February 2009 - mathfun@mailman.xmission.com

[math-fun] Wired: The Formula That Killed Wall Street
by Henry Baker 25 Feb '09

25 Feb '09

FYI -- http://www.wired.com/print/techbiz/it/magazine/17-03/wp_quant WIRED MAGAZINE: 17.03 Recipe for Disaster: The Formula That Killed Wall Street By Felix Salmon Email 02.23.09 In the mid-'80s, Wall Street turned to the quantsÂbrainy financial engineersÂto invent new ways to boost profits. Their methods for minting money worked brilliantly...until one of them devastated the global economy. A year ago, it was hardly unthinkable that a math wizard like David X. Li might someday earn a Nobel Prize. After all, financial economistsÂeven Wall Street quantsÂhave received the Nobel in economics before, and Li's work on measuring risk has had more impact, more quickly, than previous Nobel Prize-winning contributions to the field. Today, though, as dazed bankers, politicians, regulators, and investors survey the wreckage of the biggest financial meltdown since the Great Depression, Li is probably thankful he still has a job in finance at all. Not that his achievement should be dismissed. He took a notoriously tough nutÂdetermining correlation, or how seemingly disparate events are relatedÂand cracked it wide open with a simple and elegant mathematical formula, one that would become ubiquitous in finance worldwide. For five years, Li's formula, known as a Gaussian copula function, looked like an unambiguously positive breakthrough, a piece of financial technology that allowed hugely complex risks to be modeled with more ease and accuracy than ever before. With his brilliant spark of mathematical legerdemain, Li made it possible for traders to sell vast quantities of new securities, expanding financial markets to unimaginable levels. His method was adopted by everybody from bond investors and Wall Street banks to ratings agencies and regulators. And it became so deeply entrenchedÂand was making people so much moneyÂthat warnings about its limitations were largely ignored. Then the model fell apart. Cracks started appearing early on, when financial markets began behaving in ways that users of Li's formula hadn't expected. The cracks became full-fledged canyons in 2008Âwhen ruptures in the financial system's foundation swallowed up trillions of dollars and put the survival of the global banking system in serious peril. David X. Li, it's safe to say, won't be getting that Nobel anytime soon. One result of the collapse has been the end of financial economics as something to be celebrated rather than feared. And Li's Gaussian copula formula will go down in history as instrumental in causing the unfathomable losses that brought the world financial system to its knees. [ ... see the link above for the rest of the article]

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[math-fun] a table of optimal partitions
by David P. Moulton 20 Feb '09

