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October 9

US default on 17th Oct

Does the probability of "US Default on 17th Oct" exists? And if so what is the proper Mathematical way of calculating it? 202.177.218.59 (talk) 01:53, 9 October 2013 (UTC)

Yes, as argued here all probabilities are quantum mechanical in nature (even what looks like obvious counter examples like betting on digits of pi turn out not to be so upon a precise analysis). So, the probability is in principle well defined by the laws of physics and has a quantum mechanical origin. Count Iblis (talk) 15:04, 9 October 2013 (UTC)

The people making these estimates presumably use data on daily government spending, and know how much is spent in a day and how much that value varies. From there they can build statistical models of how likely the funds are to run out at any given day. Presumably this is simpler than working from first principles as Count Iblis suggests. :-) I don't want to give too much detail, because it is speculation - maybe someone can track down the source that the 17th date came from originally to see if methodology is given. Katie R (talk) 19:06, 9 October 2013 (UTC)

According to Time "Economists at JPMorgan have been more precise, estimating that the Treasury will run out of cash on the 24th—one week after Treasury says it’s extraordinary measures will be exhausted. They write that it is “extremely unlikely” the Treasury will be able to make its payments more than a few days after the 24th, and that the Treasury would most certainly have to default on some payments by November 1st, when large outlays for Social Security, Medicare, retirement benefits for military and civil services workers, and interest payments are due." Gandalf61 (talk) 20:04, 9 October 2013 (UTC)

Common factors of a sum

This might be a terribly difficult question, given how greatest common divisors behave. Suppose I have two odd integers, A and D, and two even integers, B and C, such that: gcd(A,B,C,D) = 1, but gcd(A,B) ≠ 1 and gcd(C,D) ≠ 1. Does there exist some integer K such that gcd(C + 2KA, D + 2KB) = 1? Searching for a counterexample has proved fruitless - after dozens of attempts, the gcd of the sums has been 1 for K = 1 or K = -1. It seems like there should always be such a K, but I have very little idea of how to show it. Any suggestions are greatly appreciated! Icthyos (talk) 14:02, 9 October 2013 (UTC)

Let A=3, D=5, B=6, C=10, which fulfil all of your conditions regarding A, B, C, and D.

Let K=2, so that C+2KA=22 and D+2KB=29, and gcd(C + 2KA, D + 2KB) = 1.

(Should K be relatively prime to C and D? Perhaps that helps.)Bh12 (talk) 16:01, 9 October 2013 (UTC)

Your example certainly works, but I should clarify: I'm wondering if it's possible for there to be A, B, C, D satisfying the above conditions such that there is no integer K for which gcd(C + 2KA, D + 2KB) = 1. For all the examples of A, B, C, D I pick, I can always find some K for which this gcd is 1, but I'm having a hard time proving that there always exists such a K in general. Finding examples is the easy part! Icthyos (talk) 16:15, 9 October 2013 (UTC)

So let K be prime to A, B, C, and D. Then C+2KA is even and is relatively prime to A. Also, D+2KB is odd and is relatively prime to D, K, and B.

If C+2KA and D+2KB are relatively prime, such a K does exist.

If C+2KA and D+2KB are not relatively prime, double the value of K; if C+2KA and D+2KB are now relatively prime, such a K does exist.

If C+2KA and D+2KB are still not relatively prime, again double the value of K; if C+2KA and D+2KB are now relatively prime, such a K does exist.

I need some help on this last point, but I don't think that K can be doubled more than twice without having gcd(C + 2KA, D + 2KB) = 1.Bh12 (talk) 17:29, 9 October 2013 (UTC)

What's your reason for this intuition about not having to double K more than twice? Also, I'm not sure I believe: So let K be prime to A, B, C, and D. Then C+2KA is even and is relatively prime to A. -- if A and C have some common factor bigger than 1, then A and C+2KA also have this as a common factor. (Recall, A and C can have a common factor bigger than 1, it just can't be shared in common with both B and D). Icthyos (talk) 19:45, 9 October 2013 (UTC)

So let K be prime to A,B,C,D, let C+2KA have a gcd GT 1, but that gcd is NOT shared with D+2KB because gcd(A,B,C,D)=1).

Likewise, B and D can have a gcd GT 1, but that gcd is NOT shared with A and C.

C+2KA and D+2KB should then be relatively prime - forget about what I said about doubling K - so that such a K exists.

