University:
California State University, Northridge
Course:
MATH 360 | Abstract Algebra I
Academic year:

2022
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360

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20
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darsefienseq0e

Notes on Fourier Series 1 Introduction and terminology We will be considering functions of a real variable with complex values. If f : [a, b] → C is such function, then it can be written as f = 0. This can be done because f is uniformly continuous (p. 143), that is, given ε > 0, there exists a partition x0 , x1 , · · · , xn of the interval [a, b] such that |f (x)−f (y)| < ε/3 if both x, y are in one of the subintervals [xi , xi+1 ]. It follows that if l is the function whose graph consists of the segments joining the points (xi , f (xi )) with (xi+1 , f (xi+1 )), then |f (x) − l(x)| < ε/3 for each x in [a, b]. The next step is to extend l to a function on [0, 2π] so that it becomes periodic. This is done by adding to the graph of l the segments from (0, 0) to (x0 , f (x0 )) and from (xn , f (xn )) to (2π, 0). This extended function, which we continue to call l, being piecewise linear and continuous, satisfies a Lipschitz condition. It follows that it may be uniformly approximated by its Fourier series. That is, there is N such that |SN l(x) − l(x)| < ε/3 for all x in [0, 2π]. 10 Finally, by using that einx = cos nx + i sin nx, we see that SN l is a finite linear combination of the functions sin nx and cos nx, |n| ≤ N . Since the trigonometric functions sin nx and cos nx can be uniformly approximated by their Taylor polynomials, we can find a polynomial P (x) such that |P (x) − SN l(x)| < ε/3 in [0, 2π]. Combining the three inequalities, we have |P (x) − f (x)| < ε in the smaller interval [a, b]. 7 Applications to number theory The first application concerns the nature of the values of the Riemann zeta function ∞ X 1 ζ(s) = ns n=1 at the even positive integers. It is adapted from Courant-John [5]. The Bernoulli polynomials ϕn (x), 0 ≤ x ≤ 2π, are defined recursively by the following relations: ϕ0 (x) = 1, ϕ0n (x) = nϕn−1 (x), and Z 1 ϕn (x) dx = 0 0 for n > 0. That is, knowing ϕn−1 we can calculate ϕn up to a constant, which is determined by the last condition. We see by induction that ϕn is a polynomial of degree n whose coefficients are rational numbers. For n > 1 we have Z 1 ϕn (1) − ϕn (0) = ϕ0n (x) dx = 0. 0 Therefore, if we denote by ψn (x) the periodic extension of the polynomial ϕn , then the functions ψn are continuous and satisfy a Lipschitz condition. The function ψ1 (x) = x − (1/2) is sectionally continuous. Its Fourier coefficients, as a function on [0, 1], are: ( Z 1 0, if n = 0 1 2πinx −1 ψb1 (n) = x− e dx = , if n 6= 0 2 0 2πin 11 Therefore its Fourier series is Sψ1 (x) = 1 X −1 2πinx e . 2π in n6=0 The Fourier series for the other functions ψn are obtained by successively integrating this one. We obtain: ψk (x) = k! X −1 2πinx e . (2π)k (in)k n6=0 In particular, if k is even, ψk (x) = (−1)1+(k/2) 2(k!) X 1 cos 2πnx, (2π)k n=1 nk ψk (x) = (−1)(k+1)/2 2(k!) X 1 sin 2πnx, (2π)k n=1 nk and if k is odd, These series ψk converge uniformly for all x and agree with ϕk in the interval [0, 1]. We also see that ψk (−x) = (−1)n ψk (x), Let −1/2, if k = 1 bk = ϕk (0) = ψk (0), if k > 1. These are rational numbers, and from the Fourier series we see that bk = 0 if k is odd, k 6= 1, and bk = (−1)1+(k/2) 2(k!) X 1 . (2π)k n=1 nk It follows that the values of the Riemann zeta function at the even integers are rational multiples of a power of (2π). Not much is known about the values ζ(2m + 1). Only recently Apéry [8] showed that ζ(3) is irrational. The