Examples of Fourier series Leif Mejlbro Download free books at Leif Mejlbro Examples of Fourier series Calculus 4c-1 Download free ebooks at bookboon.com Examples of Fourier series – Calculus 4c-1 © 2008 Leif Mejlbro & Ventus Publishing ApS ISBN 978-87-7681-380-2 Download free ebooks at bookboon.com Examples of Fourier series Contents Contents Introduction 5 1. Sum function of Fourier series 6 2. Fourier series and uniform convergence 62 3. Parseval’s equation 101 4. Fourier series in the theory of beams 115 Fast-track your career Masters in Management Stand out from the crowd Designed for graduates with less than one year of full-time postgraduate work experience, London Business School’s Masters in Management will expand your Please click the advert thinking and provide you with the foundations for a successful career in business. The programme is developed in consultation with recruiters to provide you with the key skills that top employers demand. Through 11 months of full-time study, you will gain the business knowledge and capabilities to increase your career choices and stand out from the crowd. London Business School Applications are now open for entry in September 2011. Regent’s Park London NW1 4SA United Kingdom Tel +44 (0)20 7000 7573 For more information visit www.london.edu/mim/ Email [email protected] email [email protected] or call +44 (0)20 7000 7573 www.london.edu/mim/ Download free ebooks at bookboon.com 4 Examples of Fourier series Introduction Introduction Here we present a collection of examples of applications of the theory of Fourier series. The reader is also referred to Calculus 4b as well as to Calculus 3c-2. It should no longer be necessary rigourously to use the ADIC-model, described in Calculus 1c and Calculus 2c, because we now assume that the reader can do this himself. Even if I have tried to be careful about this text, it is impossible to avoid errors, in particular in the first edition. It is my hope that the reader will show some understanding of my situation. Leif Mejlbro 20th May 2008 Download free ebooks at bookboon.com 5 Examples of Fourier series Sum function of Fourier series 1 Sum function of Fourier series A general remark. In some textbooks the formulation of the main theorem also includes the unnecessary assumption that the graph of the function does not have vertical half tangents. It should be replaced by the claim that f ∈ L2 over the given interval of period. However, since most people only know the old version, I have checked in all examples that the graph of the function does not have half tangents. Just in case ... ♦ n Example 1.1 Prove that cos nπ =(−1) , n ∈ N0. Find and prove an analogous expression for π π cos n and for sin n . 2 2 (Hint: check the expressions for n =2p, p ∈ N0,andforn =2p − 1, p ∈ N). Pi/2 1 (cos(t),sin(t)) 0.5 -Pi 0 –1 –0.5 0.5 1 –0.5 –1 –3/2*Pi One may interpret (cos t, sin t) as a point on the unit circle. The unit circle has the length 2π, so by winding an axis round the unit circle we see that nπ always lies in (−1, 0) [rectangular coordinates] for n odd, and in (1, 0) for n even. It follows immediately from the geometric interpretation that cos nπ =(−1)n. We get in the same way that at π 0forn ulige, cos n = 2 (−1)n/2 for n lige, and π (−1)(n−1)/2 for n ulige, sin n = 2 0forn lige. Download free ebooks at bookboon.com 6 Examples of Fourier series Sum function of Fourier series Example 1.2 Find the Fourier series for the function f ∈ K2π, which is given in the interval ]−π,π] by 0 for − π<t≤ 0, f(t)= 1 for 0 <t≤ π, and find the sum of the series for t =0. 1 –4 –2 2 4 x 1 ∗ Obviously, f(t) is piecewise C without vertical half tangents, so f ∈ K2π. Then the adjusted function f ∗(t) is defined by f(t)fort = pπ, p ∈ Z, f ∗(t)= 1/2fort = pπ, p ∈ Z. The Fourier series is pointwise convergent everywhere with the sum function f ∗(t). In particular, the sum of the Fourier series at t =0is 1 f ∗(0) = , (the last question). 