Power Page

Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green Bay
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Interesting stuff about powers of numbers

Index

About Notation

To save writing exponents, we'll denote power sums like this: terms on each side of the equality will be separated by commas, and opposite sides of the equality will be separated by a semicolon. Thus, for Pythagorean triplets, a, b; c means a2 + b2= c2. For chains of squares,
21, 22, 23, 24; 25, 26, 27 means 212+ 222+ 232+ 242=252+ 262+ 272 and similarly for sums of higher powers. Where the exponent is not clear from context, it will be noted in parentheses. Thus: 95,800, 217,519, 414,560; 422,481 (n=4) means 95,8004 + 217,5194 + 414,5604 = 422,4814.

Pythagorean Triplets

The simplest triangle with integer sides that satisfies the Pythagorean Theorem is 3, 4, 5.
That is, 32 + 42 = 52. Such a triplet is called a Pythagorean Triplet. There are infinitely many, and they are easy to generate. A classic formula, known since ancient times, can generate them at will. If the numbers in the triplet are a, b, and c, then:
a = n2 -m2, b=2mn, c=m2+n2, where m and n are two integers and m is less than n.
(Verify that a2 + b2= c2.)

We can see several things of interest:

  1. Looking at the b term, at least one number in every Pythagorean triple must be even.
  2. Looking at the a and c terms, if m and n are both even or both odd, then a and c must be even. Thus, all members of the triple are even. Since we can divide all the members of the triple by 2 to get a smaller triple, we are not really interested in these cases. So we can restrict our attention to even-odd sets of m and n.
  3. Obviously, if m and n have a common factor, so will the triple. We aren't interested in these cases either.
  4. Thus, the only interesting (relatively prime) triples are generated when m and n are relatively prime and only one is even.
  5. The c term (hypotenuse) will always be odd in any relatively prime triple.
  6. The difference between c and b is always a square: c - b = (n - m)2. Since we only get relatively prime triples when only one of (m,n) is odd, c - b is always an odd square.
  7. The sum of c and b is always a square: c + b = (n + m)2.
  8. Since a2 = c2 - b2 = (c - b)(c + b), c + b always equals a2 times an odd square. We can use this relationship to group triples into related series:
  9. Also, c - a = (n2 + m2) - (n2 - m2) = 2m2. Thus the difference between c and a is always two times a square.

We can use the relationships above to group Pythagorean triples in many ways. This set of series displays relationships 6-8 above:

c - b = 1 m= n= c - b = 9 m= n= c - b = 25 m= n=
1, 0; 1 0 1 9, 0; 9 0 3 25, 0; 25 0 5
3, 4; 5 1 2 15, 8; 17 1 4 35, 12; 37 1 6
5, 12; 13 2 3 21, 20; 29 2 5 45, 28; 53 2 7
7, 24; 25 3 4 27, 36; 45 3 6 55, 48; 73 3 8
9, 40; 41 4 5 33, 56; 65 4 7 65, 72; 97 4 9
11, 60; 61 5 6 39, 80; 89 5 8 75, 100; 125 5 10

The set of series below emphasizes relationship 9 above:

c - a = 2 m= n= c - a = 8 m= n= c - a = 18 m= n=
4, 3; 5 1 2 12, 5; 13 2 3 24, 7; 25 3 4
8, 15; 17 1 4 20, 21; 29 2 5 36, 27; 45 3 6
12, 35; 37 1 6 28, 45; 53 2 7 48, 55; 73 3 8
16, 63; 65 1 8 36, 77; 85 2 9 60, 91; 109 3 10
20, 99; 101 1 10 44, 117; 125 2 11 72, 135; 153 3 12

Almost-Isosceles Triples

The square root of two is irrational, therefore there cannot be any Pythagorean triplets a, a; c. But there are an infinite number of triplets a, a + 1; c. Here are the first ten:

0 1 1 c=
3 4 5 6*1 -1
20 21 29 6*5 -1
119 120 169 6*29 - 5
696 697 985 6*169 -29
4059 4060 5741 6*985 - 169
23660 23661 33461 6*5741 -985
137903 137904 195025 6*33461 - 5741
803760 803761 1136689 6*195025 - 33461
4684659 4684660 6625109 6*1136689-195025

The recursion relations are:

a0 b0=a0+1 c0
a1 = 3*a0 + 2*b0 + 1 b1 = 3*a0 + 2*b0 +2 c1 = 4*a0 + 3*c0 + 2
a2 = 6*a1 - a0 + 2 b2 = 6*a1 - a0 + 2 c2 = 6*c1 - c0

Almost 30-60 Triples

Like the square root of two, the square root of three is irrational, so there cannot be any Pythagorean triples that are 30-60 triangles (c = 2a). But we can come close; there are an infinite number of Pythagorean triples where c = 2a +/- 1. Here are the first 13.