20 Feb '09

Hi muffin-folk, Now I'll give the table of all the optimal partitions I've computed for the muffin problem, namely those with m and n relatively prime with mn < 128, followed by some comments. Recall that I normalize to find common partitions of m * n and n * m and that I write multiplicities after parts. Recall also that I'm lexically maximizing the list of piece sizes sorted in nondecreasing order. That is, I maximize the smallest piece, then the next smallest, etc. I give (m, n) and then the (unique) optimal partition, in increasing order. Thus the value of U(m, n) is just the first number after the colon. I repeat (m, m+1) pairs in the explicit list after the general formulas. (1, k): 1 * k (for k>=1) (2, 2k+1): 1 * 2, 2 * 2k (for k>=0) (3, 3k+-1): (3k-1)/2k * 2k, (3k+1)/2k * 2k, 3 * (k+-1) (for k>=1 odd) (3, 3k+-1): (3k-1)/2k * 2k, 3/2 * 2, (3k+1)/2k * 2k, 3 * (k-1+-1) (for k>=2 even) (4, 2k+1): (2k-1)/k * 2k, 2 * 2, (2k+1)/k * 2k (for k>=1) (3k-1, 3k): k * 6, (k+1) * 6, ..., (2k-1) * 6 (for k>=2) (3k, 3k+1): k * 2, (2k+1)/2 * 4, (k+1) * 2, ..., 2k * 2 (for k>=2) (3k+1, 3k+2): (2k+1)/2 * 4, (k+1) * 2, (2k+3)/2 * 4, ..., (4k+1)/2 * 4 (for k>=1) (5, 6): 2 * 6, 3 * 6 (5, 7): 5/3 * 3, 7/3 * 6, 8/3 * 6 (5, 8): 2 * 4, 3 * 4, 5 * 4 (5, 9): 2 * 6, 3 * 6, 5 * 3 (5, 11): 13/6 * 6, 9/4 * 4, 5/2 * 2, 11/4 * 4, 17/6 * 6 (5, 12): 2 * 4, 3 * 4, 5 * 8 (5, 13): 13/6 * 6, 22/9 * 6, 5/2 * 2, 23/9 * 6, 17/6 * 6 (5, 14): 11/5 * 10, 12/5 * 2, 13/5 * 2, 14/5 * 10, 5 * 2 (5, 16): 16/7 * 14, 17/7 * 2, 18/7 * 2, 19/7 * 14 (5, 17): 13/6 * 6, 37/15 * 10, 5/2 * 2, 38/15 * 10, 17/6 * 6 (5, 18): 9/4 * 8, 5/2 * 4, 11/4 * 8, 5 * 8 (5, 19): 16/7 * 14, 69/28 * 4, 5/2 * 2, 71/28 * 4, 19/7 * 14 (5, 21): 7/3 * 18, 8/3 * 18, 5 * 3 (5, 22): 9/4 * 8, 5/2 * 4, 11/4 * 8, 5 * 12 (5, 23): 23/10 * 10, 149/60 * 12, 5/2 * 2, 151/60 * 12, 27/10 * 10 (5, 24): 7/3 * 18, 8/3 * 18, 5 * 6 (6, 7): 2 * 2, 5/2 * 4, 3 * 2, 7/2 * 4, 4 * 2 (6, 11): 7/3 * 6, 17/6 * 4, 3 * 2, 19/6 * 4, 11/3 * 6 (6, 13): 13/5 * 10, 14/5 * 2, 3 * 2, 16/5 * 2, 17/5 * 10 (6, 17): 13/5 * 10, 44/15 * 6, 3 * 2, 46/15 * 6, 17/5 * 10 (6, 19): 19/7 * 14, 41/14 * 4, 3 * 2, 43/14 * 4, 23/7 * 14 (7, 8): 5/2 * 4, 3 * 2, 7/2 * 4, 4 * 2, 9/2 * 4 (7, 9): 5/2 * 6, 3 * 2, 7/2 * 2, 4 * 2, 9/2 * 6 (7, 10): 7/3 * 3, 19/6 * 6, 10/3 * 2, 7/2 * 2, 11/3 * 2, 23/6 * 6 (7, 11): 11/4 * 4, 13/4 * 2, 27/8 * 4, 7/2 * 2, 29/8 * 4, 15/4 * 2, 17/4 * 4 (7, 12): 3 * 12, 4 * 12 (7, 13): 8/3 * 6, 19/6 * 2, 41/12 * 4, 7/2 * 2, 43/12 * 4, 23/6 * 2, 13/3 * 6 (7, 15): 3 * 10, 4 * 10, 7 * 5 (7, 16): 3 * 12, 4 * 12, 7 * 4 (7, 17): 