1

How Bernoulli Did It:Sums of Powers, Generating Functions,

and Bernoulli Numbers

Hieu D. Nguyen(Joint work with Abdul Hassen)

Rowan UniversityRowan Math Seminar

October 11, 2006

MAA Column: Ed Sandifer’s How Euler Did it:

http://www.maa.org/news/howeulerdidit.html

2

Bernoulli Challenge

Prize Problem: Find the sum of the 10th powers ofthe first 1000 natural numbers, i.e.

10 10 10 101 2 3 ... 1000+ + + +

(Bernoulli did it “in less than half of a quarter of an hour”)

Hint: The answer is a 32-digit number.

Rules: Students only. No technology allowed.

3

Jacob (Jacques) Bernoulli

1654 - 1705

The MacTutor History of Math Archivehttp://www-history.mcs.st-and.ac.uk/index.html

4The MacTutor History of Math Archivehttp://www-history.mcs.st-and.ac.uk/index.html

Leonhard Euler

1707 - 1783

5

2 2 2 2

1

( 1)(2 1)1 2 ...6

n

k

n n nk n=

M+ += + + + = =∑

Sums of Powers

1

( 1)1 2 3 ...2

n

k

n nk n=

N+= + + + + = =∑

23 3 3 3 2

1

( 1)1 2 ...2

n

k

n nk n=

+⎡ ⎤= + + + = =⎢ ⎥ N⎣ ⎦

∑

…

Johann Faulhaber, Academia Algebrae (1631)

4 4 4 4

1

1 61 2 ...5 5

n

k

k n M=

⎛ ⎞= + + + = − +⎜ ⎟⎝ ⎠

∑ N (Haytham)

1 2 3 ... 98 99 100 50(101) 5050+ + + + + + = =

(Pythagoreans)

(Archimedes)

(Nichomacus)

2 2 2 2 2 21 2 3 ... 98 99 100 ?+ + + + + + =

6

Faulhaber’s Formulas

17 17 17 17 9 8 7

1

6 5 4

3 2

1 2 ... (1280 6720 21220

46880 72912 7422043404 10851 ) / 45

n

kk n N N N

N N NN N

=

= + + + = − +

− + −

+ −

∑

2 11 2

1

( ... ) (even powers)n

r rr

k

k M b b N b N −

=

= + + +∑

2 1 2 11 2

1

( ... ) (odd powers greater than 1)n

r rr

k

k N c c N c N+ −

=

= + + +∑

7

Jacob Bernoulli, Ars Conjectandi (1713)

3 22 2 2 2

11 2 ...

3 2 6

n

k

n n nk n=

= + + + = + +∑

2

11 2 3 ...

2 2

n

k

n nk n=

= + + + + = +∑

4 3 23 3 3 3

11 2 ...

4 2 4

n

k

n n nk n=

= + + + = + +∑5 4 3

4 4 4 4

11 2 ...

5 2 3 30

n

k

n n n nk n=

= + + + = + + −∑6 5 4 2

5 5 5 5

1

51 2 ...6 2 12 12

n

k

n n n nk n=

= + + + = + + −∑

1 1 12 2+ =

1 1 1 13 2 6+ + =

1 1 1 14 2 4+ + =

1 1 1 1 15 2 3 30+ + − =

1 1 5 1 16 2 12 12+ + − =

11 2 ... ?

np p p p

k

k n=

= + + + =∑

8

1 1 ( 1)( 2) ( 1)( 2)( 3)( 4), , , , , ,...1 2 2 2 3 4 2 3 4 5 6

p p p p p p p p pA B Cp

− − − − − −+ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

1 11: ,2 2

p =

1 1 12 : , ,3 2 6

p =

1 1 13: , ,4 2 4

p =

1 1 14 : , ,5 2 3

p =

1 1 55 : , ,6 2 12

p =

1 1 1, , ,...6 30 42

A B C= = − =

3 12 6

= ⋅

1,301,

12

4 12 6

= ⋅

5 12 6

= ⋅

2 12 6

= ⋅

, 0

, 0

, 0 , 0

4 12 30

= ⋅

5 12 30

= ⋅5 4 3 12 3 4 30⋅ ⋅

= ⋅⋅ ⋅

4 3 2 12 3 4 30⋅ ⋅

= ⋅⋅ ⋅

9

1 1 3

1

3

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

2 3 4 5 6

np p p p p

k

p

p p p pk n n An Bnp

p p p p p Cn

+ − −

=

−

− −= + + +

+ ⋅ ⋅− − − −

+ +⋅ ⋅ ⋅ ⋅

∑Bernoulli’s formula

1 1 1, , ,...6 30 42

A B C= = − =

(Original) Bernoulli numbers

1 1 20 1 2 3

1

! ! ! ! ...0!( 1)! 1! ! 2!( 1)! 3!( 2)!

