2008 MEI CONFERENCE SIX GEMS FROM C3 & C4 Nick Lord 7/4/2008
1.
Squared squares
[C3: Proof]
1.1
Problem
For which whole numbers n can a square be sub-divided into n smaller squares, not
necessarily all of the same size?
Answer
All n except for n = 2, 3, 5 .
Replacing
by
causes n → n + 3 .
So it is enough to do n = 6, 7,8 .
4 → 7 as above;
1.2
for 6 and 8.
Problem
Same problem with cubes in 3-dimensions.
Answer
All n except for n = ??? (the precise set of exceptions is not known).
Dissection/trisection of cubes causes n → n + 7, n → n + 26 .
Starting with n = 1 , the second process makes 1, 27, 53, 79, 105, 131, 157 which
leave remainders of 1, 6, 4, 2, 0, 5, 3 when divided by 7. The first process then fills in
to show all n ≥ 157 can be done.
(Then just need to hunt for lower mod 7 class representatives and prove the
exceptions can’t be done!)
2
1.3
Problem
Can a square be sub-divided into smaller squares, all of different sizes?
Answer
Conjectured to be impossible until the first example (with 55 smaller squares) was
found in 1939. The (unique) smallest example (with 21 smaller squares) was found by
Duijvestijn in 1978.
1.4
Problem
Same problem in 3-dimensions.
Answer
Impossible!
3
2.
Prince Rupert’s cube
[C4: Vectors]
2.1
Question
What is the largest hole with a square cross-section that can be drilled through a
given 1× 1× 1 cube?
(Equivalently, what is the largest cube that can be pass through a given 1× 1× 1 cube?)
Answer
First, identify a likely-looking square cross-section ABCD: then drill the hole
perpendicular to this.
B(1,1-k,1)
C(1-k,1,1)
z
D(0,k,0)
x
y
A(k,0,0)
(0,0,0)
2.2
Exercises
•
⎛ −k ⎞
⎛1 − k ⎞
⎜ ⎟
⎜
⎟
AD = BC = ⎜ k ⎟ and AB = DC = ⎜1 − k ⎟ (so that ABCD is a parallelogram).
⎜ 0 ⎟
⎜ 1 ⎟
⎝ ⎠
⎝
⎠
•
AB i AD = − k (1 − k ) + k (1 − k ) = 0 (so that ABCD is a rectangle).
•
AB = AD if
2
2
(1 − k ) 2 + (1 − k ) 2 + 12 = k 2 + k 2 ⇒ k =
4
3
4
(so that ABCD is a square).
•
Side-length of ABCD is then
( 34 ) + ( 34 )
2
2
=
9
8
≈ 1.060660... , i.e. larger than
the edge-length of the cube!
•
⎛2⎞
⎜ ⎟
Normal to ABCD, the direction of the axis of the hole, is ⎜ 2 ⎟ , so is not a
⎜ −1⎟
⎝ ⎠
⎛1⎞
⎜ ⎟
space diagonal of the cube such as ⎜ 1 ⎟ .
⎜ ⎟
⎝ −1⎠
2.3
Further exercises
A common reaction here is to assume that drilling has to be along a space diagonal.
•
Show that joining the midpoints of edges of 1× 1× 1 cube as shown produces a
regular hexagon with side-length
•
•
1
2
2.
⎛1⎞
⎜ ⎟
With axes as before, show that a normal to the plane of this hexagon is ⎜ 1 ⎟ .
⎜ ⎟
⎝ −1⎠
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 1× 1× 1 cube has side-length
(3 − 3) × 12 2 = 0.8965754... (so smaller than the edge-length of the cube).
5
3.
