Generalized multiplication tables - Williams CollegeDimitris Koukoulopoulos University of Illinois...

The multiplication table and higher dimensional analoguesGeneralized multiplication tables

Generalized multiplication tables

Dimitris Koukoulopoulos

University of Illinois at Urbana - Champaign

June 4, 2009

The multiplication table problemStatement of resultsA heuristic argument

Consider the multiplication table

1 2 3 4 · · · N1 1 2 3 4 · · · N2 2 4 6 8 · · · 2N3 3 6 9 12 · · · 3N4 4 8 12 16 · · · 4N...

......

...N N 2N 3N 4N · · · N2

Question (Erdos, 1955)How many distinct integers does this table contain?

We may ask the same question for products of 3 integers, inwhich case we have a multiplication ‘box’, or 4 integers, and soon. To this end, define

Ak+1(N) := |{n1 · · · nk+1 : ni ≤ N (1 ≤ i ≤ k + 1)}|.

Then the problem is equivalent to estimating Ak+1(N).

The key to understanding the combinatorics of themultiplication table is the function

Hk+1(x ,y , z) := |{n ≤ x :∃d1 · · · dk |n such thatyi < di ≤ zi (1 ≤ i ≤ k)}|,

where y = (y1, . . . , yk ) and z = (z1, . . . , zk ).

We may ask the same question for products of 3 integers, inwhich case we have a multiplication ‘box’, or 4 integers, and soon. To this end, define

Ak+1(N) := |{n1 · · · nk+1 : ni ≤ N (1 ≤ i ≤ k + 1)}|.

Then the problem is equivalent to estimating Ak+1(N).

The key to understanding the combinatorics of themultiplication table is the function

Hk+1(x ,y , z) := |{n ≤ x :∃d1 · · · dk |n such thatyi < di ≤ zi (1 ≤ i ≤ k)}|,

where y = (y1, . . . , yk ) and z = (z1, . . . , zk ).

The transition from Hk+1(x ,y , z) to Ak+1(N) is achieved via theelementary inequalities

2k ,(N

2, . . . ,

), (N, . . . ,N)

)≤ Ak+1(N)

≤∑

2mi≤√

N1≤i≤k

2m1+···+mk,

2m1+1 , . . . ,N

2m1, . . . ,

+ Nk+1/2.

Note that it suffices to study Hk+1(x ,y , z) whenz = 2y ;the numbers log y1, . . . , log yk all have the same order ofmagnitude.

The transition from Hk+1(x ,y , z) to Ak+1(N) is achieved via theelementary inequalities

2k ,(N

2, . . . ,

), (N, . . . ,N)

)≤ Ak+1(N)

≤∑

2mi≤√

N1≤i≤k

2m1+···+mk,

2m1+1 , . . . ,N

2m1, . . . ,

+ Nk+1/2.

Note that it suffices to study Hk+1(x ,y , z) whenz = 2y ;the numbers log y1, . . . , log yk all have the same order ofmagnitude.

Set Q(u) := u log u − u + 1.

Then the following estimate holds.

Theorem (Ford (2004), K. (2008))Let k ≥ 1, 0 < δ ≤ 1 and c ≥ 1. Assume that x ≥ 1 and3 ≤ y1 ≤ y2 ≤ · · · ≤ yk ≤ yc

1 with xy1···yk

≥ max{2k , yδ1}. Then

Hk+1(x ,y ,2y) �k ,δ,cx

(log y1)Q( k

log(k+1))(log log y1)3/2

Consequently

Corollary

Ak+1(N) �kNk+1

(log N)Q( k

log(k+1))(log log N)3/2

(N ≥ 3).

Set Q(u) := u log u − u + 1. Then the following estimate holds.

1 with xy1···yk

Hk+1(x ,y ,2y) �k ,δ,cx

(log y1)Q( k

Consequently

Corollary

Ak+1(N) �kNk+1

(log N)Q( k

(N ≥ 3).

Set Q(u) := u log u − u + 1. Then the following estimate holds.

1 with xy1···yk

Hk+1(x ,y ,2y) �k ,δ,cx

(log y1)Q( k

Consequently

Corollary

Ak+1(N) �kNk+1

(log N)Q( k

(N ≥ 3).

