removing parentheses - humboldt state university · \kvant" (\quantum" in russian) ... a...

Removing Parentheses

Tyler J. Evans

Fifty-Seventh Annual Redwood Area Mathematics Tournament

April 30, 2016

T.J. Evans (HSU) Parentheses 1 / 29

Overview

1 Kvant Selecta

2 Euler’s Identity

3 Euler’s Identity and the Partition Function

4 A Partition Calculator

5 Proof of Euler’s Identity

6 Power’s of Euler’s Function and Missed Opportunities

What is Kvant?

“Kvant” (“Quantum” in Russian) is a Russian mathematics magazinepublished for a highschool student audience.

Kvant was first published in the former Soviet Union in 1970.

Kvant was initiated by a group of prominent Soviet physicists andmathematicians, and it’s editorial board and list of contributingauthors consisted of the very best Russian academics in these fields.

Kvant was distributed by subscription only, and in the 1970s it had inexcess of 300,000 subscribers!

A typical issue of Kvant consisted of four feature articles, two inphysics and two in mathematics; there would also be one or twoshorter articles called “Physics Labs” and “Mathematical Circles.”

Since 1991, an English version, Quantum, has been published bySpringer-Verlag.

What is Kvant?

Kvant Selecta

In 1999, the AmericanMathematical Societypublished a series oftranslations of articlesfirst published in Kvant.

The articles were dividedinto categories: algebraand analysis, geometryand topology and discretemathematics.

Kvant Selecta

In 1999, the AmericanMathematical Societypublished a series oftranslations of articlesfirst published in Kvant.

The articles were dividedinto categories: algebraand analysis, geometryand topology and discretemathematics.

Kvant Selecta

The Russian original ispublished in Kvant 1981, no.8, pp. 12-20. The Englishtranslation is published inKvant Selecta: Algebra andAnalysis, II, MathematicalWorld, vol. 15, AMS, pp. 39-49, 1999.

Today we will look at thecontents of a beautifulalgebraic article by D. B.Fuchs entitled “On theremoval of parentheses,on Euler, Gauss, andMacDonald, and onMissed Opportunities.”

The organization of thematerial presented herebelongs to D.B. Fuchs;any mistakes ormisunderstandings belongto the speaker!

Kvant Selecta

The Russian original ispublished in Kvant 1981, no.8, pp. 12-20. The Englishtranslation is published inKvant Selecta: Algebra andAnalysis, II, MathematicalWorld, vol. 15, AMS, pp. 39-49, 1999.

Today we will look at thecontents of a beautifulalgebraic article by D. B.Fuchs entitled “On theremoval of parentheses,on Euler, Gauss, andMacDonald, and onMissed Opportunities.”

The organization of thematerial presented herebelongs to D.B. Fuchs;any mistakes ormisunderstandings belongto the speaker!

Removing Parentheses

You are no doubt familiar with the multiplication of polynomials.

(a+ b)(a+ b) = a2 + 2ab+ b2,

(a+ b)(a− b) = a2 − b2,

(1− a)(1 + a+ · · ·+ an) = 1− an+1.

Fuchs refers to the expansion of these products, and the gathering of liketerms “removing the parentheses.”

Euler’s Identity

In the middle of the 18th century (in approximately 1748), the great Swissmathematician Leonhard Euler wrote about “removing the parentheses” inthe polynomial expression

ϕn(x) = (1− x)(1− x2)(1− x3) · · · (1− xn).

Euler’s Identity

Let’s do some of the removal ourselves.

ϕ1(x) = 1− x,

ϕ2(x) = 1− x− x2 +x3,

ϕ3(x) = 1− x− x2 + x4 +x5 − x6,

ϕ4(x) = 1− x− x2 +2x5 − x8 −x9 + x10,

ϕ5(x) = 1− x− x2 +x5 + x6 +x7 − x8 −x9 − x10 · · · ,

ϕ6(x) = 1− x− x2 +x5 +2x7 −x9 − x10 · · · ,

ϕ7(x) = 1− x− x2 +x5 +x7 + x8 −x10 · · · ,

ϕ8(x) = 1− x− x2 +x5 +x7 +x9 · · · ,

ϕ9(x) = 1− x− x2 +x5 +x7 +x10 · · · ,

ϕ10(x) = 1− x− x2 +x5 +x7 · · · ,

The dots refer to terms of degree higher than 10 - there’s no room here! For example,

the polynomial ϕ10(x) has degree 55.