0 1 1
3 4 5
8 15 17
33 56 65
120 209 241
451 780 901
1680 2911 3361
6273 10864 12545
23408 40545 46817
87363 151316 174725
326040 564719 652081
1216801 2107560 2433601
4541160 7865521 9082321

The recursion relations are:

a0 b0 c0 = 2*a0 +/- 1
a1 = b0 + c0 +/- 1 b1 = c0 + 2*b0 - a0 c1 = 2*b0 + 2*c0 +/- 1

Chains of Consecutive Squares

The Pythagorean triplet 3,4;5 is the first example of another interesting pattern. The first few examples are:
3, 4; 5
10, 11, 12; 13, 14
21, 22, 23, 24; 25, 26, 27

For any value of a, there is a chain of a squares (n-a), (n-a+1) ..... (n-1),n; (n+1), n+2)....(n+a)
The value of n is given by n=2a(a+1).

Proof: Pair up terms across the equality as shown:
...... (n-3)2 + (n-2)2 + (n-1)2 + n2 = (n+1)2 + (n+2)2 + (n+3)2 .... or,
.......n2 - 6n + 9 + n2 - 4n + 4 + n2 - 2n + 1 + n2 = n2 + 2n + 1 + n2 + 4n + 4 + n2 + 6n + 9 ....
All the squares of integers cancel, as do all the n2 terms except one. We get:
...... -2an - ..... - 6n - 4n - 2n + n2 = 2n + 4n + 6n + ..... 2an. Dividing by n and rearranging:
n = 4(1 + 2 + 3 + 4 + 5 ...... a). The sum of integers from 1 to a is a(a+1)/2 (see Sums of Powers of Consecutive Integers below). Thus n = 2a(a+1). There is a chain of a+1 squares left of the equality and a to the right. The examples above are for a = 1, 2, and 3. For a = 4, we find n = 40 and the resulting chain is: 36, 37, 38, 39, 40; 41, 42, 43, 44

Pythagorean Quartets

There are an infinite number of quartets of squares a, b, c; d. There are so many that they are not all that interesting, except for one point. The three-dimensional version of the Pythagorean Theorem is a2 + b2 + c2 = d2. In other words, in three-dimensional Cartesian Coordinates, point (a,b,c) is distance d from the origin. (Proof: Consider the x-y plane. Point (a,b,c) - call it Q - projects to (a,b,0). The distance from this point - call it P - to the origin is given by a2 + b2 = r2. Now look at the vertical plane through Q and P. The distance from Q to the origin is r2 + c2 = d2, and since a2 + b2 = r2, a2 + b2 + c2 = d2.) Thus Pythagorean quartets are the coordinates of points in three dimensions that lie an integer distance from the origin.

A formula that generates Pythagorean quartets is:
a = m2, b = 2mn, c = 2n2; d = (m2 + 2n2) = a + c. Also note that b2 = 2ac. When m =1 and n = 1, we get 1, 2, 2; 3, the simplest example.

Fermat's Last Theorem

There are an infinite number of Pythagorean triplets. Are there any triplets for higher powers? That is, are there any integers for which an + bn = cn, where n is a power higher than 2? The French mathematician Pierre Fermat claimed about 1637 that there were none. He wrote in the margin of a book that he had a marvelous proof too complex to fit in the margin. For centuries, mathematicians wished Fermat had been reading a bigger book. The only reasons for taking such an unsupported conjecture seriously are (a) the proofs Fermat did furnish were brilliant, (b) in every other case where Fermat claimed to have proven something profound without actually providing a proof, a proof was later found and (c) Fermat's Last Theorem survived every assault made on it. He has credibility.

