§5.1. FOURIER级数 16/67回 2 2n丌 cos T xdx 应用积分公式: r cos Px dr n cos Px+n sin Px 2 CoS SIn T 0 2n丌 COS 3T(2nT 0 =2 2n丌 2n丌 f(t)sint dt o 3-t sin xdx ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 16/67 = 2 T Z T 0 1 3 x cos 2nπ T xdx , A^È©úªµ Z x cos Px dx = 1 P2 cos Px + x P sin Px an = 2 T · 1 3 " 1 2nπ T 2 cos 2nπ T x + x 2nπ T sin 2nπ T x #T 0 = 2 3T T 2nπ 2 cos 2nπ T x + 2nπ T x sin 2nπ T x T 0 = 0 , bn = 2 T Z T 0 f(t) sin 2nπ T t dt = 2 T Z T 0 1 3 x sin 2nπ T x dx
§5.1. FOURIER级数 7/67圆 2 2 not T coS 0 2n丌2 2n丌 SIn r cos 3T\ 2nT T T 3n丌 T 2n丌 f(x)= n 3 T S例3(P92,4(1)将函数f(x)=cos3x展为 Fourier级数 解:可按书上的标准方法展开.此外,还可令t=e后把f(x 化为t的有理分式,展为 Taylor级数后再将变数换回x e cos x ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 17/67 = 2 T · 1 3 " 1 2nπ T 2 sin 2nπ T x − x 2nπ T cos 2nπ T x #T 0 = 2 3T T 2nπ 2 sin 2nπ T x − 2nπ t x cos 2nπ T x T 0 = − T 3nπ , f(x) = T 6 − X ∞ n=1 T 3nπ sin 2nπ T x . ✿ ~®(P.92§4(1))ò¼ê f(x) = cos3 x. ÐFourier?ê© )µUÖþIO{Ðm©d §- t = e ix r f(x) z t kn©ª§Ð Taylor ?ê2òCê£ x© f(x) = cos3 x = e ix + e −ix 2 3
§5.1. FOURIER级数 18/67回 e3x+ 3er+3e-ix +e-13x 83-43-4 十e 2 cos x + cos 3x 注:本题其实就是三倍角公式: cos 3x 4 cos-3 cos x 所以 3 f(r)= cos'x=cos x + cos 3x S例4(P:92,4(3)在(-x,x)这个周期上,∫(x)= cos ar,(非整 数).将其展为 Fourier级数 ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 18/67 = 1 8 e i3x + 3e ix + 3e −ix + e −i3x = 3 4 · e ix + e −ix 2 + 1 4 · e i3x + e −i3x 2 = 3 4 cos x + 1 4 cos 3x. 5µ KÙ¢Ò´núªµ cos 3x = 4 cosx −3 cos x ¤± f(x) = cos3 x = 3 4 cos x + 1 4 cos 3x. ✿ ~¯(P.92§4(3))3 (−π, π) ù±Ïþ§f(x) = cos αx§(α ê)©òÙÐFourier?ê©
§5.1. FOURIER级数 19/67回 解:因为f(x)是偶函数,所以bk=0 f(x)=a ak cos kx sina丌 00-2 cos as ds sin as 2a丌 2 k cos as cos ks dE cos(k+a)5 ds +-/cos(k-a)5ds sin(k +a)5 sin(k-a)5 k+a k sin(k+a)丌 * on(k-a)T k+a k L.[sin kT cos aT-cos k sin aT] ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 19/67 )µÏ f(x) ´ó¼ê§¤± bk = 0§ f(x) = a0 X ∞ k==1 ak cos kx. a0 = 1 2π Z π −π cos αξ dξ = 1 2απ sin αξ π −π = sin απ απ . ak = 2 π Z π 0 cos αξ cos kξ dξ = 1 π Z π 0 cos(k + α)ξ dξ + 1 π Z π 0 cos(k − α)ξ dξ = 1 π sin(k + α)ξ k + α + sin(k − α)ξ k − α π 0 = 1 π sin(k + α)π k + α + sin(k − α)π k − α = 1 π · 1 k + α [sin kπ cos απ − cos kπ sin απ]
§5.1. FOURIER级数 20/67圆 1-元 1 sin k COS OT-cos k sin aT] 1 CoS KT SIn (TT k +a k-a (-1) sin aT k2 (-1)+1 s 所以 2sina丌 Q(-1)k+1 f(r cos Kr k=1 ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 20/67 + 1 π · 1 k − α [sin kπ cos απ − cos kπ sin απ] = 1 π cos kπ sin απ 1 k + α − 1 k − α = 1 π (−1)k sin απ · −2α k 2 − α2 = (−1)k+1 π sin απ · 2α k 2 − α2 . ¤± f(x) = 2 sin απ π " 1 2α + X ∞ k=1 α(−1)k+1 k 2 − α2 cos kx#