Or can you provide a counterexample?Bh12 (talk) 20:20, 9 October 2013 (UTC)

Take A = 5, B = 20, C = 38, D = 19 and K = -1. Then C + 2KA = 28 and D + 2KB = -21, which have gcd 7. The problem is that new divisors can appear in the sums that are not divisors of any of A, B, C or D. Icthyos (talk) 21:13, 9 October 2013 (UTC)

If you want "some integer K" such that the required gcd=1, then in your example take K=2, so that C+2KA=58 and D+2KB=99. What diference would it make that for K=-1 the gcd is not 1?Bh12 (talk) 00:01, 10 October 2013 (UTC)

Yes, but the example demonstrates that not any old K will work. A priori, it's entirely possible that there exists A, B, C, D satisfying the conditions, for which there is no K giving the gcd in question to be 1. I want to prove that there is always such a K. Icthyos (talk) 12:35, 10 October 2013 (UTC)

I think a more general form would be more tractable. Suppose the following statement is true:

Given a, b, c, d, with GCD(a, b, c, d)=1 and ad-bc≠0, then GCD(c+ka,d+kb)=1 for some k.

Let a=2A, b=2B, c=C, d=D in the above problem. D is odd so 2 cannot be a common factor of a, b, c, d, and neither can any other prime else GCD(A, B, C, D) is not 1. Also A and D are odd, and B and C are even, so AD-BC≠0, so ad-bc=2(AD-BC)≠0. So the conditions for the above statement hold which would imply that there is a k so that GCD(c+ka,d+kb)=GCD(C+2kA,D+2kB)=1.

I believe I have a proof of the statement given and I'll post that in a bit. --RDBury (talk) 22:24, 9 October 2013 (UTC)

(Part 2): To prove the above statement, I claim it's sufficient to prove the special case a=0. Suppose true for a=0, i.e.:

Given b, c, d, with GCD(b, c, d)=1 and b, c≠0, then GCD(c,d+kb)=1 for some k.

First:

Lemma: If ps-qr=1 then GCD(m,n)=GCD(pm+qn,rm+sn) for any m, n.

Proof: If u divides m and n then u divides pm+qn and rm+sn, so GCD(m,n)|GCD(pm+qn,rm+sn). Also s(pm+qn)-q(rm+sn)=m and -r(pm+qn)+p(rm+sn)=n. So if u divides pm+qn and rm+sn then u divides m and n, so GCD(pm+qn,rm+sn)|GCD(m,n). Therefore GCD(m,n)=GCD(pm+qn,rm+sn) as required.

Let a, b, c, d be given with GCD(a, b, c, d)=1, and ad-bc≠0. Let GCD(a,b)=e, there are x and y so that ax+by=e. Then (a/e)x+(b/e)y=1. Apply the lemma to get GCD(c+ka,d+kb)=GCD((b/e)(c+ka)-(a/e)(d+kb),x(c+ka)+y(d+kb))=GCD((ad-bc)/e,xc+yd+k(ax+by))=GCD((ad-bc)/e,xc+yd+ke). Let c'=(ad-bc)/e, d'=xc+yd, b'=ax+by. GCD(c',d')=GCD(c,d) by the lemma, and b'=e=GCD(a,b). So GCD(b',c',d')=GCD(GCD(a,b),GCD(c,d))=GCD(a,b,c,d)=1. Both b'=e and c'=(ad-bc)/e are nonzero so there is a k so that GCD(c',d'+kb')=GCD(c+ka,d+kb)=1.

If remains so show the a=0 which I will post next. --RDBury (talk) 00:47, 10 October 2013 (UTC)

Wow, thanks for taking the time to explain this so carefully. I followed this up until the final step. We know that GCD(b',c',d') = 1, so we can find integers f, g, h so that fb' + gc' + hd' = 1, and so 1 = h(d' + (f/h)b') + gc'. Taking k = f/h, we then know GCD(c',d'+kb') = 1...but how do we know that f/h is an integer? Or are you arguing this step in another way? Icthyos (talk) 12:35, 10 October 2013 (UTC)

Oh wait, I see, we're using our assumption that it's true when a = 0. Clever! Icthyos (talk) 12:46, 10 October 2013 (UTC)

(Part 3). It remains to show, if GCD(b,c,d)=1, and b, c≠0, then GCD(c,d+kb)=1 for some k. First, let c' be the product of the prime factors of c. Then c' divides c so GCD(b,c',d)=1 and if p divides c and d+kb then it divides c' and d+kb. So if there is a k so that GCD(c',d+kb)=1 then GCD(c,d+kb)=1. So we can assume without loss of generality that c is square free.

