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Ramanujan sums - TTK

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitRamanujan sumsLászló Tóthhttp://www.ttk.pte.hu/matek/ltothInstitute of Mathematics and InformaticsUniversity of PécsHungaryMay 2007Dumlupinar University, KütahyaTurkey

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit1. Periodic functionsAn arithmetic function f : N → C is called periodic if there is a number r ∈ Nsuch that f(n + r) = f(n) for all n ∈ N. This is equivalent to the condition:f(n 1 ) = f(n 2 ) for all n 1 , n 2 ∈ N, n 1 ≡ n 2 (mod r). Here r is called the period ofthe function f and we say that f is periodic (mod r).If f is periodic (mod r), then it can be extended to a function defined on Z, denotedalso by f, such that f(n 1 ) = f(n 2 ) for every n 1 , n 2 ∈ Z, n 1 ≡ n 2 (mod r).Examples. 1) The function f(n) = (n, r), where (n, r) is the gcd of n and r, isperiodic (mod r).2) e(n) := exp(2πin/r) = cos(2πn/r) + i sin(2πn/r) is periodic (mod r). Herethe values of e(n) are the roots of unity of order r. More generally, e k (n) :=exp(2πikn/r) is also periodic (mod r), where k ∈ N is fixed and e k (n) are the k-thpowers of the roots of unity of order r.3) The function c r (n) := ∑exp(2πikn/r), called Ramanujan sum, is alsok (mod r)(k,r)=1periodic (mod r), here the sum is over a reduced residue system (mod r).

If r is a period of the function f, then each multiple of r is also a period. Thereexists a least (positive) period. What is the set of all periods?The answer is given byTheorem 1.1. If r 1 and r 2 are periods of the function f, then the gcd (r 1 , r 2 )is also a period. The least period divides each period, therefore the set ofperiods is the set of multiples of the least period.Proof. There exist u, v ∈ Z such that (r 1 , r 2 ) = ur 1 + vr 2 . Using that r 1 and r 2are periods,f(n + (r 1 , r 2 )) = f(n + ur 1 + vr 2 ) = f(n + ur 1 ) = f(n)for all n. If r 0 is the least period and r is an other period, then (r 0 , r) is also one,and since r 0 ≤ (r 0 , r) we obtain r 0 | r. •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTheorem 1.2.∑k (mod r)e k (n) =∑k (mod r)exp(2πikn/r) ={r, if r | n,0, if r ∤ n.Proof. Let ε = exp(2πin/r). By the formula for the sum of geometric sequences,{∑∑r−1exp(2πikn/r) = ε k r, if ε = 1,== 0, if ε ≠ 1,k (mod r)where ε = 1 iff r | n. k=0Remark. A usual notation is e(x) := exp(2πix), x ∈ Q. If a ≡ b (mod r), thene( a r ) = e( b r), and this is the property that gives number theoretic importance tocertain trigonometric sums, involving e(x).We show that every periodic function (mod r) can be written as a sum of thefunctions e k (n) = exp(2πikn/r). More exactly,ε r −1ε−1

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTheorem 1.3. If the function f is periodic (mod r), then f can be written asr∑(1) f(n) = g(k) exp(2πikn/r), n ∈ N,k=1where the values g(k) are unique and are given by(2) g(k) = 1 r∑f(j) exp(−2πijk/r), 1 ≤ k ≤ r.rj=1Proof. For all n ∈ N, F (n) :==r∑k=11r⎛⎝= 1 rr∑g(k) exp(2πikn/r) =k=1⎞r∑f(j) exp(−2πijk/r) ⎠ exp(2πikn/r) =j=1r∑f(j)j=1r∑exp(2πik(n − j)/r).k=1By Theorem 1.2. the inner sum is r if r | (n − j) ⇔ j ≡ n (mod r) and otherwiseit is 0. Hence F (n) = f(n).

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitNow suppose that f(n) van be written in the form (1) and also in this way:r∑(3) f(n) = g ′ (k) exp(2πikn/r), n ∈ N.k=1We show that g(j) = g ′ (j) for all j. By (1) and (3)r∑(g(k) − g ′ (k)) exp(2πikn/r) = 0, n ∈ N.k=1Multiplying by exp(−2πijn/r) and summing we have:r∑ r∑(g(k) − g ′ (k)) exp(2πin(k − j)/r) = 0,n=1k=1r∑(g(k) − g ′ (k))k=1r∑exp(2πin(k − j)/r) = 0,n=1and using again Theorem 1.2 we obtain (g(j) − g ′ (j))r = 0, consequently g(j) =g ′ (j).

