The Bernoulli Numbers

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March 16, 2021 Outline

1 Sm(n) and its History

2 The Bernoulli Method

3 Euler’s Impact and Applications

Savage The Bernoulli Numbers Chapter 1

Sm(n) and its History

Savage The Bernoulli Numbers Sums of Powers of Integers

Deﬁnition Let n, m ∈ N with n ≥ 1. The sums of powers of integers, denoted Sm(n), is

n X m m m m m Sm(n) := k = 1 + 2 + 3 + ··· + n . k=1

*Historically, ﬁnding quick ways of computing these sums were initially intriguing. *Applications in physics, engineering, and possibly in scripts for games such as Dota 2.

Savage The Bernoulli Numbers Sums of Powers of Integers

Example (Forms you may have seen)

If m = 0, then n X 0 0 0 0 S0(n) = k = 1 + 2 + ··· + n = n. k=1 If m = 1, then n X n + 1 S (n) = k1 = 1 + 2 + ··· + n = . 1 2 k=1

Savage The Bernoulli Numbers Historical Questions

Big Question: Can we ﬁnd a “nice” formula for the sums of powers of integers? That is, is there a concise way of computing

n X m m m m m Sm(n) = k = 1 + 2 + 3 + ··· + n ? k=1

Savage The Bernoulli Numbers Historical Questions

Example (Geometric Sums) Let m ≥ 1. Then for b > 0 with b 6= 1, we know that bm − 1 b0 + b1 + b2 + b3 + ··· + bm−1 = b − 1

Savage The Bernoulli Numbers Historical Questions

Big Question: Can we ﬁnd a “nice” formula for the sums of powers of integers? That is, is there a concise way of computing

n X m m m m m Sm(n) := k = 1 + 2 + 3 + ··· + n ? k=1

We have know of nice formulas for m = 0 and m = 1, but what if m > 1? Is each formula related somehow or is each independent?

Savage The Bernoulli Numbers Historical Questions

Example (Related?) We know n + 1 S (n) = n and S (n) = , 0 1 2 but are these formulas (and others) related somehow?

Savage The Bernoulli Numbers History: Archimedes (287-212 BC) of Syracuse, Sicily

Our story begins with Archimedes. He is credited with ﬁrst discovering that

n + 1 S (n) = . 1 2

It’s also believed Archimedes also discovered a nice formula for S2(n).

Savage The Bernoulli Numbers History: Aryabhata (476-550) of Kusumapura?, India

Aryabhata was an Indian astronomer. He is credited with discovering a nice formula for S3(n), namely

2 S3(n) = (1 + 2 + ··· + n) .

n + 12 n2(n + 1)2 ⇒ S (n) = = 3 2 4

Savage The Bernoulli Numbers History: Abu Ali al-Hassan ibn al Haytham (965-1039) of Basra, Iraq

Commonly referred to as Alhazan. Ar- guably discovered that 1 1 1 1 S (n) = n5 − n4 + n3 − n. 4 5 2 3 30

Savage The Bernoulli Numbers History: Thomas Harriot (1560-1621) of Oxford, U.K. and Pierre De Fermat (1601-1665) of Beaumont-de-Lomagne, France

Both independently “proved” 1 1 1 1 S (n) = n5 − n4 + n3 − n. 4 5 2 3 30

*Given Sm(n), Fermat developed a recur- sive formula using integration to compute

Sm+1(n).

Savage The Bernoulli Numbers History: Johann Faulhaber (1580-1635) of Ulm, Germany

In 1610, found nice formulas for

Sm(n) where

0 ≤ m ≤ 10.

In 1631, he published in Academia

Algebrae nice formulas for Sm(n) where

0 ≤ m ≤ 23.

Savage The Bernoulli Numbers The Faulhaber Method

n+1 Given n ≥ 1, let α = S1(n) = 2 . Theorem (Fact - Key Observation)

If m ≥ 1 is odd, then Sm(n) can be expressed as rational polynomial in α.

Savage The Bernoulli Numbers The Faulhaber Method

Example n+1 Given n ≥ 1, let α = 2 .

1 m = 1 ⇒ S1(n) = α. 2 2 2 m = 3 ⇒ S3(n) = (1 + 2 + ··· + n) = α .

3 4 3 1 2 m = 5 ⇒ S5(n) = 3 α − 3 α . 4 m = 13 ⇒ 64 80 656 944 2764 691 S (n) = α7 − α6 + α5 − α4 + α3 − α2. 13 7 3 15 21 105 105

Savage The Bernoulli Numbers The Faulhaber Method

Example (Continued) Suppose m = 13 and n = 4. Then

13 13 13 13 S13(4) = 1 + 2 + 3 + 4 = 68, 711, 380 .

