Harmonic Sums - Lecture 11

Harmonic SumsLecture 11

Justin Stevens

Justin Stevens Harmonic Sums (Lecture 11) 1 / 25

Outline

1 Harmonic SumsDivergenceEuler-Mascheroni ConstantReciprocals of Primesp-adic Valuation

2 Shuffling Deck of Cards

3 References

Divergent Harmonic Series

Definition. The harmonic series is the divergent infinite series:∞∑

1k = 1

1 + 12 + 1

3 + 14 + 1

5 + · · · .

One way to prove this series is divergent is the comparison test:

1 + 12 + 1

3 + 14 + 1

5 + 16 + 1

7 + 18 + 1

9 + · · ·

>1 + 12 + 1

4 + 14︸︷︷︸

+ 18 + 1

8 + 18 + 1

8︸︷︷︸1/2

+ 116 + · · · .

=1 + 12 + 1

2 + 12 + · · · =∞.

1k = 1

1 + 12 + 1

3 + 14 + 1

5 + · · · .

1 + 12 + 1

3 + 14 + 1

5 + 16 + 1

7 + 18 + 1

9 + · · ·

>1 + 12 + 1

4 + 14︸︷︷︸

+ 18 + 1

8 + 18 + 1

8︸︷︷︸1/2

+ 116 + · · · .

=1 + 12 + 1

2 + 12 + · · · =∞.

1k = 1

1 + 12 + 1

3 + 14 + 1

5 + · · · .

1 + 12 + 1

3 + 14 + 1

5 + 16 + 1

7 + 18 + 1

9 + · · ·

>1 + 12 + 1

4 + 14︸︷︷︸

+ 18 + 1

8 + 18 + 1

8︸︷︷︸1/2

+ 116 + · · · .

=1 + 12 + 1

2 + 12 + · · · =∞.

1k = 1

1 + 12 + 1

3 + 14 + 1

5 + · · · .

1 + 12 + 1

3 + 14 + 1

5 + 16 + 1

7 + 18 + 1

9 + · · ·

>1 + 12 + 1

4 + 14︸︷︷︸

+ 18 + 1

8 + 18 + 1

8︸︷︷︸1/2

+ 116 + · · · .

=1 + 12 + 1

2 + 12 + · · · =∞.

Ant Traveling on Rubber Band Paradox

Example.Suppose an ant crawls along an infinitely-elastic one-meterrubber band at the same time as the rubber band is uniformly stretched.If the ant travels 1 centimeter per minute and the band stretches 1 meterper minute, will the ant ever reach the end of the rubber band?

The answer, counterintuitively, is “yes". Consider the table below:

Time Length of Band Distance Traveled Ratio Traveled1s 1 meter 1 centimeter 1/1002s 2 meter 1 centimeter 1/2003s 3 meter 1 centimeter 1/3004s 4 meter 1 centimeter 1/400

After n seconds, the ratio of distance traveled by the ant to total length is

n∑k=1

Logarithms

Definition. Logs are the inverse of exponents: x = bc ⇐⇒ logb(x) = c.

Exponential Logarithmicbm · bn = bm+n logb(xy) = logb(x) + logb(y)bm/bn = bm−n logb(x/y) = logb(x)− logb(y)(bm)n = bmn logb(xy ) = y logb(x)

Definition. e ≈ 2.71828 is a constant discovered by Jacob Bernoulli.

Figure 1: e is the unique number such that the shaded area of y = 1/x equals 1.

Logarithms

Euler-Mascheroni Constant

Definition. The difference between the nth partial sum, Hn, and ln(n)converges to the Euler-Mascheroni constant, γ ≈ 0.5772156649015328.

In 1734, Leonhard Euler estimated the constant to 15 digits.

In 1790, Lorenzo Mascheroni estimated the constant to 19 digits.Modern computers have approximated the number to 100 billion digits.However, it is unknown whether the number is rational or irrational.In physics, the Euler-Mascheroni constant is used in quantumcorrections to calculate the mass of the electron or Higgs boson.The number of Mersenne primes less than x is conjectured to be about

eγlog 2 log log x .

In 1898, de la Vallée-Poussin proved γ = limn→∞

n∑k=1

(dnk e −nk ).

In 1734, Leonhard Euler estimated the constant to 15 digits.In 1790, Lorenzo Mascheroni estimated the constant to 19 digits.

