Honors Combinatorics CMSC-27410 = Math-28410 CMSC-37200 Instructor: - - PowerPoint PPT Presentation

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Honors Combinatorics CMSC-27410 = Math-28410 CMSC-37200 Instructor: - - PowerPoint PPT Presentation

Apr 28, 2023 576 likes •1.33k views

Honors Combinatorics CMSC-27410 = Math-28410 CMSC-37200 Instructor: Laszlo Babai University of Chicago Week 7, Tuesday, May 19, 2020 CMSC-27410=Math-28410 CMSC-3720 Honors Combinatorics Sum of inverse squares 1 n 2 n = 1

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Honors Combinatorics

CMSC-27410 = Math-28410 ∼ CMSC-37200 Instructor: Laszlo Babai University of Chicago Week 7, Tuesday, May 19, 2020

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 2

Sum of inverse squares

∞

  • n=1

1 n2

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 3

Sum of inverse squares

∞

  • n=1

1 n2 < 1 +

∞

  • n=2

1 n(n − 1)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 4

Sum of inverse squares

∞

  • n=1

1 n2 < 1 +

∞

  • n=2

1 n(n − 1) = 1 +

∞

  • n=2
  • 1

n − 1 − 1 n

  • = 2

Theorem (Euler, 1734)

∞

  • n=1

1 n2 = π2 6

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 5

Sum of inverse squares

∞

  • n=1

1 n2 < 1 +

∞

  • n=2

1 n(n − 1) = 1 +

∞

  • n=2
  • 1

n − 1 − 1 n

  • = 2

Theorem (Euler, 1734)

∞

  • n=1

1 n2 = π2 6 ≈ 1.645

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 6

Sum of inverse squares

∞

  • n=1

1 n2 = π2 6 Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Lemma 1 =⇒ lower bound ∞

n=1 1/n2 ≥ π2/6

Proof. 0 < α < π/2 =⇒ α < tan α

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 7

Sum of inverse squares

∞

  • n=1

1 n2 = π2 6 Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Lemma 1 =⇒ lower bound ∞

n=1 1/n2 ≥ π2/6

Proof. 0 < α < π/2 =⇒ α < tan α ∴ Lemma =⇒

m

  • k=1

(2m + 1)2 π2k 2 > m(2m − 1) 3

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 8

Sum of inverse squares

∞

  • n=1

1 n2 = π2 6 Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Lemma 1 =⇒ lower bound ∞

n=1 1/n2 ≥ π2/6

Proof. 0 < α < π/2 =⇒ α < tan α ∴ Lemma =⇒

m

  • k=1

(2m + 1)2 π2k 2 > m(2m − 1) 3

m

  • k=1

1 k 2 > π2 3 · m(2m − 1) (2m + 1)2 → π2 6

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 9

Sum of inverse squares Lemma 2

m

  • k=1

1 sin2

kπ 2m+1

= 2m(m + 1) 3

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 10

Sum of inverse squares Lemma 2

m

  • k=1

1 sin2

kπ 2m+1

= 2m(m + 1) 3 Lemma 2 =⇒ upper bound ∞

n=1 1/n2 ≤ π2/6

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 11

Sum of inverse squares

Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Proof. De Moivre formula (cos α + i sin α)n = cos(nα) + i sin(nα)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 12

Sum of inverse squares

Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Proof. De Moivre formula (cos α + i sin α)n = cos(nα) + i sin(nα) sin(nα) =

  • n

1

  • sin α cosn−1 α−
  • n

3

  • sin3 α cosn−3 α+
  • n

5

  • sin5 α cosn−5

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 13

Sum of inverse squares

Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Proof. De Moivre formula (cos α + i sin α)n = cos(nα) + i sin(nα) sin(nα) =

  • n

1

  • sin α cosn−1 α−
  • n

3

  • sin3 α cosn−3 α+
  • n

5

  • sin5 α cosn−5

cot α = cos α

sin α =⇒

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 14

Sum of inverse squares

Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Proof. De Moivre formula (cos α + i sin α)n = cos(nα) + i sin(nα) sin(nα) =

