competing interest. We do not work for any airline or have a commercial relationship with any airline or any financial interest.

## Authors’ contributions

JB developed de business logic and DE carried out the data modelling. All authors read and approved the final manuscript.

## Background

Previously, several works have addressed the issue of optimizing operations of the airline industry. In 2003, a model that predicts no-show ratio with the purpose of optimizing overbooking practice claimed to increase revenues by 0.4% to 3.2% [ 1]. More recently, in 2013, Alaska Airlines in cooperation with G.E and Kaggle Inc., launched a $250,000 prize competition with the aim of optimizing costs by modifying flight plans [ 2]. However, there are very little works on miles programs or frequent flier programs that focus on enhancing their value. One example is [ 3] from Taiwan, but they focused on segmenting customers by extracting decision-rules from questionnaires. Using a Rough Set Approach they report classification accuracies of 95%. [ 4] focused on segmenting users according to return flight and length of stays using a k-means algorithm. A few other works such as [ 5], focused on predicting the future value of passengers (Customer Equity). They attempt to predict the top 20% most valuable customers. Using SAS software they report false positive and false negative rate of 13% and 55% respectively. None of them have applied this knowledge to increase the value-proposition of miles programs. All this in spite the fact that some of these programs have become more valuable than the airline that started them. Such is the case of Aeroplan, a spin off of the miles programs started by Toronto based Air Canada. Air Canada is as of today valued at$500 M whereas Aeroplan is valued at \$2bn, four times more [ 6]. Given this it is surprising that so many works focus in ticket or “revenue management” [ 7] but few focus on their miles programs. In this case study we report the experience gained when we analyzed the data of an airline’s miles program. We show how by applying modern data mining techniques it is possible to create value for both: (1) the owner of the program and (2) future high-value passengers.
The data of the miles program was provided by Etihad Airlines and was 1.2GBytes in size when zip compressed. It contained three tables of anonymized data from year 2008 to 2012, corresponding to about 1.8 million unique passengers. The tables contain the typical air miles program data such as:
1.
Age and demographics.

2.
Loyalty program purchases etc.

3.
Flight activity, miles earned etc.

## Case Description

### Objective

After co-examining the opportunities that the data offered with the airline we decided to focus on high-value passengers with the objective to: predict and discriminate which of the new passengers who enroll the miles program will be high-value customers before it is obvious to a human expert.

### Rationale

Business justification:
1.
So that potential high-value customers are identified in order to gain their loyalty sooner.

2.
So that limited resources such as upgrades are optimally allocated to passengers with potential to become high-value.

3.
To enhance customer profiling.

### Current method: linear extrapolation

The miles program was using three main tiers to classify customer by value. Ordered form less to more value these are: basic, silver, gold. The airline was using the following method to predict future tier status: a passenger flight activity is observed on a monthly basis then, a linear extrapolation on how much they will fly in the coming month(s) is performed (Figure  1). This information was then used in interactions with customers.

### Definition of the target variable

After some discussion we decided that we are interested in transitions from basic to silver or gold. After a user flights for first time with the airline, their activity is observed for D days, (Figure  2). At the same time we mark each user with 1 or 0 depending on whether they will attain Silver or greater tier status within S additional days after day D. We will later refer to this binary variable as silver_attain, and to this model as D/S model. Then past behavior of customers is feed into the model. The model is described further in the text.

### Identifying future high value customers

When a new customer joins the program, (and D days have passed since then) the model can be used to make a prediction about the likelihood that they will become Silver in the near or far future. Figure  2 and Table  1 show examples. For business purposes we equate Silver tier status with high-value customer. However, this hypothesis was not validated in the scope of the project.
Table 1
Miles program dataset
Dataset
Fields
Rows
Cols
Passengers
Id, Date_of_birth, Nationality, City, State, Country, Interest_1, Interest_2, Interest_3, Interest_4, Interest_5, Tier
1.8 M
12
Flights
Id, Company, Activity_date, Origin, Destination, Class_code, Flt_number, Definition, Miles, Points,
12 M
10
Activity (not used)
Id, Definition, Issue_date, Miles, Redeposited_points, Flight_date, Class_code, Origin, Destination, Flight_number, ret_flight_date, Ret _flight_number, Ret_class_code, Ret_origin, Ret_destination, Product_code, Cash_before_premium
1.6 M
17
Previous to model construction we performed two actions:
1)
Cleaning of the data.

2)
Feature extraction.

### Cleaning

Initially, we had three tables: Passengers, Flights and Activity. Activity, which contains data related to miles transactions, was not used because it did not help improve results. The Airline suggested that this might be due to incompleteness of data. For example, by Airline policy all monetary transactions were unavailable to us and had been removed from datasets. Table  1 shows the list of fields for each table. First we removed two outlier cases for passengers whose future tier prediction is straightforward or has little merit:
1)
Passengers who fly less than one return trip in six years.

2)
Passengers with very high activity (number of flights greater than 500).

Additionally, before the data was handed over to us, the airline anonymized the id field. The id had been replaced by an alphanumeric hash code of length 32 characters long. We changed these cumbersome hash ids into integer numerical values to increase the performance. Big datasets made it very difficult for R Language to operate with hash ids during feature engineering.

### Feature extraction

The system works in the following way: For each passenger a vector is constructed considering data pertaining to the D period only. For example, if D = 15 days, then the flights and data we would consider to build the aforementioned vector are any flights between start_date and start_date +  D, where start_date is the first flight date of a new passenger. Note that start_date varies for every passenger. Now for each passenger we construct a vector that depends on three variables:
1)
The passenger data in the Tables.

2)
start_date.

3)
Length of D.

Each component of the vector is called a “feature”. The length of the vector is 634 features. The vector is equivalent to a digital fingerprint for each passenger considered in a given period. Some features are straightforward to calculate and others require complex calculations. Following we explain how each feature was calculated. We have divided the features in three groups: metric, categorical and cluster features.

#### Group 1 - Metric features

A metric feature is data that is already in numerical format. For example, age of a customer.

