《数学分析》课程教学课件（讲稿，打印版）第12章 Fourier分析（7/7）

Fourier È© Âñ½n Fourier C† Fourier C†5Ÿ òÈ Parseval ª A^ §12.4 Fourier È©ÚFourier C† 12.4.1 Fourier È© £ Fourier ?ê:  f(x) 3 [−π, π] þŒÈ½ýéŒÈ, K f(x)  Fourier ?ê a0 2 + X ∞ n=1 ￾ an cos nx + bn sin nx
, Ù¥ an = 1 π Z π −π f(x) cos nx dx, bn = 1 π Z π −π f(x) sin nx dx. éu f(x)  Fourier ?ê, k Dirichlet Âñ5½n. „/, f(x) ∼ a0 2 + X ∞ n=1 ￾ an cos nx + bn sin nx
. 1/34 kJ Ik J I ˆ£ ¶ '4 òÑ

Fourier È© Âñ½n Fourier C† Fourier C†5Ÿ òÈ Parseval ª A^  f(x) ´ (−∞, +∞) þýéŒÈ¼êž, XJÄ f(x) 3 [−π, π] ùãŠ, 2^ Fourier ?ê{5?Ø, ÒØÜ· .  N/ïÄ f(x) 3 (−∞, +∞) þ5Ÿ. •ì Fourier ?ê‰{, - a(λ) = 1 π Z +∞ −∞ f(t) cos λt dt, b(λ) = 1 π Z +∞ −∞ f(t) sin λt dt, (1) ¡ f(x)  Fourier È©Xê, È© Z +∞ 0 ￾ a(λ) cos λx + b(λ) sin λx
dλ ¡ f(x)  Fourier È©, P f(x) ∼ Z +∞ 0 ￾ a(λ) cos λx + b(λ) sin λx
dλ. (2) ¯K 1 þªmॠFourier ȩ۞Âñ? Âñž´ÄÂñ f(x)? 2/34 kJ Ik J I ˆ£ ¶ '4 òÑ

Fourier È© Âñ½n Fourier C† Fourier C†5Ÿ òÈ Parseval ª A^ Ún 1 e f(x) ´ (−∞, +∞) þýéŒÈ¼ê, K a(λ) = 1 π Z +∞ −∞ f(x) cos λx dx, b(λ) = 1 π Z +∞ −∞ f(x) sin λx dx, 3 (−∞, +∞) þëY, … λ → +∞ ž, a(λ), b(λ) Ѫu". y² Ϗ f(x) ´ (−∞, +∞) þýéŒÈ¼ê, ¤±é?¿ ε > 0,  3 A > 0 ¦ Z +∞ A |f(x)| dx + Z −A −∞ |f(x)| dx 0 ¦ |x1−x2| < δ1 ž, k | cos x1 − cos x2| < πε 2M , Ï  δ = δ1 A . K |λ1 − λ2| < δ ž, k | cos λ1x − cos λ2x| < πε 2M , (|x| 6 A). 3/34 kJ Ik J I ˆ£ ¶ '4 òÑ

Fourier È© Âñ½n Fourier C† Fourier C†5Ÿ òÈ Parseval ª A^ u´ |a(λ1) − a(λ2)| 6 1 π Z +∞ −∞ |f(x)| | cos λ1x − cos λ2x| dx = 1 π Z +∞ A |f(x)| | cos λ1x − cos λ2x| dx + 1 π Z −A −∞ |f(x)| | cos λ1x − cos λ2x| dx + 1 π Z A −A |f(x)| | cos λ1x − cos λ2x| dx 6 2 π Z +∞ A |f(x) dx + 2 π Z −A −∞ |f(x)| dx + 1 π Z A −A |f(x)| πε 2M dx 6 ε 2 + ε 2 = ε. ùÒy² a(λ) 3 (−∞, +∞) þëY. Ón, Œy b(λ) 3 (−∞, +∞) þëY. Šâ Riemann-Lebesgue ÚnŒ,  λ → +∞ ž, a(λ), b(λ) Ѫu". 4/34 kJ Ik J I ˆ£ ¶ '4 òÑ

Fourier È© Âñ½n Fourier C† Fourier C†5Ÿ òÈ Parseval ª A^ Ún 2 e f(x) ´ (−∞, +∞) þýéŒÈ¼ê, K Z u 0 Z +∞ −∞ f(t) cos λ(t − x)dt dλ = Z +∞ −∞ Z u 0 f(t) cos λ(t − x)dλ dt. y² Iy²mà2ÂÈ©Âñ†àÈ©. Ϗ     Z A −A Z u 0 f(t) cos λ(t − x)dλ dt − Z u 0 Z +∞ −∞ f(t) cos λ(t − x)dt dλ     =     Z u 0 Z A −A f(t) cos λ(t − x)dt dλ − Z u 0 Z +∞ −∞ f(t) cos λ(t − x)dt dλ     =     Z u 0 Z +∞ A + Z −A −∞ f(t) cos λ(t − x)dt dλ     6 Z u 0 Z +∞ A + Z −A −∞ |f(t)|dt dλ = u Z +∞ A + Z −A −∞ |f(t)|dt → 0 (A → +∞), ¤±Úny. 5/34 kJ Ik J I ˆ£ ¶ '4 òÑ

