Challenge of the week: mathematical paradox, $500,000 award* - AnalyticBridge2019-05-26T14:59:44Zhttps://www.analyticbridge.datasciencecentral.com/forum/topics/challenge-of-the-week-mathematical-paradox-500-000-award?feed=yes&%3Bxn_auth=noHere's an interesting comment…tag:www.analyticbridge.datasciencecentral.com,2015-04-01:2004291:Comment:3224532015-04-01T23:21:38.848ZVincent Granvillehttps://www.analyticbridge.datasciencecentral.com/profile/VincentGranville
<p>Here's an interesting comment from one of our readers:</p>
<p><br></br><em>You've just found a mathematical thing, that is called a 'chaotic attractor'. Read: Tönu Puu:"Attractors, Bifurcations, and Chaos: Nonlinear Phenomena in Economics"<a class="ot-anchor aaTEdf" dir="ltr" href="http://www.springer.com/gp/book/9783662040942" rel="nofollow" target="_blank">http://www.springer.com/gp/book/9783662040942</a></em><br></br><br></br><em>These often appear like 'natural constants' and sometimes appear in…</em></p>
<p>Here's an interesting comment from one of our readers:</p>
<p><br/><em>You've just found a mathematical thing, that is called a 'chaotic attractor'. Read: Tönu Puu:"Attractors, Bifurcations, and Chaos: Nonlinear Phenomena in Economics"<a rel="nofollow" target="_blank" href="http://www.springer.com/gp/book/9783662040942" class="ot-anchor aaTEdf" dir="ltr">http://www.springer.com/gp/book/9783662040942</a></em><br/><br/><em>These often appear like 'natural constants' and sometimes appear in context of well known, mathematical functions or irrational mathematical constants, which can also be produced by recursive functions.</em><br/><br/><em>Also see 'point clouds' and Haussdorff. Mathematics, around 100 years old.</em><br/><br/><em>Have fun!</em></p> The core of the problem is wh…tag:www.analyticbridge.datasciencecentral.com,2015-03-31:2004291:Comment:3224382015-03-31T03:28:40.409ZVincent Granvillehttps://www.analyticbridge.datasciencecentral.com/profile/VincentGranville
<p>The core of the problem is whether or not 1/(1+2/(1+3/(1+4/(1+5/...)))) = <span>0.525135276... = function of Pi, e, SQRT, erfc, can be expressed using a finite combination of elementary functions - that is, not including erfc. The interest is because most its sister continued fractions do, but this one does not seem to. </span></p>
<p>The core of the problem is whether or not 1/(1+2/(1+3/(1+4/(1+5/...)))) = <span>0.525135276... = function of Pi, e, SQRT, erfc, can be expressed using a finite combination of elementary functions - that is, not including erfc. The interest is because most its sister continued fractions do, but this one does not seem to. </span></p> I guess I'm a little bit conf…tag:www.analyticbridge.datasciencecentral.com,2015-03-31:2004291:Comment:3222152015-03-31T01:52:22.322ZGregory Hursthttps://www.analyticbridge.datasciencecentral.com/profile/GregoryHurst169
<p>I guess I'm a little bit confused because as I pointed out in my first post (with the attached pdf), f(2) does not equal <span>0.525135276..., it equals 1/a - 1.</span></p>
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<p><span>As I said before the reason excel is giving you 0.525135276 as you iterate, is that rounding error propagated and took over. Using a CAS with exact arithmetic showed me this.</span></p>
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<p><span>So with that said, the constant <span>sqrt(pi*e/2)*erfc(1/sqrt(2)) should be irrelevant to the original…</span></span></p>
<p>I guess I'm a little bit confused because as I pointed out in my first post (with the attached pdf), f(2) does not equal <span>0.525135276..., it equals 1/a - 1.</span></p>
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<p><span>As I said before the reason excel is giving you 0.525135276 as you iterate, is that rounding error propagated and took over. Using a CAS with exact arithmetic showed me this.</span></p>
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<p><span>So with that said, the constant <span>sqrt(pi*e/2)*erfc(1/sqrt(2)) should be irrelevant to the original problem, right? It has nothing to do with f(2).</span></span></p>
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<p><span><span>Also wasn't the original paradox about how you were observing 1/a - 1 = 0.525135276...? It makes sense to you would call that a paradox, but asking if erfc(1/sqrt(2)) can be expressed in terms of elementary functions doesn't seem like a paradoxical thing to ask.</span></span></p> All rational or quadratic alg…tag:www.analyticbridge.datasciencecentral.com,2015-03-30:2004291:Comment:3222122015-03-30T23:27:27.415ZVincent Granvillehttps://www.analyticbridge.datasciencecentral.com/profile/VincentGranville
<p>All rational or quadratic algebraic numbers have periodic developments as continued fractions. But there are many ways to represent a number as a continued fraction, for instance Pi has its traditional standard a-periodic development (that looks chaotic) and one that uses 1, 9, 25, 49, 81 (the odd squares) which is also a-periodic. But I'm not aware of a theorem that claims that a rational or quadratic algebraic number can not be represented by an a-periodic (obviously non standard)…</p>
<p>All rational or quadratic algebraic numbers have periodic developments as continued fractions. But there are many ways to represent a number as a continued fraction, for instance Pi has its traditional standard a-periodic development (that looks chaotic) and one that uses 1, 9, 25, 49, 81 (the odd squares) which is also a-periodic. But I'm not aware of a theorem that claims that a rational or quadratic algebraic number can not be represented by an a-periodic (obviously non standard) continued fraction. Maybe such a result exists, I'm just not aware of it. If it does exist, it would help solve the mystery regarding the nature of many numbers (is it a rational or not). I would think - maybe I'm wrong - that you can create a made-up continued, a-periodic fraction, converging to a rational number, by design.</p> The continued fraction form t…tag:www.analyticbridge.datasciencecentral.com,2015-03-30:2004291:Comment:3224372015-03-30T23:06:40.111ZDaniel Lichtblauhttps://www.analyticbridge.datasciencecentral.com/profile/DanielLichtblau
<p>The continued fraction form tells you it cannot be rational (or quadratic algebraic).</p>
<p>The continued fraction form tells you it cannot be rational (or quadratic algebraic).</p> Yes for efrc in general (as a…tag:www.analyticbridge.datasciencecentral.com,2015-03-30:2004291:Comment:3224362015-03-30T23:01:10.163ZVincent Granvillehttps://www.analyticbridge.datasciencecentral.com/profile/VincentGranville
<p>Yes for efrc in general (as a function). But here we are dealing with a specific value. Even proving that 0.5251... is irrational, is no easy task, and has probably never been done.</p>
<p>Yes for efrc in general (as a function). But here we are dealing with a specific value. Even proving that 0.5251... is irrational, is no easy task, and has probably never been done.</p> It's known that it can't be e…tag:www.analyticbridge.datasciencecentral.com,2015-03-30:2004291:Comment:3224352015-03-30T22:51:35.582ZGregory Hursthttps://www.analyticbridge.datasciencecentral.com/profile/GregoryHurst169
It's known that it can't be expressed in terms of elementary functions. I believe the Risch algorithm shows this.