4+1 In this case, α = 2 = 10 and 64 80 656 944 2764 691 (10)7 − (10)6 + (10)5 − (10)4 + (10)3 − (10)2 7 3 15 21 105 105

is indeed 68, 711, 380 .

Savage The Bernoulli Numbers Chapter 2

The Bernoulli Method

Savage The Bernoulli Numbers Jacob Bernoulli (1655-1705) of Basel, Switzerland

Eﬀectively used Fermat’s methods to compute Sm(n) for

1 ≤ m ≤ 10.

Savage The Bernoulli Numbers Bernoulli’s Sums, b = n − 1

1 2 1 1 S1(b) = 2 n − 2 n 1 3 1 2 1 1 S2(b) = 3 n − 2 n + 6 n 1 4 1 3 1 2 S3(b) = 4 n − 2 n + 4 n 1 5 1 4 1 3 1 1 S4(b) = 5 n − 2 n + 3 n − 30 n 1 6 1 5 5 4 1 2 S5(b) = 6 n − 2 n + 12 n − 12 n 1 7 1 6 1 5 1 3 1 1 S6(b) = 7 n − 2 n + 2 n − 6 n + 42 n 1 8 1 7 7 6 7 4 1 2 S7(b) = 8 n − 2 n + 12 n − 24 n + 12 n 1 9 1 8 2 7 7 5 2 3 1 1 S8(b) = 9 n − 2 n + 3 n − 15 n + 9 n − 30 n 1 10 1 9 3 8 7 6 1 4 3 2 S9(b) = 10 n − 2 n + 4 n − 10 n + 2 n − 20 n

Savage The Bernoulli Numbers 1 h 2 1i S1(b) = 2 n − 1n

1 h 3 3 2 1 1i S2(b) = 3 n − 2 n + 2 n

1 h 4 3 2i S3(b) = 4 n − 2n + 1n

1 h 5 5 4 5 3 1 1i S4(b) = 5 n − 2 n + 3 n − 6 n

1 h 6 5 5 4 1 2i S5(b) = 6 n − 3n + 2 n − 2 n

1 h 7 7 6 7 5 7 3 1 1i S6(b) = 7 n − 2 n + 2 n − 6 n + 6 n

1 h 8 7 14 6 7 4 2 2i S7(b) = 8 n − 4n + 3 n − 3 n + 3 n

1 h 9 9 8 7 21 5 3 3 1i S8(b) = 9 n − 2 n + 6n − 5 n + 2n − 10 n

1 h 10 9 15 8 6 4 3 2i S9(b) = 10 n − 5n + 2 n − 7n + 5n − 2 n

Savage The Bernoulli Numbers 1 h 2 −1 1i 2 n + (2)( 2 )n

1 h 3 −1 2 1 1i 3 n + (3)( 2 )n + (3)( 6 )n

1 h 4 −1 3 1 2i 4 n + (4)( 2 )n + (6)( 6 )n

1 h 5 −1 4 1 3 −1 1i 5 n + (5)( 2 )n + (10)( 6 )n + (5)( 30 )n

1 h 6 −1 5 1 4 −1 2i 6 n + (6)( 2 )n + (15)( 6 )n + (15)( 30 )n

1 h 7 −1 6 1 5 −1 3 1 1i 7 n + (7)( 2 )n + (21)( 6 )n + (35)( 30 )n + (7)( 42 )n

1 h 8 −1 7 1 6 −1 4 1 2i 8 n + (8)( 2 )n + (28)( 6 )n + (70)( 30 )n + (28)( 42 )n

1 h 9 −1 8 1 7 −1 5 1 3 −1 1i 9 n + (9)( 2 )n + (36)( 6 )n + (126)( 30 )n + (84)( 42 )n + (9)( 30 )n

1 h 10 −1 9 1 8 −1 6 1 4 −1 2i 10 n + (10)( 2 )n + (45)( 6 )n + (210)( 30 )n + (210)( 42 )n + (45)( 30 )n

