Waring - PowerPoint PPT Presentation

1 / 21
About This Presentation
Title:

Waring

Description:

Waring s Problem M. Ram Murty, FRSC, FNA, FNASc Queen s Research Chair Queen s University Lagrange s theorem In 1770, Lagrange proved that every natural ... – PowerPoint PPT presentation

Number of Views:107
Avg rating:3.0/5.0
Slides: 22
Provided by: RamM154
Category:
Tags: circle | theorem | waring

less

Transcript and Presenter's Notes

Title: Waring


1
Warings Problem
  • M. Ram Murty, FRSC, FNA, FNASc
  • Queens Research Chair
  • Queens University

2
Lagranges theorem
  • In 1770, Lagrange proved that every natural
    number can be written as a sum of four squares.
  • This was first conjectured by Bachet in 1621 who
    verified the conjecture for every number less
    than 326.

Joseph Louis Lagrange (1736-1813)
3
Fermat and Euler
  • Fermat claimed a proof of the four square
    theorem.
  • But the first documented proof of a fundamental
    step in the proof was taken by Euler.

Pierre de Fermat (1601-1665)
Leonhard Euler (1707-1783)
4
Edward Waring and Meditationes Algebraicae
  • In 1770, Waring wrote in his book Meditationes
    Algebraicae that every natural number can be
    written as a sum of four squares, as a sum of
    nine cubes, as a sum of 19 fourth powers and so
    on.
  • This is called Warings problem.

Edward Waring (1736-1798)
5
The problem with cubes
  • Can every natural number be expressed as a sum of
    9 cubes?
  • This was first proved in 1908 by Arthur Wieferich
    (1884-1954)
  • Wieferich was a high school teacher and wrote
    only five papers in his entire life.
  • But all of these papers were of high quality.
  • A small error in Wieferichs paper was corrected
    by A.J. Kempner in 1912.

6
What about fourth powers?
  • Theorem (Balasubramanian, Deshouillers, Dress,
    1986) Every number can be written as a sum 19
    fourth powers.

R. Balasubramanian
J.-M. Deshouillers
7
So what exactly is Warings problem?
  • For each natural number k, there is a number g
    g(k) such that every number can be written as a
    sum of g kth powers.
  • Moreover g(2)4, g(3)9, g(4)19 and so on.
  • The first question is if g(k) exists.
  • The second is what is the formula (if there is
    one) for g(k)?

8
Hilberts Theorem
  • Theorem (Hilbert, 1909) For each k, there is a
    gg(k) such that every number can be written as a
    sum of g kth powers.

David Hilbert (1862-1943)
9
What is g(k)?
  • J.A. Euler (the son of L. Euler) conjectured in
    1772 that g(k) 2k (3/2)k 2.
  • The number 2k(3/2)k-1 lt3k can only use 1s and
    2ks when we try to write it as a sum of kth
    powers.
  • The most frugal choice is (3/2)k -1 2ks
    followed by 1s.
  • This gives g(k) 2k (3/2)k 2.
  • Thus, g(1)1, g(2)4, g(3)9, g(4)19.
  • g(5)37 (J. Chen, 1964)
  • g(6)73 (S. Pillai, 1940)

J. Chen (1933-1996)
S.S. Pillai (1901-1950)
10
Pillais Theorem
  • Write 3k 2kq r, with 0 lt r lt 2k.
  • If rq 2k, then g(k) 2k (3/2)k 2.
  • Equivalent formulation if (3/2)k1-(3/4)k,
    then g(k) 2k (3/2)k 2.
  • Mahler (1957) proved that this condition holds
    for all k sufficiently large. However, his proof
    was ineffective since it uses Roths theorem in
    Diophantine approximation which is ineffective.

11
The circle method
  • In his letter to Hardy written in 1912, Ramanujan
    alluded to a new method called the circle method.

