Math 782: Analytic Number Theory (Instructor’s Math 782: Analytic Number Theory...

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Transcript of Math 782: Analytic Number Theory (Instructor’s Math 782: Analytic Number Theory...

  • Math 782: Analytic Number Theory (Instructor’s Notes)*

    Analytic Versus Elementary:

    • Terminology (Analytic Number Theory makes use of Complex Analysis and Elemen- tary Number Theory does not; but it isn’t so simple to distinguish.)

    • Writing an integer as a sum of two squares. This is the first of a few examples of how Complex Analysis can be used to answer a question seemingly unrelated to it: if a and b are sums of 2 squares, why must ab be also?

    • Coverings. A covering of the integers is a system of congruences x ≡ aj (mod mj) such that every integer satisfies at least one of these congruences. A perfect covering is one in which each integer satisfies exactly one of the congruences. Suppose we have a finite perfect covering of the integers with each modulus mj > 1. Show that the largest modulus occurs twice. Use that

    1 1 − z =

    r∑ j=1

    zaj

    1 − zmj for |z| < 1,

    where 1 < m1 ≤ m2 ≤ · · · ≤ mr and 0 ≤ aj < mj . Suppose that mr−1 6= mr. Then the right-hand side, and not the left-hand side, gets large (in absolute value) as z approaches exp(2πi/mr). The contradiction implies mr−1 = mr.

    • Preliminaries on exponentials. The following are useful formulas: n−1∑ j=0

    e(i2πk/n)j = {

    0 if k 6= 0 n if k = 0,

    ∫ 1 0

    ei2πkθ dθ = 1 2π

    ∫ 2π 0

    eikθ dθ = {

    0 if k 6= 0 1 if k = 0.

    • Putnam Problem A-6, 1985. If p(x) = a0 + a1x+ · · · + amxm is a polynomial with real coefficients aj , then set Γ(p(x)) = a20 + a

    2 1 + · · · + a2m. Let f(x) = 3x2 + 7x + 2.

    Find, with proof, a polynomial g(x) with real coefficients such that (i) g(0) = 1, and (ii) Γ(f(x)n) = Γ(g(x)n) for all positive integers n.

    There are perhaps better ways to do this problem, but we use what we just learned to establish

    Γ(p(x)) = ∫ 1

    0

    p ( e2πiθ

    ) p ( e−2πiθ

    ) dθ.

    Note that f(x) = (3x+ 1)(x+ 2). That g(x) = 6x2 + 5x+ 1 = (3x+ 1)(2x+ 1) provides a solution follows from

    Γ (f(x)n) = ∫ 1

    0

    f ( e2πiθ

    )n f ( e−2πiθ

    )n dθ

    *These notes are from a course taught by Michael Filaseta in the Fall of 1996.

    1

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    = ∫ 1

    0

    ( 3e2πiθ + 1

    )n ( e2πiθ + 2

    )n ( 3e−2πiθ + 1

    )n ( e−2πiθ + 2

    )n dθ

    = ∫ 1

    0

    ( 3e2πiθ + 1

    )n ( 2e2πiθ + 1

    )n ( 3e−2πiθ + 1

    )n ( 2e−2πiθ + 1

    )n dθ

    = ∫ 1

    0

    g ( e2πiθ

    )n g ( e−2πiθ

    )n dθ

    = Γ (g(x)n) .

    • A Geometric Problem. Let R be a rectangle, and suppose R can be expressed as a union of rectangles Rj with edges parallel to R and common points only along these edges. Suppose each Rj has at least one edge of integer length. Then R itself must have an edge of integer length.

    Let R be positioned so that it is in the first quadrant of the xy-plane, with an edge on each of the x and y axes. Let Rj have bottom left-hand corner (uj , vj), and let αj and βj be the horizontal and vertical dimensions of Rj respectively. Observe that∣∣∣∣

    ∫ w+γ w

    e2πit dt

    ∣∣∣∣ = 12π ∣∣e2πiγ − 1∣∣ = 0 ⇐⇒ γ ∈ Z.

    One uses that one of αj and βj is an integer for each j and that∣∣∣∣ ∫ ∫ R

    e2πi(x+y) dx dy

    ∣∣∣∣ = ∣∣∣∣

    r∑ j=1

    ∫ ∫ Rj

    e2πi(x+y) dx dy

    ∣∣∣∣ to obtain the desired result.

    Homework:

    (1) (a) If a and b are integers which can each be expressed in the form x2 + 5y2 for some integers x and y, explain why it is possible to express ab in this form as well.

    (b) Determine whether the following is true: if a and b are integers and ab can be expressed in the form x2 + 5y2 with x and y integers, then either a or b can be as well. Either prove the result or give a counterexample.

    (2) (Putnam Problem A-5, 1985) For m a positive integer, define

    Im = ∫ 2π

    0

    cosx cos(2x) cos(3x) · · · cos(mx) dx.

    (a) Use one of the preliminary results above (yes, you must to get credit for the problem) to determine with proof for which integers m with 1 ≤ m ≤ 10 we have Im 6= 0. (Hint: Use that cosx = (eix + e−ix)/2.)

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    (b) Generalize part (a). In other words, determine which positive integers m are such that Im 6= 0. The answer should be in the form: Im 6= 0 if and only if m ≡ u (mod v) where v is an integer and u is a list of integers. Justify your answer.