Modern computers have approximated the number to 100 billion digits.However, it is unknown whether the number is rational or irrational.In physics, the Euler-Mascheroni constant is used in quantumcorrections to calculate the mass of the electron or Higgs boson.The number of Mersenne primes less than x is conjectured to be about

n∑k=1

(dnk e −nk ).

In 1734, Leonhard Euler estimated the constant to 15 digits.In 1790, Lorenzo Mascheroni estimated the constant to 19 digits.Modern computers have approximated the number to 100 billion digits.

However, it is unknown whether the number is rational or irrational.In physics, the Euler-Mascheroni constant is used in quantumcorrections to calculate the mass of the electron or Higgs boson.The number of Mersenne primes less than x is conjectured to be about

n∑k=1

(dnk e −nk ).

In 1734, Leonhard Euler estimated the constant to 15 digits.In 1790, Lorenzo Mascheroni estimated the constant to 19 digits.Modern computers have approximated the number to 100 billion digits.However, it is unknown whether the number is rational or irrational.

In physics, the Euler-Mascheroni constant is used in quantumcorrections to calculate the mass of the electron or Higgs boson.The number of Mersenne primes less than x is conjectured to be about

n∑k=1

(dnk e −nk ).

In 1734, Leonhard Euler estimated the constant to 15 digits.In 1790, Lorenzo Mascheroni estimated the constant to 19 digits.Modern computers have approximated the number to 100 billion digits.However, it is unknown whether the number is rational or irrational.In physics, the Euler-Mascheroni constant is used in quantumcorrections to calculate the mass of the electron or Higgs boson.

The number of Mersenne primes less than x is conjectured to be about

n∑k=1

(dnk e −nk ).

In 1734, Leonhard Euler estimated the constant to 15 digits.In 1790, Lorenzo Mascheroni estimated the constant to 19 digits.Modern computers have approximated the number to 100 billion digits.However, it is unknown whether the number is rational or irrational.In physics, the Euler-Mascheroni constant is used in quantumcorrections to calculate the mass of the electron or Higgs boson.The number of Mersenne primes less than x is conjectured to be about

n∑k=1

(dnk e −nk ).

In 1734, Leonhard Euler estimated the constant to 15 digits.In 1790, Lorenzo Mascheroni estimated the constant to 19 digits.Modern computers have approximated the number to 100 billion digits.However, it is unknown whether the number is rational or irrational.In physics, the Euler-Mascheroni constant is used in quantumcorrections to calculate the mass of the electron or Higgs boson.The number of Mersenne primes less than x is conjectured to be about

n∑k=1

(dnk e −nk ).

Figure 2: The area of the blue region converges to the Euler–Mascheroni constant.Source: William Demchick

How Euler Did It

After letters from Goldbach, Euler published the paper Variae observationescirca series infinitas or Several observations about infinite series in 1744.

Theorem. The infinite product 2·3·5·7·11·13·17···1·2·4·6·10·12·16··· equals the sum of the

divergent harmonic series, where each factor is of the form p/(p − 1).

Proof.By the Fundamental Theorem of Arithmetic and distributive property,

∞∑n=1

∏p∈P

(1 + 1

p + 1p2 + 1

p3 + · · ·)

Since |1/p| < 1, we can use the infinite geometric series formula:

=∏p∈P

( 11− 1/p

How Euler Did It

∞∑n=1

∏p∈P

(1 + 1

p + 1p2 + 1

p3 + · · ·)

=∏p∈P

( 11− 1/p

How Euler Did It

∞∑n=1

∏p∈P

(1 + 1

p + 1p2 + 1

p3 + · · ·)

=∏p∈P

( 11− 1/p

How Euler Did It

∞∑n=1

∏p∈P

(1 + 1

p + 1p2 + 1

p3 + · · ·)

=∏p∈P

( 11− 1/p

Sums of Reciprocals of Primes

Taking the natural logarithm of both sides, we see

ln( ∞∑

pln( 1

1− 1/p )

p− ln(1− 1/p).

For |x | < 1, 11−x = 1 + x + x2 + · · · . Taking the definite integral,∫ 1

1− x dx =∫ (

1 + x + x2 + · · ·)dx

− ln(1− x) = x + x2

2 + x3

3 + · · ·

Therefore, ln(∞∑

1n ) =

(1p + 1

2p2 + 13p3 + · · ·

ln( ∞∑

pln( 1

1− 1/p )

p− ln(1− 1/p).