Suppose c and d have a common prime factor p. Let c=c'p. Since c is square free, p does not divide c'. Suppose also there is k so that GCD(c',d+kb)=1. If p divides k then replace k with k+c', GCD(c',d+(k+c')b)=GCD(c',d+kb+c'b)=GCD(c',d+kb)=1 so this can be done. Suppose c and d+kb have a common prime factor q. If q=p, then since p divides c and d, p divides kb. But p does not divide k so p divides b, contradicting the assumption that that GCD(b, c, d)=1. So q ≠ p. Then q divides c' and d+kb contracting GCD(c',d+kb)=1. So if there is k so that GCD(c',d+kb)=1 then there is k so that GCD(c,d+kb)=1. So we can assume that c and d are relatively prime. But in this case simply take k=0 to get the result. --RDBury (talk) 01:29, 10 October 2013 (UTC)

October 11

Does there exist a function that...?

It maps R+ to R+, and has f(f(x))<f'(x) for every x>0. I guess it is not possible, but what is the trick?--124.172.170.233 (talk) 07:20, 11 October 2013 (UTC)

Am I missing something? What's wrong with f(x)=2x? AlexTiefling (talk) 09:28, 11 October 2013 (UTC)

There's a ' on the right f.Phoenixia1177 (talk) 10:04, 11 October 2013 (UTC)

$f(x)=-{1 \over -{1 \over x}}\,\!$ perhaps? -- Toshio Yamaguchi 10:23, 11 October 2013 (UTC)

Isn't that just x?Phoenixia1177 (talk) 10:27, 11 October 2013 (UTC)

How is the derivative of that x? The inner derivative of the denominator is $1 \over x^{2}$ and then doing the outer derivative you get $1 \over {x^{4}}$, unless am I missing something here? -- Toshio Yamaguchi 10:44, 11 October 2013 (UTC)

Perhaps I should have added some parentheses. -- Toshio Yamaguchi 10:46, 11 October 2013 (UTC)

Like this: $f(x)=-{1 \over (-{1 \over x})}\,\!$ -- Toshio Yamaguchi 10:57, 11 October 2013 (UTC)

Perhaps I should try to graph that as its becoming a bit confusing. -- Toshio Yamaguchi 11:04, 11 October 2013 (UTC)

f(x) simplifies to f(x) = x. You've made a mistake calculating the derivative: $f'(x)=-{1 \over (-{1 \over x})^{2}}*{1 \over x^{2}}=x^{2}*{1 \over x^{2}}=1,$ as expected.--80.109.106.49 (talk) 11:07, 11 October 2013 (UTC)

I don't what the proper notation for the function I have in mind would be. Consider the following. Take say $x=2$. Calculate $y=-{1 \over x}$. Then calculate $z=-{1 \over y}$. I am pretty sure z is not 2. -- Toshio Yamaguchi 11:17, 11 October 2013 (UTC)

It could be expressed as a function composition:

$f(x)=-{1 \over x}\,\!$

$g(x){=}f\circ f$

-- Toshio Yamaguchi 11:30, 11 October 2013 (UTC)

Your notation is correct for what you're trying to express. However, it does simplify to x. The negatives cancel out, and dividing by a fraction is the same as multiplying but the reciprocal. So it becomes 1*x = x.--80.109.106.49 (talk) 11:34, 11 October 2013 (UTC)

Yes, I can see it now. Sorry, don't know what I thought when writing that, I was trying to follow some weird thought that actually led nowhere. -- Toshio Yamaguchi 11:39, 11 October 2013 (UTC)

By an argument using Rolle's theorem, f must be monotonic. Clearly it cannot be decreasing, so it must be strictly increasing. If it were bounded, the derivative would converge to 0, so it must be unbounded. Thus the derivative must be unbounded. So eventually, f(x) - x > 1. By the mean value theorem, ff(x) = f(x) + f'(c)( f(x) - x), for some c greater than x. So ff(x) > f(x) + f'(x)( f(x) - x) > f'(x). Contradiction.--80.109.106.49 (talk) 10:28, 11 October 2013 (UTC)