The following heuristic was first given by Ford when k = 1.

Assume that log y1, . . . , log yk have the same order ofmagnitude. For n ∈ N write n = ab, where

a =∏

pe‖n,p≤2y1

Assume that µ2(a) = 1 and a ≤ yC1 .

Consider the set

Dk+1(a) = {(log d1, . . . , log dk ) : d1 · · · dk |a}.

Main assumption: Dk+1(a) is well-distributed in [0, log a]k .

a =∏

pe‖n,p≤2y1

Consider the set

a =∏

pe‖n,p≤2y1

Consider the set

a =∏

pe‖n,p≤2y1

Consider the set

Then we would expect that

τk+1(a,y ,2y) := |{d1 · · · dk |a : yi < di ≤ 2yi (1 ≤ i ≤ k)}|

=∣∣∣Dk+1(a) ∩

k∏i=1

(log yi , log yi + log 2]∣∣∣

≈ |Dk+1(a)|(log 2)k

(log a)k ≈(k + 1)ω(a)

(log y1)k .

This expression is ≥ 1 when

ω(a) ≥ m :=⌊ k

log(k + 1)log log y1

⌋+ O(1).

We have

|{n ≤ x : ω(a) = r}| ≈ xlog y1

(log log y1)r

=∣∣∣Dk+1(a) ∩

k∏i=1

≈ |Dk+1(a)|(log 2)k

(log a)k ≈(k + 1)ω(a)

(log y1)k .

ω(a) ≥ m :=⌊ k

⌋+ O(1).

We have

|{n ≤ x : ω(a) = r}| ≈ xlog y1

(log log y1)r

=∣∣∣Dk+1(a) ∩

k∏i=1

≈ |Dk+1(a)|(log 2)k

(log a)k ≈(k + 1)ω(a)

(log y1)k .

ω(a) ≥ m :=⌊ k

⌋+ O(1).

We have

|{n ≤ x : ω(a) = r}| ≈ xlog y1

(log log y1)r

Therefore

Hk+1(x ,y ,2y) ≈ xlog y1

∑r≥m

(log log y1)r

(log y1)Q( k

This is slightly too big. The problem arises from the fact thatDk+1(a) is usually not well-distributed, but instead it has manyclumps.

To see this define

Lk+1(a) := Vol( ⋃

d1···dk |a

[log(d1/2), log d1)×· · ·×[log(dk/2), log dk )),

which is a quantitative measure of how well-distributed Dk+1(a)is.

Therefore

∑r≥m

(log log y1)r

(log y1)Q( k

To see this define

Lk+1(a) := Vol( ⋃

d1···dk |a

Therefore

∑r≥m

(log log y1)r

(log y1)Q( k

To see this define

Lk+1(a) := Vol( ⋃

d1···dk |a

If a = p1 · · · pm, where p1 < · · · < pm ≤ 2y1, then we expect that

log log pj ∼jm

log log y1 = jlog(k + 1)

k+ O(1).

But with probability tending to 1 there is a j so that

log log pj ≤ jlog(k + 1)

k− (log log y1)

which implies that

Lk+1(a) ≤ τk+1(pj+1 · · · pm)Lk+1(p1 · · · pj)

. (k + 1)m exp{−(log log y1)1/3}.

This is much less than τk+1(a) = (k + 1)m and so most of thetime Dk+1(a) contains large clusters of points.

log log pj ∼jm

k+ O(1).

k− (log log y1)

which implies that

Lk+1(a) ≤ τk+1(pj+1 · · · pm)Lk+1(p1 · · · pj)

. (k + 1)m exp{−(log log y1)1/3}.

log log pj ∼jm

k+ O(1).

k− (log log y1)

which implies that

Lk+1(a) ≤ τk+1(pj+1 · · · pm)Lk+1(p1 · · · pj)

. (k + 1)m exp{−(log log y1)1/3}.

log log pj ∼jm

k+ O(1).

k− (log log y1)

which implies that

Lk+1(a) ≤ τk+1(pj+1 · · · pm)Lk+1(p1 · · · pj)

. (k + 1)m exp{−(log log y1)1/3}.