next application (Exercise 10.5 of Mardsen [6]) is a representation of the function sin πx as an infinite product that resembles the factorization of a polynomial. This was already known to Euler, of course by other means. We consider the function f (x) = cos λx, −π ≤ x ≤ π, where λ is a nonintegral real number. The function f (−π) = f (π) so it can be extended to a periodic continuous function. To compute its Fourier series we use the interval 12 [−π, π]. We have fb(n) = = = Z π 1 cos λx e−inx dx 2π −π Z π 1 (eiλx + e−iλx )e−inx dx 4π −π (−1)n 2λ sin λx. 2π λ2 − n2 Since the function f satisfies a Lipschitz condition, its Fourier series converges to f (x) at all points. Hence, for −π ≤ x ≤ π, cos λx = ∞ sin λπ X 2λ einx . (−1)n 2 2π n=−∞ λ − n2 In particular, for x = π we have sin πλ cos πλ = π Therefore π tan πλ − 1 X 2λ + λ n=1 λ2 − n2 ! . X 2λ 1 = , λ n=1 λ2 − n2 if λ is not an integer. The series on the right converges uniformly for 0 ≤ λ ≤ λ0 < 1. The function on the left is integrable because π tan πλ − (1/λ) → 0 as λ → 0. By integrating, X sin πλ λ2 log = log 1 − 2 πλ n n=1 for |λ| < 1. That is Y λ2 sin πλ = πλ 1− 2 . n n=1 These product formula is also valid for λ = ±1, and then for all real λ because the expression on the right defines a periodic function of period 2. This product formula is interesting because it exhibits directly that the function sin πλ vanishes at the integer values of λ. In this respect it corresponds to the factorization of a polynomial when its zeros are known. If we take λ = 1/2, we obtain Wallis’ product formula for π/2: π 22446688 = ··· . 2 13355779 13 8 The isoperimetric inequality This application of Fourier series will show that among all simple closed plane curves of a given perimeter, the circle is the one that encloses the largest area. By a plane curve we mean a continuous function z : t ∈ [0, 2π] 7→ z(t) ∈ C. It is closed if z(0) = z(2π), and it is simple if the function z is one to one on [0, 2π). We assume that the curves considered here have continuous derivative. The length L of the curve z(t), 0 ≤ t ≤ 2π is Z 2π L= |z 0 (t)| dt 0 as is described in the Appendix to Chapter 13. We assume that L = 2π, and that the curve is parametrized by arc-length: |z 0 (t)| = 1 for all t. The area A enclosed by the curve z is seen to be Z 1 2π z(t)z 0 (t) dt. A= 2i 0 The hypothesis imposed on the function z imply that it can be represented by its Fourier series: ∞ X z(t) = zb(t)eint . −∞ By replacing z(t) by z(t) − zb(0) (which is a translation of the plane, so it does not change the quantities A and L), we may assume that zb(0) = 0. The Fourier coefficients of z and z 0 are easily computed: Z 2π 1 b z(n) = z(t)e−int dt = zb(−n), 2π 0 and zb0 (n) = inb z (n). Hence A Z ∞ ∞ X 1 2π X zb(−n)eint im zb(m)eimt dt 2i 0 n=−∞ m=−∞ X 2 = π n|b z (n)| . = n6=0 On the other hand, for the length we have (recall the assumption L = 2π) Z 2π 2π = dt 0 Z = = 2π |z 0 (t)|2 dt 0 X 2π n2 |b z (n)|2 . n6=0 14 By comparing term by term we obtain the isoperimetric inequality 0 ≤ π − A. Equality holds if and only if X (n2 − n)|b z (n)|2 = 0, n6=0 that is, if and only if |b z (n)|2 = 0 for n = −1, ±2, ±3, · · · . In this situation, z(t) = zb(1)eit , which is a circle of radius |b z (1)|. Proposition 9. If z(t) is a curve of length L enclosing an area A, then 4πA ≤ L2 , with equality if and only if z(t) is a circle. Proof. The curve w(t) = (2π/L)z(t) has lenght 2π and encloses an area of (2π/L)2 A. By the discussion above, 0≤π− 2π L 2 A or 4πA ≤ L2 . 