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April 20[edit]

Surprising periodic series[edit]

I'm looking for a property, such that the existense of series having this property - is not surprising, yet the existence of periodic (or even just cyclic) series having this property - is very surprising, whereas... the mathematicians have really discovered such a periodic (or cyclic) series having this property ! HOOTmag (talk) 20:18, 20 April 2015 (UTC)

Surprising is in the eye of the beholder. For me, when I first encountered the Gibbs phenomenon, it was surprising. I could believe that a function with a jump discontinuity could be represented by a infinite series of some basis functions (e.g., the Dirac delta), but that approximating a square wave with a Fourier periodic series would result in an extra spike jump discontinuity was counterintuitive. And yet, there it is. --Mark viking (talk) 00:55, 22 April 2015 (UTC)
Thank you so much.
Your example inspired me to think about many other examples of series whose periodicity is surprising. For example, until De-Moivre's and Euler's era, everyone thought that no smooth non-permanent exponential function - can be periodic (by "exponential function" I mean a function whose value for any sum of two numbers is the product of the values of the function for both of them). Then De-Moivre and Euler surprised the world by coming up with De-Moivre's formula and Euler's formula, that involve some kind of exponential function which - despite its being smooth and non-permanent - it was (surprisingly) periodic, so that one can now define a series whose periodicity is surprising - by using (trivially) the surprising periodicity of the function defined in De-Moivre's formula and in Euler's formula ...
Actually, it's unnecessary to go up to De-Moivre's and Euler's modern age: Even the early man, who only knew natural numbers, could easily set up a series by stipulating its periodicity in some surprise. For example, let P be any proposition which - surprisingly - turns out to be true (e.g. "the animal sitting behind me is a tiger"). Now we define the "property" of the series S, as follows: If P is false then S is not periodic (but say converges with the identity function), while if P is true then S is periodic (say converges with the constant number zero) ...
HOOTmag (talk) 08:21, 22 April 2015 (UTC)

April 22[edit]

Euclid's theorem[edit]

Go to Euclid's theorem. I think it would be interesting if someone could add a proof that uses the fact that ln(0) is negative infinity. In other words, it contains a formula that would show that ln(0) is a finite negative number if there were a finite number of primes. Georgia guy (talk) 00:20, 22 April 2015 (UTC)

Boxed expression in LaTeX[edit]

See what I discovered:

\begin{array}{|c|} \hline EXPRESSION \\ \hline \end{array}

See 'A box' section in Talk:Minkowski's question mark function for some details. :) CiaPan (talk) 11:57, 22 April 2015 (UTC)

Hm... why doesn't \fbox (or \framebox) work in math mode on WP? I get parse errors when I try it here, but on my home installation, I can use it both within and outside of math mode, and even nest in various ways , e.g. \fbox{$X_{n}^{m} \fbox{foo $X_{m}^{n}$}$} SemanticMantis (talk) 16:04, 22 April 2015 (UTC)
Because wiki math markup is only a fairly small subset of LaTeX. Maybe not even a subset — I'm not at all sure that everything that works in wikimath also works in LaTeX. --Trovatore (talk) 23:37, 22 April 2015 (UTC)
Fair enough. Part of me wonders why it can't be full LaTeX, or why not at least a proper subset, but I suspect the technical details might not be that interesting :) SemanticMantis (talk) 14:36, 23 April 2015 (UTC)

zeroes of infinite sums[edit]

Can someone refer me to some material on how to find the zeroes of functions that are defined as infinite sums? RJFJR (talk) 23:18, 22 April 2015 (UTC)