S. Ramanujan (1887-1920)
This was developed by Hardy and Littlewood in
several papers and the method is now called the
Hardy-Littlewood method.
12
The function G(k)
  • Define G(k) as follows. For each k there is an
    no(k) such that every n no(k) can be written as
    a sum of G(k) kth powers.
  • Clearly G(k) g(k).
  • Note that g(k) 2k (3/2)k 2 implies that
    g(k) has exponential growth.
  • Using the circle method, Vinogradov in 1947
    showed that G(k)k(2log k 11)
  • Hardy Littlewood conjectured that G(k)lt4k and
    this is still an open problem.

13
Schnirelmans theorem
  • Let A be an infinite set and set A(n) be the
    number of elements of A less than or equal to n.
  • If B is another infinite set, then what can we
    say about the set AB ab a e A, b e B?
  • Here we allow for the empty choice.
  • For example, is there a relation between A(n),
    B(n), and (AB)(n)?

L. Schnirelman (1905-1938)
14
Schnirelmans density
  • Define the density of A as d(A) infn1 A(n)/n.
  • Thus, A(n)dn for all values of n.
  • Note the density of even numbers is zero
    according to this definition since A(1)0. The
    density of odd numbers is ½.
  • d(A)1 if and only if A is the set all natural
    numbers.
  • Theorem (Schnirelman, 1931)
  • d(AB)d(A)d(B)-d(A)d(B).

15
An elementary approach
  • Let A(n)s, B(n)t. Write a1 lt a2 lt ... lt as n.
  • Let ri B(ai1 ai -1) and write these numbers
    as b1 lt b2 lt ... lt bri
  • Then ai lt ai b1 lt ai b2 lt ... lt ai brilt
    ai1
  • Therefore, (AB)(n) is at least
  • A(n) r1 r2 ... rs-1 B(n-as) B(a1-1)
  • A(n)
  • d(B)((a1-1) (a2-a1-1) ... (as-as-1-1)
    (n-as) )
  • A(n) d(B)(n-s) (1-d(B))A(n)d(B)n
  • (1-d(B))d(A)n d(B)n.

16
An application of induction
  • So we have d(AB) 1-(1-d(A))(1-d(B)).
  • By induction, we have
  • d(A1 ... As) 1 (1-d(A1))...(1-d(As))
  • Notation 2A AA, 3A AAA, etc.
  • Corollary. If d(A)gt0, then for some t, we have
    d(tA) gt ½.
  • Proof. By the above, d(tA) 1-(1-d(A))t.
  • If d(A)1, we are done. Suppose 0ltd(A)lt1.
  • Then, (1-d(A))t tends to zero as t tends to
    infinity.

17
What happens if d(A)gt1/2?
  • Then A(n) gt n/2.
  • This is the size of A a e A, an.
  • Consider the set B n a a e A, a n.
  • This set has size A(n) gt n/2.
  • If B and A are disjoint, we get more than n
    numbers which are n, a contradiction.
  • Thus, every number can be written as a sum of two
    elements of A.

18
Consequence of Schnirelmans Theorem
  • If A has positive Schnirelman density, then for
    some t, we have tA is the set of natural numbers.
  • In other words, every number can be written as a
    sum of at most t elements from the set A.
  • Let us apply this observation to count the number
    of numbers n which can be written as a sum of t
    kth powers.

19
X1k X2k ... Xtk n
  • Let A be the set of numbers that can be written
    as a sum of t kth powers.
  • Key lemma The number of solutions is at most
    A(n)nt/k-1.
  • On the other hand, a lower bound is given by
    (n/t)1/kt .
  • This gives that A(n) has positive Schnirelman
    density.

U.V. Linnik (1915-1972)
20
Some open problems
  • Can every sufficiently large number be written as
    a sum of 6 cubes? (Unknown)
  • The conjecture is that G(3)4. What we know is
    that 4G(3)7.
  • Hypothesis K (Hardy and Littlewood) Let rg,k(n)
    be the number of ways of writing n as a sum of g
    kth powers. Then, rk,k(n) O(ne) for any egt0.
  • G(k)max(k1, ?(k)) where ?(k) is the smallest
    value of g predicted by local obstructions.

21
THANK
YOU!
Write a Comment
User Comments (0)
About PowerShow.com