Series: Oresme to Euler to $1,000,000

© Joe ConradSolano Community College

December 8, 2012CMC3 Monterey Conferencejoseph.conrad@solano.edu

Series

= 0.3 + 0.03 + 0.003 + 0.0003 + …

= 0.3333…

=

3 3 3 310 100 1000 10000

13

Series

Harmonic Series:

Nicole Oresme (ca. 1323 – 1382)

1

1

n n

1

22n

n

n

Pietro Mengoli (1626 – 1686)

1

( 1) ln2n

nn

2

1

1 ?n n

Jacob Bernoulli (1654 – 1705)

p-Series:

1

1p

n n

Basel Problem

“If anyone finds and communicates to us that which thus far has eluded our efforts, great will be our gratitude.”

- Jacob Bernoulli, 1689

21

1 ?n n

1.6449

Enter Euler! Euler (1707 - 1783)

in 1735 computed the

sum to 20 decimal places.

“Quite unexpectedly I have

found an elegant formula involving the quadrature of the circle.”

2

21

16n n

Euler’s First “Proof”

Recall that if P(x) is a nth degree polynomial with roots a1, a2, …, an, then P(x) can be factored as

for some constant A.

1 2( ) ( )( )(...)( )nP x A x a x a x a

Euler let P(x) be

Note: xP(x) = sin(x), so

So if a is a root of P(x), then sin(a) = 0

which implies that a = ±, ±2, ±3, …

2 4 6 8

( ) 13! 5! 7! 9!x x x xP x

sin( )( )

xP x

x

So, we can factor P(x) as

Letting x = 0, we get B = 1.

2 4 6 8

( ) 13! 5! 7! 9!x x x xP x

( ) ( )( )( 2 )( 2 )P x A x x x x 2 2 2 2 2 2 2 2( )( 4 )( 9 )( 16 )A x x x x

2

21 xB

2 2 2

2 2 21 1 1

4 9 16x x x

2 4 6 8

( ) 13! 5! 7! 9!x x x xP x

2 2 2 2

2 2 2 2( ) 1 1 1 1

4 9 16x x x xP x

2 1 1 1 16 1 4 9 16

12

2x

2 2 2

2 2 24 9 16x x x

13!

2 2 2 21 1 1 1

4 9 16

Extending this argument, Euler got:

In 1750, he generalized this to …

44

1

190n n

66

1

1945n n

2626

1

1315862111094481976030578125n n

But, first!

1 2 3 n ( 1)2

n n

2 2 2 2 ( 1)(2 1)1 2 36

n n nn

2 23 3 3 3 ( 1)1 2 3

4n nn

Bernoulli discovered how to compute these in general:

1

01

111

pnp p j

jjk

pk B njp

1 1 1 12 6 30 42

{ } {1, , ,0, ,0, ,0, }jB

4 3 20 1 2 3

3

1

4 4 4 40 1 2 3

14

n

kB n B n B n B nk

4 3 21 24

n n n

2 2( 1)4

n n

4 3 212

1 11 1 4 6 4 04 6

n n n n

“…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

91 409 924 241 424 243 424 241 924 242 500.”

What about ?

The first 20 Bernoulli numbers:

51 1 1 1 11, , ,0, ,0, ,0, ,0, ,2 6 30 42 30 66

691 7 3617 438670, ,0, ,0, ,0, ,02730 6 510 798

{ }nB

What did Euler know and when?

He knew Bernoulli’s work.

He knew his p-series sums (1735).

He knew the Euler-MacLaurin formula (1732):

11

( ) ( )n n

kf k f x dx

(1) ( )2

f f n

( , )n f pR ( 2 1) ( 2 1)

1

( ) (1)p

k kn

k

A f n f

He knew the Taylor series for many functions.

Somehow, he noticed that the Bernoulli numbers tied these things together.

Appear in Taylor series:

01 !

nn

xn

B xxe n

Euler-Maclaurin became:

11

( ) ( )n n

kf k f x dx

(1) ( )2

f f n

(2 1) (2 1)2

1(1)( )

(2 )!( , )k kk

n

p

k

Bf n f

kf pR

1 2 1 22

21

( 1) 21(2 )!

k k kk

kn

Bkn

1 1 1 2 22( 1) 2

Check 1:(2 1)! 6

Bk

42 1 3 4 1 4304 ( )8( 1) 2

2 :(2 2)! 24 90

Bk

What about ?

Nobody knows the exact sum!

Roger Apéry (1916 – 1994) proved this is irrational in 1977.

31

1n n

31

1 1.202056903n n

Where to next?

Being calculus, we define a function:

This function is defined for all

real x > 1.

1

1( ) xn

xn

Bernhard Riemann (1826 – 1866)

Define a function:

where s complex. 1

1( ) sn

sn

This function can be extended to all the complex numbers except s = 1.

Riemann’s Functional Equation:

Note: , n a natural number

1( ) 2(2 ) ( )! (1 )sin( ), 02

s ss s s s

( 2 ) 0n

Question: Are there any other zeros?

Riemann found three:

½ + 14.1347i

½ + 21.0220i

½ + 25.0109i

The Riemann Hypothesis

All the nontrivial zeros of the zeta

function have real part equal to ½.

Carl Siegel

(1896 – 1981)

What is known?• All nontrivial zeros have 0 < Rez < 1.• If z is a zero, then so is its conjugate.• There are infinitely many zeros on the

critical line.• At least 100 billion zeros have been

found on the critical line.• The first 2 million have been calculated.• This verifies the RH up to a height of

about 29.5 billion.

What is known?

• The 100,000th is ½ + 74,920.8275i. • The 10,000,000,000,000,000,010,000th is

½+1,370,919,909,931,995,309,568.3354i

Andrew Odlyzko

In 2000, the Clay Institute of Mathematics offered a prize for solving the Riemann Hypothesis:

$1,000,000

Main Sources

Julian Havil, Gamma, Princeton University Press, Princeton, NJ, 2003.

William Dunham, Euler: The Master of Us All, MAA, 1999.

Ed Sandifer, How Euler Did It: Bernoulli Numbers, MAA Online, Sept. 2005.