I don't think there is any general method. Can you give a little more context? Are you looking for exact solutions, or approximations within some tolerance, or approximations within an arbitrarily small tolerance? Are the sums power series or trigonometric series or something else? --Trovatore (talk) 23:27, 22 April 2015 (UTC)
It seems like you try to decide Riemann Hypothesis, don't you? HOOTmag (talk) 23:33, 22 April 2015 (UTC)
Assuming that you have uniform estimates on the truncation error (e.g., for a uniformly convergence series), you can use the method of bisection to locate individual zeros. In more favorable cases, standard numerical methods like Newton's method can be used, but one needs to be careful that the series converges in an appropriately strong sense. Sławomir Biały (talk) 23:39, 22 April 2015 (UTC)
A power series can be treated like this. Truncate the series and compute the zeroes of the resulting polynomial by the the Durand-Kerner method. Bo Jacoby (talk) 09:17, 23 April 2015 (UTC).
That result is probably not useful for the full series, since analytic functions in general need not have roots at all, such as with the exponential function, nor have a uniform limit for the position of the zeros.--Jasper Deng (talk) 09:08, 24 April 2015 (UTC)
The result is useful when the series converges around a root. Bo Jacoby (talk) 07:02, 24 April 2015 (UTC).

I was hoping for an answer like "read chapter 5 of book XYZZY for a general overview". I'm not sure what is knowable. They tell me there is a zero to zeta around .5 + 14.***i (or something) but I don't know how they found it. And when I try to think about finding the zeroes of complex valued functions of a complex argument my intuition gets confused. I think you pointed me in the right direction: I need to go back and look in my old numerical analysis book about how if you truncate the series you can bound how close the zero is. Thank you. RJFJR (talk) 17:47, 23 April 2015 (UTC)

For the zeta function specifically, the complex argument of zeta is under control on the critical line, so the problem of finding zeros is effectively reduced to finding zeros of a single real function. See Titchmarsh "The Zeros of the Riemann zeta-function", 1935, Proceedings of the Royal Society, Vol 151, No 873 pp. 234-255. Parts of that paper are readable. Sławomir Biały (talk) 20:50, 23 April 2015 (UTC)
Our Z function and Riemann–Siegel formula and links therein may help, and they cite how people found such zeros. The Z function is rigged to be real on the critical line by the functional equation, so intuition won't get so confused. :-)John Z (talk) 06:05, 25 April 2015 (UTC)

Thank you, everyone. (I think I need to get out my old text books and review some.) RJFJR (talk) 22:54, 25 April 2015 (UTC)

Need help estimating the radius of a planet from its gravity[edit]

So I've been trying to figure out a way to estimate certain variables about a planet (primarily its radius) from the force of its gravity, and I seem to be getting it wrong. (And no, this is not a homework problem. Just something I'm trying to figure out / trying to remember from high school physics in my spare time.) Could someone help me out?

So I start out with this standard take on Newton's law of universal gravitation (b stands for the hypothetical Planet B, since that's as good a letter as any):

g_{b} = \frac{GM_{b}}{{r_{b}}^{2}}

Since we know what the planet's g will be, solve for radius:

{r_{b}}^{2} = \frac{GM_{b}}{g_{b}}

Here's where things might be getting dicey -- since this is going to be a rocky planet, like Earth, I assume that its mass and radius will be proportional to Earth's mass and gravity (i.e. if planet is bigger than Earth, its mass will be greater). Meaning:

\frac{M_{b}}{r_{b}} = \frac{M_{e}}{r_{e}}

Solve for Planet B's mass:

M_{b} = \frac{r_{b}M_{e}}{r_{e}}

Then introduce that back into the other equation.

{r_{b}}^{2} = \left(\frac{G}{g_{b}}\right)\left(\frac{r_{b}M_{e}}{r_{e}}\right)

r_{b} = \frac{GM_{e}}{g_{b}r_{e}}

So it looks like it should work. But the calculations don't seem to be working out. Can anyone tell me where I'm going wrong here? Thanks. --Brasswatchman (talk) 23:51, 22 April 2015 (UTC)