1− x dx =∫ (

1 + x + x2 + · · ·)dx

− ln(1− x) = x + x2

2 + x3

3 + · · ·

1n ) =

(1p + 1

2p2 + 13p3 + · · ·

ln( ∞∑

pln( 1

1− 1/p )

p− ln(1− 1/p).

For |x | < 1, 11−x = 1 + x + x2 + · · · . Taking the definite integral,

∫ 11− x dx =

∫ (1 + x + x2 + · · ·

− ln(1− x) = x + x2

2 + x3

3 + · · ·

1n ) =

(1p + 1

2p2 + 13p3 + · · ·

ln( ∞∑

pln( 1

1− 1/p )

p− ln(1− 1/p).

1− x dx =∫ (

1 + x + x2 + · · ·)dx

− ln(1− x) = x + x2

2 + x3

3 + · · ·

1n ) =

(1p + 1

2p2 + 13p3 + · · ·

ln( ∞∑

pln( 1

1− 1/p )

p− ln(1− 1/p).

1− x dx =∫ (

1 + x + x2 + · · ·)dx

− ln(1− x) = x + x2

2 + x3

3 + · · · .

1n ) =

(1p + 1

2p2 + 13p3 + · · ·

ln( ∞∑

pln( 1

1− 1/p )

p− ln(1− 1/p).

1− x dx =∫ (

1 + x + x2 + · · ·)dx

− ln(1− x) = x + x2

2 + x3

3 + · · · .

1n ) =

(1p + 1

2p2 + 13p3 + · · ·

Divergence of Sum of Reciprocals of Primes

ln(∞∑

1n ) =

(12 + 1

3p + 14p2 + 1

5p3 + · · ·)

(1 + 1

p + 1p2 + · · ·

( 1p2 ·

11− 1/p )

1p2 − p =

1p + O(1).

Hence the reciprocal sum of primes diverges. Let S(n) =∑

p≤n1p , then

ln ln n < S(n) < ln ln n + M + 1/(ln n)2.

The value M is known as the Meissel-Mertens constant and depends on γ.

ln(∞∑

1n ) =

(12 + 1

3p + 14p2 + 1

5p3 + · · ·)

(1 + 1

p + 1p2 + · · ·

( 1p2 ·

11− 1/p )

1p2 − p =

1p + O(1).

p≤n1p , then

ln(∞∑

1n ) =

(12 + 1

3p + 14p2 + 1

5p3 + · · ·)

(1 + 1

p + 1p2 + · · ·

( 1p2 ·

11− 1/p )

1p2 − p =

1p + O(1).

p≤n1p , then

ln(∞∑

1n ) =

(12 + 1

3p + 14p2 + 1

5p3 + · · ·)

(1 + 1

p + 1p2 + · · ·

( 1p2 ·

11− 1/p )

1p2 − p =

1p + O(1).

p≤n1p , then

ln(∞∑

1n ) =

(12 + 1

3p + 14p2 + 1

5p3 + · · ·)

(1 + 1

p + 1p2 + · · ·

( 1p2 ·

11− 1/p )

1p2 − p =

1p + O(1).

Hence the reciprocal sum of primes diverges.

Let S(n) =∑

p≤n1p , then

ln(∞∑

1n ) =

(12 + 1

3p + 14p2 + 1

5p3 + · · ·)

(1 + 1

p + 1p2 + · · ·

( 1p2 ·

11− 1/p )

1p2 − p =

1p + O(1).

p≤n1p , then

ln(∞∑

1n ) =

(12 + 1

3p + 14p2 + 1

5p3 + · · ·)

(1 + 1

p + 1p2 + · · ·

( 1p2 ·

11− 1/p )

1p2 − p =

1p + O(1).

p≤n1p , then

The value M is known as the Meissel-Mertens constant and depends on γ.Justin Stevens Harmonic Sums (Lecture 11) 10 / 25

p-adic Valuation

Let a positive integer n > 1 be written as n = pe11 pe2

2 · · · pekk .

Definition. For each prime, the p-adic valuation of n is vpi (n) = ei .

If P is the set of primes, then another way to write the factorization is

n =∏p∈P

pvp(n).

Theorem. vp (mn) = vp (m) + vp (n) and vp(nc) = cvp(n).