The possibility that there might be a simple solution to a famous problem attracted cranks by the score. (Why exactly cranks think they will become famous by solving famous problems is a mystery. How many mathematicians have megabuck endorsement contracts?) Most mathematicians now think Fermat was probably mistaken in thinking he had a proof (even Brett Favre gets sacked occasionally.) Fermat's Last Theorem was proven for specific powers of n to such large values - in the millions - that nobody would ever actually write a counter-example on paper. Finally, in 1993, the theorem was proven by Andrew Weil of Princeton University. (Weil is now endorsing sneakers and married to a supermodel - not.)

Euler's Conjecture

There are no triplets of numbers for powers larger than 2, but there are longer sums. It is an interesting coincidence that we have 3, 4; 5 for squares and 3, 4, 5; 6 for cubes (The pattern does not hold for larger powers.) The mathematician Leonhard Euler conjectured that it always required n terms to sum to an nth power: two squares, three cubes, four fourth powers, and so on. Nice guess, but wrong. In 1966, L. J. Lander and T. R. Parkin found the first counterexample: four fifth powers that summed to a fifth power: 27, 84, 110, 133; 144.

In 1988 Noam Elkies of Harvard found a counterexample for fourth powers:
2,682,440, 15,365,639, 187,960; 20,615,673. Roger Frye of Thinking Machines Corporation did a computer search to find the smallest example:
95,800, 217,519, 414,560; 422,481

Sums of Cubes (Cubic Quartets)

Even though 3, 4, 5; 6 for cubes is as pretty a mathematical formula as anyone could hope to find, sums of cubes do not yield to formulas anything like as neat as those for Pythagorean triplets. One additional complication is that cubes can be negative. We could ignore negative numbers in dealing with Pythagorean triplets, but we cannot in dealing with cubes. Also, Pythagorean triplets generally have terms that are more or less the same size. The largest triplet with a term less than 10 is 9, 40; 41. Any term larger than 41 will differ from all other squares by more than 81. But there is much less limitation on cubic quartets. For example:

The list above resulted from a search where b and c ranged up to 10,000, so it appears the cutoff at b = 2676 is a real limit. These are "almost" exceptions to Fermat's Last Theorem. Since cubes can be negative, we also find this set of "almost" exceptions to Fermat's Last Theorem. 

Again, since the search ran to b = c = 10,000, the cutoff at 3753 is probably the real limit.

Polynomial Generators

There are quite a few polynomials known that generate cubic quartets. Here are some:

Vieta, 1591

In 1591, Francois Vieta developed the formula below. We can generate numbers a, b, c; d for cubes as follows:

a = m(m3 - 2n3), b = n(2m3 - n3), c = n(m3 + n3), d = m(m3 + n3), where m and n are any two numbers. If m = 2 and n =1 we get 12, 15, 9; 18. If m = 3 and n = 2 we get 33, 70, 92; 105.

Unfortunately, the sum of two squares has no simple factors, but the sum of two cubes does. Vieta's formulas yield large numbers that are often not relatively prime. Elementary reasoning suggests that if it takes two variables to create a Pythagorean triple, it should take at least three to make the cubic analogue, so Vieta's solution is not complete. 

Euler, 1760

We can write A3 + B3 = D3 - C3, and then factor both polynomials. Unfortunately, this seductively symmetrical approach tends to lead to solutions with four variables. Euler's solution (1760-63) is typical. He let A,B,C and D be as follows:

A = (m-n)p + q3     B = (m+n)p - q3     C = p2 - (m+n)q     D = p2 + (m-n)q

A3 + B3 = D3 - C3 factors to: (A+B)(A2-AB+B2) = (D-C)(D2+CD+C2)

Substituting and simplifying, we have m2 +3n2 = 3pq. Obviously m is divisible by 3. Call m = 3k. Also pq = n2 + 3k2

However, Euler also proved that every divisor of n2 + 3k2, if n and k are relatively prime, is of the same form. Thus we can write

p = x2 + 3y2     q = z2 + 3w2     m = 3(yz +- xw)     n =(xz -+ 3yw)

and plug these back into the formulas for A,B,C and D. This, unfortunately, is pretty typical of most of the solutions of this problem. We end up with a chain of substitutions leading back to extremely complex formulas with no attempt to condense or simplify them. And since D is wholly dependent on A, B, and C, there should be no more than three independent variables.

