高等微積分問題

2010-06-15 7:37 am
Let f(x) = Σ(1~∞) [n^(-2)] sin(nx). Show that f is a continuous function on R and that ∫(0~π/2) f(x)dx = Σ(n=1,3,5,...) [n^(-3)] + 2Σ(n=2,6,10,...) [n^(-3)].
更新1:

兩位大師都回答得很詳細~ 所以我只好選先回答的天助大師~ Schwartz Space大師不好意思囉~

更新2:

根據Σ(1~∞) sin(nx)/n^2 conv. uniformly 就可以得到 f is a continuous function on R ?

回答 (1)

2010-06-15 7:56 am
✔ 最佳答案
Σ(1~∞) |sin(nx)|/n^2 <= Σ(1~∞) 1/n^2 conv.
so, Σ(1~∞) sin(nx)/n^2 conv. uniformly

∫(0~π/2) [Σ(1~∞) sin(nx)/n^2 ] dx
=Σ(1~∞) ∫(0~π/2) sin(nx)/n^2 dx (since Σ(1~∞) sin(nx)/n^2 conv. to f(x) uniformly)
=Σ(1~∞) [1-cos(nπ/2)]/n^3 ----(A)
if n is odd, then [1-cos(nπ/2)]=1
if n=4k, then [1-cos(nπ/2)]=0
if n=4k+2, then [1-cos(nπ/2)]= 2
so, (A)=Σ(n=1,3,5,...) 1/n^3+ 2Σ(n=2,6, 10, 14, 18,...) 1/n^3

2010-06-15 01:54:37 補充:
連續函數,且均勻收斂,則極限函數亦連續!


收錄日期: 2021-04-30 14:55:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100614000010KK09327

檢視 Wayback Machine 備份