Furthermore, p-adic numbers are well defined for fractions. In particular,

Theorem. vp(m/n) = vp(m)− vp(n).

p-adic Valuation

2 · · · pekk .

n =∏p∈P

pvp(n).

Furthermore, p-adic numbers are well defined for fractions.

In particular,

p-adic Valuation

2 · · · pekk .

n =∏p∈P

pvp(n).

Furthermore, p-adic numbers are well defined for fractions. In particular,

p-adic Sum

Theorem. Prove that if vp(a) 6= vp(b), then vp(a + b) = min(vp(a), vp(b)).

Proof.Let a = pvp(a)a′ and b = pvp(b)b′ and WLOG vp(a) < vp(b). Then,

a + b = pvp(a)a′ + pvp(b)b′

= pvp(a)(a′ + pvp(b)−vp(a)b′

Since p - a′ and vp(b)− vp(a) ≥ 1, we see that

a′ + pvp(b)−vp(a)b′ ≡ a′ 6≡ 0 (mod p).

Hence, vp(a + b) = vp(a) = min(vp(a), vp(b)).

p-adic Sum

Proof.Let a = pvp(a)a′ and b = pvp(b)b′ and WLOG vp(a) < vp(b).

a′ + pvp(b)−vp(a)b′ ≡ a′ 6≡ 0 (mod p).

p-adic Sum

a′ + pvp(b)−vp(a)b′ ≡ a′ 6≡ 0 (mod p).

p-adic Sum

a′ + pvp(b)−vp(a)b′ ≡ a′ 6≡ 0 (mod p).

p-adic Sum

a′ + pvp(b)−vp(a)b′ ≡ a′ 6≡ 0 (mod p).

p-adic Sum

a′ + pvp(b)−vp(a)b′ ≡ a′ 6≡ 0 (mod p).

p-adic Sum

a′ + pvp(b)−vp(a)b′ ≡ a′ 6≡ 0 (mod p).

Harmonic Sum is Never an Integer for n ≥ 2

Example.Prove that the harmonic sum is never an integer for n ≥ 2:

Hn = 11 + 1

2 + · · ·+ 1n .

Solution. Let aj = n!/j for 1 ≤ j ≤ n, hence we can rewrite the sum as

11 + 1

2 + · · ·+ 1n = a1 + a2 + a3 + · · ·+ an

Let the largest power of 2 less than or equal to n be r = 2s . Hence,

v2 (a1 + a2 + · · ·+ an) = v2(ar )= v2(n!)− v2(r)< v2(n!).

Since there are strictly more factors of 2 in the denominator thannumerator, Hn is never an integer for n ≥ 2.

Hn = 11 + 1

2 + · · ·+ 1n .

11 + 1

2 + · · ·+ 1n = a1 + a2 + a3 + · · ·+ an

v2 (a1 + a2 + · · ·+ an) = v2(ar )= v2(n!)− v2(r)< v2(n!).

Hn = 11 + 1

2 + · · ·+ 1n .

11 + 1

2 + · · ·+ 1n = a1 + a2 + a3 + · · ·+ an

v2 (a1 + a2 + · · ·+ an) = v2(ar )= v2(n!)− v2(r)< v2(n!).

Hn = 11 + 1

2 + · · ·+ 1n .

11 + 1

2 + · · ·+ 1n = a1 + a2 + a3 + · · ·+ an

Let the largest power of 2 less than or equal to n be r = 2s .

Hence,

v2 (a1 + a2 + · · ·+ an) = v2(ar )= v2(n!)− v2(r)< v2(n!).

Hn = 11 + 1

2 + · · ·+ 1n .

11 + 1

2 + · · ·+ 1n = a1 + a2 + a3 + · · ·+ an

v2 (a1 + a2 + · · ·+ an) = v2(ar )

= v2(n!)− v2(r)< v2(n!).

Hn = 11 + 1

2 + · · ·+ 1n .

11 + 1

2 + · · ·+ 1n = a1 + a2 + a3 + · · ·+ an

v2 (a1 + a2 + · · ·+ an) = v2(ar )= v2(n!)− v2(r)

< v2(n!).

Hn = 11 + 1

2 + · · ·+ 1n .

11 + 1

2 + · · ·+ 1n = a1 + a2 + a3 + · · ·+ an

v2 (a1 + a2 + · · ·+ an) = v2(ar )= v2(n!)− v2(r)< v2(n!).

Hn = 11 + 1

2 + · · ·+ 1n .

