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September 10

Twisted Rack Cohomology

I saw a reference, in a paper on entropic algebras, about "twisted rack cohomology" with a reference to a paper I don't have access to, "Twisted quandle homology theory and cocycle knot invariants". Our article Racks and quandles mentions a quandle cohomology, but doesn't go into any details- does anyone have, or know where I can find, details on quandle cohomology and what "twisted" refers to? Does it have any relation to Twisted K-theory?Phoenixia1177 (talk) 05:25, 10 September 2013 (UTC)[reply]

Never mind, the paper is freely available! I don't know how I missed that, it's the first link that comes up in Google.Phoenixia1177 (talk) 05:36, 10 September 2013 (UTC)[reply]

If someone wants to remove this, please feel free. I'm not sure how to.Phoenixia1177 (talk) 05:36, 10 September 2013 (UTC)[reply]

nLab also has a nice discussion of twisted cohomology. --Mark viking (talk) 06:16, 10 September 2013 (UTC)[reply]

Thanks for the link:-) That's an interesting site, by the way:-)Phoenixia1177 (talk) 06:20, 10 September 2013 (UTC)[reply]

Power rule in calculus

Can someone explain to me the differences in the layout of the power rule? I have two versions of it but I don't know how or why they differ.

They are:

${\frac {d(x^{n})}{dx}}=n\cdot x^{n-1}$

And:

${\frac {d(u^{n})}{dx}}=n\cdot u^{n-1}{\frac {du}{dx}}$

I guess I don't understand the significance of the letters and what they represent because, before the equal sign, the first looks identical to the second. The only difference being the letters that are being used. Thanks, Dismas|^(talk) 05:46, 10 September 2013 (UTC)[reply]

No, if you changed the letter from x to u, you would get

{\frac {d(u^{n})}{du}}=n\cdot u^{n-1}

. Note that the left hand side has a u in the bottom, not an x as in the left hand side of your second expression. Looie496 (talk) 05:51, 10 September 2013 (UTC)[reply]

Usually, in textbooks I've seen, u denotes any differentiable function of x. The second case includes provisions for the chain rule.--Jasper Deng (talk) 05:53, 10 September 2013 (UTC)[reply]

There is a difference between the two left hand sides, both have a dx on the bottom, it's x -> u in the numerator only. In that case, it tells us that we are taking the deriv. of u with respects to x- so the first case is the second with u(x) = x.Phoenixia1177 (talk) 05:59, 10 September 2013 (UTC)[reply]

Jasper, I understood your response the best. I see what is meant now. Thank you! Dismas|^(talk) 06:21, 10 September 2013 (UTC)[reply]

Resolved

Triangular Prism Die.

I asked a more general question that got bogged down a few days ago, and I thought I'd ask it here. If a triangular prism of sides 1 (the side of the triangle) and x (the sides connecting the triangles) is thrown like a die, what is x so that all five sides are equally likely to come up?Naraht (talk) 14:41, 10 September 2013 (UTC)[reply]

I'm not certain how to go round this, but if you just want a die that gives 1 in 5 chances then you can use an icosahedron which has the advantage of being visibly fair. Dmcq (talk) 15:07, 10 September 2013 (UTC)[reply]

Yes, or a pentagonal or dodecagonal trapezohedron or pentagonal or dodecagonal bipyramid or even a Zocchihedron. I know there are other options for a die for a 1 in five chance (or even reroll a d6 if it comes up a 6). That's why I tried to ask it *very* specifically.Naraht (talk) 15:49, 10 September 2013 (UTC)[reply]

Here is a patent for such a die, with precise measurements included. Whether is it a fair die is another matter. Some say that the faces should have equal area for fairness, others say equal solid angle. But the position of the center of mass surely plays a role in the physics, too. --Mark viking (talk) 16:33, 10 September 2013 (UTC)[reply]

Yes, the fairness is claimed but not proven. Also, even if 1-5 are equiprobable, they cannot sum to 1, because the die could also land on the beveled edges labeled 22, 24. It would be unlikely, but the chance is still not zero :) SemanticMantis (talk) 18:54, 10 September 2013 (UTC)[reply]

The patent says that fairness was determined experimentally, in other words it wasn't derived using math. There is no guarantee that if the die is rolled on a a different surface or made of different material then the probabilities would not change. --RDBury (talk) 20:09, 10 September 2013 (UTC)[reply]

My own intuition is this could only be determined experimentally. Having done simulations of such problems even such a simple case like this is chaotic and so non-deterministic. It's simple enough though that it could be simulated on a computer, letting you do millions of trials and so estimate x with a high degree of precision.--JohnBlackburne^words_deeds 20:22, 10 September 2013 (UTC)[reply]

