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Analytic Number Theory

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Useful summation formulas

Analytic number theory is so abysmally complex that we need a basic toolkit of summation formulas first in order to prove some of the most basic theorems of the theory.

Abel's summation formula

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Theorem 1.1 (Abel's summation formula, also called Abel's identity):

Let be a sequence and let be a differentiable function such that is Riemann integrable. If we define

,

then we have

.

Note: We need the Riemann integrability to be able to apply the fundamental theorem of calculus.

Proof 1:

We prove the theorem by induction on .

1. :

First, we have in this case

.

Then, we have

by the fundamental theorem of calculus.

2. Induction step:

Define . We have

by the induction hypothesis. Further,

.

Putting things together, we obtain

and thus the desired formula.

The method of proof we applied here was using induction and then trying to express the terms from the induction hypothesis in terms of the terms from the desired formula.

Proof 2:

We prove the theorem by direct manipulation of the term on the left.

Define .

Proof 3:

We prove the formula by the means of the Riemann-Stieltjes integral. Indeed, by integration by parts, we have

.

Corollary 1.2:

.

Proof 1:

We deduce the formula from integration by parts for the Riemann-Stieltjes integral.

Proof 2:

We directly manipulate the LHS (left hand side).

Define and .

Two further proofs are given in exercises 1.1.1 and 1.1.5.

We note that induction and direct manipulation are quicker proofs for theorem 1.1, while corollary 1.2 is quicker proven from theorem 1.1 or Riemann-Stieltjes integration.

Exercises

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  • Exercise 1.1.1: Prove corollary 1.2 from theorem 1.1. Hint: .
  • Exercise 1.1.2: Compute . Hint: Use , , apply Abelian summation and split the resulting integral into pieces where is constant. Then apply a similar process.
  • Exercise 1.1.3: Prove that the limit exists. This limit is called the Euler–Mascheroni constant. Hint: Use and .
  • Exercise 1.1.4: Prove theorem 1.1 from corollary 1.2.
  • Exercise 1.1.5: Prove corollary 1.2 using induction on .

Euler's summation formula

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Definition 1.3:

For , we define

.

Theorem 1.4 (Euler's summation formula):

Let be a differentiable function, such that is Riemann integrable. Then

.

Proof:

We prove the theorem from Corollary 1.2, setting and using integration by parts (integration by parts is proven using the fundamental theorem of calculus).

Indeed,

,

where in the last line we used integration by parts on the integral .

Corollary 1.5:

Exercises

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  1. Prove corollary 1.5.

Euler–Maclaurin formula

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Theorem 1.6 (Euler–Maclaurin formula):

Define the functions and . Then for any twice continuously differentiable function such that is Riemann integrable, we have

.

Proof 1:

We prove the theorem by direct computation.

Proof 2:

We prove the theorem from Euler's summation formula.


The Chebychev ψ and ϑ functions

Proposition (the Chebychev ψ function may be written as the sum of Chebyshev ϑ functions):

We have the identity

.

Proposition (estimate of the distance between the Chebychev ψ and ϑ functions):

Whenever , we have

.

Note: The current proof gives an inferior error term. A subsequent version will redeem this issue. (Given the Riemann hypothesis, the error term can be made even smaller.)

Proof: We know that the formula

holds. Hence,

.

By a result obtained by Pierre Dusart (based upon the computational verification of the Riemann hypothesis for small moduli), we have

whenever . If is in that range, we hence conclude

.

By Euler's summation formula, we have

.

Certainly and . Moreover, . Now derivation shows that

is an anti-derivative of the function

of . By the fundamental theorem of calculus, it follows that

for real numbers such that . This integral is not precisely the one we want to estimate. Hence, some analytical trickery will be necessary in order to obtain the estimate we want.

We start by noting that if only the bracketed term in the integral were absent, we would have the estimate we desire. In order to proceed, we replace by the more general expression (where ), and obtain

.

The integrand is non-negative so long as

.

