euler sine formula

Chapter 6 Euler’s Formula for sin( z) 6.0. Motivation In this chapter we take a little break and show how the complex analysis we have learned so far can be used to prove Euler’s infinite-prod uct represen- tation of the sine function: sin(πz) =πz∞/productdisplay n=1/parenleftbigg 1−z2 n2/parenrightbigg . We should note that this is a special case of general results a bout factor- izations of entire functions to be discussed later; here we g ive two ad hoc proofs, one using the Residue Theorem and one by Liouville’s Theorem. How could someone come up with a formula like that in the first p lace? A person might start with the observation that sin( πz) vanishes precisely at the integers, so it could be that a formula something like sin(πz) =∞/productdisplay n=−∞(z−n) works; the product on the right apparently vanishes at the in tegers, so if sine were a polynomial this would be at worst off by a constant f actor. Unfortunately a person realizes very quickly that the produ ct/producttext∞ n=−∞(z−n) simply does not converge; the factors do not even tend to 1, which seems a likely prerequisite for convergence of an in finite product. We could try to fix this this by multiplying each factor by some constant to 87 88 6. Euler’s Formula for sin(z) make the product converge somewhere . The simplest approach would be to multiply each factor (or each factor with n/\e}atio\slash= 0) by a constant so as to make the factor equal 1 at the origin. So we want to replace the fact or (z−n) with ( z−n)/(0−n), or 1 −z/n. That gives us a tentative sin(πz) =/parenleftbigg−1/productdisplay n=−∞/parenleftBig 1−z n/parenrightBig/parenrightbigg (z)/parenleftbigg∞/productdisplay n=1/parenleftBig 1−z n/parenrightBig/parenrightbigg . It ispossible that that is within a constant factor of being correct. If so then what would the constant be? If you divide both sides by zand then letz→0, you get π= 1. So we fix that: sin(πz) =/parenleftbigg−1/productdisplay n=−∞/parenleftBig 1−z n/parenrightBig/parenrightbigg (πz)/parenleftbigg∞/productdisplay n=1/parenleftBig 1−z n/parenrightBig/parenrightbigg . This is actually correct. Almost. The products still do not c onverge. Whether an infinite product converges has to do with how fast t he factors tend to 1, and (as we will see below) the fact that the series/summationtext∞ n=11/n diverges implies that the product/producttext∞ n=1(1−z/n) diverges. But it turns out that this product converges if we modify it by grouping ce rtain pairs of factors together: (1 −z/n) and (1 + z/n) are both too far from 1 for convergence, but their product is (1 −z2/n2), which is much closer to 1. And so we combine pairs of factors to get sin(πz) =πz∞/productdisplay n=1/parenleftbigg 1−z2 n2/parenrightbigg , which turns out to be correct. Of course we have not yet given much indication why this produ ct should actually equal sin( πz). It is not hard to see that the product converges to an entire function, and if we define P(z) =πz∞/productdisplay n=1/parenleftbigg 1−z2 n2/parenrightbigg then it is not hard to show that P(z) = 0 if and only if z∈Z, and one can even show without much trouble that P(z+ 1) = −P(z), but sin( πz) is not the only function with all these properties (for example if gis entire then the function sin( πz)eg(sin2(πz))has all the properties listed). But it seems that it could be so; now we just have to prove it. There are vari ous ways to do that. 6.1. Proof by the Residue Theorem 89 6.1. Proof by the Residue Theorem One might imagine proving that Euler’s infinite product equa ls sin( πz) by taking the logarithm of both sides — we need only show that log(sin( πz)) = log( πz) +∞/summationdisplay n=1log/parenleftbigg 1−z2 n2/parenrightbigg . Of course this cannot quite work because these functions do n ot quite have (“single-valued”) logarithms; at the very least we would ne ed to be very careful about saying exactly what branches of the logarithm we are using (note that the factors in the product have zeroes, and we are n ot going to be able to define branches of the logarithm holomorphic near tho se zeroes ...) . This begins to seem somewhat tricky. Before abandoning this idea we might note that although most holomor- phic functions do not have (holomorphic) logarithms, and in particular these ones do not, the nonexistent logarithm of a holomorphic func