It can't land on the beveled edge "22" (or the beveled edges between the "22" sides) since it would not be stable (it would fall on the triangular side), OTOH, *theoretically*, the die could stand on the "24" side. But I wonder if what required the testing was the beveled sides.Naraht (talk) 21:26, 10 September 2013 (UTC)[reply]

From reading the spec, it appears that the bevelled edges are just to "make the shape of the present invention 10 interesting and ... capture and radiate illumination which enhances the lustrous and sparkling gem-like appearance", rather than to enhance the rolling (though it's entirely possible that this is an attempt by the patent author to mislead those trying to replicate the design and inadvertently make his own patent easier to invalidate for insufficiency). For those worrying about making their own, the patent has expired due to failure to pay renewal fees 4 years ago, so it's well beyond ressurection (in the US at least, I've not checked for related patents in other jusridictions). I could rant about those claims all day... the main claim is OK, the dependents are awful... MChesterMC (talk) 08:34, 11 September 2013 (UTC)[reply]

Neither fairness nor unfairness can be proven experimentally. Fairness and probability are matters of faith. I believe that if the five sides have equal solid angle as seen from the center of mass, then the die is fair. I may change my mind when confronted with theoretical arguments or experimental facts, but even that does not prove unfairness. Bo Jacoby (talk) 08:52, 11 September 2013 (UTC).[reply]

I'm pretty certain that's wrong. If you have a hexagon and roll it along it has 1/6 chance of coming up on each side. If you cut it down to a rectangle of 1 x sqrt(3) then by that argument it should end up on the short ends one third of the time. I'm pretty certain it will now end up on one of the long sides more than 2/3 of the time even though they take up only 2/3 of the angle and less than 2/3 of the perimeter. Dmcq (talk) 13:41, 11 September 2013 (UTC)[reply]

Do you have any argument besides being 'pretty certain'? By the way, I did not say that double solid angle means double probability, but only that equal solid angle means equal probability. Are you pretty certain that's wrong? Do you have suggestions for improvement? Bo Jacoby (talk) 19:33, 11 September 2013 (UTC).[reply]

(Assuming you meant any n and not just 5:) I don't think that there is a separate "equal solid angles" rule because I can subdivide all the faces of a die into many (almost) equal-solid-angle pieces. It certainly can't just be proportional to solid angle, because a face can be impossible (by having the center of gravity project outside it) and yet still subtend solid angle from the CG. Even if we reject such dice, I think the answer requires a precise definition of the "throwing the die" process in order to define probabilities and thence fairness. Certainly, throwing a die in a very viscous fluid will tend to follow the solid angles (because there is no rotation, so all that matters is the face through which the weight vector points). --Tardis (talk) 13:02, 12 September 2013 (UTC)[reply]

Let's stick to the problem at hand, the triangular prism of sides 1 (the side of the triangle) and x (the sides connecting the triangles). Do you have a better suggestion for constructing a fair die, or are you merely providing negative criticism? Bo Jacoby (talk) 13:42, 12 September 2013 (UTC).[reply]

It seems that even if you manage to compute x such that the die is actually fair, you can't convince people that this is the case, and so it is pointless after all. Bo Jacoby (talk) 16:42, 16 September 2013 (UTC).[reply]

Question about notation for partial derivatives and integrals

If ⌈x⌉ means the ceiling of x, i.e. the smallest integer greater than or equal to x, then why does ∂f/∂xi mean the partial derivative of f with respect to xi, where f is a function on (x1, …, xn)? Also, does ∫ab f(x) dx means the signed area between the x-axis and the graph of the function f between x = a and x = b? --Marticleeë (talk) 17:58, 10 September 2013 (UTC)[reply]

I don't understand the "if ... then" part of your question, but if you mean

\int _{a}^{b}\!f(x)\,dx\,

for the "also" part then yes, see Integral. Dbfirs 21:56, 10 September 2013 (UTC)[reply]

September 11

If the plane was fly in Atmosphere by escape velocity from earth's gravity . what will happen?