9 Ergodic theory Ergodic theory is a field of mathematics which studies the long term average behavior of systems. The collection of all states of a system forms a space X. The evolution of the system is represented by a transformation f : X → X, where f (x) is taken to be the state at time 1 of a system which at time 0 is in the state x. This of course sounds very abstract. Here is a concrete example. Suppose that your state space X is the unit circle |z| = 1 in the complex plane, and that your transformation is a rotation by an angle 2πα. Thus if you are at the point z = e2πix at time 0, at time 1 you would be at the point e2πiα z = e2π(x+α) , and at e2π(x+nα) at time n. If α is a rational number of the form p/q, then you would come back to the initial state at time q. On the other hand, if α is irrational, you will never visit any of the states through which you have already passed. In this case, we may further ask how the states are distributed on the circle. 15 Another problem related to this one is the following (see Arnold-Avez [3]). Let an , n = 1, 2, · · · , denote the first digit of 2n . Let N (7, k) denote the number of 7’s included in the first k terms of the sequence an . The problem is to compute, if it exists, the limit N (7, k) . k→∞ k lim Here are the first terms of the sequence; the number 7 is quite slow in showing up. 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 7, 1, 2, 5, 1, 2, 4, 9, 1, 3, 7, 1, 2, 5, · · · Let x1 , x2 , . . . be a sequence of numbers in the interval [0, 1). Given any pair of real numbers a, b, such that 0 ≤ a < b ≤ 1, denote by ξn (a, b) the number of the first n terms of the sequence xk which are in the interval [a, b). A sequence xk is said to be uniformly distributed in the interval [0, 1) if lim n→∞ ξn (a, b) =b−a n , for each pair of numbers 0 ≤ a < b ≤ 1. For example, the sequence xk = 1 − (1/k) is not uniformly distributed in [0, 1). Proposition 10. An infinite sequence of numbers (xk ) in the interval [0, 1) is uniformly distributed if and only if n 1X f (xk ) = n→∞ n Z lim k=1 1 f (x) dx, 0 for every function f integrable on [0, 1]. Proof. Given an interval [a, b), the function 1, if a ≤ x < b χ[a,b) (x) = 0, otherwise, is integrable on [0, 1). We have Z 1 χ[a,b) = b − a, 0 and n 1X ξn (a, b) . χ[a,b) (xk ) = n n k=1 Therefore the sequence is uniformly distributed. 16 Conversely, if the sequence is uniformly distributed, then n 1X lim f (xk ) = n→∞ n 1 Z f (x) dx 0 k=1 holds when f is of the form χ[a,b) . Since any step function on [0, 1] is a linear combination of functions of the form χ[a,b) , this result also holds for them. If f is integrable on [0, 1], then, by the theorem on p.256, given ε > 0, we can find step functions s1 , s2 such that s1 ≤ f ≤ s2 , and Z 1 (s2 (x) − s1 (x)) dx < ε. 0 Since the proposition holds for s2 , n 1X s2 (xk ) = lim n→∞ n k=1 1 Z Z 1 s2 (x) dx ≤ ε + 0 f (x) dx, 0 so that if n is large enough, n 1X s2 (xk ) < 2ε + n 1 Z f (x) dx. 0 k=1 Since f ≤ s2 , this implies n 1X f (xk ) < 2ε + n k=1 Z 1 f (x) dx, 0 for n sufficiently large. Similarly, by using s1 instead of s2 , we obtain n 1X f (xk ) > n f (x) dx − 2ε. 0 k=1 Therefore n 1X f (xk ) − n 1 Z Z 1 f (x) dx < 2ε, 0 k=1 for n large enough. Proposition 11. A sequence xk is uniformly distributed if and only if n 1 X 2πimxk e = 0, n→∞ n lim k=1 for each non-zero integer m. 17 Proof. The function x 7→ e2πimx is integrable on [0, 1], and if m 6= 0 Z 1 e2πimx dx = 0. 