§5.1. FOURIER级数 5/67 这就是周期函数的f(x)的 Fourier展开式,其系数称为 Fourier系 数 函数族(51-2)满足正交归一关系 k≠n kT IT cos=rcos"xdx=〈1 k=n≠0 k=n=0 (5.1-4) n k≠n sin -x sin -xdx cosx sin-xdx =0 ●Fist●Prev·Next●Last● Go Back● Full Screen●cose●Quit• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 5/67 ù. Ò. ´. ±. Ï. ¼. ê. . f(x) . Fourier Ð. m. ª. §Ù. X. ê. ¡. . Fourier X. ê. © ~ ¼. ê. x. (5.1-2)÷. v. . . 8. . '. X. µ    Z ` −` cos kπ ` x cos nπ ` xdx =    0 k , n 1 δk` =    1 ` k = n , 0 1 2` k = n = 0 Z ` −` sin kπ ` x sin nπ ` xdx =    0 k , n 1 ` k = n Z ` −` cos kπ ` x sin nπ ` xdx = 0. (5.1-4)