If the plane was fly in Atmosphere by escape velocity from earth's gravity . what will happen? — Preceding unsigned comment added by 37.238.81.178 (talk) 08:42, 11 September 2013 (UTC)[reply]

Escape_velocity on Earth is about 40,000 km/hr. I do not know of any planes that come even close to that, so it is not possible.196.214.78.114 (talk) 09:40, 11 September 2013 (UTC)[reply]

This Escape_velocity#Orbit explains why.196.214.78.114 (talk) 09:41, 11 September 2013 (UTC)[reply]

—but I say if! — Preceding unsigned comment added by 37.238.81.178 (talk) 10:25, 11 September 2013 (UTC)[reply]

Then it would break apart due to the stress of travelling at hypersonic speeds, or burn up due to air friction. Gandalf61 (talk) 10:52, 11 September 2013 (UTC)[reply]

The cabin (because it is not designed to) would not be able to maintain pressure at those altitudes - lots of nasty things happen then. E.g. Depressurization. You'd also probably freeze to death. http://www.kansasflyer.org/index.asp?nav=Avi&sec=Alti&tab=Theory&pg=2 196.214.78.114 (talk) 12:28, 11 September 2013 (UTC)[reply]

Isn't it much more likely that you'd be burnt to a crisp. Sławomir Biały (talk) 12:40, 11 September 2013 (UTC)[reply]

Maybe on re-entry. But seriously, all this speculation of ours is based on varying altitudes, velocities, structural integrity of the plane, its propulsion system(s), and what you consider a "plane". etc. These is no way to answer this in a couple of lines. 196.214.78.114 (talk) 12:55, 11 September 2013 (UTC)[reply]

An airplane gets it's thrust from accelerating air away from the engines at high speeds. Therefore, as the air gets thinner, the thrust lessens, making it impossible to get to escape velocity. However, some type of a hybrid is possible, say where the bottom stage is an airplane, then a more conventional rocket is launched from the high point of the airplane's flight. This could make it possible to launch a given payload into space using less fuel, and thus a far smaller rocket, but the additional complexity is an argument against it. StuRat (talk) 12:47, 11 September 2013 (UTC)[reply]

But a rocket never gets close to escape velocity during launch - it doesn't need to since it is under continuous propulsion. The short answer is that no practical propulsion system could sustain a speed anywhere near escape velocity in the lower atmosphere; nothing that structurally resembled a plane (i.e. was not a solid cannonball) could maintain structural integrity at such speeds; and in any case almost all known materials would be rapidly melted by the high temperatures caused by air friction, shock waves etc. Gandalf61 (talk) 13:57, 11 September 2013 (UTC)[reply]

I never said it did. I was merely answering a more general Q of "could a plane be used to get into space". Admittedly this isn't exactly what was asked, but it seemed relevant. StuRat (talk) 11:49, 12 September 2013 (UTC)[reply]

The question, unfortunately, is a hypothetical that needs more conditions specified before any sensible answer can be given, since nature of propulsive and aerodynamic forces must be specified to distinguish wildly differing answers (fall to the ground (under any realistic assumption without power, craft staying intact), go into orbit, returning to atmosphere (neglecting air drag), be destroyed (typical aircraft construction), etc.; e.g. I've seen a proposal that would allow these speeds by explosively parting the air ahead of the craft, allowing it to travel in a near-vacuum using magnetohydrodynamics for propulsion; unexpected results are feasible, for example supersonic underwater speeds can be sustained). This question actually belongs at Wikipedia:Reference_desk/Science, not here. — Quondum 13:01, 11 September 2013 (UTC)[reply]

See also the Kármán line article. Count Iblis (talk) 14:11, 11 September 2013 (UTC)[reply]

The What-If section of xkcd has a good explination of this [1]. Its not getting into space which is different, its gaining enough horizontal velocity so that you can go into orbit which takes the energy.--User:Salix alba (talk): 19:54, 11 September 2013 (UTC)[reply]

Marvelous piece! I might quibble with one line of it, where he says "If you're in [...] Seattle [...], space is closer than the sea". That clearly implies not only that he doesn't count Puget Sound as being part of "the sea" in spite of connecting with it (I imagine a lot of people would agree with him there), but that not even the Strait of Juan de Fuca is part of "the sea", which is less intuitive to me. --Trovatore (talk) 20:38, 11 September 2013 (UTC)[reply]

It also didn't mention that while the distance to space isn't great, the problem is the air resistance on the way, which requires more fuel to overcome, and then you need a larger rocket to contain that fuel, and the added weight of that fuel and bigger rocket requires still more fuel, etc. So, if you can skip that thick atmosphere, you can get quite an advantage. StuRat (talk) 11:56, 12 September 2013 (UTC)[reply]

At the Science Desk "not feasible" might be a reasonable answer. However, since the question is at the Mathematics Desk, let us attempt an answer on the supposition that (for some sufficiently great altitude), the technology is available to travel through the atmosphere this fast. So, first note that the escape velocity starting from a particular altitude is

\scriptstyle {\sqrt {2}}

times the orbital speed at the same height. This means that travelling parallel to the surface of the earth at