Calculating π
[C3: Integration, C4: Algebra, Trigonometry]
The hunt for the digits of π
3.1
Number of DPs
calculated
•
3.2
Comment
Archimedean methods
2 (4? 6?)
by 240BC
1
by 1429
3.5 × 10
Archimedes – using inner/outer polygons
Ludoph van Ceulen
•
17th/18th Century calculus (especially series for tan-1x)
1706
John Machin
10 2
2
1874
Shanks – claimed 707 DPs: slip at 527th!)
( 5 × 10
1949
First computer calculation
2 × 103
4
1958
10
5
1961
10
6
1973
Print-out “world’s most boring book”
10
•
19th/20th Century calculus (e.g. Gauss- Salamin algorithm)
1983
107
8
1987
10
9
1989
10
10
1997
5 × 10
11
1999
2.06 × 10
•
Back to tan-1x!
1.2411× 1012
2002
Kanada – 600h on a supercomputer
The tan-1x method
We calculate
(i)
Date
1
dt in two different ways:
0 1+ t2
∫
x
Expand binomially and integrate term-by-term:
6
x
1
dt = ∫ (1 − t 2 + t 4 − ...)dt
2
0 1+ t
0
x3 x5
= x − + − ....
3 5
∫
x
Use the substitution t = tan θ :
(ii)
t = tan θ ⇒
dt
= sec 2 θ so that
dθ
tan −1 x
tan −1 x
1
1
2
−1
.sec
dt
=
θ
d
θ
=
∫0 1 + t 2 ∫0 1 + tan 2 θ
∫0 1dθ = tan x
x
Thus:
x3 x5
tan x = x − + − ... (in fact valid for −1 ≤ x ≤ 1 )
3 5
−1
Putting x = 1 gives Leibniz’s Series:
π
1 1
= tan −1 1 = 1 − + − ... .
4
3 5
Refining the method
3.3
•
•
π
4
= tan −1
1
1
(equivalent to tan −1 1 + tan −1 2 + tan −1 3 = π )
+ tan −1
2
3
(Machin’s Formula)
π
1
1
= 4 tan −1 − tan −1
4
5
239
Proof: tan θ =
1
5
by the double-angle formula for tan.
⇒ tan2θ = 125 ⇒ tan −1 4θ = 120
119
Then tan(4θ − π4 ) =
−1
tan 4θ − 1 120
= 119 120 =
1 + tan 4θ 1 + 119
1
239
.
(It takes about 1h to do 15DPs by hand!)
•
The 1961 and 1973 calculations (for 105 and 106 digits) used the pair of formulae
π
1
1
1
= 6 tan −1 + 2 tan −1
+ tan −1
,
4
8
57
239
π
1
1
1
= 12 tan −1 + 8 tan −1
− 5 tan −1
.
4
18
57
239
•
The current 2002 record (for 1.2411× 1012 digits) used the pair of formulae
7
π
1
1
1
1
,
+ 32 tan −1 − 5 tan −1
+ 12 tan −1
4
49
57
239
110443
π
1
1
1
1
= 44 tan −1
+ 7 tan −1
− 12 tan −1
+ 24 tan −1
.
4
57
239
682
12943
4.
= 12 tan −1
Radioactive decay
[C3: Functions, Calculus, C4: Trigonometry]
4.1
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:
(D1)
f (t ) > 0
(mass is positive!)
(D2)
f ′(t ) < 0
(the process is decay)
(D3)
f (t + 1) =
1
2
f (t )
(so that 1 is the half-life – equivalently, take the
half-life as the unit of time)
then is it the case that f (t ) = A.2− t ?
4.2
Analysis
Consider g (t ) = f (t ).2t .
Then g (t + 1) = f (t + 1).2t +1 =
1
2
f (t ).2.2t = g (t ) , so g (t ) is periodic with period 1.
Thus f (t ) = g (t ).2− t , where g (t ) is periodic with period 1.
If g (t ) = A , a constant, then f (t ) = A.2− t , i.e. exponential decay.
But g (t ) doesn’t have to be a constant.
