MCGILL UNIVERSITY SEPTEMBER 21, 2007 COPYRIGHT © 2007 MICHAEL I. SHAMOS Surprises in Experimental...

MCGILL UNIVERSITY

SEPTEMBER 21, 2007

COPYRIGHT © 2007 MICHAEL I. SHAMOS

Surprises inExperimental Mathematics

Michael I. ShamosSchool of Computer ScienceCarnegie Mellon University

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Mathematical Discovery

• Where do theorems come from?

• Theorem easy to conjecture, proof is hard– Fermat’s last theorem, four-color theorem

• Theorem surprising, but results from a logical investigation– Fundamental theorem of algebra

• Theorem difficult to invent, straightforward to prove– Hadamard three-circles theorem

– Where do these come from?

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Hadamard Three-Circles Theorem• If f(z) is holomorphic (complex differentiable) on the annulus

centered at the origin and

• then

)(loglog)(loglog)(loglog bMa

raM

r

brM

a

b

bza )(sup)( zfrMrz

a

b

r

• •

•)(aM

)(rM)(bM

How was this theorem ever conjectured?

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Outline• The problem

– Closed-form expression for• The approach

– Build a catalog of real-valued expressions indexed by first 20 digits– Equivalent expressions will “collide”– Look up 1.20205690315959428539

• The discoveries– The Partial Sum Theorem– Overcounting functions– How many ways can n be expressed as an integer power k

j ?

– Expression for – . . .

...2020569031.11

)3(1

3

k k

122

1

kk

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


The Problem

• Closed form expressions for values of the zeta function

• Euler found an expression for all even values of s:

• No expression is known for even a single odd value, e.g.

1

1)(

ksk

s

ssss

s

Bs 22

121

)!2(

2)1()2(

)3(

2

1;

1

1

11

1

0

BBk

n

nB

n

kkn

)2( s s2is a rational multiple of

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


The Catalog• Some values of :

2906251531329465

176411)20(;

90)4(;

6)2(

2042

2906251531329465

176411)20(4162033872790000009539.1

20

90)4(516037111381911.08232323

4

6)2(472468482264361.64493406

2

?)3(399731595942851.20205690

12

1152717616113171.20202249

Catalan

0

2)12(

)1(0541772190150.91596559

k

k

kCatalan

)(n

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Other Catalogs

• Sloane’s Encyclopedia of Integer Sequences

– Terrific, but for integer sequences, not reals

• Plouffe’s Inverter

– Huge (215 million entries), but not “natural” expressions from actual mathematical work

• Simon Fraser Inverse Symbolic Calculator

– 50 million constants

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Discovery A

primep

pk

k

k

2

12

2

)(

1

where is the number of primes k)(k

• Is this a coincidence?

• Why the factor of 2?

• Is there a general principle at work?

primepp

kk aa

a

a

k 1

1

)(

1

– In fact,

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Observation

is a partial sum function, i.e.,)(k

kp

primep

k 1)(

otherwise

primejjIwherejIk prime

k

jprime ,0

,1)(,)()(

1

11

)()()()()(k

primeprimepk

kgkIkgkfk

primep

pprimep

pk

k

k1

1 2

1

2

12

2

)(So can be rewritten as:

where f and g are “related”

More generally, is the partial sum function of theindicator function of the property “primeness”:

)(k

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


The Partial Sum Theorem• Given a sequence S of complex numbers s(k), let

be the sequence of partial sums of S.

• Given a function f, if certain convergence criteria are satisfied, then

where

(the partial tails of f) is a transform of f independent of s & t

11

)()()()(jk

jgjskfkt

jk

kfjg )()(

n

k

ksnt1

)()(

PARTIALSUMS OF s( j )

PARTIALTAILS OF f ( k )

(New)

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Partial Sum Functions

• Many sequences are partial sum functions:

n

k knH

1

1)( the harmonic function

n

ks

s

knH

1

)( 1)( generalized harmonic function

n

kn

n

1 2

121

n

k

kn1

log)(log

n

kkn

knn

1

22

22

646

• Actually, every sequence is the partial sum function of some other sequence

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Some Partial Sum Transforms

ka

1

jk

kfjg )()()(kf )( jg)(kf

1)1(

1 jaa

kk 2

1

j

1

kk 3

1

)1(2

1

jj

!

1

k

)(

)1,(1

j

je

)!2(

12

k

kk

)!1( jj

ka

ksin 1cos21

sin)1sin(21 aaa

jajj

n

i ik0 )(

11

1

11 ),()1(

1

1

n

i

ii jinSn

1

)1(2 k

k

)1(2

)1(

jj

j

kak

1

0,,1

ja

k

k21

¼½,,1,1)½(4

)(12

jjFj

jj

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Some Partial Sum Transforms

1

1)(

kka

kt

1

)()(j

jgjs

1

)()(k

kfkt

11)1(

1)(

jjaa

js

12

1)(

k kkkt

1

1)(

j jjs

23

1)(

k kkkt

2 )1(2

1)(

j jjjs

1 !