#### Group 2 - Categorical features

A categorical feature is a text variable, the content of which can only belong to a finite group of choices. The column “City” is such an example. The main problem with categorical variables is that they must be converted to numbers somehow. Since a computer does not understand “city names” per se, there are different ways to operate with such variables. One way is to encode code each name (aka level or category) of each categorical variable into a binary feature. Therefore, we opted for an interpretation of categorical variables using the dummy variable method as follows:
Let N be the total number of values for a given categorical variable, then we create N new “dummy” features. If a given record has value = i th level, then the i th dummy feature equals 1, otherwise 0.
Unfortunately, this transformation restricts range of algorithms that are effective: algorithms based on metrical approach such as SVM [ 8] yield poor results in such cases.
One of such features is: “ city” (Cities to where the passenger has flown during D, from the table flights). Following, we explain an example on how the categorical city variable was processed. Each city to where the airline flies is represented as a feature in the vector. If the given passenger did not fly even once to the given city during D, then the feature is set to 0, if it flew one or more times it is set to 1. The same process is performed for Class tickets letters. The categorical variables that are exploded to binary format using the dummy method are:
1)
Passenger Nationality.

2)
City of Passenger’s address.

3)
State of Passenger’s address.

4)
Country of Passenger’s address.

5)
Passenger’s Company (Employer).

6)
Flight Origin (Airport).

7)
Flight Destination (Airport).

8)
Ticket Class Code (Economy E, F, K…).

Note that the absolute order of the features must be always the same for all passengers. However, the chosen ordering is arbitrary and does not affect performance. We chose to use alphabetical order. Table  2 shows an example of how city was converted from categorical to binary.
Table 2
Example of conversion from categorical to binary for a given passenger
Feature label
Example value
City_1 (Amsterdam)
0
City_2 (Barcelona)
1
City_323 (Zagreb)
1
Class_code_1 (“A”)
0
Class_code_2 (“B”)
1
Class_code_3
1
Class_code_45 (“Z”)
0

#### Group 3. Cluster features

Once all categorical features are converted to binary ones, even if their quantity has increased, it is possible to apply clustering algorithms to identify groups of passengers with similar associated “vectors”. In this case, we applied the tried and proven k-means algorithms with different number of clusters (2, 3, 5, 7, 10, 15, and 20) to classify all passengers in clusters. The k denotes the number of clusters. So for example is k = 3 it means there are 3 clusters (A, B, C). Every passenger vector will be assigned to the “closest” cluster center as defined by Euclidean distance in n-dimensions, where n is the number of features of the vectors. At this point a cluster label (“A”, “B” or “C”) is assigned to each passenger vector. Then as before we use the dummy variable method to explode cluster labels into binary features. This was done in the following way: For each passenger all previous features (of Tables  3 and 2) are put in vector form. Then, if we are considering 100 passengers, and data of say 354 flights that satisfy the condition of belonging in the D period, then we produce 100 vectors. This is input into a k-means algorithm for k = 3. k denotes de number of clusters into which the vectors will be classified. The algorithm will attempt to classify each of the 100 vectors into k = 3 clusters, A or B or C. Using the dummy variable method we generate three new variables: Cluster_k3_A, Cluster_k3_B , Cluster_k3_C. These 3 variables will become new additional features instead of a categorical feature that says (“A”, “B” or “C”). Then if a vector (passenger) belongs to “A”, its Cluster_k3_A feature is set to 1 and 0 otherwise, conversely, if the vector belongs to “B” then only Cluster_k3_B is set to 1, and so on. This process is repeated for k = 2, 5, 7, 10, 15, 20. Table  4, shows the features and an example where a passenger vector has been classified into cluster “B” for k = 2, cluster “A” for k = 3 etc…
Table 3
Metric features
Feature label
Example
Age_of_passenger
43 (years)
Sum of miles
25500 (miles)
Average miles
2300
Interest_1
0 or 1
Interest_2
0 or 1
Interest_3
0 or 1
Interest_4
0 or 1
Interest_5
0 or 1
(1 = customer marked as interested in Interest_1 in the website when he registers).
Table 4
Example of creation of 60 cluster features by the dummy method
k
Example of K-means output ( categorical)
Label of the new binary features
Binary value
2
“B”
Cluster_k2_A
0
Cluster_k2_B
1
3
“A”
Cluster_k3_A
1
Cluster_k3_B
0
Cluster_k3_C
0
5
“C”
Cluster_k5_A
0
Cluster_k5_B
0
Cluster_k5_C
1
Cluster_k5_D
0
Cluster_k5_E
0
7
“C”
Cluster_k7_A
0
Cluster_k7_B
0
Cluster_k7_C
1
Cluster_k7_D
0
Cluster_k7_E
0
Cluster_k7_F
0
Cluster_k7_G
0
10
“A”
Cluster_k10_A
1
Cluster_k10_K
0
20
“A”
Cluster_k20_A
1
Cluster_k20_T
0
After all this process, for each passenger vector, we add these binary cluster features to the corresponding passenger vector. Table  5 shows an example. At this point it should be clear the label assigned to each feature is irrelevant. The order of the features is also irrelevant, only maintaining a consistent ordering of features in the vectors is important. Naturally, the more training vectors available, the more accurate the predictions can be. We used 50,000 vector examples to train the model. A benefit of feeding clustered features to a model is that it could help to find high order relationships between similar data points. We use a GBM model that accounts for up to 5-degree relations.
Table 5
Example of feature vectors generated per each passenger
Passenger Id
Example vector
1
(43, 13.54, …, 0, 0, 12, 0, …, 0, 1, 1, 0, 0, 0, …, 0, 0, 0)
2
(26, 2.9, …, 0, 0, 12, 0, …, 0, 1, 1, 0, 0, 0, …, 0, 0, 0)