  • n

1

  • sin α cosn−1 α−
  • n

3

  • sin3 α cosn−3 α+
  • n

5

  • sin5 α cosn−5

cot α = cos α

sin α =⇒

sin(nα) = sinn α

  • n

1

  • cotn−1 α −
  • n

3

  • cotn−3 α +
  • n

5

  • cotn−5 α − . . .
  • CMSC-27410=Math-28410∼CMSC-3720

Honors Combinatorics

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SLIDE 15

Sum of inverse squares

Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Proof. sin(nα) = sinn α

  • n

1

  • cotn−1 α −
  • n

3

  • cotn−3 α +
  • n

5

  • cotn−5 α − . . .
  • n := 2m + 1

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 16

Sum of inverse squares

Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Proof. sin(nα) = sinn α

  • n

1

  • cotn−1 α −
  • n

3

  • cotn−3 α +
  • n

5

  • cotn−5 α − . . .
  • n := 2m + 1

P(x) :=

  • 2m + 1

1

  • xm −
  • 2m + 1

3

  • xm−1 +
  • 2m + 1

5

  • xm−2 − . . .

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 17

Sum of inverse squares

Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Proof. sin(nα) = sinn α

  • n

1

  • cotn−1 α −
  • n

3

  • cotn−3 α +
  • n

5

  • cotn−5 α − . . .
  • n := 2m + 1

P(x) :=

  • 2m + 1

1

  • xm −
  • 2m + 1

3

  • xm−1 +
  • 2m + 1

5

  • xm−2 − . . .

sin((2m + 1)α) = sin2m+1 α · P(cot2 α)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 18

Sum of inverse squares

Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Proof. P(x) :=

  • 2m + 1

1

  • xm −
  • 2m + 1

3

  • xm−1 +
  • 2m + 1

5

  • xm−2 − . . .

sin((2m + 1)α) = sin2m+1 α · P(cot2 α) Roots of P?

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 19

Sum of inverse squares

Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Proof. P(x) :=

  • 2m + 1

1

  • xm −
  • 2m + 1

3

  • xm−1 +
  • 2m + 1

5

  • xm−2 − . . .

sin((2m + 1)α) = sin2m+1 α · P(cot2 α) Roots of P? LHS vanishes at

kπ 2m+1

∴ roots of P are rk = cot2 kπ 2m + 1 (k = 1, . . . , m)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 20

Sum of inverse squares

Lemma 1

m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 Proof. P(x) :=

  • 2m + 1

1

  • xm −
  • 2m + 1

3

  • xm−1 +
  • 2m + 1

5

  • xm−2 − . . .

sin((2m + 1)α) = sin2m+1 α · P(cot2 α) Roots of P? LHS vanishes at

kπ 2m+1

∴ roots of P are rk = cot2 kπ 2m + 1 (k = 1, . . . , m) these are all the roots P(x) = (2m + 1)(x − r1) · · · (x − rm)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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Sum of inverse squares

P(x) :=

  • 2m + 1

1

  • xm −
  • 2m + 1

3

  • xm−1 +
  • 2m + 1

5

  • xm−2 − . . .

P(x) = (2m + 1)(x − r1) · · · (x − rm) rk = cot2 kπ 2m + 1 (k = 1, . . . , m)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 22

Sum of inverse squares

P(x) :=

  • 2m + 1

1

  • xm −
  • 2m + 1

3

  • xm−1 +
  • 2m + 1

5

  • xm−2 − . . .

P(x) = (2m + 1)(x − r1) · · · (x − rm) rk = cot2 kπ 2m + 1 (k = 1, . . . , m) Coeff(xm−1) (2m + 1)(r1 + · · · + rm) = (2m+1

3 ), i.e.,

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 23

Sum of inverse squares

P(x) :=

  • 2m + 1

1

  • xm −
  • 2m + 1

3

  • xm−1 +
  • 2m + 1

5

  • xm−2 − . . .

P(x) = (2m + 1)(x − r1) · · · (x − rm) rk = cot2 kπ 2m + 1 (k = 1, . . . , m) Coeff(xm−1) (2m + 1)(r1 + · · · + rm) = (2m+1

3 ), i.e., m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 24

Sum of inverse squares

P(x) :=

  • 2m + 1

1

  • xm −
  • 2m + 1

3

  • xm−1 +
  • 2m + 1

5

  • xm−2 − . . .