Korneck, 1873

G. Korneck in 1873 derived this solution in three variables for the formula 
A3 + B3 = C3 + D3. Since cubes can be negative, this is merely a slight rearrangement of the formula above.

Table of Quartets with a and b < 100

Below is a table of cube sums for a and b less than 100. Click here for a complete listing of quartets with a, b and c up to 1000.

a b c d
1 6 8 9
1 71 138 144
2 17 40 41
3 4 5 6
3 10 18 19
3 34 114 115
3 36 37 46
4 17 22 25
4 57 248 249
5 76 123 132
5 86 460 461
6 32 33 41
7 14 17 20
7 54 57 70
9 55 116 120
9 58 255 256
11 15 27 29
12 19 53 54
12 31 102 103
12 81 136 145
12 86 159 167
13 51 104 108
13 65 121 127
14 23 70 71
15 42 49 58
15 64 297 298
15 82 89 108
16 23 41 44
16 47 108 111
16 51 213 214
17 40 86 89
17 57 177 179
18 19 21 28
19 53 90 96
19 60 69 82
19 92 101 122
19 93 258 262
20 54 79 87
21 43 84 88
21 46 188 189
22 51 54 67
22 57 255 256
22 75 140 147
23 81 300 302
23 86 97 116
23 94 105 126
25 31 86 88
25 38 87 90
25 48 74 81
26 55 78 87
27 30 37 46
27 46 197 198
27 64 306 307
28 53 75 84
29 34 44 53
29 75 96 110
31 33 72 76
31 64 137 142
31 95 219 225
32 54 85 93
33 70 92 105
34 39 65 72
35 77 202 206
36 38 61 69
36 147 341 350
38 43 66 75
38 48 79 87
38 57 124 129
42 83 205 210
44 51 118 123
44 73 128 137
45 53 199 201
45 69 79 97
46 47 148 151
47 75 295 297
47 97 162 174
48 85 491 492
49 80 263 266
49 84 102 121
50 61 64 85
50 67 216 219
50 74 97 113
51 82 477 478
53 58 194 197
54 80 163 171
56 61 210 213
57 68 180 185
57 82 495 496
58 59 69 90
58 75 453 454
59 93 148 162
61 90 564 565
64 75 477 478
65 87 142 156
66 97 632 633
69 99 146 164
71 73 138 150
71 81 384 386
72 85 122 141
86 95 97 134
88 95 412 415
94 96 99 139

Sums of Higher Powers

Sums of Fourth Powers

If we look at the fourth powers of the digits 0-9, we see an interesting fact. The fourth powers are: 04=0, 14=1, 24=16, 34=81, 44=256, 54=625, 64=1296, 74=2401, 84=4096, 94=6561. Fourth powers only end in 0,1,5 and 6. To get fourth powers that sum to a fourth power, we have to find combinations that sum to something ending in those digits. Using sums of four terms, we find only (0,0,0,1=1), (0,0,0,5=5), (0,0,0,6=6), (0,0,1,5=6), (0,0,5,5=0), (0,0,5,6=1), (0,1,5,5=1), (0,5,5,5=5), (0,5,5,6=6), (1,5,5,5=6), (5,5,5,5=0), (5,5,5,6=1). Note here that we're talking about the powers, not the numbers themselves; a fourth power ending in 6 could belong to a number ending in any even digit. Nevertheless, this improves the odds a lot over the 10,000 combinations if any combination of four ending digits were possible. Noting that Euler's conjecture isn't true, but Fermat's Last Theorem is, one of the four terms at most can be zero (although other terms can end in zero.)

Kermit Rose and Simcha Brudno, (More about four biquadrates equal one biquadrate, Mathematics of computation, vol. 27, no. 123, July 1973, p. 491-494.) found the following sets, where A4+B4+C4+D4=E4

