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2010 -2011 31 6 16 §1-§2 8:00-9:30 0. 1. y = f (x) a f (0) = 0, lim n→∞ n k=1 f ( k n 2 )= 1 2 f (0). f (x) x =0 Taylor 2. y = f (x) (−∞, +) lim x0 f (x) x = A. Φ(x)= 1 0 f (xt)dt, lim x0 Φ(x) x . lim x0 f (x) x = A, f (0) = 0, f (0) = A. Φ(x) Φ(x)= 1 x x 0 f (u)du x =0, l’Hospital 3. lim n→∞ 1 n +1 + ··· + 1 2n = ln 2. 1 1 2 + 1 3 + 1 2 + 1 n ln n = γ + o(1) n →∞ γ Euler 2 1 n +1 + ··· + 1 n + n y = 1 1+x [0, 1] n Riemann n k=1 1 1+ k n 1 n , 1 n x k , ξ k = k n , y = 1 1+x [0, 1] 1

Mathematical Analysis

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Mathematical Analysis

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Page 1: Mathematical Analysis

2010 -2011 d℄WXd^W 31 Va[� 6 e 16 _ §1-§2 8:00-9:30

0. �S�>DK�:�:+����C"(;�O�-*6�;5?�1. ` �: y = f(x) L� a ��$I F�0 f(0) = 0, P.

limn→∞

n∑

k=1

f(k

n2) =

1

2f ′(0). b�'G�: f(x) L x = 0 � Taylor N%8�

2. ` y = f(x) L (−∞, +∞) R)C�0 limx→0

f(x)

x= A. , Φ(x) =

1

0f(xt)dt,1��

limx→0

Φ(x)

x. b�< lim

x→0

f(x)

x= A, E=T f(0) = 0, 0 f ′(0) = A. 4�! Φ(x) B�

Φ(x) =1

x

x

0

f(u)du x 6= 0,G l’Hospital �M�3. f\ ��

limn→∞

( 1

n + 1+ · · · +

1

2n

)

= ln 2.ZY 1 'G7;#�1

2+

1

3+

1

2+

1

n− ln n = γ + o(1) n → ∞/R γ 9 Euler �:�ZY 2 !�8

1

n + 1+ · · · +

1

n + n�B��: y = 1

1+xL2� [0, 1] 7 n ��? Riemann ��

n∑

k=1

1

1 + k

n

1

n,/R 1

nAJ △xk, ÆR��Q3U ξk = k

n, HJ y = 1

1+xL [0, 1] &�����&�� ��

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