[分析] 高微一題

看板Math作者 (佳佳)時間14年前 (2011/11/26 22:24), 編輯推噓2(2推 0噓 10→)
留言12則, 3人參與, 最新討論串1/5 (看更多)
Let f:[-1,1] -> |R be real analytic, and f(1/k) = 0, k = 1,2,3,... then f is identically zero. 想了一陣子，在 0 那一點可算出各階的泰勒展開係數皆為 0，就不知怎麼做了。 十分感謝！ 佳佳 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.114.112.122
→
11/26 22:29, , 1F
analytic = 可泰勒展開
11/26 22:29, 1F
→
11/26 22:47, , 2F
Then let B={z in [-1,1]|f(z)=0}. Try to prove
11/26 22:47, 2F
→
11/26 22:48, , 3F
B is both closed and open, since [-1,1] is conn
11/26 22:48, 3F
→
11/26 22:48, , 4F
we have B=[-1,1]
11/26 22:48, 4F
→
11/26 22:48, , 5F
Note: B is nonempty since 1/k in B
11/26 22:48, 5F
推
11/26 23:37, , 6F
Given a z in B, since f is analytic, then the
11/26 23:37, 6F
→
11/26 23:37, , 7F
zero of f is isolated
11/26 23:37, 7F
→
11/26 23:38, , 8F
hence there is open ball B(z;r) such that f=0
11/26 23:38, 8F
→
11/26 23:38, , 9F
on B(z;r), which impies open
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推
11/26 23:42, , 10F
改一下好了 B={z|f(z)=0 is some nhb of z}
11/26 23:42, 10F
→
11/26 23:42, , 11F
這樣比較正確
11/26 23:42, 11F
→
11/26 23:46, , 12F
十分感謝～
11/26 23:46, 12F
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文章代碼(AID): #1EqFQ3Cc (Math)
‣ 更多 tiwsjia 的文章
文章代碼(AID): #1EqFQ3Cc (Math)