It's known that it can't be expressed in terms of elementary functions. I believe the Risch algorithm shows this. The value for f(2) is derived…tag:www.analyticbridge.datasciencecentral.com,2015-03-30:2004291:Comment:3224112015-03-30T22:37:43.977ZVincent Granvillehttps://www.analyticbridge.datasciencecentral.com/profile/VincentGranville
<p>The value for f(2) is derived from Wolfram Encyclopedia. It involves the erfc function, which is not an elementary transcendental function from the list that I provided. But can it be represented by these "elementary" functions, using finite combinations? Other similar continued fractions converge to simple expressions such as e, Pi, etc. Why not this one? Do we really need to use erfc or Gamma or Bessel functions to express it? In fact, this 0.525135276... might even be a rational number.…</p>
<p>The value for f(2) is derived from Wolfram Encyclopedia. It involves the erfc function, which is not an elementary transcendental function from the list that I provided. But can it be represented by these "elementary" functions, using finite combinations? Other similar continued fractions converge to simple expressions such as e, Pi, etc. Why not this one? Do we really need to use erfc or Gamma or Bessel functions to express it? In fact, this 0.525135276... might even be a rational number. Nobody knows, but I guess it is not. This is the 0.5 million dollar question - unanswered yet.</p> Hi Jordan,
How much is the r…tag:www.analyticbridge.datasciencecentral.com,2015-03-30:2004291:Comment:3222092015-03-30T22:33:07.382ZLuis V. Montiel Cendejashttps://www.analyticbridge.datasciencecentral.com/profile/LuisVMontielCendejas
<p>Hi Jordan,</p>
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<p>How much is the rent on that new house you own in Seattle? </p>
<p>Hi Jordan,</p>
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<p>How much is the rent on that new house you own in Seattle? </p> [Not to worry, the house is s…tag:www.analyticbridge.datasciencecentral.com,2015-03-30:2004291:Comment:3222062015-03-30T22:25:55.507ZDaniel Lichtblauhttps://www.analyticbridge.datasciencecentral.com/profile/DanielLichtblau
<p>[Not to worry, the house is safe so far as this post is concerned.]</p>
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<p>the limiting behavior should be a root of f_infinity=1/(1 + n*f_infinity). This is a quadratic and one gets solutions (-1 - sqrt(1 + 4 n))/(2 n) and (-1 + ...)/...</p>
<p>That first one is the "correct" one in terms of the sequence having an asymptotic behavior, at least for initial values I have tried.</p>
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<p>If one does a forward iteration say from 1 to 50 with whatever initial value 'a' for f(1)…</p>
<p>[Not to worry, the house is safe so far as this post is concerned.]</p>
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<p>the limiting behavior should be a root of f_infinity=1/(1 + n*f_infinity). This is a quadratic and one gets solutions (-1 - sqrt(1 + 4 n))/(2 n) and (-1 + ...)/...</p>
<p>That first one is the "correct" one in terms of the sequence having an asymptotic behavior, at least for initial values I have tried.</p>
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<p>If one does a forward iteration say from 1 to 50 with whatever initial value 'a' for f(1) then, iterating backwards from f(50) should NOT approximate your f(2). But perturbing slightly from f(50) and iterating backwards will give an approximation to that value. "Slightly" is loose wording: make it small enough and again you do not recover that f(2) value.</p>
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<p>I think what happens is along these lines. If your initial value for f(large n) is sufficiently close to (-1 - sqrt(1 + 4 n))/(2 n), then it will correspond to a value obtained by iterating forward from some seed value 'a' for f(1). In that case, iterating backwards will simply give approximately (1 - a)/a. Away from (-1 - sqrt(1 + 4 n))/(2 n), backwards iterations will approach the value you found. There is a midrange for perturbations, wherein one approaches neither of these.</p>
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<p>Since the iteration for nth term is defined explicitly in terms of n this seems to be a generalization of the dynamics that involve Fatou and Julia sets. Might be related though (I simply do not know enough about that stuff to say).</p>
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