11/4 * 4, 17/5 * 10, 69/20 * 2, 7/2 * 2, 71/20 * 2, 18/5 * 10, 17/4 * 4 (7, 18): 3 * 6, 4 * 6, 7 * 12 (8, 9): 3 * 6, 4 * 6, 5 * 6 (8, 11): 8/3 * 6, 11/3 * 6, 4 * 6, 13/3 * 6 (8, 13): 13/4 * 8, 7/2 * 2, 15/4 * 2, 4 * 2, 17/4 * 2, 9/2 * 2, 19/4 * 8 (8, 15): 3 * 6, 4 * 6, 5 * 6, 8 * 6 (9, 10): 3 * 2, 7/2 * 4, 4 * 2, 9/2 * 4, 5 * 2, 11/2 * 4, 6 * 2 (9, 11): 13/4 * 4, 7/2 * 2, 15/4 * 4, 9/2 * 2, 21/4 * 4, 11/2 * 2, 23/4 * 4 (9, 13): 3 * 3, 4 * 6, 5 * 6, 9 * 6 (9, 14): 7/2 * 4, 17/4 * 8, 9/2 * 4, 19/4 * 8, 11/2 * 4 Recall the upper bound I mentioned previously: Corollary. For m < n with m not dividing 2n we have U(m, n) <= m/3 for 4m/3 <= n < 3m/2 U(m, n) <= min( n/([2n/m]+1), m - n/[2n/m] ) otherwise. This upper bound seems to be sharp in many cases. In particular, among all the values in the table above, the only cases when it isn't sharp are (m, m+1) (with m mod 3 not 2), (5, 11), and (9, 11). I imagine the counterexamples will be more common for larger values, and yes, I do think it's a coincidence that the two sporadic counterexamples so far have n=11. Anyway, I thought I would give proofs for those two pairs. For (5, 11), we may, as usual, assume that each 5 splits into exactly 2 pieces, so that there are 22 pieces in all. Splitting an 11 into at least 6 pieces gives parts that are too small, and splitting an 11 into only 3 pieces gives parts whose complements in 5 are too small. So each 11 splits into 4 or 5 pieces, and of the five 11s, there are 3 that split into 4 pieces and 2 that split into 5 pieces. Now all 22 pieces pair up to make the 5s, and since there are 12 pieces coming from 11s split into 4 parts, but only 10 pieces from 11s split into 5 parts, there must be two of the former that pair up. But now consider the 8 pieces making up the two corresponding 11s split into 4 parts. These 8 pieces sum to 11 * 2 = 22, but 2 of them sum to 5, so the remaining 6 sum to 17, and one of them must be at least 17/6. Finally, the complement of this piece in its 5 is at most 13/6. Now consider (9, 11). Again, we may assume that each 9 splits into exactly 2 parts, so there are 22 pieces in all. Each 11 splits into 2 or 3 parts, so 5 of them split into 2 parts, and the other 4 split into 3 parts. Again, two pieces from two of the 11s that split into 3 must be paired to give a 9. Then the other 4 pieces of those 11s sum to 11 * 2 - 9 = 13, and the smallest of them is at most 13/4. Okay, now that I look at the proofs, I think it's not so much coincidence, since it's almost the same argument with the same numbers. David P. Moulton

