New PDF release: Advanced Analysis

By Min Yan

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Xn 2. x1 = 1, xn+1 = x2n + 2 . 2. 17. Discuss the convergence of the sequence defined by x1 = α, xn+1 = γ β + γxn . Do the same to the sequence defined by x1 = α, xn+1 = β + . 18. Let a, b > 0. Define sequences by a1 = a, b1 = b, an = an−1 + bn−1 a+b 2an−1 bn−1 2ab . Use , bn = ≥ to prove the sequences con2 an−1 + bn−1 2 a+b verge. Moreover, find the limits. 19. Let xn = 1+ 1 n n and yn = 1+ 1 n n+1 . 1. Use induction to prove (1+x)n ≥ 1+nx for x > −1 and any natural number n. 2. By showing decreasing.

5) lim xα = 0 for α < 0. 3. LIMIT OF FUNCTION 45 Note that only +∞ is considered here because xα may not be defined for negative x and non-integer α. Of course in the special case n is a natural number, the same argument leads to 1 lim = 0 for natural number n. 9. 3) is lim αx = 0 for 0 < α < 1. 8) x→+∞ 1 The proof, however, needs to be modified. Again let = 1 + β. Then β > 0. For α 1 any > 0, take N = + 1. 12. 7. Rigorously verify the limits. where the inequality 1. limx→∞ 2. limx→∞ x2 x = 0. +1 sin x = 0.

Therefore max{x1 , x2 , . . , xN , xN +1 + 1} is an upper bound for the sequence, and min{x1 , x2 , . . , xN , xN +1 − 1} is a lower bound. 8, there is a subsequence {xnk } converging to a limit l. Thus for any > 0, there is K, such that k > K =⇒ |xnk − l| < . 2 On the other hand, since {xn } is a Cauchy sequence, there is N , such that m, n > N =⇒ |xm − xn | < . 2 Now for any n > N , we can easily find some k > K, such that nk > N (k = max{K, N } + 1, for example). Then we have both |xnk − l| < and 2 |xnk − xn | < .

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Advanced Analysis by Min Yan


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