# PRIME NUMBERS AND THE RIEMANN HYPOTHESIS

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1 PRIME NUMBERS AND THE RIEMANN HYPOTHESIS CARL ERICKSON This minicourse has two main goals. The first is to carefully define the Riemann zeta function and exlain how it is connected with the rime numbers. The second is to elucidate the Riemann Hyothesis, a famous conjecture in number theory, through its imlications for the distribution of the rime numbers.. The Riemann Zeta Function Let C denote the comlex numbers. They form a two dimensional real vector sace sanned by and i where i is a fixed square root of, that is, C = {x + iy : x, y R}. Definition. The Riemann zeta function is the function of a comlex variable ζ(s) := + s 2 + s 3 + = n s, s C. s It is conventional to write s = σ + it where s = σ + it and σ, t R While this definition may seem straightforward, erhas you ve not thought of what comlex owers are before. We are robably comfortable with integral owers of a comlex number we do this in PROMYS at least in the case of Z[i] C but it is by no means obvious what a comlex ower of a real number should be. While more could certainly said, we ll aeal to standard rules of exonentials and the Euler identity e iθ = cos θ + i sin θ, for θ R. Then for n R we have that n s = n σ it = n σ n it = n σ e it log n = n σ( cos( t log n) + i sin( t log n) ). We note that by standard trigonometric identities, the quantity in the rightmost arentheses has absolute value. Thus the mantra is the the real art of the exonent controls the magnitude of an exonential, while the imaginary art of the exonent controls the angle of the ray drawn from the origin to the result. This is like the olar form of comlex numbers, i.e. an element of C may be written as n= re iθ, where r R 0, θ [0, 2π). This way of writing a comlex number is unique unless r = 0. Now that we understand what each term in the Riemann zeta function s definition means, the next thing that we must ask is whether or for which s C the sum defining ζ(s) converges. A useful definition for us is the following Date: July 23, 2008 (last corrected January 4, 203).

2 Definition 2. Let (a i ) C be a sequence of comlex numbers. We say that the sum converges absolutely or is absolutely convergent rovided that the sum a i converges in R, where denotes the comlex absolute value. i= i= Examle 3. The sequence (a i ) i defined by a i = ( ) i+ /i converges in C but is not absolutely convergent. Absolute convergence is useful because of the following Proosition 4. If a sum is absolutely convergent, then it is convergent. a i Proof. Exercise. Alying the idea of absolute convergence and the fact from calculus class that for σ R the sum n n σ converges for σ >, we deduce that ζ(s) converges for σ >. We might ask if ζ(s) converges when σ =, or erhas for some σ <. However, for s =, we find that the zeta function gives the harmonic series, which does not converge. Therefore the statement that ζ(s) converges for σ > is best ossible. 2. The Riemann Zeta Function and the Primes Why would one think that the rimes had anything to do with this function? Well, Euler was fond of writing such equalities as = ( ) ( ) ( ) + as a roof of the infinitude of the rimes. We can quickly make this more rigorous. As a consequence of unique rime factorization, we can write the zeta function (in its region of absolute convergence) as ζ(s) = + s 2 + s 3 + s 4 + s 5 + s 6 + s 7 + s = ( ) ( ) ( ) + s s s s s s = = k=0 ks ( s ), 2

3 where the last equality alies the geometric series formula i=0 ri = ( r) for r < and signifies a roduct over the rimes in N. As the equality on the second line follows from unique rime factorization, we can say that the equation ζ(s) = ( ) s is an analytic statement of unique rime factorization. It is known as the Euler roduct. This gives us a first examle of a connection between the zeta function and the rimes. Now lets take the limit of ζ(s) as s +, that is, as s aroaches from the right along the real line. Clearly this limit does not converge, as we know that the harmonic series diverges. Yet the limit of each factor in the Euler roduct is the real number ( ) =. Since the roduct of these real numbers over the rimes diverges, we may then conclude the following Theorem 5. There are infinitely many rimes. Thus using calculus concets and the unique rime factorization theorem, the zeta function has yielded another roof of the most ancient theorem about the rimes. This is another sign that the zeta function and the rimes are connected. The Euler roduct may also be alied to solve this roblem. Exercise 6. Call a ositive integer m nth ower free rovided that for all l Z +, l n m = l =. Now for x R + and integers n 2 define f n (x) as Prove that for integers n 2, f n (x) = #{m Z + : m x and m is nth ower free}. f n (x) lim x x = ζ(n). Ok, what else can we find out about the rimes by alying calculus concets to the Euler roduct? Take the (modified) Taylor series for the log function at, that is, the series log( x) = n= x n n x2 = x 2 x3, for x C, x <. 3 Alying this to the Euler roduct for the zeta function we find that log ζ(s) = log ( ) = log( s ) s = = m ms m= s + ( m=2 3 ). m ms

