since a p 1 1 (mod p). x = 0 1 ( 1) p 1 (p 1)! (mod p) (p 1)! 1 (mod p) for p odd and for p = 2, (2 1)! = 1! = 1 1 (mod 2).

Transkript

1 5 Sice φ = ϕ is multiplicative, if = m j= pα j j is the stadard factorisatio, m φ() = φ(p α j j ). j= Theorem so ( φ(p α ) = p α ) p φ() = ( ). p p Proof. Cosider the atural umbers i the iterval j p α. There are p α = p α p multiples of p ad the rest are coprime with p, (j, p) = hece (j, p) =. Therefore φ(p α ) = p α p α = p α ( ). p Ex φ(00) = φ( ) ( = 00 ) ( ) 2 5 ( ) ( ) 4 = = 40 40% are coprime with 00 Theorem 2 (Wilso) If p P, (p )! (mod p). Proof. I Z p, f(x) = x p has degree p ad roots [ ] p, [ 2 ] p,..., [ p ] p sice a p (mod p). x = 0 ( ) p (p )! (mod p) (p )! (mod p) for p odd ad for p = 2, (2 )! =! = (mod 2). Note The coverse also holds. Note o Fermat Numbers These ca be defied as F 0 = 3 F + = F 2 2F + 2, 0

2 6 sice the F = (F ) 2 = (F 2 ) 22. = (F 0 ) 2 = 2 2 so F = Propositio 23 (F, F m ) = m.

3 7 3 Möbius Fuctio ad Möbius Iversio (Mathematica: MoebiusMu[]) Defiitio if = µ() = ( ) m if is a product of m distict p P 0 if p P with p 2 Ex µ() =, µ(2) =, µ(6) = ( ) 2 =, µ(p) =, µ(4) = 0, µ(2) = µ(2 2 3) = 0 Propositio 24 µ is multiplicative. Proof. Let (a, b) =, a = m i= pα i i, b = j= qβ j j. If α i or β i 2 the µ(ab) = 0 ad µ(a) or µ(b) = 0 so µ(ab) = 0 = µ(a)µ(b). If ot µ(ab) = ( ) +m = ( ) m ( ) = µ(a)µ(b) so µ is multiplicative. Defiitio I() = { if = 0 if >. Propositio 25 If f() is multiplicative ad ot idetically zero, the f() =. Proof. (, a) = f( a) = f()f(a) so f(a) = f()f(a). If we choose a so f(a) 0 the = f(). Theorem 3 Let g() ad h() be multiplicative. The the fuctio f() = ( ) g(d)h d d is also multiplicative. Proof. Let (a, b) =. The f(ab) = ( ) ab g(d) h d d ab ( ) ab = g(d) h d = u a = u a d = uv u a, v b ( ) ab g(uv) h uv v b v b ( a g(u) g(v) h h u) ( ) b v

4 8 sice (u, v) = ( a, b u v) =. f(ab) = ( ( ) a b g(u) h g(v) h u) v u a v b = ( a ) g(u) h ( ) b g(v) h u v u a v b = f(a)f(b). Propositio 26 Let f be multiplicative ad ot idetically zero. The µ(d)f(d) = ( f(p)) () p d where the product icludes oe term for each prime divisor of. Proof. Let g() = µ()f() ad h() = i Theorem 3. The LHS of equatio () is d g(d)h ( d ) so is multiplicative. The RHS of () is also multiplicative sice if = ab the (a, b) =, p p a or p b. At = LHS = µ()f() = RHS = empty product = (by defiitio). At = p α LHS = µ(d)f(d) d p α = µ()f() + µ(p)f(p) + µ(p 2 )f(p 2 ) + = + ( )f(p) = f(p) RHS = p p α ( f(p)) = f(p) = LHS Hece they are equal, sice multiplicative fuctios are determied by their values at ad prime powers.

