Giáo trình giải tích 2 part 2 pdf

0 n →∞ f(x)=Tf(x)  e x =1+x + 1 2! x 2 + ···+ 1 n! x n + ··· cos x =1− 1 2! x 2 + 1 4! x 4 + ···+ (−1) n (2n)! x 2n + ··· sin x = x − 1 3! x 3 + 1 5! x 5 + ···+ (−1) n (2n +1)! x 2n+1 + ··· 1 1 − x =1+x + x + ···+ x n + ··· , |x| < 1 ln(1 + x)=x − 1 2 x 2 + 1 3 x 3 + ···+ (−1) n+1 n xan + ··· , |x| < 1 (1 + x) α =1+αx + α(α − 1) 2! x 2 + ···+ α(α − 1) ···(α − n +1) n! x n + ··· , |x| < 1 (x)=  x 0 e −t 2 dt e x x = −t 2 e −t 2 =1− t 2 + 1 2! t 4 + ···+ (−1) n n! t 2n + ··· erf(x)=x − x 3 3 + x 2 2!5 + ···+ (−1) n n!(2n +1) x 2n+1 + ···= ∞  k=0 (−1) k k!(2k +1) x 2k+1 x ∈ R (x)=  x 0 sin t t dt sin x Si(x)=  x 0 (1− 1 3! t 2 + 1 5! t 4 +···+ (−1) n (2n +1)! t62n+···)dt = ∞  k=0 (−1) k (2k + 1)!(2k +1) x 2n+1 ln 2 ln(1 + x) ln(1 −x)=x + 1 2 x 2 + 1 3 x 3 + ···+ x n n + ··· , |x| < 1 ln(1 + x) −ln(1 −x) ln  1+x 1 − x  =2(x + 1 3 x 3 + ···+ x 2n+1 2n +1 + ···), |x| < 1 x = 1 3 ln 2 = 2( 1 3 + 1 3.3 3 + ···+ 1 (2n +1)3 2n+1 )+R n R n =  k>n 1 (2k +1)3 2k+1 < 1 3(2n +3)  k>n 1 9 k = 1 3(2n +1) (1/9) n 1 − 1/9 = o( 1 9 n ) sin cos a 0 2 + ∞  k=1 (a k cos kx + b k sin kx) f T ϕ(x)=f( T 2π x) 2π 2π [−π, π] <f,g>=  π −π f(x)g(x)dx f,g ∈ C[−π, π] 1, cos x, sin x, cos 2x, sin 2x, ···, cos nx, sin nx, ··· 0  π −π cos kx cos lxdx =0 k = l  π −π sin kx sin lxdx =0 k = l  π −π cos kx sin lxdx =0 ∀k, l  π −π dx =2π,  π −π cos 2 kxdx =  π −π sin 2 kxdx = πk=1, 2, ··· f f(x)= a 0 2 + ∞  k=1 (a k cos kx + b k sin kx),x∈ [−π,π] f(x)coslx = a 0 2 cos lx + ∞  k=1 (a k cos kx cos lx + b k sin kx cos lx) f(x)sinlx = a 0 2 sin lx + ∞  k=1 (a k cos kx sin lx + b k sin kx sin lx) a k = 1 π  π −π f(x)coskxdx, k =0, 1, 2, ··· b k = 1 π  π −π f(x)sinkxdx, k =1, 2, ··· f f [−π, π] f Ff(x)= a 0 2 + ∞  k=1 (a k cos kx + b k sin kx) a k ,b k f • f f(−x)=f(x) f(x)sinkx b k =0 Ff(x)= 1 2 a 0 + ∞  k=1 a k cos kx • f f(−x)=−f(x) f(x)coskx a k =0 Ff(x)= ∞  k=1 b k sin kx • F (af + bg)=aF f + bF g f, g a, b ∈ R f(x), |x|≤π Ff(x) x 4 π ∞  k=0 sin(2k +1)x 2k +1 x 2 ∞  k=1 (−1) k+1 sin kx k x 2 π 2 3 +4 ∞  k=1 (−1) k cos kx k 2 Ax 2 + Bx + C A π 2 3 + C +4A ∞  k=1 (−1) k cos kx k 2 +2B ∞  k=1 (−1) k+1 sin kx k Ff(x)=f(x) Ff(x) 2π Ff(x) Ff(x) = f(x) Ff(x)=f(x) Ff(x)=f(x) n f F n f(x)= a 0 2 + n  k=1 (a k cos kx + b k sin kx) F n f F n f(x)= a 0 2 + n  k=1 (a k cos kx + b k sin kx) = 1 2π  π −π f(u)du + n  k=1 1 π  π −π f(u)(cos ku cos kx +sinku sin kx)du = 1 π  π −π f(u)  1 2 + n  k=1 cos k(u − x)  du g T  a+T a g(t)dt =  T 0 g(t)dt t = u − x T =2π a = −π −x F n f(x)= 1 π  π −π f(x + t)  1 2 + n  k=1 cos kt  dt =  π −π f(x + t)D n (t)dt D n (t)= 1 π  1 2 + n  k=1 cos kt  2sin t 2 cos kt =sin(k + 1 2 )t − sin(k − 1 2 )t D n (t)= 1 π sin 2n +1 2 t 2sin t 2 D n 2π  π −π D n (t)dt =1 g [a, b] lim λ→+∞  b a g(t)cosλtdt = lim λ→+∞  b a g(t)sinλtdt =0 g lim λ→+∞  b a g(t)cosλtdt = g(t)sinλt λ     b a − 1 λ  b a g  (t)sinλtdt g  → 0 λ → +∞ g g  g >0 s  π −π |g −s| <  b a g(t)cosλtdt =  b a (g(t) −s(t)) cos λtdt +  b a s(t)cosλtdt s |cos λx|≤1 lim λ→+∞       b a g(t)cosλtdt      ≤  b a |g(t) −s(t)|dt <  lim λ→+∞  b a g(t)cosλtdt =0  f [a, b] a = a 0 <a 1 < ··· <a s = b f (a i−1 ,a i ) lim x→a + i f(x)=f(a + i ) lim x→a − i f(x)=f(a − i ) i =0, ··· ,s f x f  + (x) = lim t→0 + f(x + t) − f(x6+) t ,f  − (x) = lim t→0 + f(x − t) − f(x − ) t , f(x)=|x| 0 f  + (0) = 1,f  − (0) = −1 f(x)= x 0 f(0 + )=1,f(0 − )=−1 f  (0 + )=f  − (0) = 0 f 2π [−π, π] f  + (x),f  − (x) F n f(x) f x Ff(x)= 1 2 (f(x + )+f(x − )) f x Ff(x)=f(x) A f (x)= 1 2 (f(x + )+f(x − )) D n F n f(x) − A f (x)=  π −π (f(x + t) − A f (x))D n (t)dt =2  π 0  f(x + t)+f(x −t) 2 − A f (x)  D n (t)dt =2  π 0 g(t)sin(n + 1 2 )tdt g(t)= f(x + t) − f(x + )+f(x −t) −f(x − ) t t 2π sin t 2 f  + (x),f  − (x) lim t→0 + g(t)= 1 π (f  + (x) − f  − (x)) g 0 n →∞ F n f(x) → A f (x) n →∞  x = 4 π ∞  k=0 sin(2k +1)π 2k +1 0 < |x| <π x =0, −π, π 1 2 ( (x + )+ (x − )) = 0 x = π/2 ∞  k=0 (−1) k 2k +1 = π 4 1 − x 2 π 2 = 2 3 − 4 π 2 ∞  k=1 (−1) k cos kx k 2 |x|≤π x = ±π x = π ∞  k=1 1 k 2 = π 2 6 x =0 ∞  k=1 (−1) k k 2 = − π 2 12 ∞  k=1 1 (2k −1) 2 = 1 2  ∞  k=1 1 k 2 − ∞  k=1 (−1) k k 2  = π 2 8 f 2 [π, π] a 2 0 2 + ∞  k=1 (a 2 k + b 2 k ) ≤ 1 π  π −π f 2 (x)dx  π −π (f(x)−F n f(x))F n f(x)dx =0  π −π (F n f(x)) 2 dx = π  a 2 0 2 + n  k=1 (a 2 k + b 2 k )   π −π f 2 (x)dx =  π −π (f(x) − F n f(x)+F n f(x)) 2 dx =  π −π (f(x) − F n f(x)) 2 dx +  π −π (F n f(x)) 2 dx +2  π −π (f(x) − F n f(x))F n f(x)dx =  −π 6π(f(x) −F n f(x)) 2 dx + π( a 2 0 2 + n  k=1 (a 2 k + b 2 k )) a 2 0 2 + n  k=1 (a 2 k + b 2 k ) ≤  π −π f 2 (x)dx n → +∞  f 2π f  [−π, π] Ff f R F n f(x) f(x) a  k ,b  k f  a k = 1 π  π −π f(x)coskxdx = 1 π  f(x) sin kx k | π −π − 1 k  π −π f  (x)sinkxdx  = − 1 k b  k b k = 1 π  π −π f(x)sinkxdx = 1 π  −f(x) cos kx k | π −π + 1 k  π −π f  (x)coskxdx  = 1 k a  k |a k cos kx + b k sin kx|≤|a k | + |b k |≤ 1 2 (b  2 k + 1 k 2 )+ 1 2 (a  2 k + 1 k 2 ) ∞  k=0 (a  2 k + b  2 k ) ∞  k=1 1 k 2 Ff  • f(x) T x = T 2π X f(x)=f( T 2π X) 2π X X a 0 2 + ∞  k=1 ( a k cos kX + b k sin kX ) a k = 1 π  π −π f( T 2π X)coskXdX, b k = 1 π  π −π f( T 2π X)sinkXdX X = 2π T x a 0 2 + ∞  k=1 ( a k cos 2kπ T x + b k sin 2kπ T x ) f a k = 2 T  T/2 −T/2 f(t)cos 2kπ T tdt, k =0, 1, 2, ··· b k = 2 T  T/2 −T/2 f(t)sin 2kπ T tdt, k =1, 2, ··· • f [a, b] f ˜ f R T ≥ b − a ˜ f(x + kT)=f(x),x∈ [a, b],k ∈ Z ˜ f • cos sin f [0,l] f(x) cos f (−l, l] f ( x)=f(−x) x ∈ (−l, 0) f(x) sin f (−l, l] f(x)=−f(−x) x ∈ (−l, 0) [−π, π] 2π f(x)= x, x ∈ [−π, π] Ff(x)= 4 π ∞  k=0 sin(2k +1)x 2k +1 ✲ x ✻ y ✲ ✲ rr ✲ ✲ rr ✲ ✲ rr ✲ ✲ rr ✲ ✲ rr −ππ f(x)=x, x ∈ [−π, π] Ff(x)=2 ∞  k=1 (−1) k+1 sin kx k ✲ x ✻ y     ✒ r     ✒ r     ✒ r     ✒ r     ✒ r −ππ f(x)=x 2 ,x∈ [−π, π] Ff(x)= π 2 3 +4 ∞  k=1 (−1) k cos kx k 2 ✲ x ✻ y rr −ππ [0, 2π] 2π f(x), 0 ≤ x<2π Ff(x) x π −2 ∞  k=1 sin kx k x 2 4 3 π 2 +4 ∞  k=1 cos kx k 2 − 4π ∞  k=1 sin kx k Ax 2 + Bx + C A 4 3 π 2 + Bπ + C +4A ∞  k=1 cos kx k 2 − (4πA − 2B) ∞  k=1 sin kx k Ff(x)=x, 0 <x<2π ✲ x     ✒ r     ✒ r     ✒ r     ✒ r     ✒ r 02π Ff(x)=x 2 , 0 <x<2π ✲ x ✕ r ✕ r ✕ r ✕ r ✕ r 02π f(x) f(x)=x, x ∈ [−π,π] f(x)=x, x ∈ [0, 2π] 2π f(x)=x, x ∈ [0,π] f(x) cos f f(x)=|x|,x∈ [−π, π] f |x| = π 2 − 4 π ∞  k=1 cos(2k +1)x (2k +1) 2 , −π ≤ x ≤ π ✲ x ✻ y   ❅ ❅ ❅   ❅ ❅ ❅   ❅ ❅ ❅   ❅ ❅ ❅   ❅ ❅ ❅ −ππ f(x) sin f f(x)=x, x ∈ [−π, π] f x =2 ∞  k=1 (−1) k+1 sin kx k , −π<x<π ✲ x ✻ y   ✒    r   ✒    r   ✒    r   ✒    r   ✒    r −ππ . π ∞  k=1 1 k 2 = π 2 6 x =0 ∞  k=1 (−1) k k 2 = − π 2 12 ∞  k=1 1 (2k −1) 2 = 1 2  ∞  k=1 1 k 2 − ∞  k=1 (−1) k k 2  = π 2 8 f 2 [π, π] a 2 0 2 + ∞  k=1 (a 2 k + b 2 k ) ≤ 1 π  π −π f 2 (x)dx  π −π (f(x)−F n f(x))F n f(x)dx. b k sin kx|≤|a k | + |b k |≤ 1 2 (b  2 k + 1 k 2 )+ 1 2 (a  2 k + 1 k 2 ) ∞  k=0 (a  2 k + b  2 k ) ∞  k=1 1 k 2 Ff  • f(x) T x = T 2 X f(x)=f( T 2 X) 2 X X a 0 2 + ∞  k=1 ( a k cos kX +. ) a k = 1 π  π −π f( T 2 X)coskXdX, b k = 1 π  π −π f( T 2 X)sinkXdX X = 2 T x a 0 2 + ∞  k=1 ( a k cos 2kπ T x + b k sin 2kπ T x ) f a k = 2 T  T /2 −T /2 f(t)cos 2kπ T tdt, k =0, 1, 2, ··· b k = 2 T  T /2 −T /2 f(t)sin 2kπ T tdt,