For example, choose g (t ) = A ⎡⎣1 + k sin ( 2π t ) ⎤⎦ so f (t ) = A ⎡⎣1 + k sin ( 2π t ) ⎤⎦ 2− t .
Can we choose k > 0 to satisfy the properties D1, D2, D3?
(D3)
is satisfied by construction,
(D1)
is certainly satisfied if A > 0, 0 < k < 1,
(D2)
needs more work!
8
f ′(t ) = A.2π k cos ( 2π t ) 2 − t + A ⎡⎣1 + k sin ( 2π t ) ⎤⎦ 2− t. − ln 2
(because
( )
(
)
d −t
d − t ln 2
e
2 =
= − ln 2e − t ln 2 = − ln 2.2− t )
dt
dt
= A.2π k cos(2π t )2−t − Ak ln 2.sin(2π t )2− t − A ln 2.2− t
ln 2 ⎤ −t
⎡
= kA ⎢ 2π cos ( 2π t ) − ln 2.sin ( 2π t ) −
2
k ⎥⎦
⎣
ln 2 ⎤ −t
⎡
= kA ⎢ 4π 2 + (ln 2) 2 cos ( 2π t + ε ) −
2
k ⎥⎦
⎣
ln 2
< 0 if 4π 2 + (ln 2) 2 <
k
or k <
5
ln 2
4π + (ln 2) 2
2
≈ 0.1096 .
y
4
3
2
1
x
1
2
3
Graph shows half-life 1 with A = 5, k = 0.1
9
4
5
6
4.3
An interesting graph
It is well worth asking students to plot y = 2
5.
−1
x
or y = 10
−1
x
….
Pythagoras causes chaos!
[C3: Numerical methods, C4: Trigonometry]
Can we get from C3 & C4 to some contemporary mathematics?
5.1
Iterating Pythagorean triples
For whole numbers a, b as, the formula a 2 − b 2 , 2ab, a 2 + b 2 always gives a
Pythagorean triple because (a 2 − b 2 ) 2 + (2ab) 2 = (a 2 + b 2 ) 2
What happens if you repeatedly feed Pythagorean triples into this formula, so that
a 2 − b 2 , 2ab are used as the next a, b values?
Example: 3, 4, 5 → 7, 24, 25 → 527, 336, 625 → 164833, 354144, 390625 → ...
There quickly arise overflow problems, so it’s best to work with the ratios
a
b
A = , B = (with A2 + B 2 = 1 ).
c
c
The “shape” of the triangle produced = smallest angle = min{tan −1 ( BA ) , tan −1 ( BA )} .
TABLE : Angles generated from a 3-4-5 triangle
n
⎛A ⎞
tan −1 ⎜ n ⎟
⎝ Bn ⎠
⎛B ⎞
tan −1 ⎜ n ⎟
⎝ An ⎠
θ n =smallest
angle=shape
1
36.8699
53.1301
36.8699
2
16.2602
73.7398
16.2602
3
57.47959
32.52041
32.52041
4
24.95918
65.04082
24.95918
5
40.08164
49.91836
40.08164
6
9.836725
80.16328
9.836725
7
70.32655
19.67345
19.67345
10
8
50.6531
39.3469
39.3469
9
11.3062
78.6938
11.3062
10
67.38759
22.61241
22.61241
11
44.77519
45.22481
44.77519
12
0.449621
89.55038
0.449621
What patterns are produced?
•
Shape column is chaotic – literally!
•
If α < β are the angles in one row, then 2α , β − α are the angles in the next row.
Proofs and a picture
5.2
•
Case 1: a < b
Then α = tan −1 ( ba ) ⇒ tan 2α =
•
2 ba
2ab
.
= 2
a2
1 − b2 b − a 2
Case 2: a > b
Then α = tan −1 ( ba ) ⇒ tan 2α =
2 ba
2ab
.
= 2
b2
1 − a2 a − b 2
In either case, 2α will be in the next row.
