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Continued Fractions John D Barrow. Headline in Prairie Life. Decimals. = 3.141592… = i a i 10 -i = (a i ) = (3,1,4,1,5,9,2,…). But rational fractions like 1/3 = 0.33333.. do not have finite decimal expansions Why choose base 10? Hidden structure?.
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Continued Fractions John D Barrow
Decimals • = 3.141592… • = i ai 10-i • = (ai) = (3,1,4,1,5,9,2,…) But rational fractions like 1/3 = 0.33333.. do not have finite decimal expansions Why choose base 10? Hidden structure?
A Different Way of Writing Numbers x2 – bx – 1 = 0 x = b + 1/x Substitute for x on the RH side x = b + 1/(b +1/x) Do it again…and again… b = 1 gives the golden mean x = = ½(1 + 5) = 1·6180339887..
William Brouncker First President of the Royal Society Introduced the ‘staircase’ notation (1620-84) by using Wallis’ product formula for John Wallis Wallis: ‘continued fraction’ (1653-5) (1616-1703)
Euler’s Formula Log{(1+i)/(1-i)} = i/2 i = -1
Avoiding the Typesetter’s Nightmare x [a0 ; a1, a2, ……] cfe of x Rational numbers have finite cfes Take the shortest of the two possibilities for the last digit eg ½ = [0;2] not [0;1,1] Irrational numbers have a (unique) infinite cfes
Pi and e = [3;7,15,1,292,1,1,3,1,14,2…..] e = 2.718…. = [2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,….] Cotes (1714) = [1;1,1,1,1,1,1,1,……..] golden ratio 2 = [1;2,2,2,2,2,2,2,2,2,2,….] 3 = [1;1,2,1,2,1,2,1,2,1,2,1,.] ‘Noble’ numbers end in an infinite sequence of 1’s
Rational Approximations for Irrational Numbers Ending an infinite cfe at some point creates a rational approximation for an irrational number = [3;7,15,1,292,1,1,…] Creates the first 7 rational approximations for labelled pn/qn 3, 22/7, 333/106, 355/113, 103993/33102, 104384/33215, 208341/66317,… A large number (eg 292) in the cfe expansion creates a very good approx
Better than Decimals Truncating the decimal expn of gives 31415/1000 and 314/100 The denominators of 314/100 and 333/106 are almost the same, but the error in the approximation 314/100 is 19 times as large as the error in the cfe approx 333/106. As an approximation to , [3; 7, 15, 1] is more than one hundred times more accurate than 3.1416.
= (2143/22)1/4 is good to 3 parts in 104 ! Ramanujan knew that 4 = [97;2,2,3,1,16539,1,…] Note that the 431st digit of is 20776
Minding your p’s and q’s • As n increases the rational approximations to any irrational number, x, get better and better • x – pn/qn 0 In the limit the best possible rational approx is x – p/q <1/(q25) qk> 2(k-1)/2 The golden ratio is the most irrational number: it lies farthest from a rational approximation 1/(q25) Approximants are 5/3, 8/5, 13/8, 21/13,… They all run close to this boundary Same is true for all (a + b)/(c + d) with ad – bc = + 1
Getting Your Teeth Into Gears The ratio of the numbers of teeth on two cogs governs their speed ratio. Mesh a 10-tooth with a a 50 tooth and the 10-tooth will rotate 5 times quicker (in the opposite direction). What if we want one to rotate 2 times faster than the other. No ratio will do it exactly. Cfe rational approximations to 2 are 3/2, 7/5, 17/12, 41/29, 99/70,… So we could have 7 teeth on one and 5 on the other (too few for good meshing though) so use 70 and 50. If we can use 99 and 70 then the error is only 0.007%
Scale Models of the Solar System