I guess I'm assuming f'(x) is monotonic at the end. We can fix that. Eventually, f(x) - x > 2. Since $\liminf f'(x)=\infty$, choose x such that for all c > x, f'(c) > f'(x)/2. Note that this does not require f'(x) be continuous. Now we have ff(x) = f(x) + f'(c)( f(x) - x) > f'(x)/2 *2 = f'(x).--80.109.106.49 (talk) 10:43, 11 October 2013 (UTC)

I disagree that, just because in the bounded case f' goes to zero, f must be unbounded; even though f(x) keeps getting larger and hence greater than f', it could be that f(f(x)) keeps getting smaller and always stays smaller than f'.

How about this example?: Let f(x) = k₁x for 0 < x < 1/k₁, f(x) = k₂x for 1/k₁ ≤ x < 1/k₂ (with k₂ < k₁), .... Here {k_i} is a decreasing sequence that goes asymptotically to zero. In the range 1/J < x < 1/K we have f(x) = Kx and f(f(x)) = K²x and f'(x) = K, so f'(x) > f(f(x)). This function has an infinite number of kink points in which the left derivative f' is not defined, but I wonder if this can be fixed by smoothing out the kink somehow. Duoduoduo (talk) 16:13, 11 October 2013 (UTC)

every base is base 10 -- isn't calling it that kind of dumb?

in the sense that "every base is base 10", isn't calling it that kind of dumb? Wouldn't it be much more descriptive to call base 10 'base 9', call octal 'base 7', call binary 'base 1' and call hexedecimal 'base f' (base 15)?

I mean what kind of sense does it make to use two positions to name our base - a 1 and a 0? 212.96.61.236 (talk) 22:30, 11 October 2013 (UTC)

Nothing 'dumb' about it at all, given that (unless indicated to the contrary), written numbers are normally given in base ten: so '10' is clearly intended to mean 'ten', rather than 'the digits 1 and 0 in some unknown base'. AndyTheGrump (talk) 22:42, 11 October 2013 (UTC)

isn't calling it that kind of dumb? — Perhaps it is... but -after all- we're not geniuses, we're just mathematicians... :-) — 79.113.213.168 (talk) 00:47, 12 October 2013 (UTC)

How is "every base base 10"? Bubba73 ^{You talkin' to me?} 01:09, 12 October 2013 (UTC)

The OP's point is that if you write the base in the number system itself then it always becomes "10" = 1×base + 0. But base ten is implied so when we write "base 10" we mean ten and when we write "base 16" we mean sixteen. PrimeHunter (talk) 01:42, 12 October 2013 (UTC)

OK, now I understand. Bubba73 ^{You talkin' to me?} 01:56, 12 October 2013 (UTC)

Our system is called base 10 rather than base 9 because when we write multi-digit numbers, the position of the digits indicates multiples of powers of 10. Like 4562 means $4\times 10^{3}+5\times 10^{2}+6\times 10^{1}+2\times 10^{0}$. Each term is a constant times an exponential function with base 10. Base 9 would be representing numbers in terms of powers of 9. Staecker (talk) 01:30, 12 October 2013 (UTC)

Also, we have 10 digits, not 9. As for why other bases are named in base 10 (why we would say base 12, 12 in decimal) is because we use decimal usually. Notation is all that number bases are, the point of notation is communication- it is not foundational, so it's not circular to denote the base in decimal, it's convenient.Phoenixia1177 (talk) 05:37, 12 October 2013 (UTC)

If we have ten digits, why do you have to spell 'ten' 1-0, instead of being able to write the tenth digit? As far as I can tell we have 9 and a positional null. 212.96.61.236 (talk) 13:57, 12 October 2013 (UTC)

0 counts as a digit. so in base ten we have ten digits to hold place value, 0-9. compare to binary, 2 digits: 0 and 1. likewise, the number two in binary is '10' .. so perhaps you're somehow conflating the issue of digits with counting, since we normally learn to start at 1, only later learning that 0 is actually the first digit if we're ordering them 0-9. 76.17.125.137 (talk) 05:51, 17 October 2013 (UTC)

Sure, you can make up your own definitions if you like. Doesn't change the fact that you're wrong, of course...Sebastian Garth (talk) 16:29, 12 October 2013 (UTC)

Thank you! 212.96.61.236 (talk) 00:30, 13 October 2013 (UTC)

