SIX GEMS FROM C3 & C4

Transcription

1 008 MEI CONFERENCE SIX GEMS FROM C3 & C4 Nick Lord 7/4/008

2 . Squared squares [C3: Proof]. Prolem For which whole numers n can a square e su-divided into n smaller squares, not necessarily all of the same size? Answer All n except for n =,3,5. Replacing y causes n n+ 3. So it is enough to do n = 6,7, as aove; for 6 and 8.. Prolem Same prolem with cues in 3-dimensions. Answer All n except for n =??? (the precise set of exceptions is not known). Dissection/trisection of cues causes n n+ 7, n n+ 6. Starting with n =, the second process makes, 7, 53, 79, 05, 3, 57 which leave remainders of, 6, 4,, 0, 5, 3 when divided y 7. The first process then fills in to show all n 57 can e done. (Then just need to hunt for lower mod 7 class representatives and prove the exceptions can t e done!)

3 .3 Prolem Can a square e su-divided into smaller squares, all of different sizes? Answer Conjectured to e impossile until the first example (with 55 smaller squares) was found in 939. The (unique) smallest example (with smaller squares) was found y Duijvestijn in Prolem Same prolem in 3-dimensions. Answer Impossile! 3

4 . Prince Rupert s cue [C4: Vectors]. Question What is the largest hole with a square cross-section that can e drilled through a given cue? (Equivalently, what is the largest cue that can e pass through a given cue?) Answer First, identify a likely-looking square cross-section ABCD: then drill the hole perpendicular to this. C(-k,,) B(,-k,) z D(0,k,0) y (0,0,0) x A(k,0,0). Exercises k k AD= BC = k and AB= DC = k (so that ABCD is a parallelogram). 0 ABi AD = k( k) + k( k) = 0 (so that ABCD is a rectangle). AB = AD if ( k) + ( k) + = k + k k = (so that ABCD is a square)

5 3 3 9 Side-length of ABCD is then ( ) ( ) + = , i.e. larger than the edge-length of the cue! Normal to ABCD, the direction of the axis of the hole, is, so is not a space diagonal of the cue such as..3 Further exercises A common reaction here is to assume that drilling has to e along a space diagonal. Show that joining the midpoints of edges of cue as shown produces a regular hexagon with side-length. With axes as efore, show that a normal to the plane of this hexagon is. Show that the largest square that fits inside a regular hexagon of side-length d has side-length (3 3)d. Deduce that the largest square cross-section of a cue has side-length (3 3) = (so smaller than the edge-length of the cue). 5

6 3. Calculating π [C3: Integration, C4: Algera, Trigonometry] 3. The hunt for the digits of π Numer of DPs Date Comment calculated Archimedean methods (4? 6?) y 40BC Archimedes using inner/outer polygons y 49 Ludoph van Ceulen 7 th /8 th Century calculus (especially series for tan - x) John Machin ( Shanks claimed 707 DPs: slip at 57 th!) First computer calculation Print-out world s most oring ook 9 th /0 th Century calculus (e.g. Gauss- Salamin algorithm) Back to tan - x! Kanada 600h on a supercomputer 3. The tan - x method x We calculate dt in two different ways: 0 + t (i) Expand inomially and integrate term-y-term: 6

7 x dt = ( t + t 4... ) dt + t 3 5 x x = x x 0 0 (ii) Use the sustitution t = tan : t dt d = tan = sec so that dt d d x + t + tan x tan x tan x =.sec = = tan Thus: x x = + (in fact valid for x ) tan x x... Putting x = gives Leiniz s Series: π = tan = Refining the method π = tan + tan (equivalent to 4 3 tan tan tan 3 π + + = ) (Machin s Formula) π = 4tan tan Proof: Then tan = tan = tan 4 = y the doule-angle formula for tan π tan tan4 + 9 tan(4 ) = = = (It takes aout h to do 5DPs y hand!) The 96 and 973 calculations (for and 0 digits) used the pair of formulae π = 6tan + tan + tan, π = tan 8tan 5tan. The current 00 record (for.4 0 digits) used the pair of formulae 7

8 π = π = tan 3 tan 5 tan tan, 44 tan 7 tan tan 4 tan. 4. Radioactive decay [C3: Functions, Calculus, C4: Trigonometry] 4. Question Is exponential decay the only decay process that has a half-life? Equivalently, if f () t has domain [0, ) and has the following three properties: (D) f() t > 0 (mass is positive!) (D) f () t < 0 (the process is decay) (D3) f ( t+ ) = f( t) (so that is the half-life equivalently, take the half-life as the unit of time) then is it the case that f( t) = A. t? 4. Analysis Consider gt () = f(). t t. Then t g( t+ ) = f( t+ ). = f( t).. = g( t), so gtis () periodic with period. t+ Thus f() t = g(). t t, where gtis () periodic with period. If gt () = A, a constant, then f( t) = A. t, i.e. exponential decay. But gtdoesn t () have to e a constant. For example, choose gt () = A + ksin( π t) so f t = A + k ( πt ) Can we choose k > 0 to satisfy the properties D, D, D3? t () sin. (D3) is satisfied y construction, (D) is certainly satisfied if A> 0, 0 < k <, (D) needs more work! 8