P(x) = (2m + 1)(x − r1) · · · (x − rm) rk = cot2 kπ 2m + 1 (k = 1, . . . , m) Coeff(xm−1) (2m + 1)(r1 + · · · + rm) = (2m+1

3 ), i.e., m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 QED[Lemma 1]

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 25

Sum of inverse squares

P(x) :=

  • 2m + 1

1

  • xm −
  • 2m + 1

3

  • xm−1 +
  • 2m + 1

5

  • xm−2 − . . .

P(x) = (2m + 1)(x − r1) · · · (x − rm) rk = cot2 kπ 2m + 1 (k = 1, . . . , m) Coeff(xm−1) (2m + 1)(r1 + · · · + rm) = (2m+1

3 ), i.e., m

  • k=1

cot2 kπ 2m + 1 = m(2m − 1) 3 QED[Lemma 1] Source: Matoušek – Nešetˇ ril: Invitation to Discrete Mathematics

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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Generating functions: Number partitions

5 = 5 5 = 4 + 1 5 = 3 + 2 5 = 3 + 1 + 1 5 = 2 + 2 + 1 5 = 2 + 1 + 1 + 1 5 = 1 + 1 + 1 + 1 + 1 p(n) = number of partitions of the number n

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 27

Generating functions: Number partitions

5 = 5 5 = 4 + 1 5 = 3 + 2 5 = 3 + 1 + 1 5 = 2 + 2 + 1 5 = 2 + 1 + 1 + 1 5 = 1 + 1 + 1 + 1 + 1 p(n) = number of partitions of the number n E.g., p(5) = 7

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 28

Generating functions: Number partitions

5 = 5 5 = 4 + 1 5 = 3 + 2 5 = 3 + 1 + 1 5 = 2 + 2 + 1 5 = 2 + 1 + 1 + 1 5 = 1 + 1 + 1 + 1 + 1 p(n) = number of partitions of the number n E.g., p(5) = 7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 2 3 5 7 11 15 22 30 42 56 77 101 135

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 29

Generating functions: Number partitions

Theorem (Hardy–Ramanujan 1917) p(n) ∼ 1 4 √ 3n eπ

√ 2n/3

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 30

Generating functions: Number partitions

Theorem (Hardy–Ramanujan 1917) p(n) ∼ 1 4 √ 3n eπ

√ 2n/3

A weak consequence: log-asymptotics ln p(n) ∼ π

  • 2n/3

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 31

Generating functions: Number partitions

Theorem (Hardy–Ramanujan 1917) p(n) ∼ 1 4 √ 3n eπ

√ 2n/3

A weak consequence: log-asymptotics ln p(n) ∼ π

  • 2n/3

We shall prove: ln p(n) ≤ π

  • 2n/3

Asymptotically tight upper bound!

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 32

Number partitions: warm-up

1 + x + x2 + · · · =

∞

  • n=0

xn = 1 1 − x

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 33

Number partitions: warm-up

How many ways to pay $1.30 in nickels, dimes, and quarters?

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 34

Number partitions: warm-up

How many ways to pay $1.30 in nickels, dimes, and quarters? 130 = 5a + 10b + 25c (1 + x5 + x10 + x15 + . . . )(1 + x10 + x20 + x30 + . . . )(1 + x25 + x50 + x75 + . . . )

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 35

Number partitions: warm-up

How many ways to pay $1.30 in nickels, dimes, and quarters? 130 = 5a + 10b + 25c (1 + x5 + x10 + x15 + . . . )(1 + x10 + x20 + x30 + . . . )(1 + x25 + x50 + x75 + . . . ) Solution: coeff of x130 kn = # solutions to 5a + 10b + 25c = n f(x) = ∞

n=0 knxn = (∞ a=0 x5a) ·

∞

b=0 x10b

· ∞

c=0 x25c

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 36

Number partitions: warm-up

How many ways to pay $1.30 in nickels, dimes, and quarters? 130 = 5a + 10b + 25c (1 + x5 + x10 + x15 + . . . )(1 + x10 + x20 + x30 + . . . )(1 + x25 + x50 + x75 + . . . ) Solution: coeff of x130 kn = # solutions to 5a + 10b + 25c = n f(x) = ∞