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitRemarks. 1. Formula (1) is called the finite Fourier expansion of f and thevalues g(k) given by (2) are the Fourier coefficients of f.2. The function g given by (2) is also periodic (mod r).3. Let P r denote the set of periodic functions (mod r), which is a complex linearspace with the addition of functions and pointwise multiplication. Moreover, P r isisomorphic to the euclidean space C r , hence its dimension is r and the usual innerproduct of C r can be written here in this form:〈f, g〉 = 1 ∑f(k)g(k),rk (mod r)where the sum is over a complete system of residues (mod r) and g(k) is the complexconjugate of g(k).According to Theorem 1.2., the functions e j , 1 ≤ j ≤ r, given by e j (n) =exp(2πijn/r) form an orthonormal system and the Fourier expansion of f isr∑f = g(k)e k , where g(k) = 〈f, e k 〉 = 1 r∑f(j) exp(−2πijk/r),rk=1which is Theorem 1.3, this is in fact the same proof as the proof of above.j=1

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitExercises1.1. a) The Cauchy product of the functions f, g : N 0 = {0, 1, 2, ...} → C isgiven byn∑(f ⊗ g)(n) = f(k)g(n − k) = ∑f(k)g(l)k=0(where the sum has n + 1 terms).k+l=nShow that the set C N 0 of functions f : N 0 → C forms an integral domain withrespect to addition and Cauchy product. When has a function an inverse withrespect to the Cauchy product?b) The Cauchy product of the periodic functions (mod r) f, g is defined by∑(f ⊙ g)(n) =f(k)g(l),k+l ≡ n (mod r)where the sum is over the solutions (mod r) of the congruence k + l ≡ n (mod r),and the sum has r terms.Observe that f ⊙ g is also periodic (mod r).operation?What are the properties of this

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit1.2. Show that the Cauchy product of the exponential functions e s (k) =exp(2πisk/r) and e t (l) = exp(2πitl/r) is{∑re s (n) = r exp(2πisn/r), if s ≡ t (mod r) ,e s (k)e t (l) =0, otherwise.k+l ≡ n (mod r)

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit2. Ramanujan sumsRamanujan sums (S. Ramanujan, 1918) are defined byc r (n) =∑exp(2πikn/r),k (mod r)(k,r)=1where k runs through a reduced system of residues (mod r) and r, n ∈ Z, r ≥ 1.Remarks. 1) The notation c r (n) = c(n, r) is also used.2) If k ≡ k ′ (mod r), then exp(2πikn/r) = exp(2πik ′ n/r), hence this definition iscorrect, it does not depend on the choosed reduced system of residues.3) c r (n) is the sum of n-th powers of the r-th primitive roots of unity.4) If k runs through a complete system of residues (mod r), then by Theorem 1.2,{∑r, if r | n,(1)exp(2πikn/r) =0, if r ∤ n,k (mod r)5) For n = 0, c r (0) = φ(r) is the Euler function.6) For n = 1, c r (1) is the sum of the r-th primitive roots of unity, and this is exactlythe Möbius function µ(r), as it can be seen from the next result.

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTheorem 2.1. For all r, n ∈ Z, r ≥ 1 we havec r (n) = ∑dµ(r/d).d|(n,r)Moreover, all the values of c r (n) are real integers, thereforec r (n) =∑cos(2πkn/r).Proof. c r (n) =∑k (mod r)k (mod r)(k,r)=1exp(2πikn/r) ∑d|(k,r)µ(d), where d | (k, r) ⇔ d | k andd | r, and using the notation k = dj, c r (n) == ∑ ∑µ(d) exp(2πijn/(r/d)) = ∑ {rµ(d)d , if r d | n0, otherwise = ∑ µ(d) r d ,d|r j (mod r/d)d|rd|rrd |nusing (1). Now let r d = δ. Then c r(n) = ∑ δµ( r δ ) = ∑δµ( r δ ). δ|nδ|(n,r)δ|r

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitRemarks. 1. From Theorem 2.1 we have {∑ r, if r | n,c d (n) =0, if r ∤ n,d|rwhich can be shown also directly.2. |c r (n)| ≤ min(σ(n), φ(r)) for all n, r ≥ 1, where σ(n) is the sum of divisors ofn and φ(r) is the Euler function.Indeed, by Theorem 2.1, |c r (n)| ≤ ∑ d|(n,r) d ≤ ∑ d|nd = σ(n), and by definition|c r (n)| ≤ φ(r) ≤ r.