tiondoeshave a derivative! Seriously: Two branches of log( f) differ by a constant, and that constant goes away when we differentiate. So, in spite of our problems with interpreting the previous formula, if we differentiate it we get some- thing that might make sense, might be true, and might then lea d to Euler’s formula for sin( πz). Noting that the derivative of log( f) isf′/f, we take the derivative of all the terms in the last formula and we get πcot(πz) =1 z+∞/summationdisplay n=1−2z/n2 1−z2/n2=1 z+∞/summationdisplay n=1/parenleftbigg1 z−n+1 z+n/parenrightbigg . Could this be right? Yes: Note that the sum on the right has pol es at the integers and residue 1 at each pole, just like the function πcot(πz). The actual proof starts here: We will show using the Residue T heorem that the function πcot(πz) is given by the infinite sum above, and we will deduce Euler’s formula for sin( πz) (after a few preliminaries about infinite products and the derivatives of nonexistent logarithms). The infinite-sum representation for the cotangent follows f rom a Cauchy’s Integral Formula–Residue Theorem hybrid: Theorem 6.1.0 (Cauchy Integral Formula for functions with s im- ple poles). Suppose that V⊂Cis open and Γis a cycle in Vsuch that Ind(Γ,a) = 0for all a∈C\V. Suppose that Sis a(relatively )closed subset ofV,S∩Γ∗=∅, and every point of Sis isolated. Suppose that f∈H(V\S) andfhas a simple pole or removable singularity at every point of S. Then forz∈V\(S∪Γ∗)we have Ind(Γ,z)f(z) =1 2πi/integraldisplay Γf(w) w−zdw+/summationdisplay p∈SInd(Γ,p)Res(f,p) z−p. 90 6. Euler’s Formula for sin(z) (Note that the sum has only finitely many nonzero terms, as in T heorem 4.17. Note also that the validity of the theorem depends on th e fact that f has at worst a simple pole at each point of S.) Proof. Fixz∈V\(S∪Γ∗) and let g(w) =f(w) w−z. Nowgis holomorphic in Vexcept for (possible) simple poles at zand the points of S. Exercise 4.7(ii) shows that Res( g,z) =f(z) and Res( g,p) = Res(f,p)/(p−z) forp∈S. So the Residue Theorem (Theorem 4.17) shows that 1 2πi/integraldisplay Γg(w)dw= Ind(Γ ,z)Res(g,z) +/summationdisplay p∈SInd(Γ,p)Res(g,p) = Ind(Γ ,z)f(z) +/summationdisplay p∈SInd(Γ,p)Res(f,p) p−z. /square For our application we need to know that the cotangent functi on remains bounded if we stay away from its poles: Lemma 6.1.1. There exists a constant Msuch that |cot(πz)| ≤M whenever |Im (z)| ≥1orRe(z) =n+ 1/2 (n∈Z). Proof. Ifz=x+iythen cot(πz) =ie−2πye2πix+ 1 e−2πye2πix−1=i1 +e2πye−2πix 1−e2πye−2πix. The first expression shows that cot( πz)→ −iuniformly in xasy→ ∞, while the second shows that cot( πz)→iuniformly in xasy→ −∞ . So there exists a number Asuch that |cot(πz)| ≤2 (|y| ≥A). Let K={x+iy: 0≤x≤1,1≤ |y| ≤A} ∪ {1 2+iy:|y| ≤1}. Since Kis compact and the function cot( πz) is continuous on K, there exists M≥2 such that |cot(πz)| ≤Mforz∈K. Since cot( π(z+1)) = cot( πz), it follows that |cot(πz)| ≤Mforz∈/uniontext n∈Z(n+K) and hence for z∈ {x+iy: |y| ≥A} ∪/uniontext n∈Z(n+K). /square 6.1. Proof by the Residue Theorem 91 Definition. Ifz∈Cands >0 then the boundary of the square with lower-left corner zand side length sis the cycle [z, z+s]˙+[z+s, z+s+is]˙+[z+s+is, z+is]˙+[z+is, z]. Theorem 6.1.2. Ifz∈C\Zthen πcot(πz) =1 z+∞/summationdisplay n=1/parenleftbigg1 z−n+1 z+n/parenrightbigg . Proof. ForNa positive integer let QNbe the boundary of the square with lower-left corner −(N+1 2)−i(N+1 2) and side length 2 N+ 1. An application of Theorem 6.1.0 shows that if z∈C\Z,|Re (z)|< N+1 2and |Im (z)|< N+1 2then πcot(πz) =N/summationdisplay n=−N1 z−n+1 2πi/integraldisplay QNπcot(πw) w−zdw, so we need only show that lim N→∞/integraldisplay QNπcot(πw) w−zdw= 0. Now Lemma 6.1.1 shows that |cot(πz)| ≤MonQ∗ N, but this is not quite enough; a direct application of the ML inequality show s only that the integral is bounded, not that it tends to 0. But the ML inequal ity is a rather crude device for estimating the size of integrals, because i t ignores the fact that the absolute value of the integrand may not be constant a nd also ignores any possible cancellation. It turns out that our integral is much smaller than the ML inequality would appear to indicate, because of cance llation: We notice that the integral of an evenfunction over QNmust equal 0. Now, cot( πw)/(w−z) is not an even function of w, but if zis fixed and w is large then it