Euler’s Identity

ϕ1(x) = 1− x,

ϕ2(x) = 1− x− x2 +x3,

ϕ3(x) = 1− x− x2 + x4 +x5 − x6,

ϕ4(x) = 1− x− x2 +2x5 − x8 −x9 + x10,

ϕ5(x) = 1− x− x2 +x5 + x6 +x7 − x8 −x9 − x10 · · · ,

ϕ6(x) = 1− x− x2 +x5 +2x7 −x9 − x10 · · · ,

ϕ7(x) = 1− x− x2 +x5 +x7 + x8 −x10 · · · ,

ϕ8(x) = 1− x− x2 +x5 +x7 +x9 · · · ,

ϕ9(x) = 1− x− x2 +x5 +x7 +x10 · · · ,

ϕ10(x) = 1− x− x2 +x5 +x7 · · · ,

Euler’s Identity

ϕ1(x) = 1− x,

ϕ2(x) = 1− x− x2 +x3,

ϕ3(x) = 1− x− x2 + x4 +x5 − x6,

ϕ4(x) = 1− x− x2 +2x5 − x8 −x9 + x10,

ϕ5(x) = 1− x− x2 +x5 + x6 +x7 − x8 −x9 − x10 · · · ,

ϕ6(x) = 1− x− x2 +x5 +2x7 −x9 − x10 · · · ,

ϕ7(x) = 1− x− x2 +x5 +x7 + x8 −x10 · · · ,

ϕ8(x) = 1− x− x2 +x5 +x7 +x9 · · · ,

ϕ9(x) = 1− x− x2 +x5 +x7 +x10 · · · ,

ϕ10(x) = 1− x− x2 +x5 +x7 · · · ,

Euler’s Identity

ϕ1(x) = 1− x,

ϕ2(x) = 1− x− x2 +x3,

ϕ3(x) = 1− x− x2 + x4 +x5 − x6,

ϕ4(x) = 1− x− x2 +2x5 − x8 −x9 + x10,

ϕ5(x) = 1− x− x2 +x5 + x6 +x7 − x8 −x9 − x10 · · · ,

ϕ6(x) = 1− x− x2 +x5 +2x7 −x9 − x10 · · · ,

ϕ7(x) = 1− x− x2 +x5 +x7 + x8 −x10 · · · ,

ϕ8(x) = 1− x− x2 +x5 +x7 +x9 · · · ,

ϕ9(x) = 1− x− x2 +x5 +x7 +x10 · · · ,

ϕ10(x) = 1− x− x2 +x5 +x7 · · · ,

Euler’s Identity

Notice that as n increases, the coefficients in ϕn(x) stabilize - thecoefficient of xk stops changing beginning with a certain value of n.

ϕn(x) = ϕn−1(x)(1− xn)

In moving from ϕn−1(x) to ϕn(x), we do not change the coefficients of1, x, . . . , xn−1 so that for k < n, the coefficient of xk in ϕn(x) does notdepend on n.

The coefficients displayed before will not change in ϕ11(x), ϕ12(x), . . . .

Setϕ(x) = (1− x)(1− x2)(1− x3) · · · (1− xn) · · ·

Euler’s Identity

ϕn(x) = ϕn−1(x)(1− xn)

Setϕ(x) = (1− x)(1− x2)(1− x3) · · · (1− xn) · · ·

Euler’s Identity

ϕn(x) = ϕn−1(x)(1− xn)

Setϕ(x) = (1− x)(1− x2)(1− x3) · · · (1− xn) · · ·

Euler’s Identity

ϕn(x) = ϕn−1(x)(1− xn)

Setϕ(x) = (1− x)(1− x2)(1− x3) · · · (1− xn) · · ·

Euler’s Identity

Let us write the power series

ϕ(x) = a0 + a1x+ a2x2 + a3x

3 + a4x4 + · · ·

Our work above shows

a0 = a5 = a7 = 1

a1 = a2 = −1

a3 = a4 = a6 = a8 = a9 = 0.