4 Now try you hand at bounding sums in the following Exercise 7. The sum on in the arentheses in the final equality converges for s. This has a very leasant consequence. Corollary 8. The sum of the recirocals of the rimes diverges. Proof. Consider the equation for log ζ(s) above as s +. We know that log ζ(s) + because lim s + ζ(s) = +. However, the quantity on the right hand side in arentheses remains bounded as s +. Therefore the sum s becomes arbitrarily large as s +, allowing us to conclude that the sum does not converge. Using more subtle roerties of the zeta function involving techniques like those that will be black boxed later in this talk, one can rove that log log x. x 3. The Proerties of the Zeta Function In the revious talk this evening, Josh discussed on the Chebyshev bounds on the rime counting function π(x) := x where varies over rime numbers no more than x, and the related function ψ(x) := n x log where ranges over rimes and n ranges over ositive integers such that n x. We wil rewrite the second function as ψ(x) = n x Λ(n), where Λ is the von Mangoldt function defined by { log if n = Λ(n) = m for some m 0 otherwise. Be sure to observe that the two formulations of ψ(x) are equivalent. Mathematicians in the nineteenth century worked toward the rime number theorem, which is a statement about the asymtotic behavior of π(x). We often use the symbol to describe asymtotics. It has the following meaning: let f(x) and g(x) be two comlex functions of a real variable x. We say that rovided that f(x) g(x) lim x + f(x) g(x) =. 4

5 With this notation in lace we can state the rime number theorem. Theorem 9 (Prime Number Theorem). Let π(x) be the rime counting function defined above. Then π(x) x log x. Other ways of stating the rime number theorem are that the robability of a randomly chosen ositive integer no more than x being rime aroaches / log x, or that the robability of a randomly chosen ositive integer near x being rime is / log x. Note that these robabilities are not trivially the same - these quantities are similar because of the log function s differential roerties. The rime number theorem was roven in 896 by Hadamard and Vallée Poussin indeendently. Each comleted the roof by constraining the comlex numbers s such that ζ(s) = 0. The next main ideas develo several main ideas having to do with these zeros, leaving comlex analysis as a black box. The zeta function is often called the Riemann zeta function because Riemann instigated serious study of it. He roved the following items, in bold. ζ(s) has an analytic continuation to the rest of C, which is comlex differentiable excet for the simle ole at s =. An analytic continuation is an imortant concet in comlex analysis, but one has already encountered this function in calculus class. Consider the function + x + x 2 + x 3 +, which converges only in the disk {z C : z < } and on some oints on the border of the disc. This is the Taylor exansion at 0 for the function x, which we know erfectly well can be defined for x outside the disk, even though its series reresentation cannot. Analytic continuation is much like this: it extends a function that is a riori defined on a constrained domain to a larger domain of definition in a sensible way. However, our intuition from real functions will fail us here, because there are many real differentiable functions like { x 2 if x 0 f(x) := x 2 if x < 0 which are once or twice differentiable, but not smooth (i.e. infinitely differentiable). However, comlex differentiable functions have the amazing roerty that they are infinitely differentiable whenever they are once differentiable. Any differentiable (thus infinitely differentiable) comlex function is termed holomorhic or analytic. This secial roerty of comlex differentiable functions is critical for showing that functions such as ζ(s) have unique analytic continuations. The continuation of ζ(s) is differentiable at every s C excet s = ; here it looks locally like, a situation which is called having s a simle ole at s =. We won t talk about what the analytic continuation of the zeta function looks like on much of C, because of the next item that Riemann roved limits zeros rather seriously. 5