is very close to the even function cot( πw)/w. So we estimate our integral by comparing it to the integral of cot( πw)/w, as follows: /integraldisplay QNcot(πw) w−zdw=/integraldisplay QNcot(πw)/parenleftbigg1 w−z−1 w/parenrightbigg dw=z/integraldisplay QNcot(πw) w(w−z)dw. The ML inequality suffices to show that this last integral tend s to 0: Note that |w|> Nforw∈Q∗ N. Hence if N >2|z|we have |w−z|> N/2 and hence /vextendsingle/vextendsingle/vextendsingle/vextendsingle1 w(w−z)/vextendsingle/vextendsingle/vextendsingle/vextendsingle<2 N2 92 6. Euler’s Formula for sin(z) forw∈Q∗ N. So the ML inequality and Lemma 6.1.1 show that /vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay QNcot(πw) w(w−z)dw/vextendsingle/vextendsingle/vextendsingle/vextendsingle<2M N2(8N+ 4) forN >2|z|; hence lim N→∞/integraldisplay QNcot(πw) w(w−z)dw= 0. /square Note that the proof shows that πcot(πz) = lim N→∞N/summationdisplay n=−N1 z−n. It would be a bad idea to rewrite this as πcot(πz) =∞/summationdisplay n=−∞1 z−n, because that might be taken to mean πcot(πz) = lim N,M→∞N/summationdisplay n=−M1 z−n, and this last equation is not correct, because the indicated limit does not exist! •Exercise 6.1. Show that the series ∞/summationdisplay n=01 z−n diverges for all z∈C\Z. Hint: If/summationtextandiverges and/summationtext(an−bn) converges then/summationtextbndiverges. •Exercise 6.2. Show that the series ∞/summationdisplay n=1/parenleftbigg1 z−n+1 z+n/parenrightbigg converges absolutely for all z∈C\Z. We need to say a little bit about infinite products. The main te chnical detail follows: 6.1. Proof by the Residue Theorem 93 Lemma 6.1.3. Ifz1,... ,z n∈Cand/summationtextn−1 j=1|1−zj|<1/2then /vextendsingle/vextendsingle/vextendsingle/vextendsingle1−n/productdisplay j=1zj/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤2n/summationdisplay j=1|1−zj|. Proof. The proof is by induction on n; the case n= 1 is clear. (By con- vention a sum of the form/summationtext0 j=1is equal to 0, being the sum of no terms.) Suppose we know that/vextendsingle/vextendsingle/vextendsingle/vextendsingle1−/producttextn j=1zj/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤2/summationtextn j=1|1−zj|whenever /summationtextn−1 j=1|1−zj|<1/2 and suppose as well that/summationtextn j=1|1−zj|<1/2. Then /vextendsingle/vextendsingle/vextendsingle/vextendsinglen/productdisplay j=1zj/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤1 +/vextendsingle/vextendsingle/vextendsingle/vextendsingle1−n/productdisplay j=1zj/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤1 + 2n/summationdisplay j=1|1−zj| ≤2, and so /vextendsingle/vextendsingle/vextendsingle/vextendsingle1−n+1/productdisplay j=1zj/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle1−n/productdisplay j=1zj+n/productdisplay j=1zj−n+1/productdisplay j=1zj/vextendsingle/vextendsingle/vextendsingle/vextendsingle ≤/vextendsingle/vextendsingle/vextendsingle/vextendsingle1−n/productdisplay j=1zj/vextendsingle/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsingle/vextendsinglen/productdisplay j=1zj/vextendsingle/vextendsingle/vextendsingle/vextendsingle|1−zn+1| ≤2n/summationdisplay j=1|1−zj|+ 2|1−zn+1| = 2n+1/summationdisplay j=1|1−zj|. /square Lemma 6.1.4. (i)Ifz1,...∈Cand/summationtext∞ j=1|1−zj|<∞then ∞/productdisplay j=1zj= lim n→∞n/productdisplay j=1zj exists;furthermore,/producttext∞ j=1zj/\e}atio\slash= 0unless zj= 0for some j. (ii)If(fj)is a sequence of complex-valued functions on some set Sand the sum/summationtext∞ j=1|1−fj|converges uniformly on SthenPn=/producttextn j=1fjtends to P=/producttext∞ j=1fjuniformly on S;ifz∈SthenP(z)/\e}atio\slash= 0unless fj(z) = 0 for some j. 94 6. Euler’s Formula for sin(z) Note. It is possible for an infinite product to equal 0 even if none of the factors vanish (consider/producttext∞ j=11/2). Some authors say that such a sum “di- verges to 0”. (And other authors say it converges to 0; for thi s reason we will try to avoid saying that an infinite product converges or diverges.) Proof. (i) Choose Nso that/summationtext∞ j=N+1|1−zj|<1/2. To show that limn→∞/producttextn j=1zjexists it is sufficient to show that ( pn) is a Cauchy sequence, where pn=N+n/productdisplay j=N+1zj. But Lemma 6.1.3 shows that |pn| ≤2, and hence another application of the lemma shows that if n > m then |pn−pm|=|pm|/vextendsingle/vextendsingle/vextendsingle/vextendsingle1−N+n/productdisplay