For planets of the same density (all rocky planets should have roughly the same average density), the mass is proportional to the cube of the radius. That makes a pretty big difference. --Trovatore (talk) 23:54, 22 April 2015 (UTC)
It would, wouldn't it? :) Okay. Let's see if I can work out the rest of this, then...
{r_{b}}^{2} = \left(\frac{G}{g_{b}}\right)\left(\frac{M_{e}{r_{b}}^{3}}{{r_{e}}^{3}}\right)
{r_{b}}^{-1} = \frac{GM_{e}}{g_{b}{r_{e}}^{3}}
\frac{1}{r_{b}} = \frac{GM_{e}}{g_{b}{r_{e}}^{3}}
r_{b} = \frac{g_{b}{r_{e}}^{3}}{GM_{e}}
Does that look right? (Numbers at least are looking more in line with what I'd expect.) --Brasswatchman (talk) 00:15, 23 April 2015 (UTC)
Yes, or simpler still  r_b = \frac{g_b}{g_e} r_e. . Abecedare (talk) 00:38, 23 April 2015 (UTC)
Excellent. Thank you both. --Brasswatchman (talk) 00:42, 23 April 2015 (UTC)
And while you are at it, if the densities don't match but somehow you have an estimate of the planet's density: r_b = \frac{g_b}{g_e} \cdot \frac{\rho_e}{\rho_b} r_e. Abecedare (talk) 00:49, 23 April 2015 (UTC)

Measure the gravitational acceleration on the ground (g at r) and on the top of a tall building (g+dg at r+dr). Then the radius of the earth is r = − 2 g dr / dg. Bo Jacoby (talk) 06:44, 23 April 2015 (UTC).

April 23[edit]

I think there might be an error in the Lagrange Multiplier article[edit]

There might be an error in example 2 in the "Lagrange Multiplier" Wikipedia article. The article says that the maximum occurs at (±sqrt(2), 1). This point falls on the circle (x^2 + y^2 = 3) and gives a value x^2 * y = sqrt(2). But the point (sqrt(3/2) , sqrt(3/2)) also meets the constraint and gives a value x^2*y = 1.5. This is larger than sqrt(2). Doesn't this show that the answer given in the example is not a maximum? — Preceding unsigned comment added by Wthurt (talkcontribs) 21:12, 23 April 2015 (UTC)

You substituted in the values incorrectly. x^2y has value \frac{3}{2}\sqrt{\frac{3}{2}} at your suggested critical point, but at the maximums the function attains the value 2, which is greater than at the point you suggested.--Jasper Deng (talk) 00:05, 24 April 2015 (UTC)

April 25[edit]

Is 1/0.5/0/0 a universal logic gate?[edit]

Suppose I have a logic gate with the following truth table:

Input 1 Input 2 Output
0 0 1
0 1 1/2
1 0 0
1 1 0

...where "1/2" means randomly either 0 or 1. Would this be universal logic gate like NOR? Do not be silly - what I mean is of course an approximation of one, due to the random element. PretendAuthority (talk) 11:02, 25 April 2015 (UTC)

Well, perhaps a faulty NOR gate. What can we say? 0,1 should map to 0, but in your implementation it sometimes does, sometimes does not. When it does not, it is not acting like a NOR gate. --Tagishsimon (talk) 11:19, 25 April 2015 (UTC)
Note that you have two possibilities:
Input 1 Input 2 Output
0 0 1
0 1 0
1 0 0
1 1 0
Input 1 Input 2 Output
0 0 1
0 1 1
1 0 0
1 1 0
The first is a NOR gate, while the second is "NOT Input1". StuRat (talk) 16:13, 25 April 2015 (UTC)
Let's try to rephrase this more precisely. There are two versions of this that I consider interesting:
  • Is it possible to find a construction in which the output has a probability approaching 1 (as number of gates increases) of being that of a universal gate for each input combination (assuming statistical independence of behaviour of the gates, and with the random entry being probability 0.5 of a 1)?
  • Is it possible to construct a universal gate (e.g. NOR) from a finite number of these gates?
As a start, we can construct an inverter: simply tie input 2 to 0, or tie the two inputs together. This is useful building block. We can use this to form an imperfect OR ("iOR") by inverting the output of an imperfect NOR. Now chain a series of iORs on input 1 with the first input 1 driven by A, and tie every input 2 to B. This produces a truth table, where entries give probability of a 1 (n0: 0 iORs, n1: 1 iOR, etc.):
A B n0 n1 n2 n3
0 0 0 0 0 0
0 1 0 1/2 3/4 7/8
1 0 1 1 1 1
1 1 1 1 1 1
We can see that as we add iORs to the chain, the truth table asymptotically approaches that of an OR gate. Remove the last inverter, and we have what could be considered an approximate universal gate, in the sense of my first bullet. So the answer to my first bullet (and thus to the OP's question) is "yes". So now the challenge would be to answer my second bullet. —Quondum 17:59, 25 April 2015 (UTC)
The first question is what the OP wanted to ask. The answer to the second question is obviously negative - suppose you built a circuit which outputs NOR with probability 1. Let f(x) represent the circuit's output for input x in the case that all gates behave as "NOT Input1". f is not NOR (otherwise "NOT Input1" would have been a universal gate), so there is some x for which f(x)\neq\mathrm{NOR}(x). There is a positive probability that the circuit will output f(x) for this input x, contradiction. -- Meni Rosenfeld (talk) 19:36, 25 April 2015 (UTC)
Dang. And here I thought that I had a counterexample, but on closer examination, your logic holds. (But until you pointed it out, it wasn't that obvious to me ...) —Quondum 20:17, 25 April 2015 (UTC)
I noticed you've tried to remove one of the answers here. It is considered extremely rude and unacceptable to do so, even if it is a poor answer (as is, in general, removing comments from talk pages). -- Meni Rosenfeld (talk) 19:36, 25 April 2015 (UTC)