5 9 Propositio 27 If > 0, µ(d) = I() = d { if = 0 if >. Proof. Let f(d) = i Propositio 26 ad ote µ(d) = µ() =. d Theorem 4 φ(d) = d Proof. Let S = {, 2,, }. If d let A(d) = {k : (k, ) = d, k }. The S = d A(d) (i.e. disjoit uio ) #S = d #A(d) or = d f(d) where f(d) = #A(d). But ( k (k, ) = d d, ) = ad d 0 < k 0 < k d d so if q = k d there is a - correspodece betwee q N satisfyig 0 < q d ad ( q, d ) =. i.e. f(d) = φ( d ) Hece = d φ( d ) But as d rus through the divisors of, so does d. Hece = d φ(d). Ex Divisors of 6 are {, 2, 3, 6} ad φ() + φ(2) + φ(3) + φ(6) = + (2 ) + (3 ) + 6 = = 6 ( ) ( ) 2 3

6 20 Dirichlet Multiplicatio Defiitio If f ad g are two real fuctios o N the defie their Dirichlet product (or covolutio) h() as h() = ( ) f(d)g = (f g)(). d d Propositio 28 I f = f I = f where { if = I() = 0 if > Propositio 29 f g = g f (commutative law) (f g) k = f (g k). (associative law) Defiitio The fuctio u() = N. The for Propositio 27: µ(d) = I() is µ u = I (2) d ad for Theorem 4: φ(d) = is φ u = N d where N() = is the idetity. If f() 0 there is a uique fuctio f with f f = f f = I. Ex By (2) u = µ, u = µ. Theorem 3 says if f ad g are multiplicative the so is f g, their Dirichlet product. Theorem 5 (Möbius Iversio Formula) f() = d g(d) g() = d ( ) µ(d)f d

7 2 Proof. ( ) f = g u f µ = (g u) µ = g (u µ) = g I = g. ( ) g = f µ g u = (f µ) u = f (µ u) = f I = f. Ex Theorem 4: φ(d) = d φ u = N φ() = (µ N)() = d µ(d) d = d µ(d) d Liouville s Fuctio Defiitio = m i= p α i i λ() = ( ) m α i The λ is completely multiplicative. Theorem 6, λ(d) = d { if is a square 0 otherwise.

8 22 Proof. Let g() = d λ(d). The g = λ u is multiplicative as the Dirichlet product of multiplicative fuctios. So we eed to compute g(p α ) for p P ad α =, 2, 3,... g(p α ) = d p α λ(d) = λ() + λ(p) + λ(p 2 ) + + λ(p α ) = ( ) α { 0 if α is odd = if α is eve. If = m i= pα i i ad is ot a square, the j so α j is odd, hece g() = m i= g(pα i i ) = 0 sice the j th term is zero. If is a square each α i is eve, hece g(p α i i ) = i g() =.

9 23 4 Averages of Arithmetic Fuctios Defiitio is the divisor fuctio. d() = d d = # of divisors of N The, as a fuctio of, d is very irregular. d(p) = 2 p P but d() ca be very large. Averages are smoother d() = d(j) ideed (later) Need the partial sums lim j= d() log() =. D(x) = j x where we defie D(x) = 0 for 0 < x <. So D(x) = d() + d(2) + + d( x ), x. Later we prove Dirichlet s theorem: d(j) x D(x) = x log(x) + (2γ )x + O ( x ) (γ is Euler s costat) where f(x) = O(g(x)) if x 0, M > 0 such that x x 0, f(x) Mg(x) defies the big-oh otatio, ad f(x) = h(x) + O(g(x)) f(x) h(x) = O(g(x)). Ex x = O(x 2 ), x 2 + 7x + 20 = O(x 2 ) Normally, f(x) is umber theoretic, like D(x), h(x) is ice ad smooth, g(x) is a ice power, or other simple mop up for the radom variatio i f(x) e.g. D(x) = x log(x)+ O(x). Defiitio We say f(x) is asymptotic to g(x) as x if ad write f(x) g(x), x. f(x) lim x g(x) = So D(x) x log(x) as x sice D(x) x log(x) = x log(x) ( ) (2γ )x x + x log(x) x log(x) + O x log(x)