j=N+m+1zj/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤4N+n/summationdisplay j=N+m+1|1−zj|; hence |pn−pm| →0 asn,m→ ∞. If none of the zjvanish then/producttextN j=1zj/\e}atio\slash= 0, and Lemma 6.1.4 shows that |1−/producttext∞ j=N+1zj|<1, so/producttext∞ j=N+1zj/\e}atio\slash= 0; hence/producttext∞ j=1zj/\e}atio\slash= 0. (ii) The proof of the second part of the lemma is the same as the proof of the first part. /square Now the sum ∞/summationdisplay n=1/vextendsingle/vextendsingle/vextendsingle/vextendsinglez2 n2/vextendsingle/vextendsingle/vextendsingle/vextendsingle converges uniformly on compact subsets of the plane; thus Le mma 6.1.4(ii) shows that the partial products PN(z) =πzN/productdisplay n=1/parenleftbigg 1−z2 n2/parenrightbigg converge uniformly on compact subsets of the plane to P(z) =πz∞/productdisplay n=1/parenleftbigg 1−z2 n2/parenrightbigg , an entire function which vanishes only at the integers. Note that we also have P′ N→P′, by Proposition 3.5. 6.1. Proof by the Residue Theorem 95 In general, if fis a differentiable function, we define the logarithmic derivative L(f) by L(f) =f′ f, at least at points where fis nonzero. The logarithmic derivative of fis the derivative of the logarithm of fwhen fhasa logarithm; this makes the formula L(fg) =L(f) +L(g) plausible. In fact the product rule shows that L(fg) =L(f) +L(g) for any differentiable nonvanishing functions fandg. It follows that L/parenleftbiggN/productdisplay n=1fn/parenrightbigg =N/summationdisplay n=1L(fn) on the set where none of the fnvanish. (This sometimes gives a convenient and efficient way to calculate derivatives of products of seve ral functions, by the way.) We will use the following continuity property of L: Proposition 6.1.5. Suppose that Vis a connected open set, f1,f2,...∈ H(V)andfn→funiformly on compact subsets of V. Suppose that fis not identically zero. Then L(fn)→L(f) uniformly on K, ifKis any compact subset of Von which fhas no zero. Proof. This is immediate from Proposition 3.5. /square And we will use the fact that L(f) determines f, at least up to a constant factor: Lemma 6.1.6. Suppose that Vis a connected open subset of C,fandg are holomorphic functions in Vneither of which vanishes identically, and L(f) =L(g) on the set where neither fnorgvanishes. Then f=cgfor some constant c. Proof. We may suppose that neither fnorghas a zero in V(because the set where they are both nonzero is in general a dense connecte d open subset ofV). Now the fact that L(f) =L(g) shows that L(f/g) = 0, and hence (f/g)′= 0; thus f/gis constant. /square 96 6. Euler’s Formula for sin(z) We can finally prove Euler’s formula. Assume first that z∈C\Z. Note that L/parenleftbigg 1−z2 n2/parenrightbigg =L/parenleftBig 1−z n/parenrightBig +L/parenleftBig 1 +z n/parenrightBig =1 z−n+1 z+n. IfPNandPare as above then it follows that L(PN)(z) =N/summationdisplay n=−N1 z−n, and now Proposition 6.1.5 shows that L(P) = lim N→∞N/summationdisplay n=−N1 z−n=1 z+∞/summationdisplay n=1/parenleftbigg1 z−n+1 z+n/parenrightbigg . But now Theorem 6.1.2 shows that L(P)(z) =πcot(πz) =L(sin(πz)), so that P(z) =csin(πz) by Lemma 6.1.6. Dividing by πzwe see that csin(πz) πz=∞/productdisplay n=1/parenleftbigg 1−z2 n2/parenrightbigg (z/\e}atio\slash= 0), and letting z→0 here shows that c= 1. This proves Euler’s formula for z∈C\Z; the case z∈Zfollows by continuity. /square 6.2. Estimating Sums Using Integrals In the second proof of Euler’s formula we will need to use an in tegral to estimate a certain infinite sum. Let us say a little bit about v arious ways one might do that. In this section we will assume that φ: (0,∞)→(0,∞) is continuous, and we will consider the question of what we ca n say about the sum ∞/summationdisplay n=1φ(n) in terms of integrals. 