2 You’re full of energy and ideas. And that’s © UBS 2010. All rights reserved. just what we are looking for. Looking for a career where your ideas could really make a difference? UBS’s Graduate Programme and internships are a chance for you to experience for yourself what it’s like to be part of a global team that rewards your input and believes in succeeding together. Please click the advert Wherever you are in your academic career, make your future a part of ours by visiting www.ubs.com/graduates. www.ubs.com/graduates Download free ebooks at bookboon.com 7 Examples of Fourier series Sum function of Fourier series The Fourier coefficients are then 1 π 1 π a0 = f(t) dt = dt =1, π −π π 0 π π 1 1 1 π an = f(t)cosnt dt = cos nt dt = [sin nt]0 =0,n≥ 1, π −π π 0 nπ π π n 1 1 1 π 1−(−1) bn = f(t)sinnt dt = sin nt dt = − [cos nt]0 = , π −π π 0 nπ nπ hence 2 1 b2n =0 og b2n+1 = · . π 2n +1 The Fourier series is (with = instead of ∼) ∞ ∞ ∗ 1 1 2 1 f (t)= a0 + {an cos nt + bn sin nt} = + sin(2n +1)t. 2 n=1 2 π n=0 2n +1 Example 1.3 Find the Fourier series for the function f ∈ K2π, given in the interval ]− π,π] by ⎧ ⎨ 0 for − π<t≤ 0, f(t)=⎩ sin t for 0 <t≤ π, and find the sum of the series for t = pπ, p ∈ Z. 1 –4 –2 2x 4 1 ∗ The function f is piecewise C without any vertical half tangents, hence f ∈ K2π.Sincef is contin- uous, we even have f ∗(t)=f(t), so the symbol ∼ can be replaced by the equality sign =, ∞ 1 f(t)= a0 + {an cos nt + bn sin nt}. 2 n=1 It follows immediately (i.e. the last question) that the sum of the Fourier series at t = pπ, p ∈ Z,is given by f(pπ) = 0, (cf. the graph). The Fourier coefficients are π π 1 1 1 π 2 a0 = f(t) dt = sin tdt= [− cos t]0 = , π −π π 0 π π π 1 · 1 2 π a1 = sin t cos tdt= sin t 0 =0, π 0 2π Download free ebooks at bookboon.com 8 Examples of Fourier series Sum function of Fourier series 1 π 1 π an = sin t · cos nt dt = {sin(n +1)t − sin(n − 1)t}dt π 0 2π 0 π 1 1 1 = cos(n − 1)t − cos(n +1)t 2π n − 1 n +1 0 1 1 1 1 1+(−1)n = (−1)n−1 − 1 − (−1)n+1 − 1 = − · for n>1. 2π n − 1 n +1 π n2 − 1 Now, 2forn even, 1+(−1)n = 0forn odd, hence a2n+1 =0forn ≥ 1, and 2 1 a2n = − · ,n∈ N, (replace n by 2n). π 4n2 − 1 Analogously, π π 1 2 1 1 2 2 1 b1 = sin tdt= · {cos t +sin t}dt = , π 0 π 2 0 2 and for n>1weget 1 π 1 π bn = sin t · sin nt dt = {cos(n − 1)t − cos(n +1)t}dt =0. π 0 2π 0 Summing up we get the Fourier series (with =, cf. above) ∞ ∞ 1 1 1 2 1 f(t)= a0 + {an cos nt + bn sin nt} = + sin t − cos 2nt. 2 − 2 n=1 π 2 π n=1 4n 1 Repetition of the last question.Wegetfort = pπ, p ∈ Z, ∞ 1 2 1 f(pπ)=0= − , 2 − π π n=1 4n 1 hence by a rearrangement ∞ 1 1 = . 2 − n=1 4n 1 2 We can also prove this result by a decomposition and then consider the sectional sequence, N N 1 1 sN = = 2 − − n=1 4n 1 n=1 (2n 1)(2n +1) N 1 1 1 1 1 1 = − = 1 − → − 2 n=1 2n 1 2n +1 2 2N +1 2 Download free ebooks at bookboon.com 9 Examples of Fourier series Sum function of Fourier series for N →∞, hence ∞ 1 1 = lim sN = . 2 − N→∞ n=1 4n 1 2 Example 1.4 Let the periodic function f : R → R,ofperiod2π, be given in the interval ]− π,π] by ⎧ ⎪ ∈ − − ⎪ 0, for t ] π, π/2[ , ⎨⎪ ∈ − f(t)=⎪ sin t, for t [ π/2,π/2] , ⎪ ⎩⎪ 0 for t ∈ ]π/2,π] . Find the Fourier series of the function and its sum function. 1 0.5 –3 –2 –1 1 x 23 –0.5 –1 1 ∗ The function f is piecewise C without vertical half tangents, hence f ∈ K2π. According to the main theorem, the Fourier theorem is then pointwise convergent everywhere, and its sum function is ⎧ ⎪ π ⎨⎪ −1/2fort = − +2pπ, p ∈ Z, ∗ π 2 f (t)=⎪ 1/2fort = +2pπ, p ∈ Z, ⎩⎪ 2 f(t)ellers. Since f(t) is discontinuous, the Fourier series cannot be uniformly convergent. Clearly, f(−t)=−f(t), so the function is odd, and thus an =0foreveryn ∈ N0,and 2 π 2 π/2 1 π/2 bn = f(t)sinnt dt = sin t · sin nt dt = {cos((n − 1)t) − cos((n +1)t)}dt.