\scriptstyle {\frac {1}{\sqrt {2}}}

times (i.e. about 71% of) the escape velocity, you will be in orbit and will need no aerodynamic lift to maintain your altitude. If you will countenance explanation in terms of inertial forces, then the centrifugal force due to following the curvature of the earth, balances the gravitational attraction of the earth. If you wish to travel any faster and still follow the curvature of the earth, then an additional downward force will be required, and this could be provided aerodynamically. The most comfortable and convenient method would be to fly the plane upside-down so that its normal aerodynamic lift supplements gravity and presses the plane towards the earth. Since the necessary centripetal force varies like the square of the tangential speed, and the target speed is

\scriptstyle {\sqrt {2}}

times the orbital speed, then the necessary force will be twice that provided by gravity - so as much again as is provide by the weight of the plane. This means that the plane will need to provide a lift equal to its weight (as is usually the case with aircraft!) but towards the earth rather than away from it. The pilot will experience a g-force of +1 (i.e. her normal weight), pressing her into her seat. --catslash (talk) 18:40, 14 September 2013 (UTC)[reply]

Regarding Goldbach's Conjecture

I have been looking around for some feedback - i.e., opinions, comments, or criticsm - on the following; could you folks be so good as to help.

LEMMA. Goldbach conjecture's can be split into three separate but similar problems: 2N mod 6 = 4; 2N mod 6 = 2; 2N mod 6 = 0.

For 2N mod 6 = 4, by inspection the primes for no solution can be 6j+1, therefore they must have the form 6j-1. For 2N mod 6 = 2, by inspection the primes for no solution can be 6j-1, therefore they must have the form 6j+1. For both 2N mod 6 = 4 and 2N mod 6 = 2, one solution may be possible for each 2N from 3 and 2N-3 if N is prime. From the above, we see that for both 2N mod 6 = 4 and 2N mod 6 = 2, the two primes for any solution differ by 12k+6 if N is even and by 12k if N is odd. where k is non-negative.

For 2N mod 6 = 0, we see by inspection that the primes for all solutions are such that one prime must have the form 6j+1 and the other prime must have the form 6j-1. Also, there is one solution from 3 and 2N-3 for 2N=6. For 2N mod 6 = 0, the two primes for any solution differ by 4k. where k is positive.

THEOREM. Statistical (heuristic?) proof of Goldbach's conjecture.

For 2N mod 6 = 0, for sufficiently large N, consider Pn, the largest prime less than 2N-1 and having the required form (see above lemma). There is always a complementary number A such that A+Pn=2N, where the probablity of A being prime is 1/ln(A). We do the same with Pm, the largest prime less than Pn: there is a complementary number B such that B+Pm=2N, and the probability of B being prime is 1/ln(B). We continue down to Ph, the smallest prime larger than N: there is a complementary number G such that G+Ph=2N, and obtain a probability of 1/ln(G).

The probabilities of A,B,...G being prime range from 1/ln(A), a relatively large number, through 1/ln(G). a relatively small number barely larger than 1/ln(N). We note that the probability of the complementary number in the middle of 1 through N of being prime is 1/ln(N/2): the "upper" half of the logarithms are between 1/ln(N) and 1/ln(N/2), while the "lower" half of the logarithms are between 1/ln(N/2) and 1/ln(A), where A is defined in 8 above. Consequently, we can take as a minimum the average probability of a complementary number A,B,...,G being prime as 1/ln(N/2).

But, since A,B, etc. cannot be divisible by 2 or 3, thus the probability of A,B, etc. being prime is actually larger than 1/ln(A), say 1/ln(A/3). Thus, we take as a minimum the average probability of a complementary number A,B,...,G being prime as 1/ln(N/6), and NOT 1/ln(N/2)..

From N to 2N, there are 2N/ln(2N) - N/ln(N) primes; namely, the number of primes from 3 through 2N minus the number of primes from 3 through N. The (number of primes) multiplied by (the minimum average probability of the complementary number being prime) equals (2N/ln(2N) - N/ln(N)) * (1/ln(N/6)). This can be further approximated as N/((ln(N))^2); do note that this minimum number of solutions for Goldbach's conjecture is for 2N, and NOT for N.

For 2N mod 6 = 4 and 2N mod 6 = 2, N may be a prime and as such is a solution to Goldbach's conjecture (for N+N), but it is only one vanishing small possibility.. The calculations for 2N mod 6 = 4 and 2N mod 6 = 2 are similar to those for 2N mod 6 = 0, except they use only one half of the primes, so that the minimum number of solutions for Goldbach's conjecture is one half of that for 2N mod 6 = 0, or (N/2)/((ln(N))^2).