ABCDE
30120315272353
240340430599651
242071043513842487
23651190113014322501
2745101085015462829
24602345227031523723
3395323035016523973
2650106020540944267
36703545175013944333
425028407006994449
1880166038049074949
50801120100032335281
5055391041014125463
5400177095526345491
540016803030435543
51504355181013545729
5695428027705426167
50008855059846609
61854790149034686801
53652850139063687101
2790134516071667209
6635544080030527339
69955620223031967703
56705500445071238373
75655230473048068433
7630592549105248493
78156100344016428517
82302905105052368577
57803695345080128637
8570618032858169137
6435287068086189243
78206935580051929431
87606935149013949519
8570705030552649639
88356800549029229797
64855660484088649877
887086351620229410419
914585305300593610939
1049086355300355611681
1145562004490147611757
8735817011801014412019
1172072703710283312167
936086557480886212259
8925441034501123412287
113908045320735212439
1243561905780161612759
10310687029351067812771
1284559502870593413137
1121075907025971213209
1304049751700789613521
12035361034401073813637
1341064201275827814029
1374079206660392914297
1440526302103414409
1335580101530949814489
1390020401920921914531
1376010245800468214751
1481589404250251215309
11110680038901457915829
11815564028801459816027
1578047904140770116049
159406670543013716113
14320131102275108816359
14890883012201210716643
15160110151085041216891
11810235018451577616893
153751105066901165817381
13060849512201564417519
1640565009501189617521
16215128505450180217661
10660323532201706817693
1732098601945725617881
1751083402760942318077
16805136605270589818477
1536512430114101266818701
1656083556101590619483
13940930544601772619493
175951344053701277219871
1925530907801270220111
11980897510901924420131
19670100301880957920253
19480755016601296920469
1810013690121401180120699
13970885587201914220719
177401652512070336221013
13915595054202480225427
162601286085453417834803
184030690410008992991179

Sums of Fifth and Higher Powers

Amazingly enough, there is a formula for fifth powers that generates an infinite number of solutions:
(75y5-x5), (x5+25y5), (x5-25y5), (10x3y2), (50xy4); (x5+75y5)

Intuition suggests there have to be a lot more than two variables needed for a complete solution, so this formula, though interesting, is only a special case.

There are no examples or formulas known for powers above 5.

Sums of Powers of Consecutive Integers

What is the general formula for the sum 1n + 2n + 3n + 4n + ..... xn for any value of n? This turns out to be a tougher and deeper problem than one might guess. It will be useful to write the sum as a function Sn(x), that is, the sum of the nth powers of consecutive integers up to x. For n=0, the answer is trivial: S0(x) = x.

For n=1, a little trick helps. Pair up the first and last terms in the series, the second and next to last, and so on. We have 1 + x, 2 + (x-1), 3+ (x-2) and so on. If x is even, we have x/2 pairs each summing to x+1, so the total is (x/2)(x+1). Thus S1(x) = x(x+1)/2. If x is odd, leave off the last term x for a moment. Now there is one less pair plus the leftover term x, so S1(x) = (x-1)/2)(x) + x or S1(x) = x(x+1)/2, again.

n = 2:

For higher powers we have to get trickier. Consider n=2. We assume S2(x)=ax3 + bx2 + cx + d. Since S2(0) = 0, d=0. This will be true for any Sn(x); there is no constant term.

Now calculate S2(x+1)-s2(x): We have:

S2(x+1)-S2(x)=a(x+1)3 + b(x+1)2 + c(x+1) - (ax3 + bx2 + cx) =
ax3 + 3ax2 + 3ax + a - ax3
+ bx2 + 2bx + b - bx2
+ cx + c - cx
or
S2(x+1)-S2(x) = 3ax2 + 3ax + a + 2bx + b + c
but also
S2(x+1)-S2(x) = (x+1)2 = x2 + 2x + 1
Now, when two polynomials are equal for any x, their coefficients must be equal. Thus: 3a = 1 and a = 1/3; 3a + 2b = 2, thus b = 1/2; a + b + c = 1, thus c = 1/6. Thus:
S2(x)=(1/3)x3 + (1/2)x2 + (1/6)x = (2x3 + 3x2 + x)/6 = x(x+1)(2x+1)/6

n=3:

Now that we have a general algorithm, we can apply it to higher powers.
Assume S3(x)=ax4 + bx3 + cx2 + dx. Write:

S3(x+1)-S3(x)=a(x+1)4 + b(x+1)3 + c(x+1)2 + dx - (ax4 + bx3 + cx2 + d) =
ax4 + 4ax3 + 6ax2 + 4ax + x - ax4
+ bx3 + 3bx2 + 3bx + b - bx3
+ cx2 + 2cx + c - cx2
+ dx + d - dx
or
S3(x+1)-S3(x) = 4ax3 + 6ax2 + 4ax + a + 3bx2 + 3bx + b + 2cx + c + d =
(x+1)3 = x3 + 3x2 + 3x + 1
Hence:
4a = 1 and a = 1/4; 6a + 3b = 3 and b = 1/2; 4a + 3b + 2c = 3 and c = 1/4;
a + b + c + d =1 and d= 0
Combining terms and simplifying: S3(x) = x2(x + 1)2/4 or S3(x) = (S1(x))2.
It looks like we might be starting a nice pattern.

n=4:

Using the routine above, assume S4(x)=ax5 + bx4 + cx3 + dx2 + ex.

S4(x+1)-S4(x)=
ax5 + 5ax4 + 10ax3 + 10ax2 + 5ax + a - ax5 + bx4 + 4bx3 + 6bx2 + 4bx + b - bx4
+ cx3 + 3cx2 + 3cx + c - cx3
+ dx2 + 2dx + d - dx2
+ ex + e - ex
or
S4(x+1)-S4(x) = 5ax4 + (10a + 4b)x3 + (10a + 6b + 3c)x2 + (5a + 4b + 3c + 2d)x + (a + b + c + d + e)
= x4 + 4x3 + 6x2 + 4x + x. Thus:
5a =1 and a = 1/5; 10a + 4b = 4 and b = 1/2; 10a + 6b + 3c = 6 and c=1/3;
5a + 4b + 3c + 2d = 4 and d = 0; a + b + c + d + e = 1 and e = -1/30. Clearing fractions, we get:
S4(x) = (6x5 + 15x4 + 10x3 - x)/30 = x(x + 1)(2x + 1)(3x2 + 3x -1)/30.
So much for our elegant pattern.

The coefficients for powers up to 12 are shown below.

x x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12
n=1 1/2 1/2
n=2 1/6 1/2 1/3
n=3 0 1/4 1/2 1/4
n=4 -1/30 0 1/3 1/2 1/5
n=5 0 -1/12 0 5/12 1/2 1/6
n=6 1/42 0 -1/6 0 1/2 1/2 1/7
n=7 0 1/12 0 -7/24 0 7/12 1/2 1/8
n=8 -1/30 0 2/9 0 -7/15 0 2/3 1/2 1/9
n=9 0 -3/20 0 1/2 0 -7/10 0 3/4 1/2 1/10
n=10 5/66 0 -1/2 0 1/1 0 -1/1 0 5/6 1/2 1/11
n=11 0 5/12 0 -11/8 0 11/16 0 -11/8 0 11/12 1/2 1/12

Readers with some advanced math background may recognize the first column as Bernoulli Numbers (with alternating signs) for the even powers. The patterns in the table are easiest to see along the diagonals. From right to left, the terms in each diagonal are as follows (n is the row number)

Factored polynomials for each n are:
n=0: x
n=1: x(x + 1)/2
n=2: x(x + 1)(2x + 1)/6
n=3: x2(x + 1)2/4
n=4: x(x + 1)(2x + 1)(3x2 + 3x - 1)/30
n=5: x2(x + 1)2(2x2 + 2x - 1)/12

Multigrades

One might guess that it is very unlikely to have sums of powers that hold for several different exponents. In fact, there are many such equations, called multigrades. For example:

It is a remarkable fact that we can add any integer to all the terms of the multigrade and it will still hold. Adding 1 to the example above, we get (2, 9, 11, 18); (3, 6, 14, 17). (n=1,2,3). For clarity we group the terms on each side of the equality in parentheses.

A few remarkable high-order multigrades are:

(1, 50, 57, 15, 22, 71); (2, 45, 61, 11, 27, 70); (5, 37, 66, 6, 35, 67) (n=1,2,3,4,5)

(1, 9, 25, 51, 75, 79, 107, 129, 131, 157, 159, 173);
(3, 15, 19, 43, 89, 93, 97, 137, 139, 141, 167, 171) (n=1, 3, 5, 7, 9, 11, 13)

References

Most of the historical results are from:

Dickson, Leonard E., History of the theory of numbers,  New York, Chelsea Pub. Co., 1952.

This is a rich but maddeningly disorganized book, with no attempt to use consistent notation or format.


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Created 2 January 1997: Last Update 14 December 2009

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