6 The zeta function obeys a functional equation, namely where ζ( s) = 2(2π) s cos(πs/2)γ(s)ζ(s), Γ(s) := and the cos function is defined on C by 0 x s e x dx for σ > 0. cos(s) = eis + e is. 2 Note that the restriction on the convergence of the Γ-function does not sto us from relating any two values of the zeta function (however, the Γ-function also admits an analytic continuation!). This functional equation relates the zeta functions value at a comlex number s to its value at the oint given by reflecting s across the oint /2 C. Thus as we know the behavior of the zeta function for σ > relatively well, we also know its behavior for σ < 0. We will demonstrate this by characterizing all of the zeros of ζ(s) just as Riemann did. Consider the functional equation and let s vary over A = {s C : R(s) > }. Let us find those values of s where the right side of the functional equation has a zero. First of all, ζ(s) 0 for all s A. This is the case because the logarithm of ζ(s) converges for all s A, which, since x 0 in C imlies that the real art of log x aroaches, means that ζ(s) cannot be zero. All of the remaining factors are never zero for any s in C. Thus the right side of the functional equation is zero only when cos(πs/2) is zero, that is, when s is an odd integer at least 3. Therefore we have that ζ( s) is zero for all odd integers that are at least three, i.e. ζ(s) = 0 and R(s) < 0 = s = 2, 4, 6, 8,.... These zeros at the negative even integers are called the trivial zeros of ζ(s). Now we have limited the unknown zeros of ζ(s) to the region 0 R(s). This region is called the critical stri. We will see the arithmetic significance of this shortly, but having gone this far in characterizing the zeros, it is time to state the Riemann Hyothesis. Conjecture 0 (Riemann Hyothesis). The non-trivial zeros of ζ(s) have real art onehalf, i.e. ζ(s) = 0 and 0 R(s) = R(s) = 2. Let s make a few observations about the zeros of ζ(s) in the critical stri. Note that via our comutations of comlex exonentials at the outset, the comlex conjugate n s of n s is equal to n s. Therefore ζ(s) = ζ(s) for all s C. Therefore if s = σ + it is a zero of zeta in the critical stri, its comlex conjugate s = σ it is a zero as well. Recall also that zeros in the critical stri will reflect over the oint /2. Therefore if the Riemann hyothesis is false and ρ = β + iγ is a zero in the critical stri but off the critical line R(s) = /2, there are zeros to the right of the critical line, two of them being in the list of distinct zeros ρ = β + iγ; ρ = β iγ; ρ = ( β) iγ; ρ = ( β) + iγ. 6

7 Many many zeros of the Riemann zeta function have been calculated, and all of them are on the critical line. We should also remark that all zeros of the Riemann zeta function are simle zeros (i.e. they look like the zero of f(x) = x at zero, not the the zero of f(x) = x n where n > ), but this is not worth getting into. But now, what can we do to connect the rimes to the zeros of the zeta function? The key is comlex analysis. 4. The Zeros of the Zeta Function and Prime Distribution Comlex analytic tools are good at finding oles - after all, oles are the interrutions in the strong condition of comlex differentiability. Since we have been claiming that the zeros of the zeta function are significant for rime distribution, it makes sense that we will further examine the log of the zeta function. The log has oles where zeta has zeros or oles. More recisely, we will examine its logarithmic derivative, i.e. the derivative of its log, which also will have oles where zeta has oles or zeros. We calculate d ds (log ζ(s)) = d ds = = log( s ) ( s ) s log log ms. Recalling the definition of the von Mangoldt function Λ(n) from the beginning of section 3, we find that the last equality can be immediately translated into m d ds log ζ(s) = n= Λ(n) n s. Now comes in the critically imortant tool from comlex analysis, which we must leave as a black box. In English, the idea is this: given a reasonably well behaved (e.g. not exonentially growing) function f : N C, the value of F (x) = n x f(n) is intimately tied to the oles of via Perron s formula. n= f(n) n s In our case, we want to discuss ψ(x) = n x Λ(n). Recall that the rime number theorem states that ψ(x) x. In fact, Perron s formula will even interolate the oints of discontinuity of ψ(x), that is we will get a formula for { ψ(x) if ψ is continuous at x ψ 0 (x) := ψ(x + )+ψ(x ) if ψ is not continuous at x. 2 7