1)(

k kkt

1 )(

)1,(1)(

j j

jejs

1

2

)!2(

1)(

k k

kkkt

1 )!1()(

j j

jjs

1

sin)(

kka

kkt

121 1cos21

sin)1sin()(

jj aaa

jajjs

1 0 )(

)(

k

n

i ik

kt

1

1

1

11 ),()1(

1

)(

j

n

i

ii jinSn

js

22 1

)1()(

k

k

kkt

2 )1(2

)1()(

j

j

jjjs

1

1)(

kkak

kt

1

0,,1

)(j

ja

js

1 21

)(k

k

kkt

112 ¼½,,1,1

)½(4

)()(

jj

jjFj

jjs

1

)()(j

jgjs

1

)()(k

kfkt

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


...00000

...00000

...)5()5(0000

...)5()4()4()4(000

...)5()3()4()3()3()3(00

...)5()2()4()2()3()2()2()2(0

...)5()1()4()1()3()1()2()1()1()1(

fs

fsfs

fsfsfs

fsfsfsfs

fsfsfsfsfs

Partial Sum Theorem (Proof)• Consider the upper triangular matrix :),()(, jijfism ji

HEADS OF s= t(i)

TAILS OF f= g(j)

Column sums are )()( kfkt

Rowsumsare

)()( jgjs

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


The Convergence Criteria

1. All sums g(k) converge

2. converges; and

3.

11

)()()()(kk

kgkskfkt iff

1

)()(k

kgks

0)()(lim

ngntn

Proof: By Markoff’s theorem on convergence of double series

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Further Applications• The number of perfect n

th powers k is

• The number of positive integers powers of a k is

• Therefore, by inspection,

1

111log

122

aankk

k

jj

aofpowerpositiveank

a

primepk ppkk

k

)1(

1)(2

23

)(11

112

mjnkk

k

jm

powermperfectank

m

th

n k

kalog

)2(log

)(1

1

)()12(

12

222

kk

k

pkk

kk

primep kk

* (Old)

primepk pk

kk

!

1

)!1(

)(

2

**

*

*

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


The Inverse Transform

• Given g(j), how can we find f(k)?

• Since g is a sum of f’s, f is the sequence of finite differences of g :

• Subtracting,

11

)()()()(jk

jgjskfkt

)1()(1

jgkfjk

)1()()( jgjgjf

)()( jgkfjk

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Some Inverse Transforms

aj

1

)( jg )1()()( kgkgkf )( jg )(kf

aa

aa

kk

kk

)1(

)1(

ja

11

1

ka

a

1

1

j )2)(1(

1

kk

!

1

j )!1( kk

!

1

jj )!1)(1(

12

kkk

kk

!!

1

jj )!1()!1(

22

kk

kk

)1(

)1(

jj

j

)2(

)1(2

kk

k

jaj

11)1(

kakk

akka

jF

1

1

1

kk

k

FF

F

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Some Inverse Transforms

1

1)(

jaj

js

1

)()(j

jgjs )()(1

kfktk

1 )1(

)1()(

kaa

aa

kk

kkkt

1

1)(

jja

js

11

1)(

kka

akt

1 1

1)(

j jjs

1 )2)(1(

)(

k kk

kt

1 !

)(

j j

js

1 )!1()(

k k

kkt

1 !

)(

j jj

js

1

2

)!1)(1(

1)(

k kkk

kkkt

1 !!

)(

j jj

js

1

2

)!1()!1(

2)(

k kk

kkkt

1 )1(

)1()(

j

j

jjjs

1 )2(

)()1(2

k

k

kk

kt

1

)(

jjaj

js

11)1(

)(k

kakk

akkakt

1

)(

j jF

js

1 1

1)(k kk

k

FF

Fkt

1

)()(j

jgjs )()(1

kfktk

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


The Partial Integral Theorem• Given a function s(x), let t(y) be the “left integral” of s :

• Given a function f(y), if certain convergence criteria are satisfied, then

where

(the right integral of f) is a transform of f independent of s & t

dxxgxsdyyfyt

00

)()()()(

dxyfxgx

)()(

y

dxxsyt0

)()(

LEFT INTEGRAL OF s(x)

RIGHT INTEGRAL OF f ( y )

*

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Example

22 )1(

1)(;