And, once 2α is there, so too is 90° − 2α = (α + β ) − 2α = β − α .
If θ n =shape of triangle in nth row, 0° ≤ θ n ≤ 45° , then θ n +1 = min {2θ n ,90° − 2θ n } .
11
y
40
30
20
10
x
20
40
60
Graph of y = min {2 x,90 − 2 x} = 45 − 45 − 2 x
Both fixed points are unstable ( gradient = 2 there). The “tent map” is one of the paradigm
examples of a chaotic dynamical system.
6.
The first integral?
[C4: Trigonometry, Integration]
… was arguably
∫
θ
0
sec λ d λ = ln tan ( π4 + θ2 ) !
I like this tale, because it emphasises the arbitrariness of module specifications and the
richness of a historical perspective.
6.1
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 be 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 by a factor of cos λ . To appear as parallels, lines of longitude have to be
1
stretched horizontally by a factor of
= sec λ at latitude λ .
cos λ
12
80
To preserve angles, lines of latitude have to be stretched vertically by “ ∑ sec λ.δλ ” (approx.)
or
∫
θ
0
sec λ d λ (exactly).
1569: Gerardus Mercator’s first world map – but he gave no formula for the
“progressive increases” of parallels of longitude ( at that time there were no complete
trig tables).
1599 Edward Wright in his Certaine errors in navigation … explained the
sec λ factor and argued that the vertical spacing is given by “perpetuall addition of the
secantes”; he also gave a table of
∑ sec λ.δλ
with δλ = 1' for λ ≤ 75° to enable
accurate Mercator maps to be drawn.
By 1620-ish, there were published tables of Naperian logs (roughly ln), lgs, logsin,
logtan.
1645 Henry Bond, “teacher of navigation, survey & other parts of mathematics”
noticed that Wright’s table closely agreed with a table of logtan!! i.e.
θ
that ∫ sec λ d λ = ln tan ( π4 + θ2 ) .
0
Justifying “whether the Artificial [logarithmic] Tangent-line be the true Meridianline” became an outstanding open problem – still live in 1666 (Nicolaus Mercator)
because it was suspected that there might be deviations for far North.
1647 Cavalieri did
∫ x dx and Gregory of St Vincent did ∫ x
n
−1
dx .
1668 James Gregory did ∫ sec λ d λ , but his proof was long and hard to follow.
1670 Barrow gave an easier proof (below).
(1873 James Thomson introduced the term “radian” in a university exam paper!)
… GoogleEarth uses a Mercator projection, up to 85.05113 degrees North!
6.2
Analysis
•
sec 2 λ + sec λ tan λ
∫0 sec λ d λ = ∫0 sec λ + tan λ d λ = ln ( secθ + tan θ )
(This “rabbit out of a hat” corresponds to the substitution t = sec λ + tan λ .)
θ
θ
•
(Barrow in 1670: the first use of partial fractions in integration.)
θ
θ cos λ
θ
θ cos λ
cos λ
cos λ
⎛ 1 + sin θ ⎞
∫0 sec λ d λ = ∫0 cos2 λ d λ = ∫0 1 − sin 2 λ d λ = 12 ∫0 1 − sin λ + 1 + sin λ d λ = 12 ln ⎜⎝ 1 − sin θ ⎟⎠
•
“The world’s sneakiest substitution” t = tan λ2 gives directly
∫
θ
0
⎛ 1 + tan θ2 ⎞
sec λ d λ = ln ⎜
= ln tan ( π4 + θ2 )
θ ⎟
1
tan
−
2 ⎠
⎝
Reconciling these answers makes a nice trigonometrical identities exercise:
sec θ + tan θ ≡
1 + sin θ 1 + tan θ2
≡
≡ tan ( π4 + θ2 )
θ
1 − sin θ 1 − tan 2
(Why are there so few names in school maths textbooks? One reason is that the history of the
subject is so tangled!)
14
15