9 t ( ) ( ) t f () t = A.πkcos πt + A + ksin πt. ln d t d tln tln t e ln e = = = ln. ) dt dt (ecause ( ) ( ) = A.πk cos( πt) Ak ln.sin( πt) Aln. t t t ln = ka π cos( πt ) ln.sin ( πt ) k ln = ka + ( t + ) k ln < 0 if 4 π + (ln ) < k t 4 π (ln ) cos π ε t or ln k < π + (ln) 5 y x Graph shows half-life with A= 5, k = 0. 9

10 4.3 An interesting graph It is well worth asking students to plot x y = or y = 0 x. 5. Pythagoras causes chaos! [C3: Numerical methods, C4: Trigonometry] Can we get from C3 & C4 to some contemporary mathematics? 5. Iterating Pythagorean triples For whole numers aas,, the formula a, a, a + always gives a Pythagorean triple ecause ( a ) + ( a) = ( a + ) What happens if you repeatedly feed Pythagorean triples into this formula, so that a,a are used as the next avalues?, Example: 3,4,5 7,4,5 57,336, ,35444, There quickly arise overflow prolems, so it s est to work with the ratios a A =, B c = c (with A + B = ). min{tan, tan }. A B The shape of the triangle produced = smallest angle = ( ) ( ) TABLE : Angles generated from a triangle B A n angle=shape tan A n B n tan B n An n =smallest

11 What patterns are produced? Shape column is chaotic literally! If α < β are the angles in one row, then α, β α are the angles in the next row. 5. Proofs and a picture Case : a< a a = = = a a Then α tan ( ) tan α Case : a> a a Then α tan ( ) tan α a = a = = a a In either case, α will e in the next row... And, once α is there, so too is 90 α = ( α + β) α = β α. If =shape of triangle in nth row, 0 45 n, then { } n = n min n,90 +. n

12 y x y = min x,90 x = x Graph of { } Both fixed points are unstale ( gradient = there). The tent map is one of the paradigm examples of a chaotic dynamical system. 6. The first integral? [C4: Trigonometry, Integration] was argualy sec d π λ λ = ln tan ( + ) 0 4! I like this tale, ecause it emphasises the aritrariness of module specifications and the richness of a historical perspective. 6. The tale: Mercator s projection How do you plot longitude and latitude lines on a map so that angles are preserved and thus compasses can e used for route-tracing? If the earth s radius is R then, at latitude λ, a circle of latitude has radius R cos λ, so arc lengths are shrunk y a factor of cos λ. To appear as parallels, lines of longitude have to e stretched horizontally y a factor of = sec λ at latitude λ. cos λ

13 To preserve angles, lines of latitude have to e stretched vertically y sec λ. δλ (approx.) or 0 sec λ d λ (exactly). 569: Gerardus Mercator s first world map ut he gave no formula for the progressive increases of parallels of longitude ( at that time there were no complete trig tales). 599 Edward Wright in his Certaine errors in navigation explained the secλ factor and argued that the vertical spacing is given y perpetuall addition of the secantes ; he also gave a tale of sec λ. δλ with δλ = ' for λ 75 to enale accurate Mercator maps to e drawn. By 60-ish, there were pulished tales of Naperian logs (roughly ln), lgs, logsin, logtan. 645 Henry Bond, teacher of navigation, survey & other parts of mathematics noticed that Wright s tale closely agreed with a tale of logtan!! i.e. that d π λ λ = ( + 4 ) 0 sec ln tan. Justifying whether the Artificial [logarithmic] Tangent-line e the true Meridianline ecame an outstanding open prolem still live in 666 (Nicolaus Mercator) ecause it was suspected that there might e deviations for far North. 647 Cavalieri did n x dx and Gregory of St Vincent did x dx.

14 668 James Gregory did secλdλ, ut his proof was long and hard to follow. 670 Barrow gave an easier proof (elow). (873 James Thomson introduced the term radian in a university exam paper!) GoogleEarth uses a Mercator projection, up to degrees North! 6. Analysis sec + sec tan λ λ λ 0 0 secλ+ tan λ (This rait out of a hat corresponds to the sustitution t = sec λ + tan λ.) secλdλ = dλ = ln ( sec + tan) (Barrow in 670: the first use of partial fractions in integration.) cos λ cos λ cos λ cos λ + sin secλdλ = 0 dλ = dλ 0 0 dλ 0 ln cos λ = + = sin λ sin λ + sin λ sin λ The world s sneakiest sustitution t = tan gives directly = + tan = + sec λdλ ln ln tan 0 tan π ( ) 4 Reconciling these answers makes a nice trigonometrical identities exercise: + sin + tan + + sec tan tan sin tan π ( ) (Why are there so few names in school maths textooks? One reason is that the history of the suject is so tangled!) 4 4

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