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[math-fun] Chocolate bar puzzle
by Dan Asimov 19 Feb '09

19 Feb '09

Speaking of muffinsectors, here is a problem about chocosectors: You are given a rectangular chocolate bar that is scored in the usual way into KxL small squares. Repeat this step: Choose one of the pieces (at first, there's only the whole bar) and break it all the way along one of its scorings . . . until all the pieces are the small squares. Show that no matter how this is done, it takes the same number of steps. No fair if you already know this puzzle (which I read in a Peter Winkler book). --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] a family of muffin problem solutions
by David P. Moulton 19 Feb '09

19 Feb '09

Last Thursday I wrote > Lemma 1: > Given m>=3, let e be 1 if m is odd and 2 if m is even. > For k>=1 we have > > U(m, (km-1)(m-e)/2) >= (km-1)/2k = m/2 - 1/2k. Now I have both a counterexample and a generalization! > Proof of Lemma 1: > We use the multiset > > (m/2 - 1/2k) * ek(m-e), m/2 * (km-2ek-1)(m-e), (m/2 + 1/2k) * ek(m-e). Well, notice that the middle multiplicity is (k(m-2e)-1)(m-e). For m=4 (and e=2), this is negative! Indeed, taking k=1 would give U(4, 3) >= 3/2, while we already know U(4, 3) = 1. This isn't a problem for my asymptotic theorem, though, since we've already got an explicit formula for U(4, 2k+1), and it tends to 2. But I can allow a more general value of e and a plus or minus sign after "km". This gives Proposition. Take m>=3, k>=1, and e>=1 with k(m-2e)+-1>=0. If (km+-1)(m-e) is even, we have U(m, (km+-1)(m-e)/2) = (km-1)/2k = m/2 - 1/2k. If (km-1)(m-e) is odd, we have U(2m, (km+-1)(m-e)) = (km-1)/k = m - 1/k. Proof. If (km+-1)(m-e) is odd, everything goes through, but then one of the inputs to U is a half-integer, and we can just rescale by 2. So we'll assume it's even. The upper bound is a special case of the bound I just posted in my previous note. First, we have [(km+-1)(m-e)/m] = [k(m-e) +- 1 -+ e/m] = k(m-e) for "+" or k(m-e)-1 for "-". For "+" we have U(m, (km+1)(m-e)/2) <= m - ( (km+1)(m-e)/2 ) /(k(m-e)) = m - (km+1)/2k = (km-1)/2k, while for "-" we have U(m, (km-1)(m-e)/2) <= ( (km-1)(m-e)/2 ) /(k(m-e)-1 + 1) = (km-1)/2k. The proof of the lower bound is exactly the same as last week's proof, mutatis mutandi, though I'll reproduce it. We use the multiset (m/2 - 1/2k) * ek(m-e), m/2 * (k(m-2e)+-1)(m-e), (m/2 + 1/2k) * ek(m-e). It is clear that the smallest part has the desired size and that the parts can be paired up to give (km+-1)(m-e)/2 copies of m. Also, all the parts of size m/2 +- 1/2k) can be added together to give e copies of (km+-1)(m-e)/2. Finally, we have m/2 * (km-2ek+-1) + (m/2 -+ 1/2k) * ek = km^2/2 - ekm/2 +- m/2 -+ e/2 = (km+-1)(m-e)/2, so the rest of the pieces can be combined to make m-e more copies of (km+-1)(m-e)/2, and this is a valid common refinement. This proves the lower bound. QED Again, it is easy to show that the partition I give is optimal, in the sense of lexically maximizing the sorted list of sizes, once you combine as many m/2's as possible into m's. In fact, if km is odd, you can combine all of them, and the optimal partition is (m/2 - 1/2k) * ek(m-e), (m/2 + 1/2k) * ek(m-e), m * (k(m-2e)-1)(m-e)/2, while if km is even, you have to leave (m-e) of them unpaired, and the optimal partition is (m/2 - 1/2k) * ek(m-e), m/2 * (m-e), (m/2 + 1/2k) * ek(m-e), m * (k(m-2e)-2)(m-e)/2. David P. Moulton

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Re: [math-fun] Interprimes, an error, and 444 digits of accuracy
by Steve Witham 17 Feb '09

17 Feb '09

Ed-- Either this is not mysterious or I'm missing something. First of all, 5/2 is the average of 2 and 3. If the sum went from k=2 to oo, instead of 1 to oo, so that the p[k] were just the odd primes, then the 5/2 crossing would be left out, and I don't think the other crossings would be changed much. Second, it looks like that formula will have zeroes near the averages of adjacent elements of *any* sorted sequence. Have you tried substituting (e.g.) the Fibonacci sequence starting at the second 1? You should get crossings at... 3/2, 5/2, 8/2, 13/2, 21/2 ... (kinda interesting itself). I would guess the accuracy (-log of error) goes up like the distance between neighbors? Or maybe distance squared? Have you plotted that? (I don't have Mathematica.) --Steve >From: Ed Pegg Jr <ed(a)mathpuzzle.com> > >Daniel Tisdale (daniel6874(a)gmail.com) sent MathWorld an interesting item. > >"The sum $f(x)=sum_(k=1)^infty(-1)^ke^(-(x-p_k)^2)$ has as zeros precisely the >interprimes, with the single additional point 5/2" > >The plot agreed precisely, so I added it.?? >http://mathworld.wolfram.com/Interprime.html > >John Pardon (jpardon(a)princeton.edu)?? pointed out that it's wrong... >that it gives a series of highly accurate approximations instead. > >How accurate? >f[x_] := N[Sum[(-1)^k*Exp[-(x-Prime[k])^2],{k,1,2500}],1000] >Table[Floor[Log[10, Abs[1/f[Prime[n]/2 + Prime[n + 1]/2]]]], {n, 1, 400}] > >2, 1, 3, 15, 35, 43, 21, 6, 10, 156, 10, 6, 73, 6, 21, 10, > 52, 10, 6, 10, 10, 43, 21, 27, 6, 21, 111, 21, 6, 73, 27, 10, 21, > 125, 21, 10, 21, 444 ... > >The accuracies to the 400th Interprime have the following distribution: >{{1, 1}, {2, 1}, {3, 1}, {6, 44}, {10, 61}, {15, 1}, {21, 70}, >{27, 47}, {35, 23}, {43, 24}, {52, 32}, {62, 6}, {73, 23}, >{85,13}, {111, 14}, {125, 19}, {156, 10}, {173, 1}, {210, 1}, >{229, 2}, {271, 2}, {340, 1}, {417, 1}, {444, 2}} > >Just the accuracies, in numbers of digits >{1, 2, 3, 6, 10, 15, 21, 27, 35, 43, 52, 62, 73, 85, 111, 125, 156, 173, >210, 229, 271, 340, 417, 444} > > > >Differences: 1, 1, 3, 4, 5, 6, 6, 8, 8, 9, 10, 11, 12, 26, 14, 31, 17, >37, 19, 42, 69, 77, 27 > >For starters, 444 digits of accuracy has occurred more than once.? >52 digits of accuracy has occurred 32 times. Something seems to >be happening here.? > >Ed Pegg Jr