6.2. Estimating Sums Using Integrals 97 The simplest approach is something found in a lot of calculus books under the heading “Integral Test”. Suppose that φis nonincreasing. This shows that φ(n) =/integraldisplayn n−1φ(n)dt≤/integraldisplayn n−1φ(t)dt, and hence that ∞/summationdisplay n=1φ(n)≤/integraldisplay∞ 0φ(t)dt. This can sometimes be useful even if it does not appear that wa y at first. For example, if we want to know that ∞/summationdisplay n=11 n2 converges then a direct application of the inequality above does not help, because/integraltext∞ 0dt/t2=∞. But of course the same argument shows that ∞/summationdisplay n=1φ(n) =φ(1) +∞/summationdisplay n=1φ(n)≤φ(1) +/integraldisplayn 1φ(t)dt, which shows that/summationtext∞ n=11/n2converges. Noting that a similar argument shows that/integraldisplay∞ 1φ(t)dt≤∞/summationdisplay n=1φ(n) we obtain the “integral test”: If φ: [1,∞)→(0,∞) is continuous and nonincreasing then/summationtext∞ n=1φ(n) converges if and only if/integraltext∞ 1φ(t)dt <∞. If we just want to check whether a sum converges then that will often suffice. However, sometimes we need more precise information about the sizeof our sum than is given by the integral test. If you think abou t it for a second, you decide that/integraltextn n−1φ(t)dtis probably a fairly poor approximation toφ(n); taking the integral from n−1 2ton+1 2should often give a better approximation. Of course now the problem of estimating the e rror in this approximation arises. •Exercise 6.3. Suppose that φ′′is continuous on [ −1/2,1/2]. (i) Show that /integraldisplay1/2 −1/2φ(t)dt−φ(0) =1 2/integraldisplay1/2 −1/2/parenleftbigg |t| −1 2/parenrightbigg2 φ′′(t)dt. 98 6. Euler’s Formula for sin(z) (ii) Deduce that /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay1/2 −1/2φ(t)dt−φ(0)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤1 8/integraldisplay1/2 −1/2|φ′′(t)|dt. Hint: For the first part, write 1 2/integraldisplay1/2 −1/2/parenleftbigg |t| −1 2/parenrightbigg2 φ′′(t)dt=1 2/integraldisplay1/2 0/parenleftbigg t−1 2/parenrightbigg2 (φ′′(t) +φ′′(−t))dt and integrate by parts until done. The exercise shows that /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay∞ 1/2φ(t)dt−∞/summationdisplay n=1φ(n)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤1 8/integraldisplay∞ 1/2|φ′′(t)|dt. This can be useful, although dealing with the error term1 8/integraltext∞ 1/2|φ′′|can be inconvenient. It is too bad that we do not have a nice simple inequality likeφ(n)≤/integraltextn+1/2 n−1/2φ, so we could get a bound for our sum in terms of an integral with no error term, as in our first integral test abov e... But wait. If φ′′≥0 then the first part of the exercise above shows that wedohave φ(n)≤/integraldisplayn+1 2 n−1 2φ(t)dt; we have been forgetting about the basic principle that conve xity is where inequalities come from! Or in any case Iforgot this in the first version of this proof; maybe some of us have never seen it stated explicitly. Here it is: Convexity is the source of many useful inequalities. Now you’ve seen it; try to remember this the next time you have an inequality you wish you could prove. We should mention a technicality: By definition the continuo us function φisconvex if φ(x)≤φ(x+t) +φ(x−t) 2 for all x,tsuch that the interval [ x−t,x+t] is contained in the domain of φ. Ifit happens that φ′′is continuous, thenφis convex if and only if φ′′≥0. 6.3. Proof Using Liouville’s Theorem 99 But convexity is actually a weaker condition: The function φ(x) =|x|shows that a convex function need not have a second derivative. And in fact convexity is exactly the condition we need here, n otφ′′≥0; ifφis convex then φ(n) = 2/integraldisplay1 2 0φ(n)dt≤2/integraldisplay1 2 0φ(n+t) +φ(n−t) 2dt=/integraldisplayn+1 2 n−1 2φ(t)dt, a much simpler argument than the previous exercise. Let us summarize the things we have proved here for future ref erence; they can all be useful in various places: Theorem 6.2.0. Suppose that φ: (0,∞)→(0,∞)is continuous. (i)Ifφis nonincreasing then /integraldisplay∞ 1φ(t)dt≤∞/summationdisplay n=1φ(n)≤/integraldisplay∞ 0φ(t)dt. (ii)Ifφ′′is continuous then /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay∞ 1/2φ(t)dt−∞/summationdisplay n=1φ(n)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤1 8/integraldisplay∞ 1/2|φ′′(t)|dt. (iii)Ifφis convex then ∞/summationdisplay n=1φ(n)≤/integraldisplay∞ 1/2φ(t)dt. We will see in the next section that the difference between the/integraltext∞ 0φin part (i) and the/integraltext∞ 1/2φin part (iii) can make a big difference. 6.3. Proof Using Liouville’s Theorem One can use Liouville’s Theorem to prove Euler’s formula; we define P(z) =πz∞/productdisplay n=1/parenleftbigg 1−z2 n2/parenrightbigg , and it follows from Exercise 3.8 in Chapter 3 that we need only show that Pis an entire function such that P(z+ 1) = −P(z) and |P(z)| ≤ceπ|Im(z)|. 