COROLLARY. The minimum number of solutions for Goldbach's conjecture for 2N mod 6 = 4 and 2N mod 6 = 2 is half of that for 2N mod 6 = 0.

This follows immediately from the above theorem, where the minimum number of solutions for 2N mod 6 = 0 is N/((ln(N))^2), while the the minimum number of solutions for 2N mod 6 = 4 and 2N mod 6 = 4 is (N/2)/((ln(N))^2. (This difference between the behavior of 2N mod 6 = 0 and the other two cases can be seen from the graph (goes to 2,000) in the Wikipedia article titled Goldbach's Comet, as well in the graph (goes to 1,000,000) in the Wikipedia article titled Goldbach's Conjecture, where this latter graph appears to be an extension of the former.)

COMPARISONS. Two comparisons of theory with reality.

For 2N=2000, the Goldbach comet number of solutions for 2N mod 6 = 0 is at least 90, and for 2N mod 6 = 4 and 2N mod 6 = 2 is at least 40. For 2N=2000, N=1000, ln(1000) is very close to 6.9, so that (ln(1000))^2 is around 47.7. This yields 21 for 2N mod 6 = 0, and 10.5 for 2N mod 6 = 4 and 2N mod 6 = 2, or about 1/4 of the actual values in both cases.

For 2N=1,000,000, the Goldbach comet number of solutions for 2N mod 6 = 0 is around 8,000, and for 2N mod 6 = 4 and 2N mod 6 = 2 is around 4,000. For 2N=1,000,000, N=500,000, ln(500,000) is a bit more than 13.1, so that (ln(500,000))^2 is a bit more than 172. This yields 2,900 for 2N mod 6 = 0, and 1,450 for 2N mod 6 = 4 and 2N mod 6 = 2, or somewhat more than 1/3 of the actual values in both cases. Bh12 (talk) 13:35, 11 September 2013 (UTC)[reply]

The difficulty with this is that it's a probabilistic argument. It essentially takes the expected number of primes in (N, 2N) times the probability that the complement of each is also prime; this product is the expected number of ways that 2N is the sum of two primes. The problem with this is that it does not preclude a smaller than expected number of ways; any counterexample to the Goldbach Conjecture is necessarily going to be a low probability event. Duoduoduo (talk) 20:41, 11 September 2013 (UTC)[reply]

Agreed. But what about the other two points mentioned above; viz., the breakdown of Goldbach's conjecture into three parts depending on the value of 2N mod 6, and that the minimum number of solutions for 2N mod 6 = 0 is twice that for each of the other two cases? Are these points new or not, and if they are new, do they have any significance?Bh12 (talk) 01:40, 12 September 2013 (UTC)[reply]

Not new. See, for instance, Mathematical Mysteries, by Calvin Clawson, p. 243. Bubba73 ^{You talkin' to me?} 02:06, 12 September 2013 (UTC)[reply]

Sigh... So it's back to the drawing board. — Preceding unsigned comment added by Bh12 (talk • contribs) 07:32, 12 September 2013 (UTC)[reply]

Question: Should the fact about there being three types of 2N mod 6, as well as the fact that the number of solutions for one of them is twice that of either of the others, be mentioned in the two Wikipedia articles titled Goldbach's Conjecture and Goldbach's Comet? I think they should - they are not at all new (known since at least 1996), and (in my opinion) they are important - but I'm biased in the matter.Bh12 (talk) 08:06, 12 September 2013 (UTC)[reply]

Neither belongs in the articles. Any problem involving natural numbers could be split into numbers that are 0, 1 or 2 modulo 3. For even numbers that means 0, 2 or 4 modulo 6. This is a trivial observation and we don't know whether it will help solve the conjecture. There are infinitely many other ways to partition the even numbers into subsets.

Your "minimum number of solutions" is based on very loose estimates which only consider divisibility by 3. The articles already have far better estimates. Any odd prime factor p in N will increase the expected number of solutions by a factor (p-1)/(p-2). This general factor is in both Goldbach's conjecture#Heuristic justification and Goldbach's comet#Anatomy of the Goldbach Comet. p=3 has the largest effect because it's the smallest odd prime, but apart from that we don't know whether there is anything special about 3. The total expected effect of {5, 7, 11, 13, 19} is larger than that of {3, 587}, so we would expect more solutions for 190190 = 2×5×7×11×13×19 than for 190188 = 2²×3⁴×587. Indeed, 190190 has 2095 solutions while 190188 has 2069. PrimeHunter (talk) 15:54, 12 September 2013 (UTC)[reply]

To PrimeHunter. I omitted a word in the above question - it should have said "the MINIMUM number of solutions for one of them is twice that of either of the others." And divisibility by 3 DOES play an impotant role in the minmum (and average) number of solutions, as witnessed by the two graphs mentioned above: numbers divisible by 3 have twice the minimum (and average) number of solutions, your counterexample not withstanding. Saying that "the MINIMUM number of solutions for one of them is twice that of either of the others" does NOT imply that ALL solutions for a number divisible by 3 must be larger than ALL solutions for a number of similar size that is not divisible by 3.