Savage The Bernoulli Numbers 1 h 2 −1 1i 2 1n + (2)( 2 )n

1 h 3 −1 2 1 1i 3 1n + (3)( 2 )n + (3)( 6 )n

1 h 4 −1 3 1 2i 4 1n + (4)( 2 )n + (6)( 6 )n

1 h 5 −1 4 1 3 −1 1i 5 1n + (5)( 2 )n + (10)( 6 )n + (5)( 30 )n

1 h 6 −1 5 1 4 −1 2i 6 1n + (6)( 2 )n + (15)( 6 )n + (15)( 30 )n

1 h 7 −1 6 1 5 −1 3 1 1i 7 1n + (7)( 2 )n + (21)( 6 )n + (35)( 30 )n + (7)( 42 )n

1 h 8 −1 7 1 6 −1 4 1 2i 8 1n + (8)( 2 )n + (28)( 6 )n + (70)( 30 )n + (28)( 42 )n

1 h 9 −1 8 1 7 −1 5 1 3 −1 1i 9 1n + (9)( 2 )n + (36)( 6 )n + (126)( 30 )n + (84)( 42 )n + (9)( 30 )n

1 h 10 −1 9 1 8 −1 6 1 4 −1 2i 10 1n + (10)( 2 )n + (45)( 6 )n + (210)( 30 )n + (210)( 42 )n + (45)( 30 )n

Savage The Bernoulli Numbers First six rows

Row # n1 n2 n3 n4 n5 n6 n7 (1) 2 1 (2) 3 3 1 (3) 6 4 1 (4) 5 10 5 1 (5) 15 15 6 1 (6) 7 35 21 7 1

*Looks like Pascal’s triangle! *(1655)

Savage The Bernoulli Numbers First six rows completed

Row # n1 n2 n3 n4 n5 n6 n7 (1) 2 1 (2) 3 3 1 (3) 4 6 4 1 (4) 5 10 10 5 1 (5) 6 15 20 15 6 1 (6) 7 21 35 35 21 7 1

Savage The Bernoulli Numbers Bernoulli Sums with Pascal numbers

1 h 2 −1 1i 2 1n + (2)( 2 )n

1 h 3 −1 2 1 1i 3 1n + (3)( 2 )n + (3)( 6 )n

1 h 4 −1 3 1 2 1i 4 1n + (4)( 2 )n + (6)( 6 )n + (4)(0)n

1 h 5 −1 4 1 3 2 −1 1i 5 1n + (5)( 2 )n + (10)( 6 )n + (10)(0)n + (5)( 30 )n

1 h 6 −1 5 1 4 3 −1 2 1i 6 1n + (6)( 2 )n + (15)( 6 )n + (20)(0)n + (15)( 30 )n + (6)(0)n

1 h 7 −1 6 1 5 4 −1 3 2 1 1i 7 1n + (7)( 2 )n + (21)( 6 )n + (35)(0)n + (35)( 30 )n + (21)(0)n + (7)( 42 )n

1 h 8 −1 7 1 6 5 −1 4 3 1 2 1i 8 1n + (8)( 2 )n + (28)( 6 )n + (56)(0)n + (70)( 30 )n + (56)(0)n + (28)( 42 )n + (8)(0)n

1 h 9 −1 8 1 7 6 −1 5 4 1 3 2 −1 1i 9 1n + (9)( 2 )n + (36)( 6 )n + (84)(0)n + (126)( 30 )n + (126)(0)n + (84)( 42 )n + (36)(0)n + (9)( 30 )n

1 h 10 −1 9 1 8 7 −1 6 5 1 4 3 −1 2 1i 10 1n + (10)( 2 )n + (45)( 6 )n + (120)(0)n + (210)( 30 )n + (252)(0)n + (210)( 42 )n + (120)(0)n + (45)( 30 )n + (10)(0)n *Each red and yellow number is a Pascal number.

Savage The Bernoulli Numbers Bernoulli Sums with Binomial Coeﬃcients

1 h2 2 2 −1 1i 2 0 (1)n + 1 ( 2 )n

1 h3 3 3 −1 2 3 1 1i 3 0 (1)n + 1 ( 2 )n + 2 ( 6 )n

1 h4 4 4 −1 3 4 1 2 4 1i 4 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n

1 h5 5 5 −1 4 5 1 3 5 2 5 −1 1i 5 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n

1 h6 6 6 −1 5 6 1 4 6 3 6 −1 2 6 1i 6 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n

1 h7 7 7 −1 6 7 1 5 7 4 7 −1 3 7 2 7 1 1i 7 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n

1 h8 8 8 −1 7 8 1 6 8 5 8 −1 4 8 3 8 1 2 8 1i 8 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n + 7 (0)n

1 h9 9 9 −1 8 9 1 7 9 6 9 −1 5 9 4 9 1 3 9 2 9 −1 1i 9 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n + 7 (0)n + 8 ( 30 )n

1 h10 10 10 −1 9 10 1 8 10 7 10 −1 6 10 5 10 1 4 10 3 10 −1 2 10 1i 10 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n + 7 (0)n + 8 ( 30 )n + 9 (0)n *These were not the numbers Bernoulli was concerned with.