Moreover, if is strictly within that range, we obtain

.

We now introduce a constant and obtain the integrals

and .

The first integral majorises the integral

,

whereas the second integral majorises the integral

.

We obtain that

.

Now we would like to set . To do so, we must ensure that is sufficiently large so that resp. is strictly within the admissible interval.

The two summands on the left are now estimated using our computation above, where is replaced by for the first computation: Indeed,

and

.

Putting the estimates together and setting , we obtain

whenever

and .

We now choose the ansatz

and

for constants and . These equations are readily seen to imply

and .

Note though that and is needed. The first condition yields

.

The equations for and may be inserted into the above constraints on and ; this yields

and , that is, and .

If all these conditions are true, the ansatz immediately yields

.

We now amend our ansatz by further postulating

.

This yields

and

.

From this we deduce that in order to obtain an asymptotically sharp error term, we need to set . But doing so yields the desired result.


Arithmetic functions

In this chapter, we shall set up the basic theory of arithmetic functions. This theory will be seen in action in later chapters, but in particular in chapter 9.

Definitions

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Definition 2.1:

An arithmetical function is a function .

Definition 2.2 (important arithmetical functions):

  1. The Kronecker delta:
  2. Euler's totient function:
  3. Möbius' -function:
  4. The von Mangoldt function:
  5. The monomials:
  6. The number of distinct prime divisors: ,
  7. The sum of prime factors with multiplicity: ,
  8. The Liouville function:

Exercises

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  • Exercise 2.1.1: Compute , and .
  • Exercise 2.1.2: Compute . Hint: .
  • Exercise 2.1.3: Compute up to three decimal places. Hint: Use a Taylor expansion.
  • Exercise 2.1.4: Prove that for each and .

The convolution and the ring of arithmetic functions

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Definition 2.3:

Let be arithmetical functions. Then the convolution of and is defined to be the function

.

In the following theorem, we show that the arithmetical functions form an Abelian monoid, where the monoid operation is given by the convolution. Further, since the sum of two arithmetic functions is again an arithmetic function, the arithmetic functions form a commutative ring. In fact, as we shall also see, they form an integral domain.

Theorem 2.4 (Abelian monoid properties of the arithmetical functions):

  1. The convolution is commutative, i. e. .
  2. The convolution is associative, i. e. .
  3. The function from definition 2.2 is an identity for the convolution, i. e. .

Proof:

1.:

,

where is a bijection from the set of divisors of to itself.

2.:

,

where the last equality follows from the identity function

being a bijection. But

and hence associativity.

3.:

Theorem 2.5:

The ring of arithmetic functions is an integral domain.

Proof: Let be arithmetic functions, and let be minimal such that , . Then

.

We shall now determine the units of the ring of arithmetic functions.

Theorem 2.6:

Let be an arithmetic function. Then is invertible (with respect to convolution) if and only if .

Proof:

Assume first . Then for any arithmetic function , .

Assume now . Then , given by the recursive formula

,
,

is an inverse (and thus the inverse) of , since and for inductively

Exercises

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  • Exercise 2.2.1:
  • Exercise 2.2.2:

Multiplicative functions

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Definition 2.7:

An arithmetical function is called multiplicative iff it satisfies

  1. , and
  2. .

Theorem 2.8:

Let be multiplicative arithmetical functions. Then is multiplicative.

Proof:

Let . Then

,

since the function is a bijection from the divisors of to the Cartesian product of the divisors of and the divisors of ; this is because multiplication is the inverse:

, .

To rigorously prove this actually is an exercise in itself. But due to the multiplicativity of and ,

.

Furthermore, .

Since is multiplicative, we conclude that the multiplicative functions form an Abelian submonoid of the arithmetic functions with convolution. Unfortunately, we do not have a subring since the sum of two multiplicative functions is never multiplicative (look at ).

Theorem 2.9:

Let be a multiplicative function such that converges absolutely. Then

.