You haven't explained what you mean by "MINIMUM" in "the MINIMUM number of solutions". The actual number of solutions varies a lot and is sometimes larger for 6N-2 or 6N+2 than for 6N. As mentioned, any odd prime factor p of N changes the expected number of solutions by a factor (p-1)/(p-2). For p=3 this is a factor 2 as you say, but you ignore all other odd p. p=5 gives a factor 4/3, p=7 gives 6/5, and so on. p=3 is a smaller prime so it gives a larger factor, but there is nothing unique about 3 changing the expectation. If we partitioned the even numbers into residue classes modulo 510510 instead of your modulo 6, then some of the classes not divisible by 3 would be expected to on average give more solutions than some of the classes divisible by 3. This is because 510510 = 2×3×5×7×11×13×17, and the total effect of {5, 7, 11, 13, 17} is around a factor 2.07, while the effect of 3 is a factor 2. PrimeHunter (talk) 23:42, 12 September 2013 (UTC)[reply]

To Duoduoduo. Did I agree too quickly? For one, I gave a statistic/heuristic proof for the MINIMUM number of expected solutions, NOT the average number. But let's assume that a glitch can wipe out a minimum number the same as it can an average number (it's a bigger/stronger/less common glitch). Also, even if this glitch does give a lower than expected number, so long as this number is not 0. who cares. But let's assume that it can give 0. Seeing as how the minimum number of solutions to Goldbach's Conjecture takes on increasing values (even if not monotonically), a counterexample would indeed have to be a very, very, very low probability event. Okay, so I do agree with you.84.109.186.188 (talk) 12:15, 12 September 2013 (UTC)[reply]

You are missing a step before you can begin to make a real probabilistic argument. You can argue that each N has a small and generally decreasing chance of being a counter example, but you still need to argue that the total chance over all N is small. Suppose we studied another problem where the chance of N being a counter example was 1/N, and we had already tested there were no small examples. The harmonic series 1 + 1/2 + 1/3 + 1/4 + ... has infinite sum, so we would expect infinitely many counter examples no matter how far we had already searched without finding one. The corresponding sum for Goldbach's conjecture is convergent, and very small when starting from the current test limit of 4×10¹⁸ [2], so we expect no counter examples. This is not in any way a proof. The expectation is based on unproven assumptions about independence between possible solutions for a given N. And even if these assumptions are valid, there will still be a small non-zero chance of counter examples. PrimeHunter (talk) 15:54, 12 September 2013 (UTC)[reply]

I don't follow - what is the "corresponding sum for Goldbach's conjecture" that is convergent?

There are different ways to estimate the terms but the sum is convergent in each case. If we for simplicity don't even consider the parity of the numbers then here is a very rough estimate of the probability that n is a counter example: (1-1/log(n)²)^n/2. 1/log(n)² is the rough chance that a random sum a+b = n will have both a and b prime. 1-1/log(n)² is the chance that at least one of them is composite, and we need that n/2 times to get a counter example. The sum of (1-1/log(n)²)^n/2 for n going from the current search limit 4×10¹⁸ to infinite is then an estimate of the number of counter examples to Goldbach's conjecture. This sum is convergent with a tiny positive value. A better estimate would change the sum by a constant factor but it would still be tiny. PrimeHunter (talk) 23:42, 12 September 2013 (UTC)[reply]

September 12

Speculations and Suspicions relating to the Factorial or Gamma function

Any input is welcome ! ( Yes, I am well aware that the statement below hasn't even been proven for all rationals in particular, let alone for all algebraics in general, but at this point I welcome any educated guesses, gut instincts, and informed opinions, or perhaps even possible counter examples I may not yet be aware of ).

