Wikipedia:Reference desk/Archives/Mathematics/2013 October 12

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October 12

Symmetrical Equation in Three Unknowns

$\left(x+{\frac {1}{2x}}-1\right)\left(y+{\frac {1}{2y}}-1\right)\left(z+{\frac {1}{2z}}-1\right)=\left(1-{\frac {yz}{x}}\right)\left(1-{\frac {xz}{y}}\right)\left(1-{\frac {xy}{z}}\right)$

Obviously, x = y = z = ⁠1/2⁠ are a solution... But are there others ? I've tried solving it by equating each symmetrical term from one side with the equivalent one from the left... but that path ultimately lead me nowhere... What approach should one generally take when faced with such an identity ? Any suggestions ? — 79.113.213.168 (talk) 00:42, 12 October 2013 (UTC)

I'm assuming you're working in the complex numbers. Multiply your equation through by xyz and simplify, you'll find that the solutions are just roots of a polynomial. Specifically, let A(y,z) = (y² - y + 0.5)(z² - z + 0.5), then the coefficients starting at the cubic working down are: yz, -y² - z² - y²z² - A, yz + yz³ + y³z + A, and -y²z² - 0.5A. Then, if you feel ambitious, you can plug those into the cubic formula, see [1], to get a formula for solutions. Or you can put in specific y, z and use [2] to get a solution. For example, 1.5, 0.5, 0.5 is also a solution, as is 1.77792355523206, 1, 1. If you simply need some solutions, or several, the simplest way is to set up an excel sheet that can do the calculating for you given y, z as input cells.

If you're looking for other routes, something more qualitative, you could examine what groups/semigroups preserve solutions- for example, you already know that permutations of x, y, and z do. You could look at linear operators next, or if that yields nothing interesting, try more esoteric setups. I'd be happy to help you more in this sort of direction, too, but it can be tedious, so I wanted to see if the above solution sufficed first. :-)Phoenixia1177 (talk) 06:39, 12 October 2013 (UTC)

Thanks! :-) (Silly me: I did think of multiplying with xyz... I just never thought of collecting the coefficients with regards to a variable... Which should've been obvious...) The only other thing worth mentioning is that all values of x, y, and z should be from within the open interval (0, 1)... Is there some general method for "binding" the values of these interdependent variables to a specific interval ? — 79.113.213.168 (talk) 07:09, 12 October 2013 (UTC)

There's nothing pretty that I can think of. You could try setting y as a function of z and x as a function of z, assume that y is increasing and in [0, 1]; then, solve for x' in terms of x, y, y', and z, then look at forcing x' >= 0 and 0 <= x(0) and x(1) <= 1. Or, for what 0 <= c < d <= 1 you'd have 0 <= x(c) < x(d) <= 1 and x' > 0 on [c, d]. Essentially, you'd be solving for what curves y in [0, 1] give you the solutions you want. Another approach would be to look at f map solutions to solutions, then find such an f that allows you to "reduce" solutions into [0, 1], if you don't mind tedious algebra, there might be linear f that does this; there might not be, though. A third method: look at the geometry of the solution set, you may be able to arrange it so tangents, or some such, through one solution net you another solution; you may be able to do some form of reduction that way to get your solutions where you want. Where this equation arise from? Depending on the context, there may be another method using the original source that makes this a bit nicer. I'm sure there are other methods, but these are things coming to mind off the top of my head; unless you get a better answer, I'll keep thinking and see if I can't find something better than the above (all of which will be very nasty and might fail...). :-)Phoenixia1177 (talk) 07:54, 12 October 2013 (UTC)

This might work: look at the coefficients of the cubic in x, you four maps u,v in R -> R. Find a maps f, g so f(u, v) and g(u,v) will give the same coefficients when substituted in as u, v. Then, suppose you have a solution x, y, z with y and z in the interval of choice, y, x, z is also a solution, so is y, f(x, z), g(x, z); if you can get f and g so f, g map R x [0, 1] into [0, 1], then you'll get a solution set in [0, 1]. I don't know if such f and g exist, but it might be the simplest to just work out and see what happens- again, lot's of algebra.Phoenixia1177 (talk) 08:17, 12 October 2013 (UTC)

Try this. Multiply through by xyz as suggested above to get

(x²-x+1/2)(y²-y+1/2)(z²-z+1/2)=(x-yz)(y-xz)(z-xy).

