K 2 2 equation holds 191 equation holds for n1

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k 2 2 equation holds. 1.91 Equation holds for n=1, assuming equation holds for n=k, then for n=k+1 x 2k 1 y 2k 1 = x 2k 1 x 2 y 2k 1 x 2 y 2k 1 y 2k 1 = x 2 x 2k 1 y 2k 1  y 2k 1 − x 2 y 2 from the assumption that equation holds for n=k, so the first item could be divided by x y , also, the second item could be divided by x y , so equation holds. 1.93 Equation holds for n=1, assuming equation holds for n=k, then for n=k+1 1 2 cosx cos2 x  coskx cos k 1 x = sin k 1 2 x 2sin 1 2 x cos k 1 x = sin k 1 2 x 2sin 1 2 x cos k 1 x 2sin 1 2 x = sin k 1 2 x sin k 3 2 x – sin k 1 2 x 2sin 1 2 x = sin k 3 2 x 2sin 1 2 x equation holds. 2.35 3 n 2 n – 1 3 = 5 n– 1 ⇒ n 1 5 (a) when = 0.01 , n=502 (b) when = 0.001 , n=5002 (b) when = 0.0001 , n=50002 2.44
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a 3 2 . Numerator and denominator are divided by n 2 . b 1 2 . Numerator and denominator are divided by n. c 3 2 . Numerator and denominator are divided by n. d 15 . Numerator and denominator are divided by 10 2n or 10 2n 1 e ∞ . f 3 . 3 n 2 n 3 n 3 n 3 n ⇒  3 n 1 n ≤ 2 n 3 n 1 n ≤ 3 n 3 n 1 n 3 ≤ 2 n 3 n 1 n 3 2 1 n 3 lim n ∞ 2 n 3 n 1 n 3 lim n ∞ 2 1 n 3 lim n ∞ 2 n 3 n 1 n 3 lim n ∞ 2 n 3 n 1 n = 3 2.48 if P u n 2 P u n u n u n P u n 2u n u n 1 u n so u n is monotonically increasing. From assumption P u n 2 , this sequence is upper bounded, so it has a limit. By same method we can prove if P u n 2 , u n is monotonically decreasing, from u n 1 = 1 2 u n P u n  P this sequence is lower bounded, so it still has a limit. Finally, no matter what P is, limit always exits.
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  • Fall '08
  • Feinberg,E
  • Division, lim, lim lim

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