\Gamma (\mathbb {A} \smallsetminus \mathbb {N} )\ \subset \ \mathbb {T} \qquad ;\qquad \Gamma ^{^{-1}}{\Big (}\mathbb {A} \smallsetminus \Gamma (\mathbb {N} ){\Big )}\ \subset \ \mathbb {T}

where $\mathbb {A}$ and $\mathbb {T}$ stand for algebraics and transcendental numbers, respectively. — 79.113.213.211 (talk) 15:47, 12 September 2013 (UTC)[reply]

I suggest you find a math-related forum somewhere else. This is a reference desk where we provide sourced answers to specific questions. We don't do "educated guesses, gut instincts, [or] informed opinions". Rojomoke (talk) 12:18, 13 September 2013 (UTC)[reply]

I am not necessarily suggesting that Wikipedia's "virtual librarians" which usually provide answers on this page should do the research and speculating themselves(!) —though frankly I wouldn't mind that either, since they are quite intelligent people— but perhaps someone might be able to bring to my attention objective information concerning (unproven) conjectures and/or (ongoing) mathematical research of other serious mathematicians... which I honestly don't think would go against the Wiki-guide-lines you mentioned... and the answer right below I think is proof of that. — 79.113.226.249 (talk) 06:10, 14 September 2013 (UTC)[reply]

From out article Gamma Function: " It has been proved that Γ(n+r) is a transcendental number and algebraically independent of π for any integer n and each of the fractions r = 1/6, 1/4, 1/3, 2/3, 3/4, and 5/6.". From the citation that gives that result, [3], and our page on periods, Ring of periods: Γ(p/q)^q is a period for natural numbers p and q. This paper, which appears to be from 2007, [4] on transcendental values of the Digamma function, the Logarithmic derivative of the Gamma function, indicates that the first part of your problem is still open, and is a long standing open problem. This excerpt, [5], mentions a general point regarding Hilbert's seventh problem, namely the about the expectation that Transcendental functions should take transcendental values at irrational algebraic arguments- the paper itself may be of interest to you. The article Particular values of the Gamma function may be of service to you also.Phoenixia1177 (talk) 13:30, 13 September 2013 (UTC)[reply]

Generally speaking,

{\tfrac {1}{n}}!\ =\int _{0}^{\infty }{e^{-x^{n}}}\ dx

so the transcendence of fractional factorials seems rather straightforward... — 79.113.226.249 (talk) 06:10, 14 September 2013 (UTC)[reply]

Also, from the reflection formula

\left(-{\tfrac {k}{n}}\right)!\ {\tfrac {k}{n}}!\ =\ {\frac {{\tfrac {k}{n}}\cdot \pi }{\sin \left({\tfrac {k}{n}}\cdot \pi \right)}}\ \in \ \mathbb {T}

it would logically follow that for all rational ⁠kn⁠ at least one of the two numbers Γ(1 ± ⁠kn⁠) is transcendental, since their product obviously is, because Sin(⁠kn⁠ π) is algebraic for all rational ⁠kn⁠ — See [6] [7] — 79.113.226.249 (talk) 10:15, 14 September 2013 (UTC)[reply]

Your observation is accurate, and not without interest; however, it lies in relation to the result, "gamma is transcendental for all noninteger rationals", in the same way that the observation "for all a there is a coprime b so ax + b is prime infinitely often" relates to Dirichlet's theorem. I think a very different approach will be required for the general case- the main meat of the conjecture would seem to be in dealing with gamma(t) for t in (0, 1/2].Phoenixia1177 (talk) 18:18, 14 September 2013 (UTC)[reply]

Perhaps combining it with Theorem 12 on page 6 (numbered 440), from the document you offered, might be a start... Actually, the fact that it is already known for a fact that that integral is transcendental was somewhat surprising, since its value represents the area of any given geometric curve described by the binomial equation xⁿ + y^m = 1... which was pretty much what triggered my initial curiosity about the nature of the various values of the factorial or gamma function all along... Now all I need is to know whether the arc length is also transcendental:

\int _{0}^{1}{{\sqrt {1+\left({\frac {d}{dx}}{\sqrt[{m}]{1-x^{n}}}\right)^{2}}}\ dx}\ \in \ \mathbb {T} \ ?\ ,\ \forall \ m,\ n\ \in \ \mathbb {N} >1

— 79.113.226.248 (talk) 19:02, 14 September 2013 (UTC)[reply]

I'll be honest, this is out of my usual areas of interest (so I don't know much about it), so this will probably be my last contribution (it has been fun, though). But relating to B(a,b) being transcendental- if you could show that {B(1/r, s/r) : s = 1,...r-2};{c = pi/sin(pi/r)} is algebraically independent, then gamma(1/r) ^ r = cB(1/r,1/r)...B(1/r,(r-2)/r) should be transcendental. But, like I said, I'm out of my element; so I seriously doubt there is any real merit to that approach. At any rate, good luck, update me on my talk page if you get any interesting results:-)Phoenixia1177 (talk) 20:35, 14 September 2013 (UTC)[reply]

Your contributions have been very helpful. Thank you. — 79.113.226.248 (talk) 20:49, 14 September 2013 (UTC)[reply]

September 13

random lookup

Having an array of N distinct items and trying to find one by randomly probing K×N times (with no guarantee that all indices tested are distinct).
How big should K be?

