[分析] 高微一題

[tiwsjia (佳佳)]

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

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◆ From: 140.114.112.122
→ wickeday    :analytic = 可泰勒展開
→ jacky7987   :Then let B={z in [-1,1]|f(z)=0}. Try to prove
→ jacky7987   :B is both closed and open, since [-1,1] is conn
→ jacky7987   :we have B=[-1,1]
→ jacky7987   :Note: B is nonempty since 1/k in B
推 jacky7987   :Given a z in B, since f is analytic, then the
→ jacky7987   :zero of f is isolated
→ jacky7987   :hence there is open ball B(z;r) such that f=0
→ jacky7987   :on B(z;r), which impies open
推 jacky7987   :改一下好了 B={z|f(z)=0 is some nhb of z}
→ jacky7987   :這樣比較正確
→ tiwsjia     :十分感謝～