Savage The Bernoulli Numbers Bernoulli Sums with Binomial Coeﬃcients

1 h2 2 2 −1 1i 2 0 (1)n + 1 ( 2 )n

1 h3 3 3 −1 2 3 1 1i 3 0 (1)n + 1 ( 2 )n + 2 ( 6 )n

1 h4 4 4 −1 3 4 1 2 4 1i 4 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n

1 h5 5 5 −1 4 5 1 3 5 2 5 −1 1i 5 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n

1 h6 6 6 −1 5 6 1 4 6 3 6 −1 2 6 1i 6 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n

1 h7 7 7 −1 6 7 1 5 7 4 7 −1 3 7 2 7 1 1i 7 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n

1 h8 8 8 −1 7 8 1 6 8 5 8 −1 4 8 3 8 1 2 8 1i 8 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n + 7 (0)n

1 h9 9 9 −1 8 9 1 7 9 6 9 −1 5 9 4 9 1 3 9 2 9 −1 1i 9 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n + 7 (0)n + 8 ( 30 )n

1 h10 10 10 −1 9 10 1 8 10 7 10 −1 6 10 5 10 1 4 10 3 10 −1 2 10 1i 10 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n + 7 (0)n + 8 ( 30 )n + 9 (0)n Instead, Bernoulli was more interested in the orange numbers.

Savage The Bernoulli Numbers Bernoulli’s Deﬁnition

Bernoulli believed that the pattern of orange numbers would continue and be consistent. Thus, Bernoulli deﬁned the Bernoulli numbers as follows: Deﬁnition (Informal - Bernoulli Numbers)

A Bernoulli number Bk is a number appearing in the sequence

(B0, B1, B2,... )

where B0 = 1 and for k ∈ N, Bk+1 is the next orange number appearing after Bk .

Savage The Bernoulli Numbers One Last Look at Bernoulli Sums

1 h2 2 2 −1 1i S1(b) = 2 0 (1)n + 1 ( 2 )n

1 h3 3 3 −1 2 3 1 1i S2(b) = 3 0 (1)n + 1 ( 2 )n + 2 ( 6 )n

1 h4 4 4 −1 3 4 1 2 4 1i S3(b) = 4 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n

1 h5 5 5 −1 4 5 1 3 5 2 5 −1 1i S4(b) = 5 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n

1 h6 6 6 −1 5 6 1 4 6 3 6 −1 2 6 1i S5(b) = 6 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n

1 h7 7 7 −1 6 7 1 5 7 4 7 −1 3 7 2 7 1 1i S6(b) = 7 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n

1 h8 8 8 −1 7 8 1 6 8 5 8 −1 4 8 3 8 1 2 8 1i S7(b) = 8 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n + 7 (0)n

1 h9 9 9 −1 8 9 1 7 9 6 9 −1 5 9 4 9 1 3 9 2 9 −1 1i S8(b) = 9 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n + 7 (0)n + 8 ( 30 )n

Savage The Bernoulli Numbers Bernoulli Sum Observations

Consider S6(b) below:

1 h7 7 7 −1 6 7 1 5 7 4 7 −1 3 7 2 7 1 1i S6(b) = 7 0 (1)n + 1 ( 2 )n + 2 ( 6 )n + 3 (0)n + 4 ( 30 )n + 5 (0)n + 6 ( 42 )n

1 Denominator of the leading fraction is m + 1.

2 m+1 The binomial coeﬃcients are of the form k where 0 ≤ k ≤ m.

3 S6(b) is a polynomial in n whose powers descend from m + 1 to 1.

Savage The Bernoulli Numbers The Bernoulli Conjecture

Conjecture (Bernoulli (1713)) If m, n are integers with m ≥ 1 and n ≥ 2, then

m 1 X m + 1 S (n − 1) = B nm+1−k m m + 1 k k k=0

th where Bk is the k Bernoulli number.

*Bernoulli died 1705. His nephew Niklaus Bernoulli published Ars Conjectandi.