Conjecture. The coefficients ak are equal to 0, 1 or −1, and most ofthem are equal to 0.

Euler’s Identity

ϕ(x) = a0 + a1x+ a2x2 + a3x

3 + a4x4 + · · ·

a0 = a5 = a7 = 1

a1 = a2 = −1

a3 = a4 = a6 = a8 = a9 = 0.

Euler’s Identity

ϕ(x) = a0 + a1x+ a2x2 + a3x

3 + a4x4 + · · ·

a0 = a5 = a7 = 1

a1 = a2 = −1

a3 = a4 = a6 = a8 = a9 = 0.

Euler’s Identity

With some hard work (or with Mathematica!) one can see that

ϕ(x) = 1− x− x2 + x5 + x7 − x12 − x15 + x22 + x26 − x35 − x40

+ x51 + x57 − x70 − x77 + x92 + x100 − · · ·

The coefficient of xk is non-zero only for k = (3n2 ± n)/2, and is equalto (−1)n in this case.

ϕ(x) = 1− x− x2 + x5 + x7 − · · ·+ (−1)nx3n2−n

2 + (−1)nx3n2+n

2 + · · ·

∞∏n=1

(1− xn) = 1 +∞∑n=1

(−1)n(x

3n2−n2 + x

3n2+n2

Euler’s Identity

ϕ(x) = 1− x− x2 + x5 + x7 − x12 − x15 + x22 + x26 − x35 − x40

+ x51 + x57 − x70 − x77 + x92 + x100 − · · ·

ϕ(x) = 1− x− x2 + x5 + x7 − · · ·+ (−1)nx3n2−n

2 + (−1)nx3n2+n

2 + · · ·

∞∏n=1

(1− xn) = 1 +∞∑n=1

(−1)n(x

3n2−n2 + x

3n2+n2

Euler’s Identity

ϕ(x) = 1− x− x2 + x5 + x7 − x12 − x15 + x22 + x26 − x35 − x40

+ x51 + x57 − x70 − x77 + x92 + x100 − · · ·

ϕ(x) = 1− x− x2 + x5 + x7 − · · ·+ (−1)nx3n2−n

2 + (−1)nx3n2+n

2 + · · ·

∞∏n=1

(1− xn) = 1 +∞∑n=1

(−1)n(x

3n2−n2 + x

3n2+n2

Euler’s Identity

ϕ(x) = 1− x− x2 + x5 + x7 − x12 − x15 + x22 + x26 − x35 − x40

+ x51 + x57 − x70 − x77 + x92 + x100 − · · ·

ϕ(x) = 1− x− x2 + x5 + x7 − · · ·+ (−1)nx3n2−n

2 + (−1)nx3n2+n

2 + · · ·

∞∏n=1

(1− xn) = 1 +

∞∑n=1

(−1)n(x

3n2−n2 + x

3n2+n2

The Partition Function

If n = 1, 2, 3, . . . is a natural number, a partition of n is a representationof n as a sum of natural numbers, repeated summands allowed. Sums withthe same terms but in different orders are considered the same partition.We let p(n) denote the number of partitions of n.

p(1) = 1 (1 = 1)

p(2) = 2 (2 = 2; 1 + 1);

p(3) = 3 (3 = 3; 1 + 2; 1 + 1 + 1);

p(4) = 5 (4 = 4; 1 + 3; 2 + 2; 1 + 1 + 2; 1 + 1 + 1 + 1);

p(5) = 7 (5 = 5; 1 + 4; 2 + 3; 1 + 1 + 3; 1 + 2 + 2);