100 6. Euler’s Formula for sin(z) (The exercise then shows that P(z) =αsin(πz) for some constant α; as noted above we can then divide both sides by zand then consider the limit asz→0 to show that α= 1.) The fact that Pis an entire function is proved as in Section 6.1; this follows from Lemma 6.1.4, since the sum/summationtext∞ n=1|z2/n2|converges uniformly forzin any compact subset of the plane. (The fact that Lemma 6.1.4 is used in both proofs is not a surprise; the lemma is simply the b asic tool one uses in checking convergence of infinite products.) The proof that P(z+ 1) = −P(z) is not hard. Let PN(z) =zN/productdisplay n=1/parenleftbigg 1−z2 n2/parenrightbigg . A little rearrangement shows that PN(z) =(−1)N (N!)2N/productdisplay n=−N(z−n). Hence PN(z+ 1) =(−1)N (N!)2N/productdisplay n=−N(z−(n−1)) =(−1)N (N!)2N−1/productdisplay n=−N−1(z−n), so that if z/\e}atio\slash∈Zwe have PN(z+ 1) PN(z)=z−(−N−1) z−N. This shows that PN(z+ 1)/PN(z)→ −1 and hence P(z+ 1) = −P(z) for z/\e}atio\slash∈Z; the fact that P(z+ 1) = −P(z) for all zfollows by continuity. The interesting part is the proof that |P(z)| ≤ceπ|Im(z)|. It seems natu- ral to take the logarithm, since sums can be easier to deal wit h than prod- ucts. So we begin by noting that log/parenleftBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞/productdisplay n=1/parenleftbigg 1−z2 n2/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenrightBigg =∞/summationdisplay n=1log/parenleftbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle1−z2 n2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenrightbigg ≤∞/summationdisplay n=1log/parenleftbigg 1 +|z|2 n2/parenrightbigg . (At first it looks like we threw away too much in that last inequ ality, so that we could not hope to get a good enough bound starting here . But we are only going to be using this inequality for zclose to the imaginary axis; for such zthe inequality is close to equality.) Now we estimate that su m by comparing it to an integral, as in the previous section: 6.3. Proof Using Liouville’s Theorem 101 Fixz/\e}atio\slash= 0 and let φ(t) = log(1+ |z|2/t2). In the first version of the proof I used part (i) of Theorem 6.1.7: ∞/summationdisplay n=1log/parenleftbigg 1 +|z|2 n2/parenrightbigg ≤/integraldisplay∞ 0log/parenleftbigg 1 +|z|2 t2/parenrightbigg dt=π|z|. (The integral can be evaluated by elementary calculus; see b elow for details.) This shows that∞/productdisplay n=1/parenleftbigg 1 +|z|2 n2/parenrightbigg ≤eπ|z|, which looks like what we want. Unfortunately there is a missi ng factor on the left-hand side; all we have proved here is that |P(z)| ≤π|z|eπ|z|. Now, this last inequality can in fact be used to prove Euler’s formula, but the simple Exercise 3.8 does not quite apply, we need a somewh at clumsier argument. A miracle happened when I realized I should be applying the co nvexity of φ: The improvement in the estimate of the sum by the integral wa s exactly enough to take care of that extra factor of z! You can easily verify that φ′′≥0. Hence part (iii) of Theorem 6.1.7 shows that∞/summationdisplay n=1log/parenleftbigg 1 +|z|2 n2/parenrightbigg ≤/integraldisplay∞ 1/2log/parenleftbigg 1 +|z|2 t2/parenrightbigg dt. Now a simple integration by parts shows that /integraldisplay log/parenleftbigg 1 +|z|2 t2/parenrightbigg dt=tlog/parenleftbigg 1 +|z|2 t2/parenrightbigg + 2|z|arctan/parenleftbiggt |z|/parenrightbigg , at least for t∈(0,∞). We need to figure out how this antiderivative behaves ast→+∞. Since log(1) = 0 and log′(1) = 1, it follows that log(1+ |z|2/t2) is approximately |z|2/t2for large t, so that tlog(1 + |z|2/t2)→0. It is clear that arctan( t/|z|)→π/2, so we obtain /integraldisplay∞ 1/2log/parenleftbigg 1 +|z|2 t2/parenrightbigg dt=π|z| −(log(1 + 4 |z|2)/2 + 2|z|arctan(1 /(2|z|))) ≤π|z| −log(4|z|2)/2 =π|z| −log(|z|)−log(2) < π|z| −log(|z|). 