Savage The Bernoulli Numbers Chapter 3 Continued

Bernoulli vs. Takakazu, a Leibniz vs. Newton sequel

Savage The Bernoulli Numbers Seki Takakazu (1642-1708) of Fujioka Gumma, Japan

Sometimes referred to as Seki Kowa.¯ Samurai Family

Is accredited with discovering the Bernoulli numbers in 1712.

Savage The Bernoulli Numbers Katsuy¯oSanp¯o(1712)

*Published by two of his students after death.

Savage The Bernoulli Numbers Takakazu numbers?

Publications after Death Europe - Math hub Unconventional Methods: Ruisai Shosa-ho Language/time barrier

Savage The Bernoulli Numbers Chapter 3

Euler’s Impact and Applications

Savage The Bernoulli Numbers Leonard Euler (1707-1783) of Basel, Switzerland

Sought to formalize Bernoulli’s work

Savage The Bernoulli Numbers One last Problem

In my opinion we should not credit Bernoulli with proving

m 1 X m + 1 S (n − 1) = B nm+1−k . m m + 1 k k k=0 Although Fermat’s method of integrals inductively proves

1 Sm(n − 1) will be a polynomial in n with no constant term and that

2 deg(Sm(n − 1)) = m + 1,

it was not clear that Fermat’s method does not diverge the Bernoulli (orange) numbers at any point.

Savage The Bernoulli Numbers Euler’s Formalization

Let E(x) be the real-valued function given by  1, if x = 0 E(x) = . x  ex −1 , if x 6= 0

*Continuously extending the function x/(ex − 1).

Savage The Bernoulli Numbers Euler’s Formalization

Proposition The function E(x) has a power series representation which converges for all x ∈ R. Equivalently, there exist ai ∈ R such that if x ∈ , then R ∞ X k E(x) = ak x . k=0

*a0 is indeed 1. *Euler recognized consistencies between ai above and the Bernoulli (orange) numbers from Jacob Bernoulli.

Savage The Bernoulli Numbers Example (Facts)

a0 = E(0) = 1 = 1/0! = B0/0! 0 a1 = E (0) = −1/2 = (−1/2)/1! = B1/1! 00 a2 = E (0) = 1/12 = (1/6)/2! = B2/2!

Savage The Bernoulli Numbers Euler’s Deﬁnition

Deﬁnition (Bernoulli numbers, Euler - 1755)) th Let k ∈ N. The k Bernoulli number, denoted Bk , is ak · k! where th ak is the k coeﬃcient in the power-series representation of E(x).

Note this implies ∞ X Bk E(x) = xk . k! k=0

Savage The Bernoulli Numbers Deﬁnition (Bernoulli Polynomials) th Let k ∈ N. The k Bernoulli polynomial, denoted Bk (x), is the real polynomial k X k B (x) := B xk−n. k n n n=0

Example

B0(x) = 1 1 B (x) = x − 1 2 1 B (x) = x2 − x + 2 6 3 1 B (x) = x3 − x2 + x 3 2 2

Savage The Bernoulli Numbers Bernoulli Polynomials

Notice that

0 B1(x) = 1 = B0(x) 0 B2(x) = 2x − 1 = 2B1(x) 1 B0 (x) = 3x2 − 3x + = 3B (x) 3 2 2

Savage The Bernoulli Numbers Interesting Result

Theorem (Proposition 7.2 in [1])

Let gk (x) ∈ R[k][x] (The coeﬃcients of g(x) are polynomials in k). Then gk (x) = Bk (x) if and only if gk (x) satisﬁes the following properties:

1 g0(x) = 1;

2 0 k ≥ 1 ⇒ gk (x) = k · gk−1(x); 1 3 R k ≥ 1 ⇒ 0 gk (x)dx = 0.

Savage The Bernoulli Numbers Application 1: Euler-Maclaurin Summation Formula (EMSF)

*Both Euler and Maclaurin (1698-1746) derived the following formula independently around 1735. Theorem (EMSF, c.1735) Let m, a, b ∈ Z with m ≥ 1 and a < b. Let f :[a, b] → R be at least m-times diﬀerentiable. Then

b−1 Z b " m # b X X Bk (k−1) f (i) = f (x)dx + f (x) + Rm k! i=a a k=1 a where m+1 Z b (−1) (m) Rm = Bm(x − bxc)f (x)dx. m! a

Savage The Bernoulli Numbers Why do we care?

Pb−1 Consider the LHS of the EMSF: i=a f (i). Suppose a = 0, b = n ≥ 2, and f (x) = xm where m ≥ 1. Plugging these values in gives us n−1 n−1 X m X m i = i = Sm(n − 1). i=0 i=1

Savage The Bernoulli Numbers Before introducing its main application, we need two propositions. Lemma If m, k ∈ N with m ≥ k, then 1 m + 1 (m)(m − 1) ··· (m − k + 2) = . m + 1 k k!

