12-3: Fourier Series with Period 2π Prakash Balachandran Department of Mathematics Duke University April 5, 2010 Pn f (k)(a) k • So far: we’ve used Taylor polynomials Pn(x) = k=0 k! (x − a) to approximate a function f(x) near the point x = a. • Observe: this relied on f(x) being n-times differentiable at the point x = a, and even then, it only gave us a local approximation to f(x): that is, it only allowed us to approximate f(x) in an interval around x = a (remember those intervals of convergence?). • What if we’re not interested in local approximations, but global approximations? That is, approximations to the function f(x) which approximate the function equally well everywhere. Clearly, as described above, the Taylor polynomials don’t work, so we have to try something else. • Let’s focus first on periodic functions with period 2π as a starting point. These functions are repeating, and so model soundwaves, heartbeats, and electrical phenomena. • Since sine and cosine are in some sense the simplest periodic function, the idea is to try approximating a periodic function f(x) by a linear combination of sine’s and cosines: n n X X Fn(x) = a0 + ak cos(kx) + bk sin(kx): k=1 k=1 This function Fn(x) is called the Fourier Polynomnial of degree n. • For the Fourier polynomial of degree n: n n X X Fn(x) = a0 + ak cos(kx) + bk sin(kx); k=1 k=1 1 Z π a0 = f(x)dx 2π −π 1 Z π ak = cos(kx)f(x)dx π −π 1 Z π bk = sin(kx)f(x)dx: π −π R π – Remark: While the coefficients in the Fourier approximation are made with integrals −π, we can use R 7π R 0 any one complete period we choose: 5π or −2π say. • We’ll do periodic functions with period different than 2π on x13:2. 1 • Construct successive Fourier polynomials for the square wave function: (0 −π ≤ x < 0 f(x) = 1 0 ≤ x < π: 1 – a0 = 2 : 2 – a1 = 0, b1 = π : – a2 = 0, b2 = 0. 2 – a3 = 0, b3 = 3π . – Graph! Notice that while the approximations are somewhat poor, they’re equally poor everywhere (global). • Theorem 1 (Fourier) Any periodic function of period 2π can be represented by 1 X f(x) = a0 + (ak cos kx + bk sin kx) k=1 where 1 Z π a0 = f(x)dx 2π −π 1 Z π ak = cos(kx)f(x)dx π −π 1 Z π bk = sin(kx)f(x)dx: π −π – Remarks: The terms ak cos(kx) + bk sin(kx) are called harmonics. The period of the first harmonic is 2π 2π 2π, the period of the second harmonic is 2 , the period of the third harmonic is 3 and so on. – It’s the period of the first harmonic that determines the overall period of the function. – When we compute the coefficients via integrals, we can use any one complete period: Z 2π Z 7π or 0 5π for instance. – Idea: use whichever period is most convenient. If we consider the sawtooth function (y=x with period R 2π R π 2π) it’s much easier to do 0 than −π. • Find a0 and the first four harmonics of a pulse train function of period 2π: ( π 1 0 ≤ x < 2 f(x) = π 0 2 ≤ x < 2π: 1 – a0 = 4 . 1 1 – First harmonic: π cos(x) + π sin(x). 1 – Second harmonic: π sin(2x). 1 1 – Third harmonic: − 3π cos(3x) + 3π sin(3x). – Fourth harmonic: 0. 2.