#### Target variable

Once the vectors for each passenger are generated we need to define the target variable. This is the variable that we want to predict. In our case, the variable is 1 if the user became Silver tier during the S period as shown in Figure  2 and 0 otherwise. To generate training examples we use samples from past years and check what happened to them. This set of vectors and associated target variables constitutes a training set. Table  6 shows an example of passenger vectors from Table  5 with the corresponding target variable.
Table 6
Example of feature vectors and target variables
Passenger Id
Feature vector
Target variable
1
(43, 13.54, …, 0, 0, 12, 0, …, 0, 1, 1, 0, 0, 0, …, 0, 0, 0)
0 (he did not achieve silver status during S period)
2
(26, 2.9, …, 0, 0, 12, 0, …, 0, 1, 1, 0, 0, 0, …, 0, 0, 0)
1 (yes he did)
Now, we can consider all this vectors as a matrix, where rows are passengers, columns are features, and the last column is the target variable. We will use such a matrix to train a mathematical model with the purpose of predicting the target variable in new passengers. Once the model is trained, to predict if a passenger will attain silver status in a given time frame S (in the future) we only need to generate its feature vector by observing the passenger for a period of time D since their first flight. Once the vector is generated (naturally, without target variable) we can input it into the model and the model will output a number. There are no restrictions on when to ask a model for a prediction as long as the data for the given S period is available.

#### Model

The high nonlinearity of the features (meaning low correlation between target variable and features) restricts the number of algorithms we can use to predict with high accuracy. We chose to blend two algorithms, which were in our opinion the most appropriate for this dataset: the GBM (Generalized Boosting Machine package) and GLM (Generalized Linear Model, glm in R). Both models are trained with the same target variable: silver_attain, and try minimize the binomial deviance (Log Loss) of prediction error. While, we chose GLM and GBM because they produce different models, GLM can be considered as non-parametric version of GBM.

### GBM

The aforementioned training matrix was used to train a GBM model in R Language. This is convenient because an implementation of GBM is available as an open source library in R, so we don't need to code it from scratch. GBM has various parameters that can affect the performance of the model. For example, the number of trees (size) is one of them and usually the most important. Manual grid search was performed to determine the optimal parameter values. 75% of the data was used to train a model and 25% was used to test the prediction. The optimal GBM parameters were:
1)
distribution = “bernoulli”,

2)
n.trees = 2000,

3)
shrinkage = 0.01,

4)
interaction.depth = 0.5,

5)
bag.fraction = 0.5,

6)
train.fraction = 0.1,

7)
n.minobsinnode = 10.

A good description of the GBM implementation we used can be found in [ 9]. It works as follows: the training matrix is used to train the model with the aforementioned parameters. After, some minutes we have a trained model. Now to make a new prediction on a passenger vector or a number of passenger vectors, we input it in the model and the model will return a number from 0 to 1 for each passenger vector. 0 means that the model predicts 0% chance for that passenger to attain silver tier status within the S period of time. A 1 means that the model is the most confident the silver tier will be attained. Training the model takes about 1 hour, but asking the model to predict what will happen to 10,000 passengers takes just seconds.

### GLM

The other algorithm used is GLM. GLM stands for General Linear Model. As before we used the implementation in R Language provided by [ 10]. GLM is just a simple logistic regression where we optimize binomial deviance, where Error = silver_attain – prediction. Default parameters where used except for family which was set to “binomial”.

### Blending with grid search

Combining predictions is known to improve accuracy if certain conditions are met [ 11]. In general, the less correlation between the individual predictors, the higher the gain in accuracy. A way that usually lowers cross-correlation is to combine models of different “nature”. We chose to combine a decision tree based algorithm (GBM) and regression based one (GLM). For a 3/3 model the relative gain in accuracy due to blending was 3%. The final prediction was constructed as a linear combination of the output of the two. Grid search was used to find which linear combination was optimal on the training set. The optimal combination was 90% of the GBM prediction plus 10% of the GLM prediction.

### Loss function

A natural way to estimate the error in problems where the target variable is binary is the use of Precision and Recall so we can account for the number of false positives and false negatives. However, one of the exceptional properties of the target variable ( silver_attain) was the huge number of 0’s and small number of 1’s. This is hardly surprising: after all, by design, only a small percentage of the total passengers are supposed to attain a privileged tier. Another interesting fact is that the airline desired to determine future silver and gold passengers with highest possible accuracy; this is, a low false positive rate (mistakes). However, since the penalty cost for false negatives is not high, the acceptable precision level can be lower. These two observations enabled us to modify the usual definition of Precision and Accuracy in the following way:
$P = N 0 N 0 + N 1 A = N 0 N 0 + N 2$
(1)
where P is precision, A is accuracy, N 0 is a number of true positives or users predicted to become silver and that later indeed became silver, N 1 is the number of false positives or users predicted to become silver but who did not indeed become silver after S days and N 2 is the number of false negatives or number of users predicted to not become silver and who indeed became silver. In the next section we will show the calculated error for the normal Precision and Recall and the calculated error according to chosen loss functions (Eq.  1).

### Discussion and evaluation

It takes about 18 hours on a laptop to clean the data and construct the features. To build a D/S model takes about 1 hour per each D/S combination. Once a model is pre-calculated, making a prediction for one single passenger takes less than 350 ms (similar to a Google search).