Proof: Let be the ordered sequence of all prime numbers. For all we have

due to the multiplicativity of . For each , we successively take , ..., and then . It follows from the definitions and the rule that the right hand side converges to

.

We claim that

.

Indeed, choose such that

.

Then by the fundamental theorem of arithmetic, there exists an and such that

.

Then we have by the triangle inequality for , and arbitrary that

From this easily follows the claim.

It is left to show that the product on the left is independent of the order of multiplication. But this is clear since if the sequence is enumerated differently, the argument works in just the same way and the left hand side remains the same.

Definition 2.10:

An arithmetical function is called strongly multiplicative iff it satisfies

  1. , and
  2. .

Equivalently, a strongly multiplicative function is a monoid homomorphism .

Theorem 2.11:

Let be a strongly multiplicative function such that converges absolutely. Then

.

Proof:

Due to theorem 2.9, we have

.

Due to strong multiplicativity and the geometric series, the latter expression equals

.

Exercises

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  • Exercise 2.3.1: Let be an arithmetic function such that for all , and let . Prove that the function is multiplicative.

Bell series

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Definition 2.12:

Let be an arithmetic function. Then for a prime the Bell series modulo is the formal power series

.

Examples 2.13:

We shall here compute the Bell series for some important arithmetic functions.

We note that in general for a completely multiplicative function , we have

.

In particular, in this case the Bell series defines a function.

1. The Kronecker delta:

2. Euler' totient function (we use lemma 9.?):

3. The Möbius function:

4. The von Mangoldt function:

5. The monomials:

6. The number of distinct prime divisors:

7. The number of prime divisors including multiplicity:

8. The Liouville function:

Theorem 2.14 (compatibility of Bell series and convolution):

Let arithmetic functions, and be a prime. Then

.

Proof:

In case of multiplicativity, we have the following theorem:

Theorem 2.15 (Uniqueness theorem):

Let be multiplicative functions. Then

.

Proof: is pretty obvious; : as formal power series is equivalent to saying . If now , then

due to the multiplicativity of and .

In chapter 9, we will use Bell series to obtain equations for number-theoretic functions.

Derivatives

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Definition 2.16:

Let be an arithmetic function. Then the derivative of is defined to be the function

.

Theorem 2.17 (rules for the derivative):

Let arithmetic functions. We have the following rules:

  1. if invertible, i.e.

Note that is not the inverse function of (this wouldn't make much sense anyway since arithmetic can not be surjective, since is uncountable), but rather the convolution inverse.

Proof:

1. is easily checked.

2.:

3.

We have and . Hence, by 2.

.

Convolving with and using yields the desired formula.

Note that a chain rule wouldn't make much sense, since arithmetic may map anywhere but to and thus doesn't make a lot of sense in general.

Further reading

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Characters and Dirichlet characters

Definitions, basic properties

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Definition 4.1

Let be a finite group. A character of G is a function such that

  1. and
  2. .

Lemma 4.2:

Let be a finite group and let be a character. Then

.

In particular, .

Proof:

Since is finite, each has finite order . Furthermore, let such that ; then and thus . Hence, we are allowed to cancel and

.

Lemma 4.3:

Let be a finite group and let be characters. Then the function is also a character.

Proof:

,

since is a field and thus free of zero divisors.

Lemma 4.4:

Let be a finite group and let be a character. Then the function is also a character.

Proof: Trivial, since as shown by the previous lemma.

The previous three lemmas (or only the first, together with a few lemmas from elementary group theory) justify the following definition.

Definition 4.5

Let be a finite group. Then the group

is called the character group of .

Required algebra

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We need the following result from group theory:

Lemma 4.6

Let be a finite Abelian group, let be a subgroup of order , and let such that is the smallest number such that . Then the group

is a subgroup of containing of order .

Proof:

Since is the disjoint union of the cosets of , is the disjoint union , as and . Hence, the cardinality of equals .

Furthermore, if , then , and hence is a subgroup.