1 + 1 + 1 + 2; 1 + 1 + 1 + 1 + 1)

You can, with some hard work, probably compute p(10) by hand. Directcomputation of p(n) becomes exceedingly difficult as n increases!

p(1) = 1 (1 = 1)

p(2) = 2 (2 = 2; 1 + 1);

p(3) = 3 (3 = 3; 1 + 2; 1 + 1 + 1);

p(4) = 5 (4 = 4; 1 + 3; 2 + 2; 1 + 1 + 2; 1 + 1 + 1 + 1);

p(5) = 7 (5 = 5; 1 + 4; 2 + 3; 1 + 1 + 3; 1 + 2 + 2);

1 + 1 + 1 + 2; 1 + 1 + 1 + 1 + 1)

p(1) = 1 (1 = 1)

p(2) = 2 (2 = 2; 1 + 1);

p(3) = 3 (3 = 3; 1 + 2; 1 + 1 + 1);

p(4) = 5 (4 = 4; 1 + 3; 2 + 2; 1 + 1 + 2; 1 + 1 + 1 + 1);

p(5) = 7 (5 = 5; 1 + 4; 2 + 3; 1 + 1 + 3; 1 + 2 + 2);

1 + 1 + 1 + 2; 1 + 1 + 1 + 1 + 1)

Partitions and Euler

For convenience, let’s put p(0) = 1 and write

π(x) = p(0)+p(1)x+p(2)x2+p(3)x3+· · · = 1+x+2x2+3x3+5x4+7x5+· · ·

Theorem. The series ϕ(x) and π(x) are reciprocals of each other. Thatis, for all x, we have

ϕ(x)π(x) = 1.

Our proof of the theorem will use an identity for the sum of a geometricseries:

1− r= 1 + r + r2 + r3 + · · ·

whenever −1 < r < 1.

π(x) = p(0)+p(1)x+p(2)x2+p(3)x3+· · · = 1+x+2x2+3x3+5x4+7x5+· · ·

ϕ(x)π(x) = 1.

1− r= 1 + r + r2 + r3 + · · ·

π(x) = p(0)+p(1)x+p(2)x2+p(3)x3+· · · = 1+x+2x2+3x3+5x4+7x5+· · ·

ϕ(x)π(x) = 1.

1− r= 1 + r + r2 + r3 + · · ·

Proof of the Theorem

We compute

ϕ(x)=

1− x· 1

1− x2· · · · · 1

1− xk· · · ·

= (1 + x+ x2 + x3 + · · · )(1 + x2 + x4 + x6 + · · · )· · · (1 + xk + x2k + x3k + · · · ) · · ·

The expanded form of this last product is the sum of all monomial terms

xa1x2a2 · · ·xkak ,

where the integers a1, . . . , ak ≥ 0.

We compute

ϕ(x)=

1− x· 1

1− x2· · · · · 1

1− xk· · · ·

= (1 + x+ x2 + x3 + · · · )(1 + x2 + x4 + x6 + · · · )· · · (1 + xk + x2k + x3k + · · · ) · · ·

The expanded form of this last product is the sum of all monomial terms

xa1x2a2 · · ·xkak ,

where the integers a1, . . . , ak ≥ 0.

For any n ≥ 1, the term xn appears in this sum as many times as we canrepresent n as a sum of the form

n = a1 + 2a2 + · · ·+ kak.

Or, equivalently, as many times as we can write

n = 1 + · · ·+ 1︸︷︷︸a1

+2 + · · ·+ 2︸︷︷︸a2

+ · · ·+ k + · · ·+ k︸︷︷︸ak

It follows that the coefficient of xn in 1/ϕ(x) is precisely p(n) so that

ϕ(x)= π(x).

n = a1 + 2a2 + · · ·+ kak.

n = 1 + · · ·+ 1︸︷︷︸a1

+2 + · · ·+ 2︸︷︷︸a2

+ · · ·+ k + · · ·+ k︸︷︷︸ak

ϕ(x)= π(x).

n = a1 + 2a2 + · · ·+ kak.

n = 1 + · · ·+ 1︸︷︷︸a1

+2 + · · ·+ 2︸︷︷︸a2

+ · · ·+ k + · · ·+ k︸︷︷︸ak

ϕ(x)= π(x).