### Compared performance

Figure 3, Tables  7 and 8 compare key performance indicators of predictive power between the previous model used by the airline (monthly extrapolation) and the D/S model. The columns P and A of Table  7 were calculated according to Eq.  1. Table  8 shows actual numbers on one particular example for D/S = 3/3 months respectively. Table  8 also shows why the calculation of Precision and Accuracy in the usual way is not suitable to assess model performance on the given dataset because of huge number of 0's. A trade-off between Accuracy, Precision, D and S clearly exists. These parameters can be adjusted to suit various forecasting needs. Additionally, Table  7 data shows that accuracy is, roughly speaking, inversely proportional to the length of S. This is, the longer the S time span the lower A and P will be. This came a bit of a surprise to us as we expected that the averaging effects of longer S time span would to facilitate prediction, but in fact the opposite is true: shorter more limited time spans lead to more accurate predictions. On the other hand, as is the case with weather forecasts, it is easier to predict events that are close in the future than those that are farther away.
Table 7
Comparison of prediction power of extrapolation vs. D/S model
Question asked D months after 1st flight: will they become Silver within S months?
Case
D
S
Extrapolation of miles Rx 1
D/S Model Rx 2P 3A 4
0
0.5
3
0.39
0.60 81% 31%
1
1
1
0.39
0.71 87% 53%
2
1
2
0.39
0.70 89% 48%
3
3
1
0.50
0.89 97% 82% 5
4
3
3
0.39
0.67 95% 46%
5
6
3
0.51
0.83 96% 69%
Notes. 1To compare predictive power between two models we use correlation as a proxy. Rx = correlation of accumulated miles with binary variable silver_attain. 2correlation of the Model’s prediction with binary variable silver_attain. 3P = Precision, of all users who were predicted to become Silver, How many of them did indeed become silver? 4A = Accuracy: Of all users who will become silver how many did we not miss? 5: See Table 8 for details on case 3.
Table 8
Precision and accuracy of silver attainment
Will a user become Silver in 1 month?
Stats
Predicted (With 3 months of data)
What really happened
Is the prediction correct?
Numbers by case
Percent of total
No
No
Yes
54752
97.09
Yes
Yes
Yes
1309
2.32
No
Yes
No
293
0.52
Yes
No
No
37
0.07

Total
56391
100
Notes: Prediction performed for each passenger with 3 month of data since 1st flight at time point: end of the three months. Precision = 2.75%. Accuracy = 81.71%.

### Confirmation of no data leakage

However, due to the spectacular high accuracy rates obtained, the airline showed a healthy concern that the prediction might be wrong due to data leakage. Data leakage happens when some data from the training set somehow contains information about what wants to be predicted ( target_variable). The only way to proof 100% that there is no data leakage is to do predictions in the future (about data that does not exist at the time of the prediction). To address this valid concern the model was used to predict what the passengers would do in the future.
To this end on February 28th 2013 we where asked to predict what existing customers would do in the future in two weeks and 3 months in the future respectively. In particular, the company asked us to predict two questions:
1)
Which of the 49572 customers that had enrolled during the last quarter of 2012 would attain silver in the first 30 days of 2013, this is a D = 3/S = 1 model and

2)
Which of the 7890 customers that had enrolled during the last two weeks of December 2012 would attain silver during the first 90 days of 2013 a D = 0.5/S = 3 model.

The data we previously had, contained information from mid 2006 but only until December 2012. So there was no possible data leakage. We made predictions but then instead of handing over all the predictions for all the passengers, we ordered the predictions by confidence from high to low, then we cutoff the predictions at confidence 0.91 and 0.48 (there was a low number of high confidence predictions in the second case), this was about 600 and 200 passengers, respectively. We sent two lists to the Airline in mid-March 23rd. Three months later, on June 9th the company sent a list with the corresponding tiers and attainment dates for each customer. The accuracy was 100% with no false positives for both models. After this proof the company was convinced about the validity of the model and asked us to provide code to integrate it with his existing CRM system.How can this knowledge help to create value? Let’s consider a flight from ADX to TYO. There are only five available seats in business class. Let’s assume that these five seats can be used to upgrade five lucky passengers. Of the 150 passengers expected to board the flight, lets assume that 20 are eligible for upgrade. The model will take less than one second to rank all the 150 passengers by probability of becoming Silver. Most importantly, it will also rank the 20 candidates. Figure  4 shows an example app. With this ranking at hand, we can now rationally allocate the five upgrades to the five customers most likely to become Silver rather than to customers with a zero probability of becoming Silver.

## Discussion

By translating passenger data from “Airline” timeline to a timeline relative to each passenger first flight, we have shown that a D/S model yields high accuracies. Furthermore, taking advantage of recently made available data mining libraries [ 9, 10] we outperformed simple extrapolation models and previous works [ 5]. False positive rates are less than 3%. The causes of a false positive have not been investigated in the scope of this project, but can be due to either/or a combination of: (1) the predictive nature of the data is not unlimited. (2) The predictive power of the model can be improved. However, the most interesting result is that the perceived value of a miles program can be increased dramatically to the very customers that matter most to the airline: the ones with high likelihood of becoming Silver.
In our experience with previous data mining projects, rather than fine tune models, the most effective way to improve accuracy is to add new features which are as uncorrelated as possible with existing features. A good place to look for potential candidates are features derived from different data sources other than the Airline CRM database, for example publicly available social media data.

## Acknowledgments

Sajat Kamal for the Aeroplan insights. Dr. Barry Green and Roy Kinnear showing themselves to be greater than their prejudices and letting us modelling the data.
Open AccessThis article is distributed under the terms of the Creative Commons Attribution 2.0 International License ( https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

## Authors’ original submitted files for images

Below are the links to the authors’ original submitted files for images.

## Competing interests

We do not have any competing interest. We do not work for any airline or have a commercial relationship with any airline or any financial interest.

## Authors’ contributions

JB developed de business logic and DE carried out the data modelling. All authors read and approved the final manuscript.
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### Benjamin Soltoff

Father, husband, political scientist, data nerd

# A Modified Draft Pick Selection Order

In preparation for teaching a new computing course for the social sciences, I’ve been practicing building interactive websites using Shiny for R. The latest Riddler puzzle from FiveThirtyEight was an especially interesting challenge, combining aspects of computational simulation and Shiny programing:

You are one of 30 team owners in a professional sports league. In the past, your league set the order for its annual draft using the teams’ records from the previous season — the team with the worst record got the first draft pick, the team with the second-worst record got the next pick, and so on. However, due to concerns about teams intentionally losing games to improve their picks, the league adopts a modified system. This year, each team tosses a coin. All the teams that call their coin toss correctly go into Group A, and the teams that lost the toss go into Group B. All the Group A teams pick before all the Group B teams; within each group, picks are ordered in the traditional way, from worst record to best. If your team would have picked 10th in the old system, what is your expected draft position under the new system?

Extra credit: Suppose each team is randomly assigned to one of T groups where all the teams in Group 1 pick, then all the teams in Group 2, and so on. (The coin-flipping scenario above is the case where T = 2.) What is the expected draft position of the team with the Nth-best record?