April 26[edit]

Understanding the structure of order 16 groups[edit]

So I have been attempting to gain a better understanding of the 14 different groups (up to isomorphism) of order 16 and I am having some difficulties finding ways make sense of their similarities and differences. With the exception of the Alternating group on a 4-element set (A_4), I understand the structure of all groups of orders<16 well enough that I could produce an operation table for them without any references. I am wanting to gain a similar faculty with order 16 groups without resorting to something like rote memorization (which would be devoid of real understanding and kinda defeats the point of learning it for me). What sort of things should I look at beyond what I already have been using? (I have been tackling this problem by so far by using GAP to get an operation table for the group, then I proceed to manually finding the various cycles, identifying which ones are primitive (meaning not a subset of a larger cycle), and then examining how their generators behave when operated together). (talk) 10:06, 26 April 2015 (UTC)

Are you already familiar with the Sylow theorems? Burnside's_lemma might also help. SemanticMantis (talk) 14:56, 26 April 2015 (UTC)
It's unclear to me what "understanding the structure" means exactly, but this list looks like a good place to start. There are five abelian groups corresponding to the five partitions of 4. There are three non-abelian split extensions of Z8 by Z2: the dihedral, quasidihedral, and M16 groups. Two groups are gotten from the non-abelian groups of order 8 by taking direct products with Z2. There is a unique non-abelian split extension of Z4 by Z4, a split extension of Z2xZ4 by Z2, and a central product of D8 and Z4. Finally, there is the generalized quaternion group of order 16. Sławomir Biały (talk) 15:20, 26 April 2015 (UTC)
I've worked out generators and relations for groups of 16 and of order 32. To get to 64 you need more sophisticated techniques than I know about and for 128 you also need a computer. A couple of things I learned though: First, what most textbooks tell you about classifying groups of order 8 (i.e. starting with a large cyclic subgroup) doesn't scale and you need to do something else for order 16. The most important elementary theorem to know is that G/Z can't be cyclic. There is a generalization of this which gives a criterion for which groups are possible G/Z, turns out that not many are. So it's relatively easy to enumerate possible Z's and G/Z's. The next thing is to learn a bit about the machinery of central extensions. This naturally leads to more sophisticated group homology theory but you probably don't need to go that far. Also, learn what you can about Frattini subgroups and their quotients. You will need a bit more linear algebra that they normally teach undergraduates as well, specifically the classification of alternating bilinear forms. As a rule of thumb, every time you multiply the order by 2 the classification problem gets an order of magnitude more difficult. So 16 isn't just a lengthier version of 8, it's more like comparing a symphony with a pop song. (Which makes 32 Wagner's Ring cycle I guess.) --RDBury (talk) 17:44, 26 April 2015 (UTC)