A Recursion for p(n)

Using Euler’s Identity, we can write

(1− x− x2 + x5 + x7 − · · · )(p(0) + p(1)x+ p(2)x2 + · · · ) = 1.

Expand the product on the left side and set the coefficients ofx, x2, x3, . . . equal to zero.

p(1)− p(0) = 0;

p(2)− p(1)− p(0) = 0;

p(3)− p(2)− p(1) = 0;

· · · · · · · · · · · ·p(n)− p(n− 1)− p(n− 2) + p(n− 5) + p(n− 7)− · · · = 0;

The sum continues as long as n− k ≥ 0.

A Recursion for p(n)

Using Euler’s Identity, we can write

(1− x− x2 + x5 + x7 − · · · )(p(0) + p(1)x+ p(2)x2 + · · · ) = 1.

Expand the product on the left side and set the coefficients ofx, x2, x3, . . . equal to zero.

p(1)− p(0) = 0;

p(2)− p(1)− p(0) = 0;

p(3)− p(2)− p(1) = 0;

· · · · · · · · · · · ·p(n)− p(n− 1)− p(n− 2) + p(n− 5) + p(n− 7)− · · · = 0;

The sum continues as long as n− k ≥ 0.

A Paper Partition Calculator

It follows that for n ≥ 1,

p(n) = p(n− 1) + p(n− 2)− p(n− 5)− p(n− 7) + · · ·

In one of the strips you were just handed (the narrow one), we have beguna table for the values of p(n). We have filled in the values of p(n) forn ≤ 5.

The other strip has an asterisk (*) in the bottom row, and then, countingfrom the bottom, +1′s in rows 1 and 2, −1′s in rows 5 and 7, +1′s inrows 12 and 15 and so on.

p(n) = p(n− 1) + p(n− 2)− p(n− 5)− p(n− 7) + · · ·

Place the asterisk adjacent to the first empty row on the left strip (n = 6).

Sum the numbers p(n) in the left strip that are adjacent to pluses andsubtract all those p(n) that are adjacent to minuses.

Enter the result into the row adjacent to the asterisk.

This new number is, by our recursion, p(6).

Now slide the right strip down one row and repeat.

In a few minutes, you will have computed a large number of values of p(n)!

You can, in your spare time, see if you can verify that p(50) = 204226.

Proof of Euler’s Identity

We need to remove the parentheses in the infinite product

(1− x)(1− x2)(1− x3)(1− x4) · · · .

The result is an infinite series in which ±xn appears as many times as wecan represent

n = n1 + n2 + · · ·+ nk

where n1 < n2 < · · · < nk. The sign will be + if k is even, and − if k isodd.

For the rest of the lecture, we will use the word S-partition to refer to anordered sequence (n1, . . . , nk) of positive integers n1, . . . , nk withn = n1 + n2 + · · ·+ nk and n1 < n2 < · · · < nk. We call k the length ofthe S-partition.

(1− x)(1− x2)(1− x3)(1− x4) · · · .

n = n1 + n2 + · · ·+ nk

(1− x)(1− x2)(1− x3)(1− x4) · · · .

n = n1 + n2 + · · ·+ nk

S-partitions and Young Tableaux

An S-partition (n1, . . . , nk) can be visualized using Young Tableaux: atable consisting of k rows with n1 columns in the first row, n2 columns inthe second row and so on.

1 2 3 . . . n1

1 2 3 . . . . . . n2...

......

1 2 3 . . . . . . . . . . . . nk

The length of the S-partition k is the number of rows in the table. Thetotal number of squares n = n1 + · · ·+ nk.

Classifying S-partitions

For a given S-partition, let s be the largest integer such that the last selements nk−s+1, . . . , nk of the S-partition are consecutive integers.

s = 2 for the S-partition 12 = 1 + 5 + 6

s = 1 for the S-partition 12 = 2 + 4 + 6

s = 3 for the S-partition 12 = 3 + 4 + 5.