n=0 knxn = (∞ a=0 x5a) ·

∞

b=0 x10b

· ∞

c=0 x25c

f(x) = 1 1 − x5 · 1 1 − x10 · 1 1 − x25

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 37

Generating functions: Number partitions

5 = 5 = 5 · 1 + 4 · 0 + 3 · 0 + 2 · 0 + 1 · 0 5 = 4 + 1 = 5 · 0 + 4 · 1 + 3 · 0 + 2 · 0 + 1 · 1 5 = 3 + 2 = 5 · 0 + 4 · 0 + 3 · 1 + 2 · 1 + 1 · 0 5 = 3 + 1 + 1 = 5 · 0 + 4 · 0 + 3 · 1 + 2 · 0 + 1 · 2 5 = 2 + 2 + 1 = 5 · 0 + 4 · 0 + 3 · 0 + 2 · 2 + 1 · 1 5 = 2 + 1 + 1 + 1 = 5 · 0 + 4 · 0 + 3 · 0 + 2 · 1 + 1 · 3 5 = 1 + 1 + 1 + 1 + 1 = 5 · 0 + 4 · 0 + 3 · 0 + 2 · 0 + 1 · 5 n = n

k=1 k · ik = ∞ k=1 k · ik

p(n) = # solutions (i1, . . . , in) or (i1, . . . , in, 0, 0, . . . ) P(x) =

∞

  • n=0

p(n)xn = 1 1 − x · 1 1 − x2 · 1 1 − x3 · · · =

∞

  • k=1

1 1 − xk

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 38

Generating functions: Number partitions

p(n) = # solutions (i1, . . . , in) or (i1, . . . , in, 0, 0, . . . ) P(x) =

∞

  • n=0

p(n)xn = 1 1 − x · 1 1 − x2 · 1 1 − x3 · · · =

∞

  • k=1

1 1 − xk converges for |x| < 1

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 39

Generating functions: Number partitions

p(n) = # solutions (i1, . . . , in) or (i1, . . . , in, 0, 0, . . . ) P(x) =

∞

  • n=0

p(n)xn = 1 1 − x · 1 1 − x2 · 1 1 − x3 · · · =

∞

  • k=1

1 1 − xk converges for |x| < 1 Pn(x) :=

n

  • k=1

1 1 − xk

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 40

Generating functions: Number partitions

p(n) = # solutions (i1, . . . , in) or (i1, . . . , in, 0, 0, . . . ) P(x) =

∞

  • n=0

p(n)xn = 1 1 − x · 1 1 − x2 · 1 1 − x3 · · · =

∞

  • k=1

1 1 − xk converges for |x| < 1 Pn(x) :=

n

  • k=1

1 1 − xk = p(0)+p(1)x +· · ·+p(n)xn + positive higher-order terms Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < x < 1 we have p(n)xn < Pn(x)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 41

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

Pn(x) :=

n

  • k=1

1 1 − xk

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 42

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

Pn(x) :=

n

  • k=1

1 1 − xk = p(0)+p(1)x +· · ·+p(n)xn + positive higher-order terms For 0 < x < 1 we have p(n)xn < Pn(x) p(n) < Pn(x) xn

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 43

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

Pn(x) :=

n

  • k=1

1 1 − xk = p(0)+p(1)x +· · ·+p(n)xn + positive higher-order terms For 0 < x < 1 we have p(n)xn < Pn(x) p(n) < Pn(x) xn We choose x so as to minimize RHS

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 44

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

Pn(x) :=

n

  • k=1

1 1 − xk

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 45

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

Pn(x) :=

n

  • k=1

1 1 − xk For 0 < x < 1 we have p(n)xn < Pn(x) p(n) < Pn(x) xn

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 46

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

Pn(x) :=

n

  • k=1

1 1 − xk For 0 < x < 1 we have p(n)xn < Pn(x) p(n) < Pn(x) xn ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 47

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

Pn(x) :=

n

  • k=1

1 1 − xk For 0 < x < 1 we have p(n)xn < Pn(x) p(n) < Pn(x) xn ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk) − ln(1 − y) = y 1 + y 2 2 + y 3 3 + y 4 4 + . . .