11 + 1

2 + · · ·+ 1n = a1 + a2 + a3 + · · ·+ an

v2 (a1 + a2 + · · ·+ an) = v2(ar )= v2(n!)− v2(r)< v2(n!).

Outline

1 Harmonic Sums

2 Shuffling Deck of CardsThe Premo Card TrickBook Stacking

3 References

Shuffling Decks of Cards

For a standard deck of playing cards, there are 52! ≈ 8.0658 · 1067

permutations, more than the number of particles in the universe.

We can use computer simulations to determine a deck’s randomness.For a perfectly shuffled deck, the expected value of correct guesses is

E = 152 + 1

51 + 150 + · · ·+ 1

3 + 12 + 1

1 = H52 ≈ 4.538 cards.

1 2 3 4 5 6 7 8 9No cut 31.17 19.69 12.92 8.80 6.56 5.51 5.01 4.76 4.65Cut 29.45 19.09 12.69 8.70 6.50 5.46 4.97 4.73 4.63

Table 1: Number of cards guessed correctly after k shuffles of 52 cards

permutations, more than the number of particles in the universe.We can use computer simulations to determine a deck’s randomness.

For a perfectly shuffled deck, the expected value of correct guesses is

E = 152 + 1

51 + 150 + · · ·+ 1

3 + 12 + 1

1 = H52 ≈ 4.538 cards.

1 2 3 4 5 6 7 8 9No cut 31.17 19.69 12.92 8.80 6.56 5.51 5.01 4.76 4.65Cut 29.45 19.09 12.69 8.70 6.50 5.46 4.97 4.73 4.63

permutations, more than the number of particles in the universe.We can use computer simulations to determine a deck’s randomness.For a perfectly shuffled deck, the expected value of correct guesses is

E = 152 + 1

51 + 150 + · · ·+ 1

3 + 12 + 1

1 = H52 ≈ 4.538 cards.

1 2 3 4 5 6 7 8 9No cut 31.17 19.69 12.92 8.80 6.56 5.51 5.01 4.76 4.65Cut 29.45 19.09 12.69 8.70 6.50 5.46 4.97 4.73 4.63

permutations, more than the number of particles in the universe.We can use computer simulations to determine a deck’s randomness.For a perfectly shuffled deck, the expected value of correct guesses is

E = 152 + 1

51 + 150 + · · ·+ 1

3 + 12 + 1

1 = H52 ≈ 4.538 cards.

1 2 3 4 5 6 7 8 9No cut 31.17 19.69 12.92 8.80 6.56 5.51 5.01 4.76 4.65Cut 29.45 19.09 12.69 8.70 6.50 5.46 4.97 4.73 4.63

Charles Jordan Premo Card Trick

A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣

Aq 2q 3q 4q 5q 6q 7q 8q 9q 10q Jq Qq Kq

Ar 2r 3r 4r 5r 6r 7r 8r 9r 10r Jr Qr Kr

A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠

1 Have a spectator cut the deck several times and give it a riffle shuffle.2 Repeat the first step, then turn your back to the spectator.3 Instruct the spectator to lift off about a quarter of the deck and

remove the top card of the lower pile, writing it down. Replace the topquarter and put the removed card on top of the reconstructed deck.

4 The card is then burried by cutting the deck and riffle shuffling.5 The performer then turns back and deals the cards into several face-up

rows and after some careful thought, reveals the spectator’s card.

A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣

A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠

1 Have a spectator cut the deck several times and give it a riffle shuffle.

2 Repeat the first step, then turn your back to the spectator.3 Instruct the spectator to lift off about a quarter of the deck and

A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣

A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠

1 Have a spectator cut the deck several times and give it a riffle shuffle.2 Repeat the first step, then turn your back to the spectator.

3 Instruct the spectator to lift off about a quarter of the deck andremove the top card of the lower pile, writing it down. Replace the topquarter and put the removed card on top of the reconstructed deck.

A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣

A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠

remove the top card of the lower pile, writing it down.

Replace the topquarter and put the removed card on top of the reconstructed deck.

A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣

A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠

A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣

A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠

4 The card is then burried by cutting the deck and riffle shuffling.

5 The performer then turns back and deals the cards into several face-uprows and after some careful thought, reveals the spectator’s card.