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTheorem 2.2. The function c r (n) is multiplicative in r, that is c rs (n) =c r (n)c s (n) for all r, s ≥ 1, (r, s) = 1 (and for every fixed n ≥ 1). Moreover⎧⎪⎨ p a − p a−1 , if p a | n,c p a(n) = −p⎪⎩a−1 , if p a ∤ n, p a−1 | n,0, if p a−1 ∤ n.Proof. LetI r (n) ={r, if r | n,0, if r ∤ n,which is multiplicative in r (for all (r, s) = 1 we have rs | n ⇔ r | n, s | n), andby Theorem 2.1,(2) c r (n) = ∑ I d (n)µ( r d ),d|rshowing that c r (n) is multiplicative in r, being the convolution of two multiplicativefunctions (c ◦ (n) = I ◦ (n) ∗ µ). Remark. c r (n) is not multiplicative in n. Indeed, if for example p ≠ q are primes,then c p (p) = p − 1, c p (q) = −1 and c p (pq) = p − 1 ≠ −(p − 1) = c p (p)c p (q).

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTheorem 2.3. (Hölder identity) For all n, r ≥ 1,c r (n) = φ(r)µ(m) , where m = rφ(m)(n, r) .Proof. Both sides are multiplicative in r, hence it is enough to prove for r = p a ,a prime power ( Exercise!). Theorem 2.4. (Orthogonality relation) If d | r and t | r, then{1r∑φ(d), if d ≠ t,c d (m)c t (m) =r0, if d = t.m=1Proof. Let r = dd 1 , r = tt 1 . Thenr∑r∑ d∑c d (m)c t (m) =m=1=d∑m=1a=1(a,d)=1t∑a=1 b=1 m=1(a,d)=1 (b,t)=1exp(2πiam/d)t∑b=1(b,t)=1r∑exp(2πim(a/d + b/t)),exp(2πibm/t) =

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quitwhere a/d + b/t = (ad 1 + bt 1 )/r and the inner sum, referring to m, is r ifr | (ad 1 + bt 1 ) and it is 0 otherwise (Theorem 1.2.) Here 1 ≤ d 1 ≤ ad 1 ≤ dd 1 = r,1 ≤ t 1 ≤ bt 1 ≤ tt 1 = r, therefore 1 < ad 1 + bt 1 ≤ 2r and the equality is valid iffd = t = 1, since (a, d) = (b, t) = 1.If d > 1 or t > 1, then r | (ad 1 + bt 1 ) ⇔ ad 1 + bt 1 = r ⇔ add 1 t + bdtt 1 = rdt ⇔at + bd = dt ⇔ d = t and a + b = d. The proof of the last equivalence:,,⇒” d | at, t | bd ⇒ d | t, t | d since (a, d) = (b, t) = 1, hence d = t, a + b = d.,,⇐” is trivial. We obtain that for d = t the sum of above isd∑ d∑ ∑r1 = rφ(d),a=1 b=1 a+b=d(a,d)=1 (b,d)=1since there are φ(d) possible values of a, and for each there is exactly one value ofb.

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTheorem 2.5. a) If r, s ∈ N, then{∑φ(r)x + O(1), if r = s,c r (n)c s (n) =O(1), if r ≠ s.b) If r ∈ N, thenn≤x∑c r (n) =n≤x{x + O(1), if r = 1,O(1), if r > 1.Proof. a) Let r, s be fixed and [x] = (rs)q + α, where 0 ≤ α < rs. Then∑c r (n)c s (n) =n≤x(rs)q∑n=1c r (n)c s (n) +[x]∑n=(rs)q+1c r (n)c s (n) = Σ 1 + Σ 2 .Here |Σ 2 | ≤ ∑ [x]n=(rs)q+1 rs < (rs)2 is bounded in x and by Theorem 2.4Σ 1 = φ(r)[x] = φ(r)x + O(1) for r = s, and Σ 1 = 0 for r ≠ s.b) Let s = 1, then c s (n) = 1 for every n and apply Part a).

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitWe say that an arithmetic function f has a mean value if the limit M(f) =1 ∑lim f(n) exists.x→∞ xn≤xAccording to Theorem 2.5 the product c r (n)c s (n) and c r (n) possess a mean valuegiven by{{φ(r), if r = s,1, if r = 1,M(c r c s ) =M(c r ) =0, if r ≠ s,0, if r > 1.The estimate of ∑ n≤x c r(n) can be given also by the first formula of Theorem 2.1.If r = 1, then c 1 (n) = 1 and it is immediate that ∑ n≤x c 1(n) = [x] = x + R(x),where R(x) = −{x} ∈ (−1, 0]. In case r > 1 we have the following result.