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 48

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < x < 1 we have ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk) − ln(1 − y) = y 1 + y 2 2 + y 3 3 + y 4 4 + · · · =

∞

  • j=1

y j j

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-49
SLIDE 49

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < x < 1 we have ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk) − ln(1 − y) = y 1 + y 2 2 + y 3 3 + y 4 4 + · · · =

∞

  • j=1

y j j −

n

  • k=1

ln(1 − xk) =

n

  • k=1

∞

  • j=1

xkj j

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

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SLIDE 50

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < x < 1 we have ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk) − ln(1 − y) = y 1 + y 2 2 + y 3 3 + y 4 4 + · · · =

∞

  • j=1

y j j −

n

  • k=1

ln(1 − xk) =

n

  • k=1

∞

  • j=1

xkj j =

∞

  • j=1

1 j

n

  • k=1

xjk

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-51
SLIDE 51

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < x < 1 we have ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk) −

n

  • k=1

ln(1 − xk) =

n

  • k=1

∞

  • j=1

xkj j =

∞

  • j=1

1 j

n

  • k=1

xjk ≤

∞

  • j=1

1 j

∞

  • k=1

xjk =

∞

  • j=1

1 j · xj 1 − xj

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-52
SLIDE 52

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < x < 1 we have ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk) < −n ln x +

∞

  • j=1

1 j · xj 1 − xj

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-53
SLIDE 53

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < x < 1 we have ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk) < −n ln x +

∞

  • j=1

1 j · xj 1 − xj Next trick: 1 − xj = (1 − x)(1 + x + x2 + · · · + xj−1) ≥ (1 − x)jxj−1 ∴

∞

  • j=1

1 j · xj 1 − xj ≤

∞

  • j=1

1 j · xj (1 − x)jxj−1 = x 1 − x

∞

  • j=1

1 j2

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-54
SLIDE 54

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < x < 1 we have ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk) < −n ln x +

∞

  • j=1

1 j · xj 1 − xj Next trick: 1 − xj = (1 − x)(1 + x + x2 + · · · + xj−1) ≥ (1 − x)jxj−1 ∴

∞

  • j=1

1 j · xj 1 − xj ≤

∞

  • j=1

1 j · xj (1 − x)jxj−1 = x 1 − x

∞

  • j=1

1 j2 = π2 6 · x 1 − x

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-55
SLIDE 55

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < x < 1 we have ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk) < −n ln x + π2 6 · x 1 − x New variable u := x/(1 − x)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-56
SLIDE 56

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < x < 1 we have ln p(n) < −n ln x −

n

  • k=1

ln(1 − xk) < −n ln x + π2 6 · x 1 − x New variable u := x/(1 − x) so 0 < u < ∞ x = u/(1 + u) ∴ ln p(n) < n ln

  • 1 + 1

u

  • + π2

6 u

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-57
SLIDE 57

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < u < ∞ we have ln p(n) < n u + π2 6 u

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-58
SLIDE 58

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < u < ∞ we have ln p(n) < n u + π2 6 u When is RHS minimal?

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-59
SLIDE 59

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < u < ∞ we have ln p(n) < n u + π2 6 u When is RHS minimal? a + b 2 ≥ √ ab

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-60
SLIDE 60

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < u < ∞ we have ln p(n) < n u + π2 6 u When is RHS minimal? a + b 2 ≥ √ ab arithmetic mean ≥ geometric mean equality holds when

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-61
SLIDE 61

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < u < ∞ we have ln p(n) < n u + π2 6 u When is RHS minimal? a + b 2 ≥ √ ab arithmetic mean ≥ geometric mean equality holds when a = b RHS min when n/u = π2u/6, and then min RHS = 2

  • (n/u) · (π2u/6) = π

√ 2n/3

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-62
SLIDE 62

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < u < ∞ we have ln p(n) < n u + π2 6 u When is RHS minimal? a + b 2 ≥ √ ab arithmetic mean ≥ geometric mean equality holds when a = b RHS min when n/u = π2u/6, and then min RHS = 2

