Series: Oresme to Euler to $1,000,000 © Joe Conrad Solano Community College December 8, 2012 CMC 3...
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Transcript of Series: Oresme to Euler to $1,000,000 © Joe Conrad Solano Community College December 8, 2012 CMC 3...
Series: Oresme to Euler to $1,000,000
© Joe ConradSolano Community College
December 8, 2012CMC3 Monterey [email protected]
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.