One could go the analytical route to solve this

But I wanted to take a computational, brute-force approach. This type of problem is ripe for Markov chain Monte Carlo (MCMC) methods, which I’ve use before in Riddler solutions.

The main task is to write a function that calculates the new draft position for a team given their current draft pick and potential assignment into one of $K$ groups. The function I wrote is:

For each simulation, I randomly sample each team into one of n_groups, then calculate draft order from worst-to-first within each group and then between groups. From this I can then calculate the expected draft position for each team given their original draft order.

So given the original problem setup, the expected draft positions for each team given random assignment into one of two groups is:

Original Draft Position Expected Draft Position
1 8.32
2 8.76
3 9.08
4 9.73
5 10.32
6 10.79
7 11.23
8 11.87
9 12.28
10 12.67
11 13.20
12 13.61
13 14.36
14 14.58
15 15.29
16 15.79
17 16.06
18 16.82
19 17.31
20 17.57
21 18.39
22 18.87
23 19.29
24 19.74
25 20.38
26 20.68
27 21.29
28 21.72
29 22.21
30 22.78

The team originally with the 10th draft can expect to have the 13th pick under this new approach.

What turned into the more complicated part was turning this function into a working Shiny app. I encourage you to try it out, as it generalizes the problem by providing expected draft picks given N teams and K groups.


text:
# Long Range

Authors: tritoke, dd Tags: misc, radio Description:

In the Heat of the Night, Some RF signal was captured at 500.5MHz from Long Range. Attachment: sig.wav

To start we are given just a single WAV file, opening this up in audacity shows us what looks to be a few small packets of data and one larger one.

On closer inspection we can see that the packets contain Frequency Modulation (FM) data - due to the squishing and stretching of the wave.

As we are told this is RF data, it is likely that each of these channels represents on part of the complex sample, i.e. one part is the real component and the other is the imaginary component.

We can use GNU radio to read the file and visualise the complex samples.

Having a look at the constellation display shows a ring when a packet is being sent and a small cluster of dots in the center otherwise, this is more evidence that we are dealing with FM data.

Lets FM-decode the data and see how it looks in audacity.

Now this is some funky data, sawtooth waves mixed with reverse sawtooth waves, I didn’t recognise this protocol at all and it took some inspired research from my teammate zeski to find that it was LoRaWAN. This was confirmed by, umm, looking at the metadata of the file :/ This also clears up why we saw what looked like FM earlier, LoRa uses Spread Spectrum Modulation which is built on top of FM.

<?xml version="1.0"?>
<SDR-XML-Root xml:lang="EN" Description="Saved recording data" Created="04-Jan-2023 13:03">
<Definition
CurrentTimeUTC="04-01-2023 13:03:07"
Filename="04-Jan-2023 210307.838 500.512MHz 000.wav"
FirstFile="04-Jan-2023 210307.838 500.512MHz 000.wav"
Folder="G:\chaitin\rw2022_lora"
InternalTag="63B5-790B-0347"
PreviousFile=""
SoftwareName="SDR Console"
SoftwareVersion="Version 3.0.28 build 2286"
UTC="04-01-2023 13:03:07"
XMLLevel="XMLLevel003"
CreatedBy="L on FXXKER"
TimeZoneStatus="0"
TimeZoneInfo="IP7//<cut long base64 string...>"
DualMode="0"
Sequence="0"
BitsPerSample="16"
BytesPerSecond="1500000"
SampleRate="375000"
UTCSeconds="1672837387"
></Definition>
</SDR-XML-Root>

So we now know the data is LoRaWAN, a bit of googling turns up rpp0/gr-lora, lets spin up the docker container and get solving.

Using the radio center frequency from the metadata and a bit of ✨ brute force ✨ we get the flag printed the debug log in hex:

>>> s = "4e 31 70 57 65 6c 63 6f 6d 65 2c 20 42 65 20 41 20 52 57 43 54 46 65 72 21 20 72 77 63 74 66 7b 47 72 33 33 74 5f 46 72 30 6d 5f 4c 6f 52 34 5f 32 36 39 33 32 38 30 32 66 32 36 61 38 63 39 62 34 35 31 39 65 62 36 66 39 30 30 66 36 37 36 66 7d 83 c3"
>>> print(bytes.fromhex("".join(s.split())).decode())
b'N1pWelcome, Be A RWCTFer! rwctf{Gr33t_Fr0m_LoR4_26932802f26a8c9b4519eb6f900f676f}\x83\xc3'

rwctf{Gr33t_Fr0m_LoR4_26932802f26a8c9b4519eb6f900f676f}

Thanks to Real World CTF challenge, it was really interesting to learn more about LoRaWAN as it wasn’t something I’d looked at in the past.


text:
# Visual Basic for Applications Reference

Visual Studio 6.0

# Exp Function

Returns a Double specifying e (the base of natural logarithms) raised to a power.

Syntax

Exp(number)

The required number argument is a Double or any valid numeric expression.

Remarks

If the value of number exceeds 709.782712893, an error occurs. The constant e is approximately 2.718282.

Note   The Exp function complements the action of the Log function and is sometimes referred to as the antilogarithm.

Show:


text:
# Conic Section Calculator

The calculator will identify the given conic section (non-degenerate or degenerate) and find its discriminant, with steps shown. Also, it will graph the conic section.

Related calculators: Parabola Calculator, Circle Calculator, Ellipse Calculator, Hyperbola Calculator

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

Identify and find the properties of the conic section $7 x^{2} - 2 x y - 22 x + 7 y^{2} - 38 y + 67 = 0$.

## Solution

The general equation of a conic section is $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$.

In our case, $A = 7$, $B = -2$, $C = 7$, $D = -22$, $E = -38$, $F = 67$.

The discriminant of the conic section is $\Delta = 4 A C F - A E^{2} - B^{2} F + B D E - C D^{2} = -2304$.

Next, $B^{2} - 4 A C = -192$.

Since $B^{2} - 4 A C \lt 0$, the equation represents an ellipse.

To find its properties, use the ellipse calculator.