It's not hard to show that the left hand side has a global minimum of 1/64 at (1/2, 1/2, 1/2). It seems that on the cube [0,1]³ which is the region you're looking at, the right hand side has a maximum of 1/64, also at (1/2, 1/2, 1/2). (I tested this on a mesh of 1331 points using a spreadsheet, not exactly rigorous but good enough to justify this approach.) If so then the only point where the two sides can be equal is (1/2, 1/2, 1/2). The equation defines a surface of degree 6, but there is an isolated point at (1/2, 1/2, 1/2), the 3 dimensional equivalent of an acnode. --RDBury (talk) 22:33, 12 October 2013 (UTC)

The only way I can approach this mathematically is by appealing to the RHS's symmetry with regard to x,y,z , and consider that each term were of the form (t - t*t) = (t - t²), whose derivative would then be (1 - 2t), whose root is t = ⁠1/2⁠... and whose second derivative is -2 < 0... as opposed to the LHS, whose second derivative is 2 > 0... QED. Thanks ! — 79.113.215.238 (talk) 04:59, 13 October 2013 (UTC)

The following trick splits the symmetrical equation in three unknowns into two equations of lower degree. Find the three roots x,y,z of the cubic equation f(w)=0 where f(w)=(w−x)(w−y)(w−z)=w³−Aw²+Bw−C and A=x+y+z and B=xy+yz+zx and C=xyz. It is possible to express the original equation in terms of A,B,C rather than x,y,z because of the symmetry. Bo Jacoby (talk) 04:53, 13 October 2013 (UTC).

As there is only one equation for three unknowns, there is a twodimensional infinity of solutions. The equation is L=R where h=1/2 and m=−1 and

L = (xx+mx+h)(yy+my+h)(zz+mz+h)

= xxyyzz+m(xy+yz+xz)xyz+h(xxyy+yyzz+xxzz)+mm(y+x+z)xyz +mh(xyy+yyz+xxy+yzz+xxz+xzz)+hh(yy+xx+zz)+mmm(xyz)+mmh(xy+yz+xz) +mhh(y+x+z)+hhh

= C²+mBC+h(B²+m2AC)+AC+h(mAB+3C)+h²(A²+m2B)+mC+hB+mh²A+h³

= C²+mBC+hB²+mhAB+hC+h²A²+mh²A+h³

and

R = (x+myz)(y+mxz)(z+mxy)

= xyz+m(xyxy+yzyz+xxzz)+(yy+xx+zz)xyz+myzxzxy

= C+m(B²+m2AC)+(A²+m2B)C+mC²

= C+mB²+2AC+A²C+m2BC+mC²

The equation is

0 = L+mR = 2C²+BC+3hB²+mhAB+mhC+h²A²+mh²A+h³+2mAC+mA²C

Multiply by 8 to get rid of the h, and reintroduce the minus sign.

0 = 16C²+8BC+12B²−4AB−4C+2A²−2A+1−16AC−8A²C

This equation is of second degree in any of the unknowns A,B,C. You may choose any two of these unknowns freely and solve for the third. Then x,y,z are the three (complex) roots of the cubic equation f(w)=0 where f(w)=w³−Aw²+Bw−C. Bo Jacoby (talk) 14:13, 13 October 2013 (UTC).

Arithmetic–geometric mean an integer?

Can the Arithmetic–geometric mean of two distinct positive integers be an integer? I did a search of small parameters, and I got one within 2.8E-12 of an integer, but I suspect that they can't be integers. Bubba73 ^{You talkin' to me?} 13:54, 12 October 2013 (UTC)

Judging by its integral expression, it obviously can't, unless either (a) the two integers are equal in absolute value, and/or (b) at least one of them is 0. Perhaps using a symbolic mathematical software (instead of a numerical one) would be a better choice in this case, where questions of quality (nature of numbers) rather than quantity (approximative value) are concerned. BTW: Which numbers did you use as input ? — 79.113.215.238 (talk) 21:54, 12 October 2013 (UTC)

Thanks. I tested 1 <= j < i <= 1000000. The one closest to an integer is AGM(894297,534879)=703058.000000000001195... Bubba73 ^{You talkin' to me?} 23:15, 12 October 2013 (UTC)

A million operations of this kind take about 6 minutes of my computer [2x2 GHz]... and yours are up to about half of a million square... And 3 million minutes are the equivalent of two years... Even with a quad-core of 4x3=12 GHz, it would still take 24/3 = 8 months time... Am I missing something ?... :-\ — 79.113.215.238 (talk) 04:36, 13 October 2013 (UTC)

I did it on one core of a quad-core i5, while the other three cores were doing something else 100% of the time, so it is slower than it would be on one dedicated core. It took somewhere around 20 hours for (nearly) 500 billion tests. I'm compiling to a native EXE; you may not be doing that. Also I didn't use multiple-precision arithmetic. Bubba73 ^{You talkin' to me?} 04:45, 13 October 2013 (UTC)

Testing AGM(i,1000000) for i = 1 to 999998 takes about 0.123 seconds on one core of my i7. Bubba73 ^{You talkin' to me?} 18:48, 13 October 2013 (UTC)

(r to 79's first comment)

Is it obvious that it can't? From the integral expression:

$M(x,y)={\frac {\pi }{4}}(x+y){\big /}K\left({\frac {x-y}{x+y}}\right)$

one can see that if the elliptic integral takes a value that is a rational multiple of $\pi$ (say, A/B $\pi$) for some rational argument (say, a/b), then we can choose x= 2A(a+b) and y=2A(b-a) , such that M(x,y) = B b is also an integer. Now the elliptical function covers all values from 0 to infinity, but I don't know whether it satisfies the underlined property for non-trivial cases... but neither is the converse obvious. Is there something I am overlooking ? Abecedare (talk) 05:23, 13 October 2013 (UTC)

You mean apart from the obvious fact that the value of the AGM is NOT an integer for the general case you describe ?... — 79.113.215.238 (talk) 06:12, 13 October 2013 (UTC)

I am not sure I understand. Did I make an algebraic error in the comment above (quite possible!), or a more fundamental conceptual error somewhere ? Abecedare (talk) 06:24, 13 October 2013 (UTC)

Give values to the A, a, and b from your "solution", and use the link I provided above to calculate the value of the AGM in the two arguments x and y thus obtained... — 79.113.215.238 (talk) 06:29, 13 October 2013 (UTC)

... but I don't know the values of a, b, A and B for which the underlined property is satisfied. I don't even know if such values exist (except for the trivial cases corresponding to x=y, x=0,y=0) !

I am afraid we may be talking past each other somehow, and what might be obvious to you may not be obvious to me and vice versa. So if you could spell out your explanation/objection that would help. If you'd like me to be clearer in the claim I am making, I can do that too. Abecedare (talk) 06:36, 13 October 2013 (UTC)

The integral form I had in mind was the one with the sinus and the cosinus in the denominator... Take a close look at it, and then tell me if the expression under the radical can be extracted or become anything meaningful (or π-friendly) if |x| ≠ |y|, and/or both of them ≠ 0... (Please keep in mind that x and y are integers, and we are dealing with a linear combination of sin² and cos²). — 79.113.215.238 (talk) 06:48, 13 October 2013 (UTC)

I don't know one way or the other. That is the reason I am asking why the result that it cannot, should be obvious. Is there some mathematical reasoning behind that claim, or just an appeal to intuition or the lack of explicit counterexamples ? Abecedare (talk) 07:15, 13 October 2013 (UTC)

For unequal arguments, let A > 0 be the minimum of their squares, and B > 0 the absolute value of the difference between their squares. Our integral then becomes $\int _{0}^{1}{\tfrac {dx}{\sqrt {(1-x^{2})(A+Bx)}}}$ — 79.113.215.238 (talk) 07:51, 13 October 2013 (UTC)

Adjusting the calendar.

The mnemonic I learned for days in a year: "30 days have Sept, April, June, & Nov.; all the rest have 31, save Feb. 28, save for every four has 29, save for ever hundred has 30, save for every thousand has 31." At the turn of the millennium, I had expected to see February with 31 days -- what a disappointment when it didn't happen. I understand that if we had applied that mnemonic, December 25th would still be the darkest day of the year. To add confusion, I don't know where I got that complete version of that old mnemonic. Question: What happened? 162.193.210.29 (talk) 14:19, 12 October 2013 (UTC)

February never has 30 or 31 days. Every century year is an exception to the leap year rule unless it is a multiple of 400. That is, 1700, 1800, 1900, and 2100 are not leap years but 2000 is. And there is another tweak or two that I don't remember. See leap year. Bubba73 ^{You talkin' to me?} 14:48, 12 October 2013 (UTC)

This question was already asked and answered at Wikipedia:Reference Desk/Miscellaneous#A question in adjusting the calendar. Any further discussion should take place there. Duoduoduo (talk) 14:54, 12 October 2013 (UTC)

I understand that for a normal distribution about 95% of the observations fall within 2 standard deviations of the mean.

What percentage of the observations in a normal distribution fall within 1.8 standard deviations?

Thank you97.93.180.21 (talk) 20:01, 12 October 2013 (UTC)

46.41% X 2 = 92.82%. Bubba73 ^{You talkin' to me?} 20:24, 12 October 2013 (UTC)

The general formulafor a normal distribution

$\mathrm {erf} \left({\frac {a}{\sqrt {2}}}\right)\times 100$

percent of the observations lie within distance $a\sigma$ from the mean. Erf is the error function. Abecedare (talk) 20:44, 12 October 2013 (UTC)

http://www.wolframalpha.com/input/?i=100*%28erf%281.8%2Fsqrt%282%29%29%29 says 92.8139. Bo Jacoby (talk) 04:01, 13 October 2013 (UTC).

It would probably be a good idea if Normal distribution gave a table out to 3 standard deviations in increments of 0.1 or 0.05. Bubba73 ^{You talkin' to me?} 04:21, 13 October 2013 (UTC)