Gears Without Tears In 1682 Christian Huygens used 29.46 yrs for Saturn’s orbit around Sun (now 29.43) Model solar system needs two gears with P and Q teeth: P/Q 29.46 Needs smallish values of P and Q (between 20 and 220) for cutting Find cfe of 29.46. Read off first few rational approximations 29/1, 59/2, 206/7,..then simulate Saturn’s motion relative to Earth by making one gear with 7 teeth and one with 206
Carl Friedrich Gauss (1777-1855)
Probability and Continued Fractions Any infinite list of numbers defines a unique real number by its cfe There can’t be a general frequency distribution for the cfe of all numbers But for almost every real number there is ! The probability of the appearance of the digit k in the cfe of almost every number is P(k) = ln[ 1 + 1/k(k + 2) ]/ln[2] P(1) = 0.41, P(2) = 0.17, P(3) = 0.09, P(4) = 0.06, P(5) = 0.04 ln(1+x) x P(k) 1/k2 as k
Typical Continued Fractions Arithmetic mean (average) value of the k’s is k=1 k P(k) 1/ln[2] k=1 1/k Geometric mean is finite and universal for a.e number (k1........kn)1/n K= 2.68545….. as n K k=1 {1+1/k(k+2)}ln(k)/ln(2) : Khinchin’s constant Captures the fact that the cfe entries are usually small e = 2.718.. is an exception (k1........kn)1/n = [2N/3(N/3)!]1/N 0.6259N1/3
k=11/k has an Infinite Sum = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + ....... = 1 + (1/2 + 1/3) + (1/4 + 1/5 + 1/6 + 1/7) + (1/8 + 1/9 + 1/10 + 1/11 +…..1/15) +.. > 1/2 + (1/4 + 1/4) +(1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16 + 1/16 + 1/16 + ..+ 1/16 ) > 1/2 + 1/2 + 1/2 + 1/2 + ……. “Divergent series are the invention of the devil, and it is a shame to base on them any demonstration whatsoever” Niels Abel
Geometric Mean for the cfe Digits of G Mean K =2.68.. k Aleksandr Khinchin 1894-1959
Slow Convergence to K-- with a pattern ? Geo Mean Cfe geometric means for , 2, , log(2), 21/3, 31/3
Lévy’s Constant If x has a rational approx pn/qn after n steps of the cfe, then for almost every number qn < exp[An] as n for some A>0 qn1/n L = 3.275… as n Paul Lévy, 1886-1971 L for cfe of 3.275…
A Strange Series What is the sum of this series?? S(N) = p=1N 1/{p3sin2p} (Pickover-Petit- McPhedran problem) Occasionally p q so sin(n) 0 and S This happens when p/q is a rational approx to 3/1, 22/7, 333/106, 355/113, 103993/33102, 104384/33215, 208341/66317,… Dangerous values continue forever and diverge faster than 1/p3
Chaos in Numberland • Generate the cfe of • u = k + x = whole number + fractional part = [u] + x • = 3 + 0.141592.. = k1 + x1 k2 = [1/x1] = [7.0625459..] = 7 x2 = 0.0625459.. k3 = [1/x2] = [15.988488..] = 15 The fractional parts change from x1 x2 x3 .. chaotically. Small errors grow exponentially
Gauss’s Probability Distribution xn+1 = 1/xn – [1/xn] As n the probability of outcome x tends to p(x) = 1/[(1+x)ln2] : 01 p(x)dx = 1 Error is < (0.7)n after n iterations p(x) In a Letter to Laplace 30th Jan 1812 ‘a curious problem’ that had occupied him for 12 years Distribution of the fractional parts x
xn+1 = 1/xn – [1/xn] = T(xn) T(x) T(x) =1/x – k (1-k)-1<x<k-1 ldT/dxl = 1/x2 > 1 as 0 < x < 1 x n steps = initial exp[ht]: h = 2/[6(ln2)2] 3.45
The Continued-Fraction Universe u = 6.0229867.. = k + x = 6 + 0.0229867.. u 1/x = 1/0.0229867 = 43.503417 = 43 + 0.503417 u 1/0.503417 = 1.9864248 = 1 + 0.9864248 Next cycles have 1, 72, 1 and 5 oscillations respectively