$7 x^{2} - 2 x y - 22 x + 7 y^{2} - 38 y + 67 = 0$A represents an ellipse.
General form: $7 x^{2} - 2 x y - 22 x + 7 y^{2} - 38 y + 67 = 0$A.


text:
Rich text (WYSIWYG) support for django-flexible-content

## Project description

A flexible item type for django-flexible-content that implements a CKEditor

## Why?

Because django-flexible-content is designed to be as minimal as possible, its default set of item types was never meant to satisfy everyone. I designed this plugin for my own use, for obvious reasons: sometimes plain text isn’t featureful enough, and HTML isn’t a user-friendly option.

## What does it do?

It allows user-friendly input of basic text:

• paragraphs
• bold, italic, underline
• lists
• links (these can be configured)

In addition, the underlying CKEditorField itself can optionally clean the submitted HTML, removing all fields except those above. So if you want to strip outside styles and tags at all cost, it can do that. Alternatively, if you want to leave room for small HTML tweaks, you can disable this as well.

## What doesn’t it do?

It intentionally doesn’t support uploading images or files. One of the ideas behind django-flexible-content was that those things are style-able and well-managed on the back-end.

A few examples:

• If you want to display an image or download with a heading above it, a caption below it, or a description next to it, your image type can support that. A WYSIWYG editor almost never could.
• If you have the same PDF or image uploaded in 18 different places, you can choose how to handle that in your image type (offer a list of existing files; detect duplicate uploads and point all those references to a single version of the file).

Technically, you could manually link to an uploaded file, or reference an image in the raw HTML… oh well.

## How do I set it up?

1. Install the package: pip install django-flexible-content-ckeditor

2. Update the project’s settings to include the package in INSTALLED_APPS:

INSTALLED_APPS = (
# ...
'flexible_ckeditor',
)

3. Update the project’s settings to include the new CKEditor flexible item type:

FLEXIBLE_CONTENT = {
'ITEM_TYPES': (
'flexible_ckeditor.CKEditorText',

# You'll probably want to include some of the default item types,
# if you hadn't configured FLEXBILE_CONTENT['ITEM_TYPES'] yet.
'default_item_types.Image',
'default_item_types.Video',
),
}

4. Restart your server and go!

## Can I use the editor elsewhere in my project?

Yes! Though the editor may be trimmed down, it can be used to support basic formatting on arbitrary models, outside of django-flexible-content’s content areas:

from django.db import models
from flexible_ckeditor.fields import CKEditorField

class Person(models.Model):
name = models.CharField(max_length=100)
bio = CKEditorField()

Note that it’ll still abide by the settings defined in FLEXIBLE_CONTENT['ckeditor'], such as whether or not to clean the HTML upon submission.

## Project details

Uploaded source


text:
# Math Help - Boolean Expression

1. ## Boolean Expression

Hello,

Ok so I have the following equation: A'B'C + A'BC + AB'C + ABC' + ABC

I know that the most simplified form is AB + C but I cannot arrive at that solution.
Every time I try to work it out the first thing I do is Simplify the last two terms as follows: AB(C' + C) = AB

So the new equation is: A'B'C + A'BC + AB'C + AB

Now I tried removing C and getting the following: C(A'B' + A'B + AB') + AB

From here I am stuck and I have tried to solve it many other ways but always get stuck or eliminate something I don't want to. Help would be much appreciated

2. A'B'C + A'BC + AB'C + ABC' + ABC

= A'B'C + A'BC + AB'C + ABC + ABC'
= C + ABC'
Let AB = X, then we have
C + XC'
= C(X+X') + XC'
= C(X + X' + X) + XC'
= C(X + X') + X(C+C')
= C + X
= C + AB

3. The problem is that the expression by which C is multiplied, i.e., A'B' + A'B + AB' is not a tautology. We can turn it into one by adding a second copy of ABC to the original expression. The expression is then broken into two parts: ABC + ABC' = AB and A'B'C + A'BC + AB'C + ABC = C(A'B' + A'B + AB' + AB) = C.

By the way, the original formula is not an equation; an equation must have an = sign.


text:
Home > CALC > Chapter 1 > Lesson 1.5.1 > Problem1-194

1-194.

NOT AGAIN!
Theo has done it again. Before he lost his graph, he used it to determine the following properties of his motion. Help him re-create a possible graph of this motion.
DETAILS:

• He changed directions three times during his $8$-foot walk.

• His average velocity was $0$ feet per second.

• Theo walked for $6$ seconds and started $5$ feet from the motion detector.

Clue 1: A change in direction means the slope goes from positive to negative or negative to positive. An $8$-foot walk means that, when you add together all the distances traveled between each high and low point on the graph, the sum is $8$.

Clue 2: An average velocity of $0$ means that Theo should end exactly where he started.

Clue 3: If Theo walked for $6$ seconds, then the domain of the graph is restricted to $0≤x≤6$. If Theo started $5$ feet from the motion detector, then the $y$-intercept is $(0,5)$.


text:
To determine.The correct graph for the function[g(x)=-frac{1}{2}f(x)+1 is B

To determine.
The correct graph for the function $g\left(x\right)=-\frac{1}{2}f\left(x\right)+1$ is B

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Asma Vang

Vertical translation:
For ,
The graph of is the graph of $y=f\left(x\right)$ {shifted up b units}
The graph of is the graph of {shifted down b units}
Reflection:
Across the x-axis:
The graph of is the reflection of the graph of $y=f\left(x\right)$ across the x-axis
Across the y-axis:
The graph of is the reflection of the graph of $y=f\left(x\right)$ across the x-axis
Vertically Stretching or Shrinking:
The graph of can be obtained from the graph of $y=f\left(x\right)$ by
Stretching verttically for or shrinking vertically for .
For $a<0$, the graph is also reflected across the x-axis
. By the properties of transformation, the graph of $g\left(x\right)=-\frac{1}{2}f\left(x\right)+1$ is,
The transformation of $y=f\left(x\right)$ and shri
vertically by a factor of $\frac{1}{2}$
Then the graph is reflection about the x-axis and shifted up one unit.
From the given graphs, the graphs B satisfies the all condition's stated above.
Thus, the correct graph for the function $g\left(x\right)=-\frac{1}{2}f\left(x\right)+1$ is B


text:
## Calendar:

The calendar on the syllabus is the plan for this semester, this calendar reflects in detail what we actually get to cover in class and when.