np p p p p

k

p p p pk B n B n B n B np p p p

+ − −

=

= + + + ++ − −∑

Modern formula

0 1 2 3 4 51 1 11, , , 0, , 0,...2 6 30

B B B B B B= = = = = − =

1

0

!( 1)!( 1 )!

kpp

p kk

k

p B nk p k

δ + −

=

= −+ −∑

10

We will observe in passing that, many [scholars] engaged in the contemplation of figurate numbers (among them Faulhaber and Remmelin of Ulm, Wallis, Mercator, in his Logarithmotechnia and others) but I do not know of one who gave a general and scientific proof of this property.

Wallis in his Arithmetica Infinitorum investigated by means of induction the ratios that the series of squares, cubes, and other powers of natural numbers have to the series of terms each equalto the greatest term. This he put in the foundation of his method.

Figurate Numbers"It appears to me that if one wants to make progress in mathematics one should study the masters, and not the pupils.“ --- Niels Abel (1802-1829)

Passage from Ars Conjectandi (p. 95)

11

His next step was to establish 176 properties of trigonal, pyramidal, and other figurate numbers, but it would have been better and more fitting to the nature of the subject if the process would have been reversed and he would have first given a discussion of figurate numbers, demonstrated in a general and accurate way, and only then have proceeded with the investigation of the sums of powers. Even disregarding the fact that the method of induction is not sufficiently scientific and, moreover, requires special work for every new series; it is a method of common judgment that the simpler and more primitive things should precede others. Such are the figurate numbers as related to thepowers, since they are formed by addition, while the others are formed by multiplication;…

Translated from the Latin by Professor Jekuthiel Ginsburg, Yeshiva College, New York City.

12

Bernoulli’s Triangle (Pascal’s) of Figurate Numbers (Binomial Coefficients)

1 0 0 0 0 0 0 01 1 0 0 0 0 0 01 2 1 0 0 0 0 01 3 3 1 0 0 0 01 4 6 4 1 0 0 01 5 10 10 5 1 0 0... ... ... ... ... ... ... 0

...n n n n nA B C D E

1 2 1

1 2 1

1 2 1

...

...

......

n n

n n

n n

A A A BB B B CC C C D

−

−

−

+ + + =

+ + + =

+ + + =

Figurate Identities

1,1,

( 1)( 2) ,1 2

( 1)( 2)( 3) ,1 2 3

...

n

n

n

n

AB n

n nC

n n nD

=

= −

− −=

⋅− − −

=⋅ ⋅

( 1)n =

13

Bernoulli’s proof of his formula based on figurate identities:

Assume the figurate identity

2

1

1 2 3 ...2 2

n

k

n nk n=

= + + + + = +∑

1 1 1 1

2 2

( 1 1) ( 1) 1

2 2 2 2

n n n n

k k k k

k k k

n n n nn

= = = =

= − + = − +

⎛ ⎞= − + = +⎜ ⎟⎝ ⎠

∑ ∑ ∑ ∑

2( 1)0 1 2 ... ( 1)1 2 2 2

n n n nn −+ + + + − = = −

⋅Then

Case p = 1:

1 2 1... n nB B B C−+ + + = holds:

Note: Observe that Bernoulli’s proof resembles the modern one using the principle of mathematical induction even though he himself viewed induction as not “sufficiently scientific”.

14

Case p = 2:

2

1 1 1 1

( 1)( 2) 1 3 11 2 2 2

n n n n

k k k k

k k k k= = = =

− −= − +

⋅∑ ∑ ∑ ∑

3 2( 1)( 2) ( 1)( 2) 3 20 0 1 3 6 ...1 2 1 2 3 6

n n n n n n n n− − − − − ++ + + + + + = =

⋅ ⋅ ⋅

Then since

3 22 2 2 2

1

1 2 ...3 2 6

n

k

n n nk n=

= + + + = + +∑

Assume the figurate identity 1 2 1... n nC C C D−+ + + = holds:

and2

1

1

3 3 3 ,2 4 4

1

n

k

n

k

k n n

n

=

=

= +

=

∑

∑

15

3 22 2

1

3 2

1 3 2 3 32 6 4 4

6 4 12

n

k

n n nk n n n

n n n=

⎛ ⎞− + ⎛ ⎞= + + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= + +

∑

it follows that

Hence,3 2

2

1 3 2 6

n

k

n n nk=

= + +∑

Case p = 3:4 3 2

3 3 3 3

1

1 2 ...4 2 4

n

k

n n nk n=

= + + + = + +∑

Assume the figurate identity 1 2 1... n nD D D E−+ + + = holds:

16

4 3 2

( 1)( 2)( 3) ( 1)( 2)( 3)0 0 0 1 4 10 ...1 2 3 1 2 3 4

6 11 66

n n n n n n n

n n n n

− − − − − −+ + + + + + + =

⋅ ⋅ ⋅ ⋅ ⋅− + −

=

3 2

1 1 1 1 1

( 1)( 2)( 3) 1 11 11 2 3 6 6

n n n n n

k k k k k

k k k k k k= = = = =

− − −= − + −

⋅ ⋅∑ ∑ ∑ ∑ ∑

Then since

and3 2

2

1

2

1

1

,3 2 6

11 11 11 ,6 12 12

1

n

k

n

k

n

k

n n nk

k n n

n

=

=

=

= + +

= +

=

∑

∑

∑

17

4 3 2 3 23 2

1

4 3 2

1 6 11 6 11 116 6 3 2 6 12 12

24 12 24

n

k

n n n n n n nk n n n

n n n=

⎛ ⎞ ⎛ ⎞− + − ⎛ ⎞= + + + − + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

= + +

∑

it follows that

Hence,4 3 2

3

1 4 2 4

n

k

n n nk=

= + +∑

18

Proof for arbitrary p?

Bernoulli’s formula for p = 10:

( )10 11 10 9 7 5 3

1

1 6 33 55 66 66 33 566

n

k

k n n n n n n n=

= + + − + − +∑

Unfortunately Bernoulli does not provide one, but the recipe for an inductive proof is evident in his demonstration for p = 1, 2, 3.

http://www.jstor.org/view/0025570x/di021151/02p0123y/0

19

Generating Functions

2 4 61 1 1 11 ...1 2 12 720 30240t

t t t t te

= − + − + +−

Leonard Euler (1755)

0 1 2 3 4 51 1 11, , , 0, , 0,...2 6 30

B B B B B B= = − = = = − =

2 4 61 1 1 11 ...2 1! 6 2! 30 4! 42 6!

t t t t⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + + − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 4 6

0 1 2 4 6 ...1! 2! 4! 6!t t t tB B B B B= + + + + +

01 !

n

ntn

t tBe n

∞

=

=− ∑

Bernoulli numbers

20

0 1 0

( 1)! ! !

k m kt

k kk m k

t t tt e B Bk m k

∞ ∞ ∞

= = =

⎛ ⎞⎛ ⎞= − = ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠∑ ∑ ∑

Recursive formula:

0

0 if 01 if 0!( 1 )!

nk

k

nBnk n k=

>⎧= ⎨ =+ − ⎩

∑

0

01 1

0 1 22

0 : 1

11: 02 2

12 : 06 2 2 6

n B

Bn B B

B B Bn B

= =

= + = ⇒ = −

= + + = ⇒ =

Equate series coefficients of t:

1

1 0 0 0( 1 )

! ! !( 1 )!

m k nnk

km k n k

BtB t n m km k k n k

+∞ ∞ ∞+

= = = =

⎛ ⎞= = + = +⎜ ⎟+ −⎝ ⎠∑∑ ∑ ∑

21

Bernoulli Polynomials

0

( )1 !

xt n

ntn

te tB xe n

∞

=

=− ∑

0

( 1) ( )!

kxt t

kk

tte e B xk

∞

=

= − ∑

Equate series coefficients of t:

0 1 0

( ) ( )! ! !

n m k

km m k

xt tt B xn m k

+∞ ∞ ∞

= = =

=∑ ∑∑

11

0 0 0

( ) ( 1 )! !( 1 )!

n n nnk

n n k

B xx t t n m kn k n k

+∞ ∞+

= = =

⎛ ⎞= + = +⎜ ⎟+ −⎝ ⎠

∑ ∑ ∑

22

Recursive formula

0

01 1

220 1 2

2

3 23

0 : ( ) 1

( ) 11: ( ) ( )2 2

( ) ( ) ( ) 12 : ( )6 2 2 2 6

3 13: ( )2 2

n B x

B xn B x x B x x

B x B x B x xn B x x x

n B x x x x

= =

= + = ⇒ = −

= + + = ⇒ = − +

= = − +

Observe that(0)n nB B=

0

( )!( 1 )! !

nnk

k

B x xk n k n=

=+ −∑

23

-2 -1 1 2

-2

-1.5

-1

-0.5

0.5

1

1.5

2

0.2 0.4 0.6 0.8 1

-0.4

-0.2

0.2

0.4

Interval [-2,2] Interval [0,1]

1( )B x0 ( )B x

3 ( )B x2 ( )B x

24

Some Properties of Bernoulli Numbers and Polynomials

What about derivative and integral properties?