    Elementary Aspects:

    • The Fundamental Theorem of Arithmetic (the atoms of which the matter called integers is made are “primes”)

    • There are infinitely many primes (Give three proofs: (i) Euclid’s, (ii) 22n + 1 is divisible by a prime > 2n+1, and (iii)

    ∏ p(1 − 1/p2)−1 = π2/6, an irrational number.)

    • Explain the notion of rearranging the terms of a series as well as the following lemmas about absolute convergence.

    Lemma 1. If the terms of an absolutely converging series ∑∞

    n=1 an with an ∈ C are rearranged, then the value of the series remains unchanged. On the other hand, if the series

    ∑∞ n=1 an with an ∈ R converges but not absolutely, then any real value can be

    obtained from an appropriate rearrangement of the series.

    Use the example of the alternating harmonic series ∑∞

    n=1(−1)n+1/n to illustrate the second half of the lemma (and to give the first half more meaning). For the first half, let S =

    ∑∞ n=1 an and let

    ∑∞ n=1 a

    ′ n be a rearrangement of it. Let ε > 0. Fix n0 such that∑

    n>n0 |an| < ε. Let N be sufficiently large so that the terms a1, . . . , an0 appear among

    a′1, . . . , a ′ N . Then ∣∣∣∣

    N∑ n=1

    a′n − S ∣∣∣∣ ≤ ∑

    n>n0

    |an| < ε.

    The first part of the lemma follows.

    Lemma 2. The product of two absolutely converging series converges absolutely.

    The product of ∑∞

    n=1 an and ∑∞

    n=1 bn is ∑∞

    n=2 cn where cn = ∑n−1

    k=1 akbn−k. The result follows since

    ∑N n=2 |cn| is an increasing function of N bounded above by

    ( N∑ n=1

    |an| )( N∑

    n=1

    |bn| )

    ≤ ( ∞∑

    n=1

    |an| )( ∞∑

    n=1

    |bn| ) .

    In fact, ∑∞

    k=1

    ∑∞ `=1 |akb`| converges. Lemma 2 is of interest but is not really needed in

    what follows.

    Homework:

    (1) Let f : (Z+)2 7→ C. We say that the iterated series ∞∑

    n=1

    ∞∑ m=1

    f(m,n) = ∞∑

    n=1

    ( ∞∑ m=1

    f(m,n) )

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    converges if for each positive integer n, the series ∑∞

    m=1 f(m,n) converges, and if

    lim N→∞

    N∑ n=1

    ( ∞∑ m=1

    f(m,n) )

    exists. The value of this limit is called the value of the iterated series. We say that the double series ∑

    1≤n

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    for any M > 0. This is a contradiction (the left-hand side is finite).

    • Logarithms and ∏∞n=1(1 − an). Suppose that 0 ≤ an ≤ 1/2. This condition implies

    (∗) ∞∑

    k=2

    akn k

    ≤ ∞∑

    k=2

    akn ≤ 2a2n ≤ an.

    We use

    log(1 − x) = − ∞∑

    k=1

    xk

    k for |x| < 1 (actually for − 1 ≤ x < 1).

    Observe that

    log N∏

    n=1

    (1 − an) = N∑

    n=1

    log(1 − an) ≥ −2 N∑

    n=1

    an.

    The left-hand side is decreasing so that if ∑∞

    n=1 an converges, then so does ∏∞

    n=1(1− an). The inequality log

    ∏N n=1(1 − an) ≤ −

    ∑N n=1 an implies that if

    ∏∞ n=1(1 − an) converges,

    then so does ∑∞

    n=1 an.

    • The estimate ∏p≤x (

    1 − 1 p

    ) ≤ 1

    log x .

    Use ∏ p≤x

    ( 1 − 1

    p

    )−1 ≥

    [x]∑ n=1

    1 n ≥ ∫ x

    1

    1 t dt = log x.

    • Second proof that ∑p 1/p diverges. Let P (x) =

    ∏ p≤x(1 − (1/p))−1 and S(x) =

    ∑ p≤x 1/p. Then by (∗)

    logP (x) = S(x) + E where 0 ≤ E ≤ 2 ∑ p≤x

    1 p2

    ≤ 2.

    The result follows since P (x) ≥ log x. (Note: we have found a “good” lower bound for∑ p≤x 1/p.)

    • A noteworthy example. The product

    ∏ p(1 − (1/p)) and

    ∑∞ n=1 µ(n)/n are related (if you expand the product

    and rearrange the terms, you get the sum). The value of the product is 0 and the value of the sum is 0; however, it is considerably harder to show the latter. (It is equivalent to the Prime Number Theorem).

    • The notation π(x) and a proof that limx→∞ π(x)/x = 0. Show that in fact π(x) < Cx/ log log x for some constant C by using the sieve of Er-

    atosthenes. (Discuss sieves in general.)

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    Homework:

    (1) (a) Modify the previous argument to show that almost all positive integers n have a prime factor ≤ log n. More precisely (and moreover) show that

    |{n ≤ x : n does not have a prime factor ≤ logn}|

    is bounded above by Cx/ log log x for some constant C. (Hint: Most n ≤ x which do not have a prime factor ≤ log n also do not have a prime factor ≤ (log x)/2.) (b) Suppose w(n) is any function defined for n ∈ Z+ which tends to infinity with n. Prove t