• 1-30-2023 – …
• 1-23-2023 – We began by going over the syllabus, discussed how the class would run and the expectations for this class. We spent the rest of class working through the review slides (Review Material). Be sure to study the slides, and chapter 0 of your text if you have it, for the quiz at the start of class next week.

## Assignments:

#### All out of class work in this class must be typed up (double spaces, 1in margins, 12pt font) and in complete sentences.

• Chapter 1: p.83 – 1.3, 1.4c, 1.5c, 1.6ej, 1.12, 1.18ej, 1.21b, 1.22, 1.28a, 1.29b Due 2/13/2023
• Chapter 2: p. 154: 2.1, 2.2, 2.4bc, 2.5bc, 2.11, 2.13, 2.14, 2.15, 2.16 Due 3/20/2023
• Chapter 3: p.187: 3.1cd, 3.2cd, 3.8b, 3.9, 3.15bc, 3.16bc Due 4/3/2023
• Chapter 4: p.210: 4.3, 4.4, 4.6 4.7 4.11, 4.13 Due 4/17/2023
• Chapter 5: Chapter 5: p.239 – 5.1 (use Theorem 5.13 and show that $$ALL_{CFG}\leq_m EQ_{CFG}$$), 5.2 (Show, directly, that  $$\overline{EQ}_{CFG}$$ is recognizable), ?? Due 5/1/2023
• Chapter 6: p.270 – 6.1, 6.11 (look at 6.10), 6.28 (look at 6.12), ??, Due 5/12/2023 by 5pm
• Extra Credit: Turing Machine Class Due 5/12/2023 by 5pm
• Extra Credit: Exploring the Infinite Due 5/12/2023 by 5pm

• Definitions:
• ??

• Definitions:
• ??

• Definitions:
• ??


text:
1. ## cos2pi

why is cos2pi = +- 1? its not logical, since cosA = sinA/tanA and sin A = tan A = 0....
??

2. Originally Posted by furor celtica
why is cos2pi = +- 1? its not logical, since cosA = sinA/tanA and sin A = tan A = 0....
??
$2\pi$ radians is a full revolution of the unit circle. So you end up back exactly where you started at an angle of 0, and so the horizontal distance (the cosine) will be exactly the same.

What's $\cos{0}$ equal?

BTW you can't work out $\cos{2\pi} = \frac{\sin{2\pi}}{\tan{2\pi}}$ because the function $\tan{x}$ is defined by $\frac{\sin{x}}{\cos{x}}$.

It would be the same as saying you're trying to find $x$ if $x = x$...


text:
overly {dave} R Documentation

## Overly of multivariate time series

### Description

This function attempts to superimpose (overlay) multivariate time series that typically stem from different locations to form one single series. Series are combined pairwise following the arrangement in a minimum spanning tree, spantree. The result is a time series which may be longer than the contributing series, but shorter than their length total.

### Usage

overly(veg, Plot.no, y, sint, ...)
overly2(veg, Plot.no, y, sint)

## Default S3 method:
overly(veg, Plot.no, y, sint, ...)
## S3 method for class 'overly'
plot(x,...,colors=NULL,l.widths=NULL)

### Arguments

 veg This is a vegetation data frame, releves are rows, species columns. It is assumed that it constitutes two or more time series Plot.no Plot names. Rows with identical name are assumed to belong to the same time series. y Transformation of species scores: x'= x exp(y) sint Length of time interval. Affects display only. ... Parameters colors=NULL, l.width=NULL, colors and line widths in plots. colors A vector of colors, such as c(1,2,3). The entries are recycled upon printing. l.widths A vector of line widths used for plotting, e.g., c=(0.5,1,1.5,2).The entries are recycled upon printing. x An object of class "overly"

### Details

In plant ecology this procedure is also known as space-for-time substitution. See also pco,spantree.

### Value

An object of class "overly" with at least the following items:

 plot.labels Names of plots, see Plot.no above n.tseps The resulting (synthetic) number of time steps tseps A vector of time steps in time units tser.data The resulting vegetation time steps ord.scores The pco scores of the ordination of time series d.mat Euclidean distance matrix of time series vegraw Input vegetation data frame veg (see above) linex1 Starting address of the time series in the synthetic time frame linex2 End address of the time series in the synthetic time frame ltex The plot names sint The time interval (see above) vegtypes The species names involved

Otto Wildi

### References

Wildi, O. and Schuetz, M. 2000. Reconstruction of a long-term recovery process from pasture to forest. Community Ecology 1: 25–32. Wildi, O. 2017. Data Analysis in Vegetation Ecology. 3rd ed. CABI, Oxfordshire, Boston.

### Examples

o.overly<- overly(sn59veg,sn59sit\$Plot.no,y=0.5,sint=5)
# Plotting (i) minimum spanning tree, (ii) temporal arrangement of time series,
# (iii) synthetic multivariate time series
plot(o.overly,colors=NULL,l.width=NULL)

[Package dave version 2.0 Index]


text:
# Give an example of a function $f$ which has a simple pole at $z=i-3$ with residue $8(i-3)$ and an essential singularity at $z=i$ with residue $6$

Give an example of a function $f$ which has a simple pole at $z=i-3$ with residue $8(i-3)$ and an essential singularity at $z=i$ with residue $6$.

I know that the function $-24/(z+3-i)$ would have a pole at $z=i-3$ with residue $8(i-3)$ but I cannot seem to make this function also have an essential singularity. Would making this function trigonometric help.

Hint. Try something of the form $$f(z)=e^{1/(z-i)^2}+\frac{A}{z-i}+\frac{B}{z-(i-3)}$$ where $A$ and $B$ are complex numbers to be found. The essential singularity is given by $e^{1/(z-i)^2}$ which has residue $0$ at $i$ and at $i-3$.