A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣

A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠

rows and after some careful thought, reveals the spectator’s card.Justin Stevens Harmonic Sums (Lecture 11) 16 / 25

Premo First Shuffle

The spectator begins by cutting the deck and giving it a riffle shuffle:

K♠ A♣ 2♣ Jq Qq 3♣ 4♣ Kq Ar 5♣ 2r 6♣ 7♣

8♣ 3r 9♣ 10♣ J♣ Q♣ 4r 5r K♣ 6r Aq 7r 2q

8r 9r 3q 4q 10r Jr Qr 5q 6q Kr A♠ 7q 2♠

3♠ 4♠ 5♠ 8q 6♠ 7♠ 8♠ 9♠ 9q 10♠ J♠ Q♠ 10q

Definition. A rising sequence is a maximal subset of an arrangement ofcards, consisting of successive face values displayed in order. Eacharrangement of a deck of cards is the union of its rising sequences.

Premo First Shuffle

K♠ A♣ 2♣ Jq Qq 3♣ 4♣ Kq Ar 5♣ 2r 6♣ 7♣

8♣ 3r 9♣ 10♣ J♣ Q♣ 4r 5r K♣ 6r Aq 7r 2q

3♠ 4♠ 5♠ 8q 6♠ 7♠ 8♠ 9♠ 9q 10♠ J♠ Q♠ 10q

Premo First Shuffle

K♠ A♣ 2♣ Jq Qq 3♣ 4♣ Kq Ar 5♣ 2r 6♣ 7♣

8♣ 3r 9♣ 10♣ J♣ Q♣ 4r 5r K♣ 6r Aq 7r 2q

3♠ 4♠ 5♠ 8q 6♠ 7♠ 8♠ 9♠ 9q 10♠ J♠ Q♠ 10q

Definition. A rising sequence is a maximal subset of an arrangement ofcards, consisting of successive face values displayed in order.

Eacharrangement of a deck of cards is the union of its rising sequences.

Premo First Shuffle

K♠ A♣ 2♣ Jq Qq 3♣ 4♣ Kq Ar 5♣ 2r 6♣ 7♣

8♣ 3r 9♣ 10♣ J♣ Q♣ 4r 5r K♣ 6r Aq 7r 2q

3♠ 4♠ 5♠ 8q 6♠ 7♠ 8♠ 9♠ 9q 10♠ J♠ Q♠ 10q

Premo Second Shuffle

The spectator then cuts the deck again and gives it a second riffle shuffle:

8r 2♣ Jq 9r 3q Qq 4q 3♣ 4♣ 10r Kq Ar 5♣

Jr 2r 6♣ 7♣ Qr 8♣ 5q 3r 9♣ 6q Kr 10♣ J♣

A♠ 7q Q♣ 2♠ 3♠ 4r 4♠ 5r 5♠ K♣ 8q 6♠ 6r

7♠ 8♠ 9♠ 9q Aq 10♠ J♠ Q♠ 7r 10q K♠ A♣ 2q

There are now the following four rising sequences:

8r 9r 10r JrQrKrA♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠Q♠K♠2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣Q♣K♣Aq 2qJqQqKqAr 2r 3r 4r 5r 6r 7r3q 4q 5q 6q 7q 8q 9q 10q

8r 2♣ Jq 9r 3q Qq 4q 3♣ 4♣ 10r Kq Ar 5♣

Jr 2r 6♣ 7♣ Qr 8♣ 5q 3r 9♣ 6q Kr 10♣ J♣

A♠ 7q Q♣ 2♠ 3♠ 4r 4♠ 5r 5♠ K♣ 8q 6♠ 6r

7♠ 8♠ 9♠ 9q Aq 10♠ J♠ Q♠ 7r 10q K♠ A♣ 2q

8r 2♣ Jq 9r 3q Qq 4q 3♣ 4♣ 10r Kq Ar 5♣

Jr 2r 6♣ 7♣ Qr 8♣ 5q 3r 9♣ 6q Kr 10♣ J♣

A♠ 7q Q♣ 2♠ 3♠ 4r 4♠ 5r 5♠ K♣ 8q 6♠ 6r

7♠ 8♠ 9♠ 9q Aq 10♠ J♠ Q♠ 7r 10q K♠ A♣ 2q

8r 2♣ Jq 9r 3q Qq 4q 3♣ 4♣ 10r Kq Ar 5♣

Jr 2r 6♣ 7♣ Qr 8♣ 5q 3r 9♣ 6q Kr 10♣ J♣

A♠ 7q Q♣ 2♠ 3♠ 4r 4♠ 5r 5♠ K♣ 8q 6♠ 6r

7♠ 8♠ 9♠ 9q Aq 10♠ J♠ Q♠ 7r 10q K♠ A♣ 2q

Premo Card Selection

The spectator then lifts off a quarter of the cards, chooses the top card, andputs the removed card on top of the reconstructed deck. This card is thenburried by cutting the deck and riffle shuffling a third and final time:

4♠ Jr 5r 5♠ K♣ 8q 6♠ 6r 7♠ 2r 8♠ 9♠ 9q

Aq 10♠ 6♣ J♠ Q♠ 7r 10q K♠ A♣ 2q Qr 8♣ 5q

7♣ 8r 3r 9♣ 2♣ Jq 6q Kr 9r 3q Qq 10♣ J♣

4q A♠ 3♣ 4♣ 7q 10r Q♣ Kq 2♠ 3♠ Ar 4r 5♣We can detect the following nine rising sequences:

4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠Q♠K♠JrQrKrA♠ 2♠ 3♠5r 6r 7r 8r 9r 10rK♣Aq 2q 3q 4q

8q 9q 10q JqQqKqAr2r 3r 4r 5q 6q 7qA♣ 2♣ 3♣ 4♣ 5♣6♣ 8♣ 9♣ 10♣ J♣Q♣

4♠ Jr 5r 5♠ K♣ 8q 6♠ 6r 7♠ 2r 8♠ 9♠ 9q

Aq 10♠ 6♣ J♠ Q♠ 7r 10q K♠ A♣ 2q Qr 8♣ 5q

7♣ 8r 3r 9♣ 2♣ Jq 6q Kr 9r 3q Qq 10♣ J♣

4q A♠ 3♣ 4♣ 7q 10r Q♣ Kq 2♠ 3♠ Ar 4r 5♣

We can detect the following nine rising sequences:4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠Q♠K♠JrQrKrA♠ 2♠ 3♠5r 6r 7r 8r 9r 10rK♣Aq 2q 3q 4q

4♠ Jr 5r 5♠ K♣ 8q 6♠ 6r 7♠ 2r 8♠ 9♠ 9q

Aq 10♠ 6♣ J♠ Q♠ 7r 10q K♠ A♣ 2q Qr 8♣ 5q

7♣ 8r 3r 9♣ 2♣ Jq 6q Kr 9r 3q Qq 10♣ J♣

4♠ Jr 5r 5♠ K♣ 8q 6♠ 6r 7♠ 2r 8♠ 9♠ 9q

Aq 10♠ 6♣ J♠ Q♠ 7r 10q K♠ A♣ 2q Qr 8♣ 5q

7♣ 8r 3r 9♣ 2♣ Jq 6q Kr 9r 3q Qq 10♣ J♣

Book Stacking for n Books

Figure 3: Albert R. Meyer, MIT 6.042J

Book Stacking for n + 1 Books

Balancing Torque Equation

Book Stacking Formula

Definition. Define Bn to be the maximum overhang of n books.

Bn ={1/2, n = 1Bn−1 + 1/(2n), n ≥ 2.

We can expliticly write the function as Bn = 12

(1 + 1

2 + · · ·+ 1n

)= Hn/2.

Figure 6: The overhang for 52 cards is 2.269 times the width of one card.

Definition. Define Bn to be the maximum overhang of n books. Then,

Bn ={1/2, n = 1Bn−1 + 1/(2n), n ≥ 2.

(1 + 1

2 + · · ·+ 1n

)= Hn/2.

Bn ={1/2, n = 1Bn−1 + 1/(2n), n ≥ 2.

(1 + 1

2 + · · ·+ 1n

)= Hn/2.

Bn ={1/2, n = 1Bn−1 + 1/(2n), n ≥ 2.

(1 + 1

2 + · · ·+ 1n

)= Hn/2.

Outline

1 Harmonic Sums

2 Shuffling Deck of Cards

3 References

Relevant Links

Numberphile: The mystery of 0.577Ed Sandifer: How Euler Did ItAdrien Dudek: Divergence of ∑p

Charles Jordan: Thirty Card MysteriesAlbert R. Meyer: Book Stacking Video (MIT 6.042J)Brian Brushwood: The Leaning Tower of Cardsa

Harmonic Sums - Lecture 11

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