Fourier È© Âñ½n Fourier C† Fourier C†5Ÿ òÈ Parseval ª A^ ½n 1 e f(x) ´ (−∞, +∞) þýéŒÈ¼ê, a(λ), b(λ) ´ f(x)  Fourier È©Xê, P S(u, x) = Z u 0 ￾ a(λ) cos λx + b(λ) sin λx
dλ, K S(u, x) = 1 π Z +∞ 0 ￾ f(x + t) + f(x − t)
sin ut t dt. (3) y² ò (1) \ S(u, x) ¥, ¿dÚn 2,  S(u, x) = 1 π Z u 0 Z +∞ −∞ f(t) ￾ cos λt cos λx + sin λt sin λx
dt dλ = 1 π Z u 0 Z +∞ −∞ f(t) cos λ(t − x)dt dλ = 1 π Z +∞ −∞ Z u 0 f(t) cos λ(t − x)dλ dt. 6/34 kJ Ik J I ˆ£ ¶ '4 òÑ

Fourier È© Âñ½n Fourier C† Fourier C†5Ÿ òÈ Parseval ª A^ Ïd S(u, x) = 1 π Z +∞ −∞ f(t) sin u(t − x) t − x dt = 1 π Z +∞ −∞ f(x + t) sin ut t dt = 1 π Z +∞ 0 f(x + t) sin ut t dt + 1 π Z 0 −∞ f(x + t) sin ut t dt = 1 π Z +∞ 0 f(x + t) sin ut t dt + 1 π Z +∞ 0 f(x − t) sin ut t dt = 1 π Z +∞ 0 ￾ f(x + t) + f(x − t)
sin ut t dt. 5 d½nò Fourier È©Ü©È©=zØ¹ a(λ), b(λ) L«. ùƒu Fourier ?ê?Ø¥ò Fourier ?êÜ©ÚL«È©. 7/34 kJ Ik J I ˆ£ ¶ '4 òÑ

Fourier È© Âñ½n Fourier C† Fourier C†5Ÿ òÈ Parseval ª A^ ½n 2 (Fourier È©ÛÜz½n) e f(x) ´ (−∞, +∞) þýéŒÈ¼ ê, K f(x)  Fourier È©3,: x ´ÄÂñ, ±9ÂñŸoŠ, =† f 3 x NC¼êŠk'. y²  δ > 0. ò (3) mà¥Ü©È©Lˆª©¤üã: S(u, x) = 1 π Z δ 0 ￾ f(x + t) + f(x − t)
sin ut t dt + 1 π Z +∞ δ ￾ f(x + t) + f(x − t)
sin ut t dt = I1 + I2. Ϗ f(x) 3 (−∞, +∞) þýéŒÈ, ¤± ϕ(t) = f(x+t)+f(x−t) t 3 [δ, +∞) þýéÂñ. Šâ Riemann-Lebesgue Ún, k lim u→+∞ R +∞ δ ϕ(t) sin ut dt = 0. Ï d lim u→+∞ I2 = 0. ùÒ`² lim u→+∞ S(u, x) ´ÄÂñ=† I1 k'. I1 =† f 3 x NCŠk'. y. 8/34 kJ Ik J I ˆ£ ¶ '4 òÑ

Fourier È© Âñ½n Fourier C† Fourier C†5Ÿ òÈ Parseval ª A^ ½n 3 (Dini ½n)  f(x) ´ (−∞, +∞) þýéŒÈ¼ê, s ´?¿¢ ê. é½ x, P ϕ(t) = f(x + t) + f(x − t) − 2s. e3 δ > 0 ¦ ϕ(t) t 3 [0, δ] þŒÈ…ýéŒÈ, K f  Fourier È©3 x ?Âñu s. y² Ϗ R +∞ 0 sin ut t dt = π 2 , =, 2 π R +∞ 0 sin ut t dt = 1, ¤±d (3),  S(u, x) − s = 1 π Z +∞ 0 ￾ f(x + t) + f(x − t) − 2s
sin ut t dt = 1 π Z δ 0 ϕ(t) t sin ut dt − 2s π Z +∞ δ sin ut t dt + 1 π Z +∞ δ ￾ f(x + t) + f(x − t)
sin ut t dt = I1 + I2 + I3. 9/34 kJ Ik J I ˆ£ ¶ '4 òÑ

Fourier È© Âñ½n Fourier C† Fourier C†5Ÿ òÈ Parseval ª A^ 3½n 2 ¥®²`² lim u→+∞ I3 = 0. e3 δ > 0 ¦ ϕ(t) t 3 [0, δ] þŒÈ…ýéŒÈ, KŠâ Riemann￾Lebesgue Ún, k lim u→+∞ I1 = 0. éu I2 ¥È©, ŠC† v = ut,  Z +∞ δ sin ut t dt = Z +∞ uδ sin v v dv. Ϗ R +∞ 0 sin v v dv Âñ, ¤± lim u→+∞ Z +∞ uδ sin v v dv = 0. Ï lim u→+∞ I2 = 0. u´ lim u→+∞ S(u, x) = s. ùL« f  Fourier È©3 x ?Âñu s. y. 10/34 kJ Ik J I ˆ£ ¶ '4 òÑ