We call it this way because we use base 10. If we were a base-12 civilization we'd probably be referring to decimal as "base ᘔ". A base-6 civilization would most likely call decimal "base 14". And so on. We would all refer to our default bases (decimal, dozenal and senary here respectively) as "base 10", understanding the base "10" is written in to be implied. The base you are using by default is implied. Double sharp (talk) 03:31, 15 October 2013 (UTC)

Write 'base ten' or 'base twelve' or 'base six' or 'base sixty' or whatever. As the base number is always written '10', writing 'base-10' is useless for communication purposes. Kind of dumb? Yes! Bo Jacoby (talk) 19:03, 15 October 2013 (UTC).

No. There is a convention that when you write "base X", X is always expressed in base ten. PrimeHunter (talk) 12:09, 16 October 2013 (UTC)

Yes, and he who does not know this dumb convention is lost. Bo Jacoby (talk) 22:40, 17 October 2013 (UTC).

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)

October 13

Product of Sines

$\prod _{k=1}^{n-1}\sin {\Big (}{\tfrac {k}{n}}\cdot \pi {\Big )}\ =\ {\frac {n}{2^{n-1}}}\qquad -$ 79.113.215.238 (talk) 12:56, 13 October 2013 (UTC)

Are you asking us or telling us? AndrewWTaylor (talk) 16:21, 13 October 2013 (UTC)

(Problems getting the notation to appear correctly - seems to work only occasionally). AndrewWTaylor (talk) 16:23, 13 October 2013 (UTC)

1. I've seen it yesterday on a math site, and, albeit it looks so deceitfully simple, I've already tried two approaches and failed... 2. Yes, I know, I've been having the same problems for months now... (I think it's due to page size, I'm not sure...) — 79.113.210.178 (talk) 02:30, 14 October 2013 (UTC)

The identity is mentioned in List_of_trigonometric_identities#Identities_without_variables, which references Mathworld (maybe where you saw it?), where the only source given is "personal communication". AndrewWTaylor (talk) 08:23, 14 October 2013 (UTC)

Question about SAT scores

How rare (or how common) are "perfect" SAT scores ... in any given year, or in general? And, has that trend gone up or down, over time? And how about "perfect" ACT scores? Thanks. Joseph A. Spadaro (talk) 14:03, 13 October 2013 (UTC)

See SAT#Raw scores, scaled scores, and percentiles. Duoduoduo (talk) 16:25, 13 October 2013 (UTC)

Yes, thanks. I had seen that. But, those are merely percentiles. I am not seeking the relative number (i.e., the percentage) of students who achieve a perfect score, but rather the absolute number of students who do (e.g., "In 2011, there were 43 students who achieved a perfect SAT score." ... or some such statement.). Thanks. Joseph A. Spadaro (talk) 17:22, 13 October 2013 (UTC)

That section does say something close to what you want:

The older SAT (before 1995) had a very high ceiling. In any given year, only seven of the million test-takers scored above 1580. A score above 1580 was equivalent to the 99.9995 percentile.

This can't be exactly right, because I doubt that the number was the same in every year as in every other year. But it's probably intended to be a statement of the average number over all the years prior to 1995. Duoduoduo (talk) 18:53, 13 October 2013 (UTC)

This dotcom site says:

Perfect Scores: A perfect score is 2,400 points. Approximately 1,000,000 students take the SAT each year and on average, only 20 of them get a perfect score.

I found this by googling "perfect score SAT". You might want to look at some of the other hits that come up for this search. Duoduoduo (talk) 19:01, 13 October 2013 (UTC)

The number of students getting 36 on the ACT is on the order of several hundred a year out of millions of test-takers, according to the Wikipedia article on the ACT.--Jasper Deng (talk) 19:16, 13 October 2013 (UTC)

Thanks to all. Much appreciated! Joseph A. Spadaro (talk) 02:50, 16 October 2013 (UTC)

Identifying a unitary

Given a complex m by n matrices $X$ and $Y=XU$ for $m\gg n$ such that $U$ is uniquely defined, what is the computationally least intensive way of identifying the n by n unitary $U$. The only approach which occurs to me is svd followed by comparison of the right hand unitaries, however this seems to involve evaluating a lot of redundant information. — Preceding unsigned comment added by 81.155.161.131 (talk) 22:51, 13 October 2013 (UTC)

As a followup question it would also be useful to know about the case where $Y\approx XU$ and I would like to find the matrix $U$ which minimises the error as determined by some defensible measure (preferably the induced norm of $Y-XU$) where the restriction of $U$ to the unitaries remains. Any help greatly appreciated. — Preceding unsigned comment added by 81.155.161.131 (talk) 23:12, 13 October 2013 (UTC)

Any help at all? If someone could even provide some pointers to a useful method or relevant article it would be useful! — Preceding unsigned comment added by 85.210.44.227 (talk) 20:42, 15 October 2013 (UTC)

October 14

Duplo Train track behavior.