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTheorem 2.6. If r > 1 and x ≥ 1, then | ∑ n≤x c r(n)| ≤ ψ(r), where ψ(r) =r ∏ p|r(1 + 1/p) is the Dedekind function.Proof. Using the first formula of Theorem 2.1 we obtain∑c r (n) = ∑ dµ(r/d) = ∑ ∑dµ(r/d) 1 = ∑n≤xn≤xd|rn=dk≤x d|rd|(n,r)= ∑ d|rdµ(r/d)(x/d − {x/d}) = x ∑ d|rµ(r/d) − ∑ d|rdµ(r/d)[x/d] =dµ(r/d){x/d} == − ∑ d|rdµ(r/d){x/d} = R r (x),where |R r (x)| ≤ ∑ d|r dµ2 (r/d) = ψ(r). Other orthogonality properties of Ramanujan sums are given in the following theorem.

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTheorem 2.7. (Orthogonality properties) If r | k and s | k, then{∑k, if r = s,(3)c r (k/d)c d (k/s) =0, if r ≠ s,(4)d|k∑φ(d)c r (k/d)c s (k/d) =d|k{kφ(r), if r = s,0, if r ≠ s.Proof. We prove relation (3), where k/s has to be an integer, this is why we havecondition s | k. The other condition r | k will be used in the proof. By Theorem2.1,S := ∑ c r (k/d)c d (k/s) = ∑ ∑c d (k/s) δµ(r/δ).d|kd|kδ|(r,k/d)Let r = δl, k/d = δm, thenS = ∑ δl=rδdm=kδµ(r/δ)c d (k/s) = ∑δ|(r,k)δµ(r/δ) ∑ d| k δc d (k/s),

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quitwhere (r, k) = r and the inner sum is k δ for k δ | k sCosequently,S = ∑ δ|rs|δδµ(r/δ) k δ = k ∑ δ|rs|δ⇔ s | δ and it is 0 otherwise.µ(r/δ),here for each term s | r, therefore S = 0 if s ∤ r. If s | r, then denoting r = δu, δ =st,S = k ∑µ( r st ) = k ∑ {µ( r/st ) = k, if r/s = 1,0, otherwise.stu=r t| r sTo prove (4) we need the following: If r | k, s | k, then(5) φ(r)c s (k/r) = φ(s)c r (k/s).This is true, since by the Hölder identity (Theorem 2.3),φ(r)c s (k/r) = φ(r)φ(s)µ( s(s,k/r) )sφ((s,k/r) ) , φ(s)c s (k/s) = φ(s)φ(r)µ( r(r,k/s) )rφ((r,k/s) ) ,

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quitwheres(s, k/r) = r. Now by (5) and (3),(r, k/s)∑φ(d)c r (k/d)c s (k/d) = ∑ φ(s)c d (k/s)c r (k/d) =d|kd|k= φ(s) ∑ {kφ(r), if r = s,c r (k/d)c d (k/s) =0, if r ≠ s. d|k

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTheorem 2.8. The Dirichlet series of the Ramanujan sum c r (n), as a functionin r, is∞∑ c r (n)= σ s−1(n), n ≥ 1, Re s > 1,r s n s−1 ζ(s)n=1where σ k (n) = ∑ d|n dk and ζ is the Riemann zeta function.Proof. c r (n) is bounded in r, since |c r (n)| ≤ σ(n) (see above), the series isabsolutely convergent for Re s > 1 and c r (n) = (I n ∗ µ)(r), where I n (d) = d ford | n and I n (d) = 0 for d ∤ n (Theorem 2.2). We obtain∞∑ c r (n)∞∑ I n (r) ∑ ∞µ(r)∞∑ r==r s r s r s r · 1s ζ(s) =r=1r=1r=1= 1 ∑r 1−s = σ 1−s(n)ζ(s)ζ(s)r|nr=1r|n= σ s−1(n)n s−1 ζ(s) .

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitRemarks. 1. For s = 2 we obtainσ(n)∞∑n= π2 c r (n)=(1 π2+ (−1)n + 2 cos(2πn/3) + 2 cos(πn/2) )+ · · · ,6 r 2 6 2 2 3 2 4 2r=1where σ(n) = ∑ d|nd, showing how the values of σ(n)/n fluctuate harmonicallyabout their mean value π 2 /6.2. A Fourier analysis of arithmetical functions, with respect to Ramanujan sums,parallel to periodic and almost periodic functions, was developed by G. H. Hardy(1921), E. Cohen (1960), J. Knopfmacher (1975), A. Hildebrand (1984),W. Schwarz, J. Spilker(1994) and others.They studied expansions, convergent pointwise or in other sense, of arithmetic functionsf of the form∞∑f(n) = a r c r (n), n ∈ N,r=1where the Ramanujan coefficients a r area r = 1φ(r) M(fc r).

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit∞∑ c r (n)3. It can be shown that for s = 1, = 0. In this case ∑ ∞ µ(r)r=1rr= 0 isr=1convergent, but not absolutely convergent (this is equivalent with the prime numbertheorem).