(1 ) ( 1) ( )nn nB x B x− = −

11

0

1 ( ) (0)n

pp p

kk B n B

p

−−

=

⎡ ⎤= −⎣ ⎦∑

1 2

221

1 ( 1) (2 )(2 )2(2 )!

n n

nnk

n Bk n

πζ−∞

=

−= =∑

1

0

1 if 10 if 1

n

kk

n nB

k n

−

=

=⎛ ⎞ ⎧= ⎨⎜ ⎟ >⎝ ⎠ ⎩

∑

0 0

1 ( 1)1

n kr n

nk r

kB r

rk= =

⎛ ⎞= − ⎜ ⎟+ ⎝ ⎠∑ ∑

1.

2.

3.

4.

5.

0

( )n

n kn k

k

nB x B x

k−

=

⎛ ⎞= ⋅⎜ ⎟

⎝ ⎠∑6.

25

Appell Sequences

1 0'( ) ( 1/ 2) 1 ( )dB x x B xdx

= − = =

11 1 2

10 00

1 1 1( ) 02 2 2

B x dx x dx x x⎛ ⎞ ⎡ ⎤= − = − =⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦∫ ∫

Derivative property

Integral property:

22 1

1 1'( ) ( ) 2 1 2 2 ( )6 2

dB x x x x x B xdx

⎛ ⎞= − + = − = − =⎜ ⎟⎝ ⎠

3 2 2 23 2

3 1 1 1'( ) ( ) 3 3 3 3 ( )2 2 2 6

dB x x x x x x x x B xdx

⎛ ⎞= − + = − + = − + =⎜ ⎟⎝ ⎠

1 1 220 0

1( ) 06

B x dx x x dx⎛ ⎞= − + =⎜ ⎟⎝ ⎠∫ ∫

1 1 3 220 0

3 1( ) 02 2

B x dx x x x dx⎛ ⎞= − + =⎜ ⎟⎝ ⎠∫ ∫

26

Theorem: (Appell?) The following two definitions of Bernoulli polynomials are equivalent:

I. Generating function:

0

( )1 !

xt n

ntn

te tB xe n

∞

=

=− ∑

II. Appell sequence with zero mean:

0(i) ( ) 1B x =

1(ii) '( ) ( 1) ( )n nB x n B x+ = +

1

0

1 if 0(iii) ( )

0 if 0n

nB x dx

n=⎧

= ⎨ >⎩∫

27

Explicit construction of Bernoulli polynomialsas an Appell sequence:

1 0 1 0 1'( ) ( ) ( ) ( ) 1B x B x B x B x dx dx x C= ⇒ = = = +∫ ∫1

1 1 21 1 10 0

0

1( ) 0 ( ) 0 02

B x dx x C dx x C x⎡ ⎤= ⇒ + = ⇒ + =⎢ ⎥⎣ ⎦∫ ∫

1 1 11 102 2

C C B⇒ + = ⇒ = − =

1 20 1 2

( 1)( ) ...1 1 2

n n nn n

n n nB x B x B x B x B− −−= + + + +

⋅

1 0 1( )B x B x B∴ = +

General formula

22 0 1 2( ) 2B x B x B x B∴ = + +

3 23 0 1 2 4( ) 3 3B x B x B x B x B∴ = + + +

28

Proof of Theorem

0( )

1 !

xt n

ntn

d te d tB xdx e dx n

∞

=

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

∑

( )⇒

( )1 1 1

0 0 00 0

( )n n

n k n kn k k

k k

n nB x dx B x dx B x dx

k k− −

= =

⎡ ⎤⎛ ⎞ ⎛ ⎞= ⋅ =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦∑ ∑∫ ∫ ∫

1 1'

10 0

( ) ( )! ( 1)!

n n

n nn n

t tB x B xn n

+ +∞ ∞

+= =

⇒ =+∑ ∑

'

1

( )1 !

xt n

ntn

te tt B xe n

∞

=

⎛ ⎞⇒ =⎜ ⎟−⎝ ⎠

∑

1 '( ) ( )1

nn

B x B xn+∴ =+

0 0

! !!( )! 1 !( 1)!