Theorems about characters

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Dirichlet characters

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Dirichlet series

For the remainder of this book, we shall use Riemann's convention of denoting complex numbers:

Definition

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Definition 5.1:

Let be an arithmetic function. Then the Dirichlet series associated to is the series

,

where ranges over the complex numbers.

Convergence considerations

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Theorem 5.2 (abscissa of absolute convergence):

Let be an arithmetic function such that the series of absolute values associated to the Dirichlet series associated to

neither diverges at all nor converges for all . Then there exists , called the abscissa of absolute convergence, such that the Dirichlet series associated to converges absolutely for all , and it's associated series of absolute values diverges for all , .

Proof:

Denote by the set of all real numbers such that

diverges. Due to the assumption, this set is neither empty nor equal to . Further, if , then for all and all , since

and due to the comparison test. It follows that has a supremum. Let be that supremum. By definition, for we have convergence, and if we had convergence for we would have found a lower upper bound due to the above argument, contradicting the definition of .

Theorem 5.3 (abscissa of conditional convergence):

Formulas

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Theorem 8.4 (Euler product):

Let be a strongly multiplicative function, and let such that the corresponding Dirichlet series converges absolutely. Then for that series we have the formula

.

Proof:

This follows directly from theorem 2.11 and the fact that strongly multiplicative strongly multiplicative.


Formulas for number-theoretic functions

Formulas for the Möbius μ function

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Lemma 2.9:

.

Proof:

For a multiindex, and a vector define

,
.

Let . Then

.

Lemma 2.10:

.

Proof 1:

We prove the lemma from lemma 2.14.

We have by lemma 2.14

Proof 2:

We prove the lemma from the product formula for Euler's totient function and lemma 2.9. Indeed, for

.

Lemma 2.14:

.

Proof 1:

We use the Möbius inversion formula.

Indeed, , and hence .

Proof 2:

We use multiplicativity.

Indeed, for a prime , we have

,

and thus due to the multiplicativity of and if contains at least one prime factor. Since further the claim follows.

Proof 3:

We prove the lemma by direct computation. Indeed, if , then

.

Proof 4:

We prove the lemma from the Binomial theorem and combinatorics.

Let . From combinatorics we note that for , there exist distinct ways to pick a subset such that . Define where . Then, by the Binomial theorem

.

Formulas for Euler's totient function

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Lemma 2.11 (Gauß 1801):

.

Proof 1:

We use the Möbius inversion formula, proven below without using this lemma, and lemma 2.10.

We have and hence by the Möbius inversion formula. On the other hand,

by lemma 2.10.

Hence, we obtain , and by cancellation of (the arithmetic functions form an integral domain) we get the lemma.

Proof 2:

We use the converse of the Möbius inversion formula, proven below without using this lemma, and lemma 2.10.

Since by lemma 2.10, we obtain from the converse of the Möbius inversion formula that .

Proof 3:

We prove the lemma by double counting.

We first note that there are many fractions of the form , .

We now prove that there are also many fractions of this form. Indeed, each fraction , can be reduced to , where . is a divisor of , since it is obtained by dividing . Furthermore, for each divisor of there exist precisely many such fractions by definition of .

Proof 4:

We prove the lemma by the means of set theory.

Define . Then . Since and is the disjoint union of the sets , we thus have

.

The next theorem comprises one of the most important examples for a multiplicative function.

Theorem 2.12 (Euler 1761):

Euler's totient function is multiplicative.

Proof 1:

We prove the theorem using double counting (due to Kronecker).

By definition of , there are sums of the form

,

where both summands are reduced. We claim that there is a bijection

.

From this would follow .

We claim that such a bijection is given by .

Well-definedness: Let , be reduced. Then

is also reduced, for if , then without loss of generality , and from follows or . In both cases we obtain a contradiction, either to or to is reduced.

Surjectivity: Let be reduced. Using the Euclidean algorithm, we find such that . Then . Define , . Then

.