We recently got our granddaughter a set of Duplo train tracks. The set has numerous curves and straights - and two sets of "points" (aka "turnouts", aka "railroad switches"). These have the interesting property that when the train goes across the switch in the "wrong" direction, the switch flips over to make it "right" (real railroad switches do this - it's called "trailing point movement"). Looking at the topmost diagram, here is how the switch behaves:

• if the train enters the switch from A, then the current switch position (B or C) determines whether it goes towards B or towards C.
• if the train enters from B then it'll exit at A - and no matter how the switch was previously set, it'll now point towards B.
• if the train enters from C then it'll exit at A - and no matter how the switch was previously set, it'll now point towards C.

This has some interesting possibilities. It's like a computer with a two-bit memory.

If my track topology is a loop with a diagonal track through it (second diagram), then if the train is travelling counter-clockwise, it'll continue to do so no matter how you fiddle with the switches manually. But if it's going clockwise, then if both switches are at B then it'll continue to do go clockwise, but if either switch is set towards C, then the train will eventually go counter-clockwise and then it's stuck like that. This is really boring!

The next two diagrams are similarly boring, the train always gets "stuck" going around some subset of the track. The leftmost one is fractionally less boring because you can manually flip one of the switches to make the train go somewhere different - but the rightmost layout is another fairly boring one.

But the bottommost diagram is fascinating. The train not only travels along every inch of the track "automatically" - but it does so in both directions without manual intervention! My granddaughter seems to be fascinated by trying to guess where the train will go next. Now, that's fun!

So - my question is, are there any other two-switch topologies that show interesting behavior? Can anyone enumerate all of the logically different track layouts with two turnouts (I don't have any bridges to make figure-8 tracks...but that's another thought). What more complicated behaviors possible if I were to buy two more switches?

SteveBaker (talk) 01:01, 14 October 2013 (UTC)

See Laying Track by Ivars Peterson describing results from an article about problems like that. There were some comments on it but they've disappeared. While you're at it you might like clicking on some of the pictures at Pancursal Track Layout which uses simple svg animation.

By the way as far as I know you can't make a computer using the switches you describe but you can using ones which switch to the wrong direction if you go the wrong way over them! Dmcq (talk) 10:04, 14 October 2013 (UTC)

Thanks! The "Laying Track" article seems to cover it - so there are indeed only five different layouts. The Pancursal article is for tracks where traversing them doesn't switch them (the Brio wooden track is like that) - so his animations don't match what happens with Duplo track. SteveBaker (talk) 15:32, 14 October 2013 (UTC)

A little armchair maths:

There are basically 2 arrangments you can make with the switches. A loop-plus-diagonal (LPD) as in the first and (sufficiently generalised) second diagram, and a loop-line-loop (LLL) as in the third and fourth diagram (i.e. two loops, connected by a line).

Let's take the LLL configuration first. It can be pretty simply split into the individual loops, and just consider what happens with an incoming train or different switch orientations. If the b or c arm is on the line, then the train will enter, exit the a arm into the loop, come back to the c arm, and then continue around the loop, trapped forever. If the a arm is on the line, then the train will enter, exit whichever arm is selected (lets say b for this example), travel around the loop, and exit to the line via the c arm, setting the switch to c. Then the next train that comes along travels the opposite way. So it looks like the fourth diagram above is the only LLL configuration that does not result in the train getting stuck in a single loop.

As for the LPD configuration, the first and second diagrams show the possible conigurations where the a arm is part of the loop for both junctions, and a quick bit of thinking will show that if the a arm is part of the line, the switch will end up set to whichever arm the opposite switch directs the trains to, and the circuit will end up equivalent to the second diagram.