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitExercises2.1. Show that if n ≥ 1, d | r and e | r, then the Cauchy product of c d (k) andc e (l) is (see Exercises 1.1., 1.2.):{∑rc d (n), if d = e,c d (k)c e (l) =0, if d ≠ e.k+l ≡ n (mod r)2.2. Show that the Dirichlet series of c r (n), as a function in n, is∞∑ c r (n)= ζ(s)φn s 1−s (r), r ≥ 1, Re s > 1,n=1where φ t (r) = ∑ d|r dt µ(r/d) is the generalized Euler function.

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit3. Even functionsThe function f : N → C is called even function (mod r) if f(n) = f((n, r)) forall n, that is if the value f(n) depends only on the gcd (n, r). Hence if f is even(mod r), then it is sufficient to know the values f(d), where d | r.Every function f which is even (mod r) is periodic (mod r), since f(n + r) =f((n + r, r)) = f((n, r)) = f(n) for all n. For example the functions f(n) = (n, r)and f(n) = c r (n) are even functions (mod r).Question: How can even functions (mod r) be characterized?We show that every even function (mod r) is a linear combination of Ramanujansums. More exactly,Theorem 3.1. If f is an even function (mod r), then f can be written in theformf(n) = ∑ h(q)c q (n), n ∈ N,q|rwhere the values h(q) are uniquely determined (Fourier coefficients of f) andh(q) = 1 ∑φ(e)f(r/e)c q (r/e) = 1 ∑f(r/e)c e (r/q), q | r.rφ(q)re|re|r

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitProof. Similar to the proof of Theorem 1.3., using the orthogonality properties ofabove.Another way: Let E r denote the set of even functions (mod r), which is a complexlinear space with the addition of functions and poitwise multiplication. FurthermoreE r is isomorphic to the space C τ(r) , where τ(r) is the number of divisors of r. HenceE r is a Hilbert space of dimension τ(r) and〈f, g〉 = 1 ∑φ(d)f(r/d)g(r/d)rd|ris an inner product, φ(d) denoting the Euler function, where g(r/d) is the complexconjugate of g(r/d).The functions c ′ q(n) = √ 1 c q (n), where q | r, form an orthonormal basis. Indeed,φ(q)if q | r, t | r, then by (4),=〈c ′ q, c ′ t〉 = 1 r∑φ(d)c ′ q(r/d)c ′ t(r/d) =d|r1r √ ∑φ(d)c q (r/d)c t (r/d) =φ(q)φ(t)d|r{1, if q = t,0, if q ≠ t.

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitWe obtain that the Fourier expansion of f according to c ′ q (q | r) isf(n) = ∑ q|rj(q)c ′ q(n), n ∈ N,wherej(q) = 〈f, c ′ q〉 = 1 r∑φ(e)f(r/e)c ′ q(r/e) =e|r1r √ φ(q)∑φ(e)f(r/e)c q (r/e).e|rThereforeh(q) =1√ j(q) = 1 ∑φ(e)f(r/e)c q (r/e).φ(q) rφ(q)e|rUsing also (5),h(q) = 1 ∑φ(q)f(r/e)c e (r/q) = 1 rφ(q)re|rwhich was to be proved. ∑f(r/e)c e (r/q),e|r

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTheorem 3.2. (L. Tóth, 2004) If f is an even function (mod r), then thereexists the main value M(f) of f and it is given byM(f) = 1 (f ∗ φ)(r).rProof. According to Theorem 3.1, f can be written as f(n) = ∑ q|r h(q)c q(n),n ∈ N, whereM(f) = ∑ h(q)M(c q ) = h(1) = 1 ∑φ(e)f(r/e)c 1 (r/e) =rq|re|r= 1 ∑φ(e)f(r/e) = 1 (f ∗ φ)(r). rrExercisese|r3.1. Let f, g be even functions (mod r) with Fourier coefficients α(q) and β(q),respectively. Prove that the Cauchy product f ⊙ g is also an even function (modr) having Fourier coefficients rα(q)β(q).3.2. If f is an even function (mod r), then according to Theorem 3.2, f has a mainvalue M(f) and M(f) = 1 r (f ∗φ)(r). Estimate the difference ∑ n≤x f(n)−xM(f).