Classifying S-partitions

We divide S-partitions (n1, . . . , nk) into three classes:First Class: If n1 ≤ s with the exception of the case where n1 = s = k.

n1 = 1, s = 2, k = 3

Second Class: If n1 > s with the exception of the case where n1 = s+ 1,s = k.

n1 = 2, s = 1, k = 3

Third Class: Either of the two exceptional cases; s = k and n1 = s ors+ 1.

n1 = 3, s = 3, k = 3T.J. Evans (HSU) Parentheses 22 / 29

First and Second Class S-partitions

To a S-partition (n1, . . . , nk) of first class, we assign the second classS-partition

(n2, . . . , nk−n1 , nk−n1+1 + 1, . . . , nk + 1) if k − n1 ≥ 2,

(n2 + 1, . . . , nk + 1) if k − n1 = 1.

(3, 5, 7, 8, 9, 10)n1 = 3, s = 4, k = 6

(5, 7, 9, 10, 11)n1 = 5, s = 3, k = 5

This gives us a mapping from the set of first class S-partitions to the setof second class S-partitions.

To a S-partition (n1, . . . , nk) of first class, we assign the second classS-partition

(n2, . . . , nk−n1 , nk−n1+1 + 1, . . . , nk + 1) if k − n1 ≥ 2,

(n2 + 1, . . . , nk + 1) if k − n1 = 1.

(3, 5, 7, 8, 9, 10)n1 = 3, s = 4, k = 6

(5, 7, 9, 10, 11)n1 = 5, s = 3, k = 5

This gives us a mapping from the set of first class S-partitions to the setof second class S-partitions.

This mapping is in fact a one-to-one correspondence for if (n1, . . . , nk) ofsecond class, we assign it the first class S-partition:

(s, n1, . . . , nk−s, nk−s+1 − 1, . . . , nk − 1) if k − s ≥ 1,

(s, n1 − 1, . . . , nk − 1) if k − s = 0.

(5, 7, 9, 10, 11)n1 = 5, s = 3, k = 5

(3, 5, 7, 8, 9, 10)n1 = 3, s = 4, k = 6

The Three S-Partitions for n = 5

If n = 5, there are three S-partitions:

(1, 4) (first class)

(5) (second class)

(2, 3) (third class)

The Four S-Partitions for n = 6

(1, 2, 3)(first class)

(1, 5)(first class)

(2, 4)(second class)

(6)(second class)

The Big Cancel

Since corresponding S-partitions have lengths that differ by 1, the terms±xn associated to them all cancel leaving only terms associated to thirdclass S-partitions.

Third class S-partitions have the form

(k, k + 1, . . . , 2k − 1) or (k + 1, k + 2, . . . , 2k).

For these, we have

n = k + (k + 1) + · · ·+ (2k − 1) =3k2 − k

n = (k + 1) + (k + 2) + · · ·+ 2k =3k2 + k

The Big Cancel

(k, k + 1, . . . , 2k − 1) or (k + 1, k + 2, . . . , 2k).

For these, we have

n = k + (k + 1) + · · ·+ (2k − 1) =3k2 − k

n = (k + 1) + (k + 2) + · · ·+ 2k =3k2 + k

The Big Cancel

(k, k + 1, . . . , 2k − 1) or (k + 1, k + 2, . . . , 2k).

For these, we have

n = k + (k + 1) + · · ·+ (2k − 1) =3k2 − k

n = (k + 1) + (k + 2) + · · ·+ 2k =3k2 + k

Missed Opportunities

The Fuchs’ article from which this talk is taken has more material onpowers of Euler’s function, an incredible connection between certainpowers and a sequence of special numbers in theoretical physics, and anamazing story of the mathematicians and physicists that discovered theseconnections independently! I highly recommend you check it out!

Thank you!

removing parentheses - humboldt state university · \kvant" (\quantum" in russian) ... a...

Documents