高等微積分*-^
2009-08-28 6:46 am
Prove: If the series Σ(n=1→ ∞) M_n of constants M_n is convergent and |f_n+1(x)-f_n(x)|<=M_n for x in E, then the sequence f_n(x) is uniformly convergent for x in E.(過程請詳答)
回答 (3)
2009-08-30 6:58 am
✔ 最佳答案
任意 ε>0 ,因Σ[n=1~∞] M_n 收斂, 故存在N>0,使m>n>N時|Σ[k=n~m-1] M_k |<ε,
又| f_m(x)-f_n(x) | = | f_m(x)-f_m-1(x) |+...+| f_n+1(x)-f_n(x) |
<= M_m-1 +....+ M_n <ε (與 x無關, 只要m,n>N就可以)
則數列 {f_n(x)}為Cauchy 數列
故數列 {f_n(x)} conv. uniformly.
2009-08-28 3:18 pm
Σ(n=1→ ∞) |f_n(x)| < Σ(n=1→ ∞) M_n - f_1(x) < ∞,
by Weierstrass-M test. f_n(x) is uniform convergent.
by Weierstrass-M test. f_n(x) is uniform convergent.
2009-08-28 6:56 am
兩題都用Cauchy sequence檢驗法即得!
收錄日期: 2021-04-30 13:57:03
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090827000010KK12102