So, your list of possible tracks is:

An LPD as in the first diagram (c on the line, both switches pointed in the same direction around the loop)

An LPD as in the second diagram (c on the line, switches pointed in opposite direcions around the loop)

An LPD where one switch has a on the line, and one has c on the line (eventually ends up on half the loop plus the line, defined by where a points for the second switch)

An LPD where both switches have a on the line (eventually ends up on half the loop plus the line, defined by the position of the switch where the train first exits the line)

An LLL where both switches have c on the line (as in the third diagram)

An LLL where one switch has a on the line, and one has c on the line (ends up stuck in the c-loop)

An LLL where both switches have a on the line (as in the fourth diagram), the only "interesting" one

Anything else with 2 switches will behave as one of the above

MChesterMC (talk) 11:11, 14 October 2013 (UTC)

So you think there are seven different setups? The "laying track" article claims that there are only five. I think the three LLL cases you describe are in his article - so I guess you are thinking of two different LPD varients? I suppose the simple distinction between LLL and LPD is that in every case there are two loops - but in LLL, there is a connecting track between them where in LPD, they share a section between them. My thinking (at right) is that there ought to be three possible arrangements where there are two separate loops and a connecting track - and three more where two loops share a common track segment - but on that basis, you have one too many combinations and "Laying Track" has one too few (LSL(AA) in my diagram)? Am I missing something here? SteveBaker (talk) 15:32, 14 October 2013 (UTC)

(Correction): OK - so LSL(AA) and LSL(BB) are kinda-sorta the same thing...nevermind...just five possible layouts. SteveBaker (talk) 15:39, 14 October 2013 (UTC)

Ah, yes, my first layout is the same as my third, and my second is the same as my fourth. MChesterMC (talk) 08:09, 15 October 2013 (UTC)

Ah I knew I'd read something besides the MathTrek page about this. In Eureka 53 February 1994 they had an article Train Sets by Adam Chalcraft and Michael Greene, there's a copy at Train Sets. If you search for 'Turing Train' you'll find lots of trains chugging round the place computing things. Dmcq (talk) 12:26, 14 October 2013 (UTC)

Laplace's Method for Integrals

Practicing using Laplace's Method, and working from de Bruijin and having a little bit of difficult with the approximation on the boundary, in particular taking $\int _{0}^{\pi }x^{n}\sin(x)dx$ and applying the approximation as $n\to \infty$, notably because the maximum of n(log(x)) is at pi in this case, and so the method gives 0 because sin(pi)=0? Any suggestions? — Preceding unsigned comment added by 130.102.158.15 (talk) 02:15, 14 October 2013 (UTC)

In general, after rescaling to make the integration limits equal to 0 and 1, you can write

$\int _{0}^{1}x^{n}f(x)dx={\frac {1}{n+1}}\int _{0}^{\infty }\exp(-u)f\left[\exp \left(-{\frac {u}{n+1}}\right)\right]du$

If f(1) is not zero, the leading contribution will be f(1)/(n+1). But in this case that isn't the case, so to get the leading contribution, you need to beyond this approximation.Count Iblis (talk) 17:23, 14 October 2013 (UTC)

You can expand:

f[exp(-u/(n+1))] = f[1 - u/(n+1) + 1/2 u^2/(n+1)^2+...]

= f(1) + [-u/(n+1) + 1/2 u^2/(n+1)^2+...] f'(1) + 1/2 [u^2/(n+1)^2 + ...]f^((2))(1) + ...

Then if f(1) = 0, the leading contribution to the asymptotic expansion becomes:

-1/(n+1)^2 f'(1)

Count Iblis (talk) 17:36, 14 October 2013 (UTC)

How to turn difficult probabilities into a simple sum a non-mathematician can use

I'm looking for the way very difficult statistics are sometimes turned into easy questionnaires that have reasonably precise results.

Let's say I'm a judge and I want to know what the odds are the thief will steal again after he is released from prison. I'm not a mathematician, so I need a form with questions like "has a girlfriend -> add 2 points", "never stolen before -> add 5", "someone close says he's a pathetic lier -> subtract 4", "says he's sorry -> add 1", and after adding it all up it says "10-15 points: 30%".