This is an outline of the question asked by 195.56.96.233 at WP:RD/Computing#random_looku, please reply there. --CiaPan (talk) 19:45, 13 September 2013 (UTC)[reply]

September 14

Square Root of Two

I've been trying to prove that the decimal expansion of root two contains more than two digits infinitally often, but can't seem to get past the extension of multiplication to an infinite decimal. Has this ever been proven before, (I'm pretty sure its Normality has not) and if it has, any pointers would be helpful. Thanks :) --Gilderien Chat|List of good deeds 20:52, 14 September 2013 (UTC)[reply]

I have no answers: only questions... But sometimes asking the right questions can be just as helpful: so here goes nothing: If a number is normal in a certain base, then is it normal in all other bases ? I'm asking this because -if this were true- you could probably simplify your work a lot by expressing these numbers in binary, since it only uses two digits, and the lack of any of tne two would imply the number to be rational... and it's also probably way easier to check the frequency of only two possible digits, as opposed to ten, for instance... Do you think such an approach would prove itself helpful ? — 79.113.226.248 (talk) 21:03, 14 September 2013 (UTC)[reply]

Normality in one base does not imply normality in all other bases. This is mentioned at normal number, and at the reference: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pjm/1103038420 Staecker (talk) 16:43, 15 September 2013 (UTC)[reply]

(edit conflict) I don't know the answer, but I have suggestions. In our article Square root of 2, in the section Series and product representations, for each of the series given: for each decimal position only finitely many terms control what that digit is. If you could characterize each term's decimal expansion, you might be able demonstrate it this way. Related to that, the section Computation algorithms spells out a method that, again by characterizing the terms of the sequence involved, could give you some results about the decimal expansion. I tried poking about for results from others on this question, but I kept getting flooded with results about computations of the decimal expansion, I'll dig about more later. :-)Phoenixia1177 (talk) 21:08, 14 September 2013 (UTC)[reply]

September 16

help understanding my math homework.

Hi I'm not asking for the answer to my math homework. I just need help understanding *why* the correct answer to my homework is correct. I'm having a bit of trouble with the radical, logarithmic and exponential equations I'm being assigned.

For example, I'm told that the answer to the equation 3^(-6x-7)= the 7th root of 81 is x = -53/42. BUT WHY?

Here's as far as I get when trying to solve the equation. I know that 81 is 3^4. so there fore the 7th root of 81 can also be expressed as 3^(4/7).

On the other side I can do a bunch of things. For example, 3^(-6x-7) is another way of saying (3^-6x)/(3^-7). But I don't know what to do to get the -53/42. Any hints you can give me? — Preceding unsigned comment added by 71.199.18.97 (talk) 04:41, 16 September 2013 (UTC)[reply]

Good, you are almost there:

3^(-6x-7) = 81^(1/7)

3^(-6x-7) = 3^(4/7)

-6x-7 = 4/7

-42x-49 = 4

-42x = 53

 x = -53/42

StuRat (talk) 04:58, 16 September 2013 (UTC)[reply]

OK, thanks. I guess I wasn't thinking that if 3^(-6x-7) = 3^(4/7) I could just forget about the 3 altogether and just set the exponents equal to each other. I have a few other problems in this format, so I will use this technique. Thanks again. — Preceding unsigned comment added by 71.199.18.97 (talk) 05:55, 16 September 2013 (UTC)[reply]

A few questions which any pointers on would be very helpful.

1) Under what conditions can a rank-3 tensor $T_{ijk}$ be decomposed into three matrices such that $T_{ijk}=\sum _{n}A_{in}B_{jn}C_{kn}$ ? Is there any other way of viewing this property?

2) The identity is preserved under pre and post multiplying by a matrix $X$ and its inverse, i.e. $XI{X}^{-1}=I$ or in index notation $\sum _{nm}X_{in}\delta _{nm}(X^{-1})_{mj}=\delta _{ij}$ . What is the relationship between three matrices $X,Y,Z$ such that they preserve a $\delta _{ijk}$ (which is 1 for i=j=k and 0 otherwise)? In other words what is the relationship between three matrices such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \delta_{ijk} = \sum_{nml} X_{in} Y_{jm} Z_{kl} \delta{nml}} ?