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit3.3. Let (s, d) = 1 and let φ(s, d, n) denote the number of elements in thearithmetic progression s, s + d, ..., s + (n − 1)d which are coprime to n. This is ageneralization of the Euler function, φ(1, 1, n) = φ(n) is the Euler function. Provethati) φ(s, d, n) is multiplicative in n,ii) for every prime power p k , φ(s, d, p k ) =(these values do not depend on s),iii) ∑ φ(s, d, n) =n≤xis the Jordan function,iv) f(d) := φ(s, d, n) ={p k (1 − 1/p), (p, d) = 1,p k , p | d,3d2π 2 J(d) x2 +O(x log x), where J(d) = φ 2 (d) = d 2 ∏ p|d (1−1/p2 )φ(n)(n, d)φ((n, d))v) M(f) = n ∏ p|n(1 − 1/p + 1/p 2 ).and this is an even function (mod n) in d,

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitSolutions of the exercises1.1. a) The Cauchy product of the functions f, g : N 0 = {0, 1, 2, ...} → C isgiven byn∑(f ⊗ g)(n) = f(k)g(n − k) = ∑f(k)g(l)k=0(where the sum has n + 1 terms).k+l=nShow that the set C N 0 of functions f : N 0 → C forms an integral domain withrespect to addition and Cauchy product. When has a function an inverse withrespect to the Cauchy product?b) The Cauchy product of the periodic functions (mod r) f, g is defined by∑(f ⊙ g)(n) =f(k)g(l),k+l ≡ n (mod r)where the sum is over the solutions (mod r) of the congruence k + l ≡ n (mod r),and the sum has r terms.Observe that f ⊙ g is also periodic (mod r).operation?What are the properties of this

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitSolution. a) It is immediate that (C N 0, +) is an abelian group, ⊗ is commutative,distributive with respect to addition and ε(0) = 1, ε(n) = 0, n > 0 is the identityelement. Furthermore,((f ⊗ g) ⊗ h)(n) = (f ⊗ (g ⊗ h))(n) =∑f(k)g(l)h(m),k+l+m=nthe operation is associative. There are no divisors of zero: let f, g ≠ 0, wheref(a) ≠ 0, f(b) ≠ 0, and a, b are the least numbers with this property. Then(f ⊗ g)(a + b) = f(a)g(b) + ∑f(k)g(l) = f(a)g(b) ≠ 0,k+l=a+bk≠a,l≠bsince for k < a one has f(k) = 0, and if k > a, then l < b and g(l) = 0.f has an inverse f iff f(0) ≠ 0 and in this casef(0) = 1f(0) , f(n) = − 1 n∑f(k)f(n − k), n > 0.f(0)b) ⊙ is commutative, associative, distributive to addition. There is no identityelement, since it would be ε(0) = 1, ε(n) = 0, n > 0, but this is not periodic.k=1

If P r denotes the set of periodic functions (mod r), then (P r , +, ⊙) is a commutativering.Remark. If f, g ∈ P r , then in general f ⊗ g ≠ f ⊙ g.•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit1.2. Show that the Cauchy product of the exponential functions e s (k) = exp(2πisk/r)and e t (l) = exp(2πitl/r) is{∑re s (n) = r exp(2πisn/r), if s ≡ t (mod r) ,e s (k)e t (l) =0, otherwise.k+l ≡ n (mod r)Solution.=∑k+l ≡ n (mod r)∑k (mod r)l ≡ n−k (mod r)= exp(2πint/r)exp(2πiks/r) exp(2πilt/r) =exp(2πiks/r) exp(2πi(n − k)t/r) =∑k (mod r)exp(2πik(s − t)/r),which is r exp(2πins/r) = r exp(2πint/r) if s ≡ t (mod r) and 0 otherwise, seeTheorem 1.2.

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit2.1. Show that if n ≥ 1, d | r and e | r, then the Cauchy product of c d (k) andc e (l) is (see Exercises 1.1., 1.2.):{∑rc d (n), if d = e,c d (k)c e (l) =0, if d ≠ e.k+l ≡ n (mod r)Solution. Let r = dd 1 , r = ee 1 . Then∑k+l ≡ n (mod r)=c d (k)c e (l) =d∑e∑∑k+l ≡ n (mod r)∑a=1 b=1 k+l ≡ n (mod r)(a,d)=1 (b,e)=1=d∑e∑d∑a=1(a,d)=1exp(2πiak/d)e∑b=1(b,e)=1exp(2πiakd 1 /r) exp(2πible 1 /r) =∑a=1 b=1 ad 1 ≡ be 1 (mod r)(a,d)=1 (b,e)=1r exp(2πiad 1 n/r),exp(2πibl/e) =by Exercise 1.2. Here 1 ≤ d 1 ≤ ad 1 ≤ dd 1 = r, 1 ≤ e 1 ≤ be 1 ≤ ee 1 = r, hencead 1 ≡ be 1 (mod r) valid iff ad 1 = be 1 ⇔ ad 1 de = be 1 de ⇔ are = bde ⇔ ae = bd.

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitBut (a, d) = 1, (b, e) = 1, therefore a | b, b | a, that is a = b, d = e and obtain thatin this case the sum isd∑r exp(2πian/d) = rc d (n).a=1(a,d)=1If d ≠ e, then each term is zero and the sum is also zero.