102 6. Euler’s Formula for sin(z) That −log(|z|) saves the day, showing that |P(z)| ≤πeπ|z|. Since P(z+ 2) = P(z), it follows (see Exercise 6.4 below) that |P(z)| ≤ceπ|Im (z)|, and this proves Euler’s formula, as noted at the start of this section. Exercises 6.4.Suppose that P:C→Cis continuous, P(z+ 2) = P(z), and |P(z)| ≤ eπ|z|. Show that there exists csuch that |P(z)| ≤ceπ|Im(z)|for all z. 6.5.Prove Wallis’ formula: π 2=∞/productdisplay n=1/parenleftbigg4n2 (2n−1)(2n+ 1)/parenrightbigg =/parenleftbigg2 1/parenrightbigg/parenleftbigg2 3/parenrightbigg/parenleftbigg4 3/parenrightbigg/parenleftbigg4 5/parenrightbigg · · ·. Note. After you prove thatπ 2=/producttext∞ n=1/parenleftBig 4n2 (2n−1)(2n+1)/parenrightBig there is still a little bit of work to do in showing that/producttext∞ n=1/parenleftBig 4n2 (2n−1)(2n+1)/parenrightBig =/parenleftbig2 1/parenrightbig/parenleftbig2 3/parenrightbig/parenleftbig4 3/parenrightbig/parenleftbig4 5/parenrightbig · · ·! Yes, it is clear that 4n2 (2n−1)(2n+ 1)=2n 2n−12n 2n+ 1, but that is not quite enough by itself, there is a little argument required. For example, it is also true that 4n2 (2n−1)(2n+ 1)=4n 2n−1n 2n+ 1, but it does not follow from that that π 2=/parenleftbigg4 1/parenrightbigg/parenleftbigg1 3/parenrightbigg/parenleftbigg8 3/parenrightbigg/parenleftbigg2 5/parenrightbigg · · ·; in fact thatproduct does not converge. 6.3. Proof Using Liouville’s Theorem 103 6.6.Find/summationtext∞ n=11 n2, using the infinite series for cot( πz). The next exercise shows that, at least for continuous functi ons, the def- inition of convexity we gave above is equivalent to an appare ntly stronger condition. (We take the domain to be all of Rjust for convenience.) 6.7.Suppose that φ:R→Ris continuous and satisfies φ(x)≤φ(x+y) +φ(x−y) 2 for all x,y∈R. Show that φ(tx+ (1−t)y)≤tφ(x) + (1 −t)φ(y) whenever 0 ≤t≤1. Hint: First note that the hypothesis is equivalent to φ/parenleftbiggx+y 2/parenrightbigg ≤φ(x) +φ(y) 2, which gives the conclusion for t= 1/2. Since φis continuous, you may assume that t=k/2nfor non-negative integers n,k. Now the result follows by induction on n; for example, the identity x+y 2+y 2=1 4x+3 4y shows how the case t= 1/4 follows from the case t= 1/2: φ/parenleftbigg1 4x+3 4y/parenrightbigg =φ/parenleftBiggx+y 2+y 2/parenrightBigg ≤φ/parenleftbigx+y 2/parenrightbig +φ(y) 2≤φ(x)+φ(y) 2+φ(y) 2=1 4φ(x) +3 4φ(y). Chapter 19 Preliminaries for the Picard Theorems (excerpt) (Notations established previously: If Ω is an open subset of the plane then Aut(Ω) is the group of invertible holomorphic maps from Ω to i tself; now if p: Ω→Vis a holomorphic covering map then Aut(Ω ,p) is the subgroup {φ∈Aut(Ω) : p◦φ=p}.) [...] Theorem 19.4.3. Suppose Ωis simply connected and pj: Ω→Vjis a holomorphic covering map for j= 0,1. Then V1andV0are conformally equivalent if and only if Aut(Ω ,p0)andAut(Ω ,p1)are conjugate in Aut(Ω) . Proof. [...] /square Forr >1 define Ar={z: 1<|z|< r}. The next theorem gives a demonstration of the power of Theorem 19.4.3. Theorem 19.4.4. If1< r1< r2thenAr1andAr2are not conformally equivalent. Recall that two square matrices AandBare said to be similar if there exists a matrix Csuch that A=CBC−1. Recall as well the trivial fact that similar matrices have the same eigenvalues. And recall from Section 10.7 that Aut(Π+) is isomorphic to SL2(R)/G, where SL2(R) is the group of all real 2 ×2 matrices with determinant 1 and Gis the subgroup {I,−I}. 337 338 19. Preliminaries for the Picard Theorems (excerpt) Proof. Forδ >0 define φδ∈Aut(Π+) byφδ(z) =δz; note that the matrix inSL2(R) corresponding to φδ(or rather one of the two matrices in SL2(R) corresponding to φδ) is Mδ=/bracketleftbiggδ1/20 0δ−1/2/bracketrightbigg . Theorem 19.1.2 shows that for j= 1,2 there is a holomorphic covering map pj: Π+→Arjsuch that Aut(Π+,pj) is generated by φδj, where δj>1 andδ1/ne}ationslash=δ2. We need only show that Aut(Π+,p1) is not conjugate to Aut(Π+,p2) in Aut(Π+). Suppose to the contrary that Aut(Π+,p1) =χAut(Π+,p2)χ−1for some χ∈Aut(Π+). Then χ◦φδ2◦χ−1must be a generator of Aut(Π+,δ1), so thatχ◦φδ2◦χ−1=φδ1orχ◦φδ2◦χ−1=φ1/δ1. Hence Mδ2must be similar to one of the four matrices ±Mδ1,±M1/δ1. But this is impossible because similar matrices have the same eigenvalues. /square [...] Chapter 20 The Picard Theorems (excerpt) Notation will be as in the previous two chapters: Γ is the grou p of all φ∈Aut(Π+) which can be written in the form φ(z) = (az+b)/(cz+d) witha,b,c,d∈Zandad−bc= 1, Γ(2) is the subgroup of Γ where aand dare odd and bandcare even, and λ: Π+→C\ {0,1}is a holomorphic covering map with Aut(Π+,λ) = Γ(2). The fact that λis a holomorphic covering map from Π+ontoC\ {0,1} makes the Little Picard Theorem very simple: Theorem 20.0 (Little Picard Theorem). Iffis a nonconstant entire function then the range f(C)is either CorC\ {α}for some α∈C. In other words, if the range of an entire function omits