Proof. This follows immediately from the fact that

m + 1 (m + 1)(m)(m − 1) ··· (m − k + 2) = . k k!

Now divide both sides by m + 1.

Savage The Bernoulli Numbers Proposition 1

Proposition m If m, k ∈ N with m ≥ k ≥ 1 and f (x) = x , then k! m + 1 f (k−1)(x) = xm+1−k . m + 1 k

Moreover, f (m)(x) = m!.

Proof. One can easily verify the second statement and that

f (k−1)(x) = (m)(m − 1)(m − 2) ··· (m − k + 2)xm+1−k .

Now apply the previous lemma.

Savage The Bernoulli Numbers Proposition 2

Proposition th Let n ∈ Z and Bk (x) be the k Bernoulli polynomial. Then Z n+1 Z 1 Bk (x − bxc)dx = Bk (x)dx. n 0

*Numerically this seems to hold

Savage The Bernoulli Numbers Application 2: The Big Corollary

Corollary (Euler, Maclaurin) If m, n ∈ Z with m ≥ 1 and n ≥ 2, then m 1 X m + 1 S (n − 1) = B nm+1−k m m + 1 k k k=0

th where Bk is the k Bernoulli number.

Savage The Bernoulli Numbers Proof of Corollary

We follow the proof in [2]. Let a = 0, b = n, and f (x) = xm. Note f is inﬁnitely diﬀerentiable and that f (m)(x) = m!. Plugging these

into Rm as in the EMSF, we see that

m+1 Z b (−1) (m) Rm = Bm(x − bxc)f (x)dx m! a Z n m+1 = (−1) Bm(x − bxc)dx 0 "n−1 Z i+1 # m+1 X = (−1) Bm(x − bxc)dx i=0 i "n−1 Z 1 # m+1 X = (−1) Bm(x)dx i=0 0 = 0.

Thus, Rm = 0!

Savage The Bernoulli Numbers Proof of Corollary Continued

Applying the rest of the EMSF now shows that

n−1 Z n " m # n X m m X Bk (k−1) i = x dx + f (x) + 0. k! i=0 0 k=1 0 Thus.

n−1 Z n " m # n X m m X Bk k! m + 1 m+1−k i = Sm(n − 1)= x dx + x k! m + 1 k i=0 0 k=1 0 " m # xn+1 1 X m + 1 = B + B nm+1−k m + 1 0 m + 1 k k k=1 " m # 1 X m + 1 = B nm+1−k . m + 1 k k k=0 Changing n → n + 1, we have that

" m # 1 X m + 1 S (n) = B (n + 1)m+1−k m m + 1 k k k=0 where this holds for all m, n ≥ 1.

Savage The Bernoulli Numbers One Last Application: Bernoulli Matrices

Let n, α ∈ N with α ≥ 1. A Bernoulli Matrix of order n, denoted B, is the (n + 1) × (n + 1) matrix where  i  j Bi−j , if i ≥ j Bi,j = 0, otherwise

and 0 ≤ i, j ≤ n. *Lower-triangular matrices

Savage The Bernoulli Numbers One Last Application: Bernoulli Matrices

Example If n = 1, then

0    0 B0 0 1 0 B =   =   . 1 1 −1 0 B1 1 B0 2 1

Savage The Bernoulli Numbers One Last Application: Bernoulli Matrices

Deﬁnition Let D be the (n + 1) × (n + 1) matrix where  i (i−j)!  j (i−j+1)! , if i ≥ j Di,j = 0, otherwise

Example Let n = 1. Then   1 0 D =   1 2 1

Savage The Bernoulli Numbers One Last Application: Bernoulli Matrices

Theorem (Theorem 2 in [3]) If n ≥ 1, then BD = DB = I .

Example If n = 1,Then       1 0 1 0 0 1 BD =   ·   =   = I . −1 1 2 1 2 1 1 0

*It does follow that DB = I as well.

Savage The Bernoulli Numbers Referencesi

[1] N. Larson,“ The Bernoulli Numbers: A Brief Primer,”, 2019. [2] D. Buenger,“ WHAT ARE THE BERNOULLI NUMBERS?,” [3] N. Tuglu,“ Bernoulli Matrices and Their Some Properties,”, 2015.

Savage The Bernoulli Numbers