Similar forms could be made for a doctor with a difficult decision to operate or not, even being able to show the patient why he decided not to. I've seen some of those, but how are they constructed? I guess its got something to do with Bayes' theorem? I can imagine that question 4 might have to be like "add 3 points, unless question 2 was also positive then just add 1" for added precision, but the questionnaire shouldn't be much harder than that and still give a quite accurate estimate of the chances. Joepnl (talk) 22:39, 14 October 2013 (UTC)

I guess Bayes theorem tells you how to update probabilities. Let S be the event that the thief will steal again. Let $E$ be some evidence. Then $P(S|E)={\frac {P(E|S)P(S)}{P(E|S)P(S)+P(E|!S)P(!S)}}$. P(S|E) is the probability the thief will steal again knowing the evidence E (what we want to find). P(S) is the probability absent that evidence. P(E|S) is the probability that evidence would arise if we know the thief will steal again. Widener (talk) 23:58, 14 October 2013 (UTC)

I think you misunderstand the nature of evidence, by the way. The fact that someone has a girlfriend, has never stolen before, or someone close says he's a pathetic liar, is not evidence he will steal again (in those cases $P(E|S)=P(E)$ and therefore $P(S|E)=P(S)$). Widener (talk) 00:05, 15 October 2013 (UTC)

Logistic regression is tailor-made for questions of the form "What are the odds that ... given various explanatory variables?". Duoduoduo (talk) 11:47, 15 October 2013 (UTC)

October 15

The encrypted German Tank Problem

In German_tank_problem#Countermeasures, it describes a simple encryption where each digit is assigned a letter. Is it possible to arrive at a good estimate for the number of "tanks" if you know that such an encryption scheme is in use, but don't know the exact letter combinations (e.g. could you estimate the number of tanks from "XH, HI, IV, VL" etc.)? Obviously, there is an upper limit if all the numbers are n digit, or there is obvious zero padding (XXXXXHI), but how large a sample would you need to get close to the real value? MChesterMC (talk) 09:00, 15 October 2013 (UTC)

The Countermeasures subsection is badly off-topic. We do have an article on Cryptography however. Bo Jacoby (talk) 19:16, 15 October 2013 (UTC).

In your sample, just from the fact that there are 4 different starting digits you can deduce that there are at least 40 tanks, and from the fact the serial numbers have 2 digits you can deduce there must be less than 100. From the German point of view it would be better to just tack an extra random digit to the end of the serial number; this would have the added benefit of scaring the heck out the Allies. --RDBury (talk) 03:40, 16 October 2013 (UTC)

Bernoulli Numbers Help!

Can anyone explain 8:30-8:40 of this video?! I just don't get why that sum has to equal zero or how that leads to deriving the values 1, -0.5 etc. Someone else asked in the comments but I don't really understand the answers there either.

http://www.youtube.com/watch?v=XHQ0OzqTjd0 — Preceding unsigned comment added by 5.81.9.195 (talk) 15:39, 15 October 2013 (UTC)

You're equating two power series, meaning their coefficients need to be the same. Widener (talk) 06:18, 16 October 2013 (UTC)

Which power series? If the Bernoulli sum/expansion is one what is the other? — Preceding unsigned comment added by 86.147.189.134 (talk) 10:37, 16 October 2013 (UTC)

The other power series is the constant function $1$. The right hand side is the power series $\sum _{n=0}^{\infty }a_{n}s^{n}$ where $a_{n}={\frac {1}{(n+1)!}}\sum _{\mu =0}^{n}{\binom {n+1}{\mu }}\beta _{\mu }$. The left hand side is the power series $\sum _{n=0}^{\infty }b_{n}s^{n}$ where $b_{0}=1$ and $b_{n}=0$ for $n>0$. As these power series are equal, $a_{n}=b_{n}$ for all $n$. That is, ${\frac {1}{(n+1)!}}\sum _{\mu =0}^{n}{\binom {n+1}{\mu }}\beta _{\mu }=0$ if $n\neq 0$ (equivalently $\sum _{\mu =0}^{n}{\binom {n+1}{\mu }}\beta _{\mu }=0$). Also, ${\frac {1}{(0+1)!}}\sum _{\mu =0}^{0}{\binom {0+1}{\mu }}\beta _{\mu }=1\implies \beta _{0}=1$, and ${\frac {1}{(1+1)!}}\sum _{\mu =0}^{1}{\binom {1+1}{\mu }}\beta _{\mu }=0\implies \beta _{1}=-{\frac {1}{2}}$. Widener (talk) 13:04, 16 October 2013 (UTC)