2004 09 Toni Beardon Golden Thoughts

Information about 2004 09 Toni Beardon Golden Thoughts

Published on February 27, 2008

Author: Marietta1

Source: authorstream.com

Content

GOLDEN THOUGHTS a ‘whistle-stop’ tour of some beautiful mathematical places:  GOLDEN THOUGHTS a ‘whistle-stop’ tour of some beautiful mathematical places Toni Beardon AIMS 30 September 2004 Ruler and compass construction for  :  Ruler and compass construction for  Draw a square of side 1 unit. Bisect the base Draw the diagonal to form a triangle with sides 1 unit and 1/2 unit so that the diagonal has length 5/2 Draw the arc of the circle with radius 5/2 and find the point where this arc cuts the extended base to give one vertex of the golden rectangle with sides (1+5)/2 and 1. The Golden Ratio  (Phi):  The Golden Ratio  (Phi) Golden Rectangles and the Fibonacci Sequence:  Golden Rectangles and the Fibonacci Sequence Continued fraction for :  Continued fraction for  and so on … Values of the finite continued fractions giving the ratios of Fibonacci numbers The Golden Ratio:  The Golden Ratio The Great Golden Pyramid of Giza c. 2560 BC:  The Great Golden Pyramid of Giza c. 2560 BC The Divine Proportion in Art and Nature:  The Divine Proportion in Art and Nature Golden Ratio or Mean or Section or Number also called The Divine Proportion :  Golden Ratio or Mean or Section or Number also called The Divine Proportion The Greeks left no record of any work on algebra. The Arabs introduced algebra and preserved and built on the legacy of Greek mathematics. Al-Khwarizmi, (780-850), from Iraq, introduced problems on dividing a line of length 10 units in two parts using a quadratic equation. Abu-Kamil (850-930), from Egypt, developed similar equations arising from dividing a line in various ways. Leonardo of Pisa (Fibonacci) used many Arabic sources and he specifically referred to the golden ratio in Liber Abaci, published in 1202. Pacioli in Divina Proportione (1509) gives an account of what was known about the golden ratio. He explains his title as follows: “… it seems to me that the proper title for this treatise must be Divine Proportion. This is because there are very many similar attributes which I find in our proportion – all befitting God himself – which is the subject of our very useful discourse.” Leonardo da Vinci 1452-1519 The Last Supper Convent of Santa Maria delle Grazie (Refectory), Milan :  Leonardo da Vinci 1452-1519 The Last Supper Convent of Santa Maria delle Grazie (Refectory), Milan The Golden Ruler The Divine Proportion in Art and Nature:  The Divine Proportion in Art and Nature Salvador Dali The Last Supper:  Salvador Dali The Last Supper Salvador Dali. 1904-1989. "The Sacrament of the Last Supper". National Gallery of Art, Washington DC. 1955 The painting incorporates part of a dodecahedron which involves the golden ratio. Pentakite ABCDE is a regular pentagon. Find the length BE:  Pentakite ABCDE is a regular pentagon. Find the length BE The angles of a regular pentagon are 108° so all the angles can be found. s CDF, EDB are congruent isosceles triangles. If the sides of the regular pentagon are 1 unit and FD = x units then BE = x units. s EDB, BEF are similar, hence More Exact values of trig ratios for special angles:  Exact values of trig ratios for special angles cos36o=sin54o=/2 = (1+5)/4 sec72o=cosec18o=2 = (1+5) cos72o=sin18o=1/2  = (5 - 1)/4 Pent Show ABFE is a rhombus and find the ratio EC:EF:  Pent Show ABFE is a rhombus and find the ratio EC:EF If EB=EC=x then FC=x-1 s EBF, DCF are similar. Hence EB:EF=CD:CF So we have x2 – x – 1 = 0 From which we get x=  and EC : EF = EF : FC =  So F divides EC and BD in the golden ratio. More Golden Triangle Triangles ABC, CBD, ABD