10 24 D(x) x log(x). From this it follows that d() log(). Theorem 7 (Euler Summatio) where 0 < y < x, the S = f() = y< x y y If f has a cotiuous derivative f o [y, x] R f(t) dt + (t t )f (t) dt + f(x)( x x) f(y)( y y). (3) Proof. Let m = y, k = x. If, [y, x]: t f (t) dt = ( )f (t) dt = ( )(f() f( )) = {f() ( )f( )} f() Summig from = m + 2 to = k, the sum i braces ({ }) telescopes to give Hece k m+ t f (t) dt = kf(k) (m + )f(m + ) k S = = k f() =m+2 = kf(k) mf(m + ) f() m+ Itegratig f(t) dt (by parts) gives y y y y< x t f (t) dt + kf(k) mf(m + ) t f (t) dt + kf(x) mf(y). (4) f(t) dt = xf(x) yf(y) y tf (t)dt. (5) The (4) (5) (3). Theorem 8 where γ = ( ) = log(x) + γ + O x x t t t 2 dt = lim x ( x ) log(x).

11 25 Proof. Let f(t) = t i Theorem 7 with y = so f (t) = t 2 Now So 0< x where γ = { }. = + < x = + t dt + t t dt + x x t 2 x dt x = t t t dt + x x t 2 x ( ) t t = log(x) dt + + O t 2 x { } = log(x) + dt + 0 x x t t t 2 t t t 2 dt x x t 2 dt = x. ( ) = log(x) + γ + O x x ad f()( ) t t t 2 ( ) dt + O x Note: γ = is Euler s costat (EulerGamma i Mathematica). It could be ratioal, but probably is ot. By Theorem 8, ( ) lim x log(x) = γ + 0 = γ. Sice log(x) as x, = Theorem 9 (Dirichlet) D(x) = x =, quoted earlier. d() = x log(x) + (2γ )x + O ( x ) (6) Proof. d() = d D(x) = d() = x x Now d = qd so we ca express the double sum as d D(x) = q, d qd x

12 26

13 27 This is a sum over a set of lattice poits i the q d plae with (q, d) such that qd = ad =, 2, 3,..., x. We sum these horizotally: D(x) = d x q x d But so = x + O() (Ex) i x { x d + O() } D(x) = d x = x d + O(x) d x ( ( )) = x log(x) + γ + O + O(x) x = x log(x) + O(x) (7) This is weaker tha (6). To prove (6) we use the symmetry of the set of poits: D(x) = 2 { x } d + d x d x = 2#(below lie q = d) + #(o q = d) (8) But y R, y = y + O() so (8) D(x) = 2 { x } d d + O() + O ( x ) d x = 2x d d 2 d + O ( x ) x d x ( = 2x log ( x ) ( )) ( x + γ + O x O( x )) + O ( x ) = x log(x) + (2γ )x + O ( x ) where we have use Lemma below for the middle sum. Lemma If α 0, x α = xα+ α + + O(xα ).

14 28 Proof. I Theorem 7 (Euler Summatio), let f(t) = t α, f (t) = αt α α = + 0< x = < x α t α dt + α = xα+ α + α + + O = xα+ α + + O(xα ) t α (t t ) dt + (x x )x α ( ) α t α dt + O(x α ) Note: Improvemets i the error term O( x) i Dirichlet s theorem for d() have come at great cost: 903 Vorooi O ( x /3 log(x) ) 922 va der Corput O ( x 33/00) 969 Kolesik O ( x ε+2/37) ε > 0 95 Hardy ad Ladau O ( x θ) θ 4 The Distributio of Primes Let Li(x) = 2 dt log(t) for x 2 be the logarithmic itegral ad π(x) = #{p P : 2 p x}. Cosider the followig data: So π(x) x π(x) but π(x) Li(x) is better, ad log(x) x π(x) x Li(x). log(x) 0 as x 0 apparetly. Ideed This distributio is the subject of the famous Prime Number Theorem, which took all of the 9 th cetury to prove.

15 29 Because log(0 ) = log(0) i [2, 00] : about the umbers are prime 2 i [2, 000] : 3 i [2,, 000, 000] : etc. 6 so they progressively thi out with a local desity log(t) #{p P : a p b} = π(b) π(a) b 2 dt log t sice if a < b a 2 dt b log t = dt log(t). a Theorem 20 For 2, 8 π() / log 2. Note: This is as close as we will get to provig the Prime Number Theorem. Lemma (Chebyshev) If H() = j=2 j the 8 π()h() 6. Proof. Proof of Theorem 20 assumig Chebychev s Lemma: For 2, ( ) log = 2 2 dt t < < dt t = log().