two co mplex values then the function is constant. Proof. Suppose that fis an entire function and fomits the two values α, β∈C, with α/\e}atio\slash=β. We need to show that fis constant. Considering ( f−α)/(β−α) in place of f, we may assume that the two values omitted are 0 and 1; thus f(C)⊂C\ {0,1}. Now Theorem 19.0.5 shows that there exists ˜f:C→Π+withλ◦˜f=f. Liouville’s Theorem shows that 1 /(i+˜f) is constant; hence ˜fis constant and so fis constant. /square I think it was Littlewood who said that this would be the world ’s short- est PhD thesis. A one-line proof of a very deep theorem. (Of co urse the 357 358 20. The Picard Theorems (excerpt) existence of a covering map from the upper half-plane onto C\ {0,1}was not entirely trivial.) Now for the Big Picard Theorem. There are various statements that commonly go by this name: [...] As previously we set D′=D\ {0}. [The Big Picard Theorem] is equiva- lent to the following: Theorem 20.3 (Big Picard Theorem). Iff∈H(D′)andf(D′)⊂ C\ {0,1}thenfhas a pole or a removable singularity at 0. [...] Traditional proofs of the Big Picard Theorem are totally unl ike the usual proof of the Little Picard Theorem, but we shall see that it is possible to give a proof of the Big Picard Theorem that is very much like a gener alization of the proof of the Little Picard Theorem; where the first proof u sed nothing but the most basic properties of covering maps, the second pr oof uses the slightly more sophisticated results from Chapter 19. The idea behind the proof may be clearer if we start with two re sults dealing with more familiar notions like harmonic conjugate s and exponen- tials: Our proof of the Big Picard Theorem is to the tradition al proof of the Little Picard Theorem exactly as the proof of Theorem B below is to the proof of Theorem A: Theorem A. Ifu:C→Ris a bounded harmonic function then uis constant. Proof. Since Cis simply connected, there exists a real-valued harmonic function vsuch that u+ivis holomorphic. Let h=eu+iv. Then his an entire function, and his bounded, since |h|=eu. Thus his constant, and hence u= log |h|is constant. /square Theorem B. Iff:D′→Ris harmonic and bounded then uextends to a function harmonic in D. The problem with Theorem B is that we cannot say that uhas a har- monic conjugate, since D′is not simply connected. (This is precisely anal- ogous to the problem with Theorem 20.3: We cannot say that the re exists ˜f:D′→Π+withλ◦˜f=fbecause D′is not simply connected.) One can give a slightly informal proof of Theorem B as follows : There does exist a “multi-valued” harmonic conjugate v(this is an informal way of saying that there is a holomorphic function u+ivdefined in some disk 20. The Picard Theorems (excerpt) 359 contained in D′which admits unrestricted continuation in D′). Now it is not hard to see that in fact there exists a multi-valued harmonic conjugate v and a real number csuch that any two branches of vdiffer by a multiple of c. Ifc= 0 we are done, so we assume that c/\e}atio\slash= 0, and define h=e2π(u+iv)/c. The fact that any two branches of vdiffer by a multiple of cshows that his an actual single-valued holomorphic function. As in the pro of of Theorem A we see that his bounded; hence hhas a removable singularity, and hence uhas a removable singularity at the origin as well. If the “informal” parts of that proof bother you don’t worry, we will give a more formal version soon. The corresponding informal proo f of Theorem 20.3 is this: Although we cannot assert that there exists ˜f:D′→Π+with λ◦˜f=f, there does exist a “multi-valued function” ˜fwith this property (that is, an ˜fdefined in some disk contained in D′which admits unrestricted continuation in D′). It turns out that there exists φ∈Γ(2) such that any two branches of ˜fdiffer by a power of φ, and, since φcannot be elliptic, we know that there exists a nonconstant bounded function h∈H(Π+) such thath◦φ=h. It follows that h◦˜fis single-valued, and hence has a removable singularity at the origin. It is now quite plausible, and not hard to prove using the resu lts in the previous chapter, that in fact fhas a (possibly infinite) limit at the origin, so that in particular fdoes not have an essential singularity (for the details here see the formal proof below). The actual proof is going to use a covering-map argument in pl ace of the analytic continuation [ ...]