1

1)(

xxf

xxs

Therefore,

02

02 )1)(1()1(

arctan

xx

dxdx

x

x

1

1)()(;arctan)()(

0

xdyyfxgydxxsyt

x

y

Note: if s(x) is a probability density

then t(y) is its cumulative distribution function

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Example

Consider the special case in which )()( xgxf

xexf )(

00

)()( dxexsdxext xxThis implies . So

1cos)1(1sin)1(2

1cosarctan

1 002

sicidxxedxx

e xx

1 2 3 4 5 6

0.2

0.4

0.6

0.8

1

xe x arctan

12

x

e x

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Example

xx eexxxtx

xs

2

1sinh;arcsinh)(;

1

1)(

2

???arcsinh0

dxxe x

580.75461002)1()1(21

00

02

YHdxx

e x

Bessel functionStruve function

What about Risch’s theorem? Risch, R. “The Solution of the Problem ofIntegration in Finite Terms.” Bull. Amer. Math. Soc., 1-76, 605-608, 1970.

Mathematica gives up.

Mathematica has no problem

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Discovery B

132

1 13

11)3()2(

)(

1

kj k kkjk

•

• Is there a general principle at work?

1 13)(

1

j k jkmust exceed ,

but by how much?

1)3(1

23

k k

)()1()(

1

1 1

ssjkj k

s

– In fact,

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Overcounting Functions

1. Every term of S+ occurs at least once in S.

2. In general, S+ overcounts S, since some terms of S occur many

times in S+

3. If Kg(k) is the number of times f(k) is included in S+, then

where Kg(k) depends only on g and not on f .

where ranges over the natural numbers),( jkg

1 11

)),(()(j kk

jkgfSkfS andLet ,

11 1

)()()),((k

gj k

kfkKjkgfS

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Examples

• Let g(k, j) = k + j . How many ordered pairs (k, j) of natural

numbers give k + j = n? Answer: Kk+j (n) = n - 1

• Therefore, by inspection,

)()1(111

)(

1

11

11 1

ssnnn

n

jk nss

ns

j ks

8)(

2

1

2

)( 2

11 1

nn

j kjk

nnjk

11!

1

)!(

1

11 1

een

n

jk nj k

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Examples

• Let g(k, j) = k • j . How many ordered pairs (k, j) give k • j = n?

Answer: Kk•j (n) = d(n), the number of divisors of n .

• Therefore, by inspection

)()(

)(

1 2

11 1

sn

nd

jk ns

j ks

1 11 1

)(

1

11

j nnj

j kjk a

nd

aa

...9076269794810610197.2!

)(

)!(

1

11 1

nj k n

nd

jk

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Enumerating Non-Trivial Powers• Let g(k, j) = k

j. How many ordered pairs (k, j) give k j = n?

Or, how many ways K(n) can n be expressed as a positive integral power of a positive integer?

• Let be the prime factorization of n

• n can be a non-trivial power of an integer > 1 iff

G(n) = gcd(e1, e2, . . . ) exceeds 1; otherwise K(n) = 1.

• Suppose b > 1 divides G(n). Then ,

where each of the ei /b is a natural number, so n is the b th power

of a natural number

• Suppose c > 1 does not divide G(n). Then at least one of the

exponents ei /c is not a natural number and n is not the c th power

of a natural number. Therefore,))nofionfactorizatprimetheofexponents(gcd()( dnK jk

...21

21ee ppn

bbebe ppn .../2

/1

21

*

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


A Remarkable Series

• Let . Then

s

nk

s

nk

knknKSS jj

11

)(1)(

ssss kkkk

11

1

12

111))(()(

22

22

nnn nnn

nGd

n

n

1)(1

2

sk

Sk

s

12 211 2

1)(1

1

)(

1

nk ks

j

sj

j ksj

snk

kk

S

22

2

1)(

nss

n

s

nnnnSS

...64

3

49

1

36

1

32

1

27

1

25

1

16

2

9

1

8

1

4

11

yields

the “overcounting” function

(Old)

*

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Goldbach’s Theorem

• In 1729, Christian Goldbach proved that

...26

1

24

1

15

1

8

1

7

1

3

11

1

1

powerintegertrivialnonaq

q

1)(1

1

1

powerintegertrivialnonaq 1 1


k kk k

k

qq

42 43 44

256

3

. . .