16 30 For 4, ( ) 2 log() log. 2 Hece log() H() log() so, by the RHS of Chebychev s Lemma, 2 π() log() ad by the LHS of Chebychev s Lemma π() 2H() 2 usig Lemma 2 whe 4. 8 π()h() π() log() If = 2, π(2) = ad If = 3, π(3) = 2 ad This completes the proof of the theorem. 8 2/ log(2) }{{} / log(3) }{{} Proof of Lemma 2 : Claim : k 0, π(2 k+ ) 2 k (9) Proof: If x > 9, π(x) x sice all eve umbers greater tha 2 are composite. Sice 2 π(2 ) = = 2 0, π(4) = 2 = 2 ad π(8) = r = 2 2, () is true k 0. where H() = Claim : 2 l H(2l ) l (0) H(2 l ) = ( 2 = 3 + ) ( ) ( l ) 2 l ( ) ( ) ( ) l 2 l = l 2

17 3 ad H(2 l ) = ( 2 + ) ( ) ( l 2 + ) ( ) ( ) l 2 l 2 l l This proves the claim. If p P has < p < 2 p 2! ad p! p ( ) 2 = 2!!! <p<2 p ( ) 2 () By Lagrage, the power of p i ( ) 2 is r ( 2 m= p m ) 2 p m (2) where p r 2 < p r+ ad the sum is r sice x, 2x 2 x (See below). Hece ( 2 ) p r 2<p r+ p r By () ad (2) π(2) π() < <p<2 p ( ) 2 p r 2<p r+ p r (2) π(2) (3) Now ad ( ) 2 = so 2(2 ) ( + ) ( ) ( ) 2 ( + ) 2 = 2 2 ( = ) ( ) ( 2 + ) ( ) 2 4 (4)

18 32 Usig LHS of (3) we get π(2) π() < 2 2 ad the RHS gives 2 < (2) π(2),. Now let = 2 k, k = 0,, 2,... so these two iequalities traslate to 2 k(π(2k+ ) π(2 k )) 2 2k+, 2 2k 2 (k+)π(2k+), k 0 or Hece k(π(2 k+ ) π(2 k )) 2 k+, 2 k (k + )π(2 k+ ). (5) (k + )π(2 k+ ) kπ(2 k ) = k(π(2 k+ ) π(2 k )) + π(2 k+ ) 2 k+ + π(2 k+ ) < 2 k+ + 2 k by (9) = 3 2 k Apply this for k = 0,, 2,..., k ad add (π(2 0 ) = π() = 0): (k + )π(2 k+ ) < 3( k ) < 3 2 k+. (6) By (5) ad (6), k 0 2 k+ 2 k + π(2k+ ) < 3 2k+ k +. If N, > choose k so 2 k+ < 2 k+2. By (0) (π is icreasig) (H is icreasig) ad π() π(2 k+2 ) < 3 2k+2 k k+ H(2 k+2 ) 2 k+ π() π(2 k+ ) 2 k + = k+2 (k + ) 2 k+2 8 H(2 k+ ) 8 H() 6 H() as claimed. 8 π() /H() 6 Ex x R, 0 2x 2 x

19 33 Proof. so 0 2x 2 x x x 2 x 2x ad 2 x Z 2 x 2x If x Z, the 2x 2 x = 2x 2 x = 0. If x Z, Z so < x < + ad x = + +ε where ε <. The 2x = 2++2ε x = ε = ε = 2 + x = Hece 2x 2 x = (2 + ) 2 =. Note we have used several times the result y + = y + Z. (Ex) 5 Primes i Gaps primes ca be close together: {, 3}, {29, 3}, {0, 03},... there ca be log stretches of N with o primes: a =! + 2 a 2 =! + 3. a =! + are composite ad cosecutive umbers, so oe are prime ad ca be as large as you like. we will prove the celebrated Bertrad s Hypothesis: N, p P with p < 2. N does there exist a p P with 2 < p < ( + ) 2?

20 34

21 35