Injectivity: Let . We show ; the proof for is the same.

Indeed, from follows , and since , is invertible modulo , which is why we may multiply this inverse on the right to obtain . Since , the claim follows.

Proof 2:

We prove the theorem from the Chinese remainder theorem.

Let . From the Chinese remainder theorem, we obtain a ring isomorphism

,

which induces a group isomorphism

.

Hence, , and from follows the claim.

Proof 3: We prove the theorem from lemma 2.11 and induction (due to Hensel).

Let such that . By lemma 2.11, we have and and hence

.

Furthermore, by lemma 2.11 and the bijection from the proof of theorem 2.8,

.

By induction on we thus have

.

Proof 4: We prove the theorem from lemma 2.11 and the Möbius inversion formula.

Indeed, from lemma 2.10 and the Möbius inversion formula, we obtain

,

which is why is multiplicative as the convolution of two multiplicative functions.

Proof 5: We prove the theorem from Euler's product formula.

Indeed, if and and , then and hence

.

Theorem 2.15 (Möbius inversion formula):

Let be an arithmetical function and define

.

Then

.

Proof:

By lemma 2.14 and associativity of convolution,

.

Theorem 2.16 (Product formula for Euler's totient function):

Let , where are pairwise different prime numbers and (recall that every number has such a decomposition by the fundamental theorem of arithmetic. Then

.

Proof 1:

We prove the theorem from lemma 2.10 and the fact that is multiplicative.

Indeed, let be a prime number and let . Then , since

by lemma 2.10. Therefore,

,

where the latter equation follows from

.

Proof 2:

We prove the identity by the means of probability theory.

Let , . Choose , , . For define the event . Then we have

.

On the other hand, for each , we have

.

Thus, it follows that are independent. But since events are independent if and only if their complements are, we obtain

.

Proof 3:

We prove the identity from the Möbius inversion formula and lemmas 2.9 and 2.10.

But by the Möbius inversion formula and since by lemma 2.10 ,

.

Proof 4:

We prove the identity from the inclusion–exclusion principle.

Indeed, by one of de Morgan's rules and the inclusion–exclusion principle we have for sets

,

where we use the convention that the empty intersection equals the universal set . Let now , and define and for . Since

,

we then have

.

But for each , we have

.

It follows

,

and since

,

the theorem is proven.

Exercises

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Formulas for the von Mangoldt function

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Theorem 8.? (The Selberg identity):


Partial fraction decomposition

Existence theorem

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Theorem 2.1 (Existence theorem of the partial fraction decomposition):

Let be polynomials over a unique factorisation domain, and let , where the are irreducible. Then we may write

,

where are polynomials of degree strictly less than and is a polynomial. The term on the right hand side is called the partial fraction decomposition of .

Proof:

We proceed by induction on . For , the statement is true since by division with remainder, we may write

with to obtain

,

and we have reduced the degree of the denominator by one (the latter summand already satisfies the required condition). By repetition of this process, we eventually obtain a denominator of one and thus a polynomial.

Let now the hypothesis be true for , and assume that . Write and . By irreducibility, . Hence, we find polynomials such that . Then

.

Each of the summands of the last term can by the induction hypothesis be written in the desired form.

Technique

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No matter how complicated our fraction of polynomials may be, we can give the partial fraction decomposition in finite time, using easy techniques. The method, which for the sake of simplicity differs from the one given in the above constructive existence proof, goes as follows:

  1. Split the polynomial into irreducible factors.
  2. Using division with remainder of by , reduce to the case (the resulting polynomial is allowed in the formula of theorem 2.1).
  3. Solve the equation given in theorem 2.1 for the (this is equivalent to solving a system of linear equations; namely multiply by and then equate coefficients).

Theorem 2.2:

The algorithm given above always terminates and gives the partial fraction decomposition of .

Proof: Due to theorem 2.1, in step three we do obtain a system of linear equations which is solvable. Hence follow termination and correctness.