82

162

83

163

...64

3

49

1

36

1

32

1

27

1

25

1

16

2

9

1

8

1

4

11

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Discovery C

1


1)(11

1

k

kq

???1

powerintegertrivialnonaq q

(Old)

What is

... 94049448666.8744643681)()(1

2powerinteger

trivialnonaq

k

kkq

2


1)()(1)1(

1

k

kkqq

factorprimerepeatedahask

primesdistinctnofproductak

k

k n

,0

,)1(

1,1

)(

*

*

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Discovery D

primepp

kk

k

12

1

2

)(

1

• Since the partial sum function of is

, where is the number of distinct prime factors of k)(k

1

)1(112

kkk

k

aaa

aa

1,1

)(

1

1)(

1112

1

aaaa

ak

aa

k

kkkk

k

primepp

kk

1

1

ka

*

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Discovery E

11 1

)(1

kpk

jp c

pk

cj

For c > 1 real and p prime,

112 14

)2(

2

1

kk

j

kj

In particular,

0

21

2

112

112 2

1

2

log

12

)12(

12

)2(

2

1

kkk

kk

kk

jkj

kkk

11 14

)2(1

12

)2(1

kk

kk

kk

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Results

32

12

2

1

11 1),max(

kk

k jjk

k

The counting function Kmax(n) of max(k, j) is 2n-1. So

1!

12

)!,max(

1

11 1

ek

k

jk kk j

)3(3

12

),max(

1 2

13

1 13

kk j k

k

jk

The counting function Klcm(n) of lcm(k, j) is d(n 2). So

)2(

)()(

),(lcm

1 3

1

2

1 1 a

s

k

kd

jk ka

k js

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


The First-Digit Phenomenon

• Given a random integer, what is the probability that its leading digit is a 1?

• Answer: depends on the distribution from which k is chosen

• If k is chosen uniformly in [1, n], then let p(d, n) be the probability that the leading digit of k is d

• For n = 19+, 5/9 < p(1,n) < .579; 1/19 p(9,n) < 1/18

• For n = 9+, p(1,n) = 1/9; p(9,n) = 1/9

• The “average” is log10(1+1/d)

• {.301, .176, .124, .097, .079, .066, .058, .051, .046}

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Relative Digit Frequency (Benford’s Law)

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

1 2 3 4 5 6 7 8 9

Benford

log10(1+1/d)

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Relative Digit Frequency

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

1, 0 2, 1 3, 2 4, 3 5, 4 6, 5 7, 6 8, 7 9, 8 9

Benford

Catalog Frac

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


First-Digit Phenomenon

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0, 1 1, 2 2, 3 3, 4 4, 5 5, 6 6, 7 7, 8 8, 9 9

Benford

Log[11,1+1/d]

Catalog Frac

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Major Ideas

• For mathematicians:– How to populate the catalog

– How to generalize from discoveries

• For computer scientists:– Use in symbolic manipulation systems

• For data miners:– How to mine the catalog, i.e. how to find new relations

• For statisticians:– How to use the fact that

where P is the cumulative distribution of density p

dxxgxpdyyfyP

00

)()()()(

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


A Parting Philosophy

“The object of mathematical rigor is to sanction and legitimize the conquests of intuition, and there was never any other object for it.”

Jacques Hadamard(as quoted by Borel in 1928)

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


QA&

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Results

2 22

1)1(

)()1(

)1(

)(

k k

k

k

k

kk

k

12 1 )12(

)(

2

1

12

)(

)(2

1

kk kk kk

k

k

k

k

)6(

)3()2(

)1(

11

)(

)(

2

primepk ppkk

k

222

1)1(

)(

)1(

)(

kk k

k

kk

k

22

2

)1(

)(1)(

kk kk

kdk

2

2)(1 log

)2()1(k

nnn

k

kn! is the nearest integer to

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Hypergometric Functions

• The are solutions of the hypergeometric differential equation:

!)(

)()(;;,

012 k

z

c

bazcbaF

k

k k

kk

0])1([)1( ybayzbacyzz

)(/)()1(...)1()( akakaaaa k where

dttzttbcb

czcbaF abcb

1

0

112 )1()1(

)()(

)(;;,

zcbaF ;;,12

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Partial Sum Theorem (Proof)

• Consider the upper triangular matrix

• The sum of row i is

• The sum of column j is

• The sum of the row sums equals the sum of the columns sums

precisely when the conditions of Markoff’s theorem are satisfied.

QED

ijj

ji igisjfism )()()()(1

,

jijfism ji ),()(,

j

iiji jfjtisjfm

11, )()()()(

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


Correspondence with Plouffe

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0, 1 1, 2 2, 3 3, 4 4, 5 5, 6 6, 7 7, 8 8, 9 9

Catalog Int

Benford

Log[11,1+1/d]

Catalog Frac

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


First-Digit Phenomenon

SOURCE: SIMON PLOUFFE

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


First Digit Frequency

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0, 1 1, 2 2, 3 3, 4 4, 5 5, 6 6, 7 7, 8 8, 9 9

Catalog Int

Benford

Catalog Frac

MCGILL UNIVERSITY

SEPTEMBER 21, 2007


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3 4 5 6 7 8 9

Catalog Int

Correspondence with Plouffe

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