n nk k

k k

B Bn nk n k n k k n k= =

= =− − + − +∑ ∑

1 if 0(Recursive formula)

0 if 0nn=⎧

= ⎨ >⎩

29

Hypergeometric Bernoulli Polynomials2

0

/ 2 (2, )1 !

xt n

ntn

t e tB xe t n

∞

=

=− − ∑

0

1

22

3 23

0 : (2, ) 1

11: (2, )3

2 12 : (2, )3 18

1 13: (2, )6 90

n B x

n B x x

n B x x x

n B x x x x

= =

= = −

= = − +

= = − + +

Recursive formula

0

0 0(2, ) , (3)3 4 (3 1) 0!(3) !

nnk

mk n k

mB x xm mk n= −

=⎧= = ⎨ ⋅ ⋅ ⋅ ⋅ + − >⎩

∑

30

We can define hypergeometric Bernoulli numbers as(2) (2,0)n nB B=

-2 -1 1 2

-2

-1.5

-1

-0.5

0.5

1

1.5

2

0.2 0.4 0.6 0.8 1

-0.4

-0.2

0.2

0.4

Interval [-2,2] Interval [0,1]

1( )B x0 ( )B x

3 ( )B x2 ( )B x

31

Weighted means:

0.2 0.4 0.6 0.8 1

-0.4

-0.2

0.2

0.4

1 1 1 210 0 0

13 2

0

(1 ) (2, ) (1 )( 1/3) ( 4 /3 1/3)

/3 2 /3 /3 0

x B x dx x x dx x x dx

x x x

− = − − = − + −

⎡ ⎤= − + − =⎣ ⎦

∫ ∫ ∫

(1 ) (2, )nx B x−

32

Theorem: The following two definitions of hypergeometric Bernoulli polynomials are equivalent:

I. Generating function:

II. Appell sequence with zero first moment:

0(i) ( ) 1B x =

1(ii) '( ) ( 1) ( )n nB x n B x+ = +

1

0

1/ 2 if 0(iii) (1 ) ( )

0 if 0n

nx B x dx

n=⎧

− = ⎨ >⎩∫

2

0

/ 2 (2, )1 !

xt n

ntn

t e tB xe t n

∞

=

=− − ∑

33

Answer to Prize Problem:

= 91,409,924,241,424,243,424,241,924,242,500

( )10 11 10 9 7 5 3

1

1 6 33 55 66 66 33 566

n

kk n n n n n n n

=

= + + − + − +∑1000

10 11 10 9

1

7 5 3

1 (6 1000 33 1000 55 100066

66 1000 66 1000 33 1000 5 1000)k

k=

= ⋅ + ⋅ + ⋅

− ⋅ + ⋅ − ⋅ + ⋅

∑

1 (603305500000000006600000000000500066

66000000000033000000000)1 (6033054999934000065999967000005000)66

=

−

=

34

"With the help of this table it took me less than half of a quarter of an hour to find that the tenth powers of the first 1000 numbers being added together will yield the sum

91409924241424243424241924242500

From this it will become clear how useless was the work of Ismael Bullialdus spent on the compilation of his voluminous Arithmetica Infinitorum in which he did nothing more than compute with immense labor the sums of the first six powers, which is only a part of what we have accomplished in the space of a single page."

35

References[1] Appell, P. E. "Sur une classe de polynomes." Annales d'École Normal Superieur,

Ser. 2 9, 119-144, 1882. [2] K. Dilcher, Bernoulli numbers and confluent hypergeometric functions,

Number Theory for the Millennium, I (Urbana, IL, 2000), 343-363, A K Peters, Natick, MA, 2002.

[3] L. Euler, Institutiones calculi differentialis, St. Petersburg, 1755, reprinted in Opera Omnia Series I vol 10. English translation of Part I by John Blanton,Springer, NY, 2000.

[4] A. Hassen and H. D. Nguyen, Hypergeometric Zeta Functions, Preprint, 2005. Available at the Mathematics ArXiv: http://arxiv.org/abs/math.NT/0509637

[5] F. T. Howard, Numbers Generated by the Reciprocal of e^x-1-x, Math. Comput. 31 (1977) No. 138, 581-598.

[6] D. J. Pengelley, The bridge between the continuous and the discrete viaoriginal sources, in Study the Masters: The Abel-Fauvel Conference, Kristiansand,2002 (eds. Otto Bekken et al), National Center for Mathematics Education,University of Gothenburg, Sweden.

[7] E. Sandifer, How Euler Did It: Bernoulli Numbers, September 2005, MAA Column