8 Perron s formula states that ψ 0 (x) = x ζ(ρ)=0 0 R(ρ) x ρ ρ log 2π 2 log( x 2 ), that is, as we add non-trivial zeros ρ to the sum on the right hand side, the right hand side will coverge ointwise to ψ 0 (x) for every x R 0. This equality is known as the exlicit formula for ψ. We remark that the first term of the right side (namely x) comes from the ole of ζ(s) at, the second term comes from the non-trivial zeros, the third term comes from the roerties of ζ(s) at 0, and the final term comes from the trivial zeros. Notice that each exonent of x is the coordinates of a ole this is art of Perron s formula. Finally we ve come u with an exression for ψ(x) deending on the zeros of x. The rime number theorem states that ψ(x) x, i.e. that the first term of the right side above dominates all of the others as x. Let s check this, term by term. The rightmost term, 2 log( x 2 ), will shrink to 0 as x gets large, so we need not worry about it. The next term is a constant, and so does not matter when x gets large either. This leaves the non-trivial zeros to contend with the x term on the left. We should remark that it is not obvious that the sum over the non-trivial zeros converges, but estimates for zero density in the critical stri first done by Riemann are sufficient to show that the sum converges. Recall from the outset that when we take a real number x to the comlex ower ρ, the real art of ρ controls the size of x ρ and its imaginary art controls the angle of the ray from 0 to x ρ. In fact, because zeros of the zeta function come in conjugate airs, we may air u the zeros in the sum over non-trivial zeros; one air ρ = β + iγ and ρ where ρ = ρ e iθρ gives us x ρ ρ + xρ ρ = xβ (e iγ log x ρ + e iγ log x ρ) β 2 + γ 2 = xβ ρ (ei(γ log x θρ) i(γ log x θρ) + e = 2 xβ ρ cos(γ log x θ ρ). As θ ρ is very close to ±π/2, this result looks a lot like a constant times x β sin(γ log x), and this is a fine function to grah or look at in order to get an idea of what each conjugate air of zeros contributes to the ψ(x) function. You may find a icture of what such a function looks like at a website to be discussed below. Note also that the fact that imaginary arts cancel when we sum over conjugate airs makes a lot of sense, since ψ(x) is a real function. With this in lace, we can say what Hadamard and Vallée Poussin roved about the zeros of the zeta function in order to have the rime number theorem as a corollary: they roved that there are no zeros of the zeta function on the line R(s) =. Let s think about why this is sufficient. Let s say that ρ = β+iγ is a zero in the critical stri that is not on the critical line, that is, a counterexamle to the Riemann hyothesis. Then there exists a zero that is to the right of the critical line, i.e. with β > /2, by the discussion of the symmetry of zeros above. 8

9 Without loss of generality, say that ρ is this zero. Then ρ and its conjugate ρ generate a term of the sum which reaches a magnitude as high as x β times a constant. If somehow the four zeros generated by ρ and the zero-symmetries were the only counterexamles to the Riemann hyothesis, then there would be crazy variations in rime density: if x were such that cos(γ log x θ ρ ) is near, then by examining the exlicit formula for ψ(x) we see that there would be fewer rimes near to this x. This is the case because the wave 2 xβ ρ cos(γ log x θ ρ) generated by ρ and its conjugate will be much bigger than the waves generated by any of the other non-trivial zeros. Likewise, when x is such that cos(γ log x θ ρ ) is close to -, there will be many more rimes near x. We can see how the truth of the Riemann hyothesis makes x as good of an estimate for ψ(x) as it can be. When we consider the rime number theorem, we realize that our aberrant zero ρ will not create trouble as long as β < ; for we calculate ψ(x) lim x x = lim x 2x β cos(γ log x θ ρ )/ ρ x x rovided that β <. If there were to be a zero ρ with real art equal to, we can see via the exlicit formula that the rime number theorem would not be true. I want to conclude by giving you the idea that the Riemann hyothesis does not just make ψ(x) as close to x as ossible, but also is very beautiful. If it is true, it means that rimes are generated by a bunch of different frequencies of waves that grow at the same rate. If some waves grow faster than others, we get havoc. To observe this all in action, check out the following website: =, htt:// This website contains grahs of the comonents of the exlicit formula coming from each zero, so that you can see what a function like 2 xβ cos(γ log x θ ρ ρ) looks like. But most imortantly, it has an alet at the bottom that shows the ψ-function in yellow and shows the right side of the exlicit formula as additional non-trivial zeros are added to the sum. Check it out! 9

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