• I see that the last part of that function is where the pole would be however where is the essential singularity? – user544158 Mar 21 '18 at 14:12
• In $\exp(1/(z-i)^2)$ that has residue $0$ at $i$ – Robert Z Mar 21 '18 at 14:28
• Recall that $e^{1/w}=\sum_{k\geq 0}(1/w)^k/k!$. – Robert Z Mar 21 '18 at 14:32
• Is it clear now? Are you able to find $A$ and $B$? – Robert Z Mar 21 '18 at 14:50
• I found that if B is equal to 8(i-3), this will give me the pole i-3 with residue 8(i-3). However I am struggling to find the residue 6 for the singularity at i. – user544158 Mar 21 '18 at 14:53

Hint:

1. $e^{1/(z-i)}=1+\frac1{z-i}+\frac1{2(z-i)^2}+\dots$ has an essential singularity at $z=i$ with residue $1$.

2. $\frac1{z+3-i}$ has a simple pole at $z=i-3$ with residue $1$.

• Would you please be able to explain why hint number 1 has residue 1? – user544158 Mar 21 '18 at 15:00
• Remember that the residue at $z=i$ is the coefficient of the $\frac1{z-i}$ term in the Laurent expansion about $z=i$ – robjohn Mar 21 '18 at 16:09


text:
# Reference Request: Korevaar and Schoen Spaces

Let $$(X,d,m)$$ and $$(Y,\rho,\mu)$$ be doubling metric measure spaces. The seminal work of Korevaar and Schoen discussed generalizations of $$L^p$$ spaces for maps from $$X$$ to $$Y$$. Standard results from Analysis show that when $$X=\mathbb{R}^d$$ and $$Y=\mathbb{R}^D$$ then $$C_c(\mathbb{R}^d,\mathbb{R}^D)$$ (continuous functions with compact support) is dense in the Bochner-Lebesgue space $$L^1(\mathbb{R}^d,\mathbb{R}^D)$$.

I'm wondering, if this is true for the $$L^p(X,Y)$$ spaces of of Korevaar and Schoen and more general if there are extensions of $$L^{p(x)}$$ spaces, Musielak-Orlicz spaces, between $$(X,d,m)$$ and $$(Y,\rho,\mu)$$ for which $$C(X,Y)$$ is dense in those spaces.


text:
# Unable to connect to remoteJMX but able to Telnet on same port after setting Dcom.sun.management.jmxremote environment variables?

I am rather perplexed at this point as we have a need to connect through remotejxm to monitor the JDK on a server through a monitoring application but I am repeatedly being told when trying to connect that it is not running. We have checked the ports and the 8999 port is open but its not letting me connect through remotejxm but I am able to connect via telnet on that same port, I have the Java options set as follows (we are running Tomcat on Windows server 2003). Any advice on what else I could look into would be greatly appreciated

``````-Dcatalina.home=C:\Program Files\Apache Software Foundation\Tomcat 6.0
-Dcatalina.base=C:\Program Files\Apache Software Foundation\Tomcat 6.0
-javaagent:C:\Program Files\Apache Software Foundation\Tomcat 6.0\AppManager\WebTransactionAgent.jar
-Djava.endorsed.dirs=C:\Program Files\Apache Software Foundation\Tomcat 6.0\endorsed
-Djava.io.tmpdir=C:\Program Files\Apache Software Foundation\Tomcat 6.0\temp
-Djava.util.logging.config.file=C:\Program Files\Apache Software Foundation\Tomcat 6.0\conf\logging.properties
-verbose:gc
-XX:+PrintGCDetails
-XX:+PrintGCDateStamps
-XX:+PrintHeapAtGC
-XX:+PrintTenuringDistribution
-Xloggc:C:\Program Files\Apache Software Foundation\Tomcat 6.0\logs\heap.log
-XX:MaxPermSize=512m
-XX:+UseConcMarkSweepGC
-XX:NewSize=1024m
-Dfile.encoding=UTF-8
-Xdebug
-Dcom.sun.management.jmxremote
-Dcom.sun.management.jmxremote.port=8999
-Dcom.sun.management.jmxremote.authenticate=false
-Dcom.sun.management.jmxremote.ssl=false
``````
-


text:
View Single Post
2021-06-28, 10:34   #11
henryzz
Just call me Henry

"David"
Sep 2007
Liverpool (GMT/BST)

22·1,499 Posts

Quote:
 Originally Posted by mart_r Here's the updated table 7 for you, including the Pari program where I previously forgot to adjust the variable names to the text. And here's a decuplet too: (193,057th prime of s = 289) * 1,889 + {220, 238, 378, 624, 934, 1048, 1414, 1612, 1678, 1750} where the 193,057th prime of s = 289 is Code: 31622342567119297681076089891955816714567540280951897365387154057445488363818198016103621097928354615483039036120985961432365652251514164595044179903724253210032238290899811114046504503777054229122725692603678959245884842428958171279396512793345739562301201289205906157466250260424464347285786720763386343358970437066090084387214846787188810038323800022249321968498343734757162251150197635271251591141913929920578564323411668207833015231803312413278100502072679946962206773893320130362397698036981816468531752935920178488432680654626988222331983116303339502122373584059843024086090134096811878534903483185031614489646009414820924548284793618591561552678558457275689527829154639089241831525443509583812990724843963675707936160940952072970697432023165485601359791369860896277765474897467473349879301 Yay! *Throws up a single confetti* Or is that called a confetto? Oh! Look at how the number starts off... wait... contfrac((log(a)/log(10)) = 796; 2, 41939, 2, 101, 1, 7, ... - nice! Now at p=1931... but, you know, the year 1931 was not an especially nice one in terms of history. I'll continue to crunch the numbers down. I am unstoppable!! You know, it only recently occured to me that those numbers have a subtle crude sense of humour, looking at the point where the first sextuplet appears.
It's a shame that extending the search made so little difference lower down. From 599 to 797 the difference is only 1. I was hoping that more could be eliminated. Maybe some more could be if the easiest targets are attacked rather than everything.