We must focus on abnormal a’s for which

log log pj ≥ jlog(k + 1)

k−O(1) (1 ≤ j ≤ m).

The probability that a has this property is about

1m� 1

log log y1(Ford).

This leads to the refined heuristic estimate

Hk+1(x ,y ,2y) ≈ x

(log y1)Q( 1

log ρ)(log log y1)3/2

which is the correct one.

k−O(1) (1 ≤ j ≤ m).

1m� 1

log log y1(Ford).

Hk+1(x ,y ,2y) ≈ x

(log y1)Q( 1

k−O(1) (1 ≤ j ≤ m).

1m� 1

log log y1(Ford).

Hk+1(x ,y ,2y) ≈ x

(log y1)Q( 1

Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates

The natural generalization of the multiplication table problem isthe estimation of

Ak+1(N1, . . . ,Nk+1) := |{n1 · · · nk+1 : ni ≤ Ni (1 ≤ i ≤ k + 1)}|.

As before, to understand this question we study Hk+1(x ,y ,2y).

The difference is that we now drop the assumption that thenumbers log y1, . . . , log yk are of the same order of magnitude.

When k = 1, Ford’s result immediately implies that

CorollaryLet 3 ≤ N1 ≤ N2. Then

A2(N1,N2) �N1N2

(log N1)Q( 1

log 2 )(log log N1)3/2

Ak+1(N1, . . . ,Nk+1) := |{n1 · · · nk+1 : ni ≤ Ni (1 ≤ i ≤ k + 1)}|.

A2(N1,N2) �N1N2

(log N1)Q( 1

Ak+1(N1, . . . ,Nk+1) := |{n1 · · · nk+1 : ni ≤ Ni (1 ≤ i ≤ k + 1)}|.

A2(N1,N2) �N1N2

(log N1)Q( 1

For n ∈ N write n = a1 · · · akb, where

ai =∏

pe‖n,2yi−1<p≤2yi

Assume that µ2(ai) = 1 and ai ≤ yCi for 1 ≤ i ≤ k .

Dk+1(a) := {(log d1, . . . , log dk ) : d1 · · · di |a1 · · · ai (1 ≤ i ≤ k)}

and assume that Dk+1(a) is well-distributed in[0, log y1]× · · · × [0, log yk ].

For n ∈ N write n = a1 · · · akb, where

ai =∏

pe‖n,2yi−1<p≤2yi

Assume that µ2(ai) = 1 and ai ≤ yCi for 1 ≤ i ≤ k .

Dk+1(a) := {(log d1, . . . , log dk ) : d1 · · · di |a1 · · · ai (1 ≤ i ≤ k)}

and assume that Dk+1(a) is well-distributed in[0, log y1]× · · · × [0, log yk ].

τ(n,y ,2y) = {(d1, . . . ,dk ) : d1 · · · di |a1 · · · ai ,

yi < di ≤ 2yi (1 ≤ i ≤ k)}

=∣∣∣Dk+1(a) ∩

k∏i=1

≈∏k

i=1(k − i + 2)ω(ai )∏ki=1 log yi

Set `i := log 3 log yilog yi−1

(r1, . . . , rk ) ∈ (N∪{0})k :k∑

ri log(k−i+2) ≥k∑

`i(k−i+1)}.

Then, heuristically, τ(n,y ,2y) ≥ 1 if-f (ω(a1), . . . , ω(ak )) ∈H .

τ(n,y ,2y) = {(d1, . . . ,dk ) : d1 · · · di |a1 · · · ai ,

yi < di ≤ 2yi (1 ≤ i ≤ k)}

=∣∣∣Dk+1(a) ∩

k∏i=1

≈∏k

(r1, . . . , rk ) ∈ (N∪{0})k :k∑

`i(k−i+1)}.

Then, heuristically, τ(n,y ,2y) ≥ 1 if-f (ω(a1), . . . , ω(ak )) ∈H .

τ(n,y ,2y) = {(d1, . . . ,dk ) : d1 · · · di |a1 · · · ai ,

yi < di ≤ 2yi (1 ≤ i ≤ k)}

=∣∣∣Dk+1(a) ∩

k∏i=1

≈∏k

(r1, . . . , rk ) ∈ (N∪{0})k :k∑

`i(k−i+1)}.