Exercises

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Tools from complex analysis

Infinite products

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Lemma 5.1 (Convergence of real products):

Let be such that

converges absolutely. Then if ,

converges.

Proof: Without loss of generality, we assume for all .

Denote

.

Then we have

.

We now apply the Taylor formula of first degree with Lagrange remainder to at to obtain for

, .

Hence, we have for

, .

Hence, and thus we obtain the (even absolute) convergence of the ; thus, by the continuity of the exponential, also the converge.

Theorem 5.2 (Comparison test for complex products):

Assume that is a non-negative real sequence such that

converges. Assume further that is a sequence of complex numbers such that . Then also

converges. Furthermore, for all

.

Proof:

We define

, . We note that
.

Without loss of generality we may assume that all the products are nonzero; else we have immediate convergence (to zero).

We now prove that is a Cauchy sequence. Indeed, we have

and furthermore

and therefore

.

Since , it is a Cauchy sequence, and thus, by the above inequality, so is . The last claim of the theorem follows by taking in the above inequality.

Theorem 5.3 (Sum test for complex products):

Let be a real sequence such that

converges absolutely. Then if ,

converges, where is a complex sequence. Furthermore, for all

.

Proof 1:

We prove the theorem using lemma 5.1 and the comparison test.

Indeed, by lemma 5.1 the product

converges. Hence by theorem 5.2, we obtain convergence and the desired inequality.

Proof 2 (without the inequality):

We prove the theorem except the inequality at the end from lemma 5.1 and by using the Taylor formula on .

We define . Then since every complex number satisfies , we need to prove the convergence of the sequences and .

For the first sequence, we note that the convergence of is equivalent to the convergence of . Now for each

Theorem 5.4 (Holomorphic products):

Let be a sequence of holomorphic functions in a domain such that for each we can find a compact and a sequence such that and

converges absolutely. Then

defines a holomorphic function.

Proof:

First, we note that is well-defined for each due to theorem 5.2. In order to prove that the product is holomorphic, we use the fact from complex analysis that if a sequence of functions converging locally uniformly to another function has infinitely many holomorphic members, then the limit is holomorphic as well. Indeed, we note by the inequality in theorem 5.3, that we are given uniform convergence. Hence, the theorem follows.

Exercises

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The Weierstraß factorisation

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The following lemma is of great importance, since we can deduce three important theorems from it:

  1. The existence of holomorphic functions with prescribed zeroes
  2. The Weierstraß factorisation theorem (a way to write any holomorphic function made up from linear factors and the exponential)
  3. The Mittag-Leffler theorem (named after Gösta Mittag-Leffler (one guy))

Lemma 5.5:

Let be a sequence of complex numbers such that

and

.

Then the function

has exactly the zeroes in the correct multiplicity.

Proof:

Define for each

.

Our plan is to prove that converges uniformly in every subcircle of the circle of radius for every . Since the function is holomorphic in a unit ball around zero, it is equal to its Taylor series there, i.e.

.

Hence, for

.

Let now be given and be arbitrary. Then we have for , arbitrary

.

Now summing over , we obtain

for all . Hence, we have uniform convergence in that circle; thus the sum of the logarithms is holomorphic, and so is the original product if we plug everything into the exponential function (note that we do have even if is an arbitrary complex number).

Note that our method of proof was similar to how we proved lemma 5.1. In spite of this, it is not possible to prove the above lemma directly from theorem 5.4 since the corresponding series does not converge if the are chosen increasing too slowly.

Theorem 5.6 (Holomorphic functions with given zeroes):

Let be a sequence of complex numbers which does not have an accumulation point. Then the function

has zeroes with the right multiplicity, where the sequence are the nonzero elements of the sequence ordered ascendingly with respect to their absolute value and is the number of zeroes within the sequence .