  • (n/u) · (π2u/6) = π

√ 2n/3 ∴

ln p(n) < π

  • 2n/3

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-63
SLIDE 63

Generating functions: Number partitions

Our goal: p(n) < eπ

√ 2n/3 — an upper bound

For 0 < u < ∞ we have ln p(n) < n u + π2 6 u When is RHS minimal? a + b 2 ≥ √ ab arithmetic mean ≥ geometric mean equality holds when a = b RHS min when n/u = π2u/6, and then min RHS = 2

  • (n/u) · (π2u/6) = π

√ 2n/3 ∴

ln p(n) < π

  • 2n/3

Source: Matoušek – Nešetˇ ril

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-64
SLIDE 64

random number puzzle

Pick two positive integers, x and y at random What is the probability that gcd(x, y) = 1 ?

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-65
SLIDE 65

random number puzzle

Pick two positive integers, x and y at random What is the probability that gcd(x, y) = 1 ? What does this question mean?

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-66
SLIDE 66

random number puzzle

Pick two positive integers, x and y at random What is the probability that gcd(x, y) = 1 ? What does this question mean? limn→∞ P(gcd(x, y) = 1 | x, y ∈ [n])

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-67
SLIDE 67

random number puzzle

Pick two positive integers, x and y at random What is the probability that gcd(x, y) = 1 ? What does this question mean? limn→∞ P(gcd(x, y) = 1 | x, y ∈ [n]) limit = 1? (Yes) limit = zero? (No) inbetween? (like)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-68
SLIDE 68

random number puzzle

Pick two positive integers, x and y at random What is the probability that gcd(x, y) = 1 ? What does this question mean? limn→∞ P(gcd(x, y) = 1 | x, y ∈ [n]) limit = 1? (Yes) limit = zero? (No) inbetween? (like) Cannot be 1: P(both x and y are even) =

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-69
SLIDE 69

random number puzzle

Pick two positive integers, x and y at random What is the probability that gcd(x, y) = 1 ? What does this question mean? limn→∞ P(gcd(x, y) = 1 | x, y ∈ [n]) limit = 1? (Yes) limit = zero? (No) inbetween? (like) Cannot be 1: P(both x and y are even) =1/4

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-70
SLIDE 70

random number puzzle

Pick two positive integers, x and y at random What is the probability that gcd(x, y) = 1 ? What does this question mean? limn→∞ P(gcd(x, y) = 1 | x, y ∈ [n]) limit = 1? (Yes) limit = zero? (No) inbetween? (like) Cannot be 1: P(both x and y are even) =1/4 less than 1/2 (Yes) more than 1/2 (No) exactly 1/2 (like)

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-71
SLIDE 71

random number puzzle

Pick two positive integers, x and y at random What is the probability that gcd(x, y) = 1 ? What does this question mean? limn→∞ P(gcd(x, y) = 1 | x, y ∈ [n]) limit = 1? (Yes) limit = zero? (No) inbetween? (like) Cannot be 1: P(both x and y are even) =1/4 less than 1/2 (Yes) more than 1/2 (No) exactly 1/2 (like)

prob = 0.607927. . .

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-72
SLIDE 72

random number puzzle

Pick two positive integers, x and y at random What is the probability that gcd(x, y) = 1 ? What does this question mean? limn→∞ P(gcd(x, y) = 1 | x, y ∈ [n]) limit = 1? (Yes) limit = zero? (No) inbetween? (like) Cannot be 1: P(both x and y are even) =1/4 less than 1/2 (Yes) more than 1/2 (No) exactly 1/2 (like)

prob = 0.607927. . . = 6 π2

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics

slide-73
SLIDE 73

random number puzzle

Pick two positive integers, x and y at random What is the probability that gcd(x, y) = 1 ? What does this question mean? limn→∞ P(gcd(x, y) = 1 | x, y ∈ [n]) limit = 1? (Yes) limit = zero? (No) inbetween? (like) Cannot be 1: P(both x and y are even) =1/4 less than 1/2 (Yes) more than 1/2 (No) exactly 1/2 (like)

prob = 0.607927. . . = 6 π2 DO: Assume ∃ lim. Prove: it must be 6/π2

CMSC-27410=Math-28410∼CMSC-3720 Honors Combinatorics