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit2.2. Show that the Dirichlet series of c r (n), as a function in n, is∞∑ c r (n)= ζ(s)φn s 1−s (r), r ≥ 1, Re s > 1,n=1where φ t (r) = ∑ d|r dt µ(r/d) is the generalized Euler function.Solution. |c r (n)| ≤ r is bounded as a function in n, the series is absolutelyconvergent for Re s > 1 and by Theorem 2.1,∞∑ c r (n)∞∑= ( ∑dµ(r/d)) 1 n s n = ∑ ∞∑ 1dµ(r/d)s (dδ) = s n=1n=1 d|(n,r) d|rδ=1= ∑ d|rd 1−s µ(r/d)∞∑δ=11δ s = ζ(s)φ 1−s(r).

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit3.1. Let f, g be even functions (mod r) with Fourier coefficients α(q) and β(q),respectively. Prove that the Cauchy product f ⊙ g is also an even function (modr) having Fourier coefficients rα(q)β(q).Solution.h(n) == ∑ q|r∑k+l ≡ n (mod r)∑α(q)β(s)s|rf(k)g(l) =∑k+l ≡ n (mod r)∑ ∑α(q)c q (k) ∑ β(s)c s (l) =q|rs|rc q (k)c s (l) = ∑ ∑α(q)β(s) ∑ rc q (n) =q|r s|rq=sk+l ≡ n (mod r)= ∑ rα(q)β(q)c q (n),q|rusing the result of Exercise 3.1. Here c r (n) is an even function (mod r) and forq | r the function c q (n) is also even (mod r), hence h(n) is even (mod r) and theFourier coefficients of h(n) are exactly rα(q)β(q).

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit3.2. If f is an even function (mod r), then according to Theorem 3.2, f has a mainvalue M(f) and M(f) = 1 r (f ∗φ)(r). Estimate the difference ∑ n≤x f(n)−xM(f).Solution. Let R f (x) = ∑ n≤xf(n)−xM(f). By Theorem 3.2 and Theorem 2.5./b) with the notations C 1 = 1, C q = 0, q > 1,∑f(n) = ∑ ∑h(q)c q (n) = ∑ h(q) ∑ c q (n) = ∑ h(q)(C q x + R q (x)) =n≤x n≤xq|r n≤xq|rq|r= h(1)x + ∑ q|rh(q)R q (x),where h(1) = 1 r (f ∗ φ)(r) = M(f), |h(q)| ≤ 1 ∑r e|r |f(r/e)||c e(r/q)| ≤≤ 1 ∑r e|r |f(r/e)|e = ∑ |f(e)|e|r eand using also Theorem 2.6,|R f (x)| = | ∑ h(q)R q (x)| ≤ ∑ |f(e)| ∑|R q (x)| ≤ ∑ |f(e)| ∑ψ(q) =eeq|re|r q|re|r q|r= rF (r) ∑ e|r|f(e)|,ewhere F (r) = ∑ d|r ψ(d) = r ∏ p|r (1 + 2(1 − p−k )(p − 1) −1 ).

Remark. f is periodic, hence bounded, let |f(n)| ≤ K, n ≥ 1. Then |R f (x)| ≤KF (r)σ(r)/r. It is true that for all ε > 0, |R f (x)| ≤ KC ε r 1+ε , where C ε dependsonly on ε, see the paper of L. Tóth, 2004, Proposition 1.•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit3.3. Let (s, d) = 1 and let φ(s, d, n) denote the number of elements in thearithmetic progression s, s + d, ..., s + (n − 1)d which are coprime to n. This is ageneralization of the Euler function, φ(1, 1, n) = φ(n) is the Euler function. Provethati) φ(s, d, n) is multiplicative in n,ii) for every prime power p k , φ(s, d, p k ) =(these values do not depend on s),iii) ∑ φ(s, d, n) =n≤xis the Jordan function,iv) f(d) := φ(s, d, n) ={p k (1 − 1/p), (p, d) = 1,p k , p | d,3d2π 2 J(d) x2 +O(x log x), where J(d) = φ 2 (d) = d 2 ∏ p|d (1−1/p2 )φ(n)(n, d)φ((n, d))v) M(f) = n ∏ p|n(1 − 1/p + 1/p 2 ).and this is an even function (mod n) in d,