Proof:

We order increasingly according to the modulus and the standard greater or equal order on the real numbers. We go on to observe that then , since if it were to remain bounded, there would be an accumulation point according to the Heine–Borel theorem. Also, the sequence is zero only finitely many often (otherwise zero would be an accumulation point). After eliminating the zeroes from the sequence we call the remaining sequence . Let the number of zeroes in . Then due to lemma 5.5, the function

has the required properties.

Theorem 5.7 (Weierstraß factorisation theorem):

Let be holomorphic and not the constant zero function with zeroes , let are the nonzero elements of the sequence ordered ascendingly with respect to their absolute value, and if , let be the order of the zero of . Then there exists a holomorphic function such that

.

Proof:

First, we note that does not have an accumulation point, since otherwise would be the constant zero function by the identity theorem from complex analysis. From theorem 5.6, we obtain that the function has exactly the zeroes with the right multiplicity, where the sequence are the nonzero elements of the sequence ordered ascendingly with respect to their absolute value and is the number of zeroes within the sequence . We have that has no zeroes and is bounded and hence holomorphic due to Riemann's theorem on resolvable singularities. For, if were unbounded, it would have a singularity at a zero of . This singularity can not be essential since dividing by finitely many linear factors would eliminate that singularity. Hence we have a pole, and this would be resolvable by multiplying linear factors to . But then has a zero of the order of that pole, which is not possible since we may eliminate all the zeroes of by writing , holomorphic and nonzero at , where is the order of the zero of at .

Hence, has a holomorphic logarithm on , which we shall denote by . This satisfies

.

Corollary 5.8 (Mittag-Leffler's theorem):

Let be a sequence of complex numbers which does not have an accumulation point. Then there exists a meromorphic function which has exactly the poles , where the pole has order .

Proof:

From theorem 5.7 we obtain a function with zeroes in the right multiplicity. Set .

Exercises

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The Hadamard factorisation

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In this subsection, we strive to factor certain holomorphic functions in a way that makes them even easier to deal with than the Weierstraß factorisation. This is the Hadamard factorisation. It only works for functions satisfying a certain growth estimate, but in fact, many important functions occurring in analytic number theory do satisfy this estimate, and thus that factorisation will give us ways to prove certain theorems about those functions.

In order to prove that we may carry out a Hadamard factorisation, we need some estimates for holomorphic functions as well as some preparatory lemmata.

Estimates for holomorphic functions

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Theorem 5.9:

Let be a holomorphic function such that , and let be the sequence of zeroes of that function ordered ascendingly by absolute value. Let . If we denote the number of zeroes of inside by , then

.

Proof:

Set and define the function by

,

where the latter limit exists by developing into a power series at and observing that the constant coefficient vanishes. By Riemann's theorem on removable singularities, is holomorphic. We now have

,

and if further , then and hence we may multiply that number without change to anything to obtain for

.

Now writing and , we obtain on the one hand

and on the other hand

.

Hence,

,

which is why both and have the same distance to , since lies on the real axis.

Hence, due to the maximum principle, we have

.

Theorem 5.10:

Let , let be holomorphic within and define . Then

.

Proof:

First, we consider the case and . We may write in its power series form

,

where . If we write and , we obtain by Euler's formula

and thus

.

Since the latter sum is majorised by the sum

,

it converges absolutely and uniformly in . Hence, by exchanging the order of integration and summation, we obtain

due to

and further for all

due to

,

as can be seen using integration by parts twice and . By monotonicity of the integral, we now have

.

This proves the theorem in the case . For the general case, we define

.

Then , hence by the case we already proved

.

Theorem 5.11:

Further preparations

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Definition 5.12 (exponent of convergence):

Let be a sequence of complex numbers not containing zero such that

converges for a . Then

is called the exponent of convergence of the sequence .

Definition 5.13 (holomorphic functions of finite order):

Let be holomorphic, and define for

.

If there exists such that

for a suitable , then is said to be of finite order. In this case,

is called the order of .

Lemma 5.14:

The theorem

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Exercises

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