Then, heuristically, τ(n,y ,2y) ≥ 1 if-f (ω(a1), . . . , ω(ak )) ∈H .13

We have that

|{n ≤ x : ω(ai) = ri (1 ≤ i ≤ k)}| ≈ xlog yk

k∏i=1

So using Taylor’s theorem and Lagrange multipliers we find that

Hk+1(x ,y ,2y) ≈ xlog yk

∑r∈H

k∏i=1

� x√log log yk

∏ki=1

(log yi

log yi−1

)Q((k−i+2)α),

where α = α(k ,y) satisfiesk∑

(k − i + 2)α log(k − i + 2)`i =k∑

(k − i + 1)`i .

We have that

|{n ≤ x : ω(ai) = ri (1 ≤ i ≤ k)}| ≈ xlog yk

k∏i=1

So using Taylor’s theorem and Lagrange multipliers we find that

Hk+1(x ,y ,2y) ≈ xlog yk

∑r∈H

k∏i=1

� x√log log yk

∏ki=1

(log yi

log yi−1

)Q((k−i+2)α),

where α = α(k ,y) satisfiesk∑

(k − i + 2)α log(k − i + 2)`i =k∑

(k − i + 1)`i .

Indeed, we may prove the following estimate.

Theorem (K, 2008)

Let 3 = y0 ≤ y1 ≤ · · · ≤ yk with xy1···yk

≥ max{2k , yk}. Then

Hk+1(x ,y ,2y)

1,(log log 3yi0−1)(log 3 log yk

log yi0)

log log yk

k∏i=1

( log yi

log yi−1

)Q((k−i+2)α),

where i0 is such that `i0 = max1≤i≤k `i .

When k = 2, this upper bound is the correct order ofH3(x ,y ,2y).

Theorem (K, 2008)

Let 3 ≤ y1 ≤ y2 so that xy1y2≥ max{4, y2}. Then

H3(x ,y ,2y)

(log log 3y1)(log 3 log y2log y1

(log log y2)5/2(log y1)Q(3α)(

log y2log y1

)Q(2α),

where y = (y1, y2).

As a direct corollary we obtain the order of magnitude of thenumber of integers in a 3-dimensional multiplication table.

CorollaryFor 3 ≤ N1 ≤ N2 ≤ N3 we have that

A3(N1,N2,N3)

N1N2N3�

(log log N1)(log 3 log N2log N1

(log log N2)5/2(log N1)Q(3α)(

log N2log N1

)Q(2α).

In general, we may reduce the counting in Hk+1(x ,y ,2y) (localdistribution of factorizations) to estimating averages of

Lk+1(a) := Vol( ⋃

d1···di |a1···ai1≤i≤k

[log(d1/2), log d1)×· · · [log(dk/2), log dk ))

(global distribution of factorizations).

Proposition (Ford (2004), K. (2008))Let k ≥ 1, x ≥ 1 and 3 ≤ y1 ≤ · · · ≤ yk with

xy1···yk

Hk+1(x ,y ,2y)

k∏i=1

( log yi

log yi−1

)−(k−i+2) ∑ai∈P(yi−1,yi )

1≤i≤k

Lk+1(a)

a1 · · · ak.

In general, we may reduce the counting in Hk+1(x ,y ,2y) (localdistribution of factorizations) to estimating averages of

Lk+1(a) := Vol( ⋃

d1···di |a1···ai1≤i≤k

[log(d1/2), log d1)×· · · [log(dk/2), log dk ))

(global distribution of factorizations).

Proposition (Ford (2004), K. (2008))Let k ≥ 1, x ≥ 1 and 3 ≤ y1 ≤ · · · ≤ yk with

xy1···yk

Hk+1(x ,y ,2y)

k∏i=1

( log yi

log yi−1

)−(k−i+2) ∑ai∈P(yi−1,yi )

1≤i≤k

Lk+1(a)

a1 · · · ak.

Thank you for your attention!

Generalized multiplication tables - Williams CollegeDimitris Koukoulopoulos University of Illinois...

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