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Re: [math-fun] Geometry puzzle
by Dan Asimov 15 Feb '09

15 Feb '09

Neil wrote: << Dan's lattice "Z_n" is of course the root lattice "A_{n-1}" of determinant n so any fundamental region has volume sqrt(n). >> Oops, I inadvertently neglected to mention that anyone whose initials are N.J.A.S. is disqualified a priori (:-)>. (For everyone else, I'm looking for a proof based on first principles.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Geometry puzzle
by N. J. A. Sloane 15 Feb '09

15 Feb '09

Dan's lattice "Z_n" is of course the root lattice "A_{n-1}" of determinant n so any fundamental region has volume sqrt(n). I could have referred to "SPLAG" of course but I didn't! Neil

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[math-fun] Geometry puzzle
by Dan Asimov 15 Feb '09

15 Feb '09

In R^n, consider the integer points whose coordinates sum to 0. Call this abelian group Z_n. Then Z_n is an (n-1)-dimensional lattice in R^n, lying in a unique (n-1)-dimensional subspace S of R^n. In S, Z_n has a well-defined Voronoi cell at (say) the origin: P_n := { x in S | for all v in L, ||x|| <= ||x-v|| }. Find the (n-1)-dimensional volume V(n) of P_n: V(n) := vol_(n-1)(P_n) without using any references. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] math humor, sort of
by rcs＠xmission.com 15 Feb '09

15 Feb '09

Bill Cordwell has passed along some math humor with an inextricable political tinge. I'm posting the URL here, along with an admonition against any discussion (on the list). I found it funny, YMMV. Rich http://iowahawk.typepad.com/iowahawk/2009/02/numbers-in-the-news.html

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Re: [math-fun] Probability of a real polynomial's having all roots real
by Dan Asimov 10 Feb '09

10 Feb '09

I wrote: << Let c = (c_0,...,c_(n-1)) have the standard normal distribution on R^n, i.e., the c_k are independent real random variables, each with probability density given by d(x) = (1/(sqrt(2 pi)) exp(-x^2 / 2)) QUESTION: Find the probability f(n) that the polynomial P(x) = x^n + c_(n-1) x^(n-1) + ... + c_1 x + c_0 has all its roots real ? >> But on second thought the measure this puts on the space of nth degree real polys does not seem the most natural. Here's a nicer one: Consider all real polynomials c_n x^n + ... + c_0, and identify any two of them if one is a nonzero constant factor times the other (i.e., if they have the same roots). Then the equivalence classes form P^n, n-dimensional projective space, which has a natural measure on it. Its double cover is the usual n-sphere S^n with its usual measure, with each polynomial corresponding to a pair of antipodal points. The polynomials of degree < n form a subset of measure 0. With this measure on nth degree polynomials, it looks feasible to get the probability of having all real roots at least for some low degrees. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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