are isosceles. Find the ratio AB:BC :  Golden Triangle Triangles ABC, CBD, ABD are isosceles. Find the ratio AB:BC Hence s ABC, BCD are similar with angles 36o,72o, 72o. Let AB=p and BC=BD=DA=q then, from these similar triangles, p : q = q : p-q so writing the ratio p/q=r, r : 1 = 1 : r – 1 Thus r = , the golden ratio. Also AD : DC = BD : DC, so D divides the side AC in the golden ratio. More Automata generating Fibonacci numbers :  Automata generating Fibonacci numbers Variables: A and B Constants: none Start: A Rules: A→B, B→AB A, B, AB, BAB, ABBAB, BABABBAB, ABBABBABABBAB, BABABBABABBABBABABBAB,…… 1, 1, 2, 3, 5, 8, 13, 21,… Gnomons carpenters’ tools used in Babylonian and Greek mathematics:  Gnomons carpenters’ tools used in Babylonian and Greek mathematics Golden Powers Find a formula for n in terms of .:  Golden Powers Find a formula for n in terms of . We know 2 =+1 so n+2 =n+1+ n Example: 7 = 6+ 5 = 25+ 4 = 3  4+ 23 = 53+ 32 = 82+ 5 = 13+8 = F7 + F6 Conjecture: n = Fn + Fn-1 Proof Let n =an+bn then, multiplying by , n+1 = an2 + bn  = an(+1) + bn  = (an+bn)  + an So an+1= an + bn and bn+1=an . Then, substituting for bn , we get an+1= an + an-1 which is the recurrence relation defining the terms of the Fibonacci sequence, hence n =Fn+Fn-1 More Golden Fibs When is a Fibonacci sequence also a geometric sequence? :  Golden Fibs When is a Fibonacci sequence also a geometric sequence? The Fibonacci sequence Fn is defined by the relation Fn+2 = Fn+1 + Fn where Fo=0 and F1=1. Now suppose that we take the same relation and more general sequences Xn with any two starting values X0 and X1 .For a geometric sequence the terms must be: X0 , r X0 , r2 X0 , … rn X0 … where r is the common ratio, so for a geometric-Fibonacci sequence we have rn+2 X0 = rn+1Xo + rnXo . As Xo 0 and r0 we can divide by rnXo , so r2 = r + 1, hence r is the golden ratio . We have shown that a geometric sequence is a Fibonacci type when the ratio is golden. It can be shown that a Fibonacci sequence with the first two terms Fo and F0 is a geometric sequence with common ratio  . More Golden Thoughts The areas A1, A2 and A3 are equal. What are the ratios RX:XS and RY:YQ?:  Golden Thoughts The areas A1, A2 and A3 are equal. What are the ratios RX:XS and RY:YQ? Pythagorean Golden Means:  Pythagorean Golden Means How many solutions?:  How many solutions? One solution is x = 2. Substitute and check! There is another solution which is easy to miss! It is approximately 2.3305 . Use a numerical method rather than an analytic method to find it. More Golden Ellipse:  Golden Ellipse An ellipse with semi axes a and b fits between two circles of radii a and b. If the area of the ellipse is equal to the area of the annulus what is the ratio b:a? The areas are and so for he areas to be equal so that Darts, kites and aperiodic tiling:  Darts, kites and aperiodic tiling Recent research by Dov Aharonov, Alan Beardon and Kathy Driver:  Recent research by Dov Aharonov, Alan Beardon and Kathy Driver It has been shown recently that many of the known identities for Fibonacci numbers have counterparts for solutions of other recurrence relations with constant coefficients. The Chebyshev polynomials satisfy a recurrence relation with a variable x in the coefficients. By taking suitable values for x these polynomials reduce to the solutions of second order recurrence relations with constant coefficients. Thus by proving identities for Chebyshev polynomials the identities from Fibonacci numbers are recaptured and also, at the same time, the corresponding identities for the general constant coefficients recurrence relation. Slide27:  Thank you Toni Beardon [email protected]

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