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitSolution. More generally, let f ∈ Z[X] be a polynomial with integer coefficientsand let φ f (n) denote the number of integers x (mod n) such that (f(x), n) = 1. Iff = s + (x − 1)d, then φ f (n) = φ(s, d, n) for all n.i) We show that φ f is multiplicative. Let N f (n) denote the number of solutions(mod n) of the congruence f(x) ≡ 0 (mod n). Using the properties of the Möbiusfunction,n∑ n∑ ∑n∑ ∑φ f (n) = 1 =µ(e) = µ(e) == ∑ e|nx=1(f(x),n)=1µ(e)x=1 e|(f(x),n)∑1≤x≤nf(x) ≡ 0 (mod e)1 = ∑ e|nx=1e|f(x)e|nµ(e) n e N f(e).Therefore φ f = µN f ∗ E, and since N f (n) is multiplicative in n we obtain that φ fis multiplicative too.ii) According to this convolutional identity we have for every prime power p k therelation (*) φ f (p k ) = p k − p k−1 N f (p) = p k (1 − N f (n)/p).

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitLet f = s + (x − 1)d. How many solutions has the congruence s + (x − 1)d ≡ 0(mod p k )? This is equivalent to dx ≡ d − s (mod p k ), being a linear congruence.If (p, d) = 1, then (p k , d) = 1 and obtain that N f (p k ) = 1 for every k. If (p, d) ≠ 1,then (p k , d) = p a , a ≥ 1, and since (s, d) = 1 we obtain that p ∤ d−s, consequentlyN f (p k ) = 0. Use now relation (*).iii) For the function φ(s, d, n) using that N f is bounded,∑µ(a)N f (a) ∑= x22where C = ∏ p∤dn≤xφ(s, d, n) = ∑ ab≤xµ(a)N f (a)b = ∑ a≤x= x22∏p∞∑a=1= ∑ a≤x(1 − N f(p)p 2(1 − 1 p 2) = ∏ pµ(a)N f (a)a 2µ(a)N f (a)( x22a 2 + O(x a )) =+ O(x 2 ∑ a>x1a 2) + O(x ∑ a≤xb≤x/a1a ) =b =) + O(x) + O(x log x) = x2C + O(x log x),2(1 − 1 p 2) ∏ p|d(1 − 1 p 2)−1 = 1ζ(2) ·d 2φ 2 (d) .

•First •Prev •Next •Last •Go Back •Full Screen •Close •Quitiv) By the multiplicativity and ii) we obtain at once:φ(n)(n, d)φ(s, d, n) =φ((n, d)) = φ(n) · (n, d)φ((n, d)) ,(n,d)φ((n,d))where the variable is d, φ(n) is fixed andby definition it is an even function (mod n).v) Use Theorem 3.2 for r = n.depends only on the gcd (n, d) andRemark. The function φ(s, d, n) was investigated by P. G. Garcia and S. Ligh(1983, 1985), see also the papers of L. Tóth (1987, 1990).The function φ f (n) was investigated by P. K. Menon (1967), and by H. Stevens(1971). The asymptotic formula for φ f (n) was established by L. Tóth and J.Sándor (1989). The results of iv) and v) were proved by T. Maxsein (1990).

•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitReferencesP. J. McCarthy, Introduction to Arithmetical Functions, Springer Verlag, NewYork Berlin Heidelberg Tokyo, 1986.R. Sivaramakrishnan, Classical Theory of Arithmetic Functions, Marcel Dekker,New York-Basel, 1989.W. Schwarz, J. Spilker, Arithmetical Functions, London Math. Soc. LectureNotes Series 184, Cambridge Univ. Press, 1994.P. Haukkanen, An elementary linear algebraic approach to even functions (modr), Nieuw Arch. Wiskd. (5), 2 (2001), 29-31.P. G. Garcia, S. Ligh, A generalization of Euler’s φ-function, Fibonacci Quart.,21 (1983), 26-28.S. Ligh, P. G. Garcia, A generalization of Euler’s φ-function, II., Math. Japonica,30 (1985), 519-522.P. Kesava Menon, An extension of Euler’s function, Math. Student, 35 (1967),55-59.

H. Stevens, Generalizations of the Euler φ-function, Duke Math. J., 38 (1971),181-186.T. Maxsein, A note on a generalization of Euler’s φ-function, Indian J. PureAppl. Math., 21 (1990), 691-694.L. Tóth, A note on a generalization of Euler’s function, Fibonacci Quart., 25(1987), 241-244.L. Tóth, J. Sándor, An asymptotic formula concerning a generalized Eulerfunction, Fibonacci Quart., 27 (1989), 176-180.L. Tóth, Some remarks on a generalization of Euler’s function, Seminar Arghiriade23 (1990), 9pp.L. Tóth, Remarks on generalized Ramanujan sums and even functions, ActaMath. Acad. Paedagog Nyházi. (N. S.), electronic, 20 (2004), 233-238, seehttp://de.arxiv.org/PS cache/math/pdf/0610/0610358.pdf•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

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