0 Thus, if the sequence xk is uniformly distributed, the conclusion holds by the previous proposition. Conversely, suppose that n 1 X 2πimxk e = 0, n→∞ n lim k=1 for m 6= 0. Then, by linearity, the equation n 1X f (xk ) = n→∞ n Z lim k=1 1 f (x) dx, 0 holds for every trigonometric polynomial (it obviously holds for the constant function ≡ 1). It then follows that the equation also holds for every continuous periodic function on [0, 1], since the proof of the Weierstrass approximation theorem implies that they can be uniformly approximated by trigonometric polynomials. More precisely, given ε > 0, choose P (x) so that sup |f (x) − P (x)| < ε. 0≤x≤1 If we let f1 = P − ε and f2 = P + ε, then f1 ≤ f ≤ f2 and Z 1 (f2 (x) − f1 (x)) dx = 2ε, 0 and the same argument as in the previous proposition shows that the equation also holds for continuous periodic functions on [0, 1]. If s is a step function on [0, 1], we can find continuous periodic functions f1 and f2 such that f1 ≤ s ≤ f2 and Z 1 (f2 (x) − f1 (x)) dx < ε. 0 The same argument then shows that the equation holds for step functions, and therefore it also holds for any integrable function on [0, 1]. Proposition 12. If λ is an irrational number, then the sequence of fractional parts {kλ} is uniformly distributed in [0, 1). Proof. Clearly, e2πin{kλ} = e2πinkλ . For each integer m 6= 0, mλ is an irrational number, and we have n X 1 e2πimkλ ≤ . | sin πmλ| k=1 18 Therefore n 1 X 2πimkλ lim e = 0. n→∞ n k=1 Further details on uniform distribution of numbes can be found in Chandrashekar [4]. Next we study the problem on the sequence of first digits of powers of 2. The first digit of 2n is equal to k if and only if k10r ≤ 2n < (k + 1)10r , or, taking logarithms, r + log10 k ≤ n log10 2 < r + log10 (k + 1) Taking fractional parts, this may be written as log10 k ≤ {nλ} < log10 (k + 1), where λ = log10 2. Since λ is irrational, the sequence {nλ} is uniformly distributed in [0, 1). Therefore, by applying the definition to the interval [a, b) = [log10 k, log10 (k + 1)), we have that 1 ξn (a, b) = log10 1 + . lim n→∞ n k 10 Complex Analysis This section requires a minimum of familiarity with complex power series. The treatment in Spivak [7] suffices. Familiarity with complex analysis is helpful; a standard reference is Ahlfors [2]. ∞ X Let f (z) = an z n be a complex power series with radius of convergence n=0 R > 0. For each 0 < r < R, let fr be its restriction to the circle |z| = r, that is fr : t ∈ [0, 2π] 7→ fr (t) = f (reit ). Each fr can be represented by its Fourier series, which is obtained directly from the power series expression: X ft (t) = an rn eint . n=0 that is, fbr (n) = an rn if n ≥ 0, fbr (n) = 0 if n < 0. If we use the formula for the Fourier coefficients, Z 2π 1 b fr (t)e−int dt fr (n) = 2π 0 19 Let M (r) = sup |f (z)|, we have |z|=r |fbr (n)| ≤ 1 2π Z 2π |fr (t)| dt ≤ M (r). 0 Proposition 13. If the power series f (z) = X an z n has radius of convergence n=0 R = ∞, and M (r) = 0, rk then f is a polynomial of degree ≤ k. lim r→∞ Proof. Clearly, |an | ≤ M (r)/rn . Thus an = 0 if n > k. This result is a generalization of Liouville’s theorem of complex analysis (see Ahlfors [2]), which has the fundamental theorem of algebra as a consequence. Another important fact of complex analysis is Cauchy integral formula which we now derive for complex power series. Consider radii r1 < r2 . Then X f (r1 eit ) = an r1n eint n=0 The relation between the Fourier coefficients of fr1 and fr2 says that Z 2π 1 f (r2 eis )e−ins ds, an r2n = 2π 0 so that we can write f (r1 eit ) = = n Z 2π 1 X r1 it e f (r2 eis )e−ins ds 2π n=0 r2 0 n ) Z 2π ( X r1 i(t−s) 1 e f (r2 eis ) ds. 2π 0 r 2 n=0 The series inside the integral is a geometric series whose sum is n −1 X r1 r1 eit i(t−s) e = 1− , r2 r2 eis n=0 so that f (r1 eit ) = it 1 2π Z 2π 0 is −1 r1 eit f (r2 eis ) 1 − ds. r2 eis If we now let z = r1 e , w = r2 e , we can rewrite the above expression as Z f (w) 1 dw, f (z) = 2πi |w|=r2 w − z which is know as the Cauchy integral formula in complex analysis. 20

Notes on Fourier Series

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