→ cgkm :{|f_k-f|<4(δ^2)ε}should be {|f_k-f|>(δ^2)ε/4} 11/08 01:22
※ 編輯: JASS0213 來自: 61.229.44.118 (11/08 14:53)
→ JASS0213 :thanks 11/08 14:53
推 q0300768 :懂了 感謝^^ 11/08 16:26
※ 引述《q0300768 (NANA真好看~^^)》之銘言：
: m 1 m 1
: 若已知f ---->f ，則該如何證明— ----->—
: k f f
: k
: f－f
: 1 1 k
: {∣—－—∣>ε}＝{∣———∣>ε}
: f f f f
: k k
: 要如何証分母的部份，使得當k→∞，他的絕對值會是0
given a>0, choose δ>0 such that
|{|f|<δ/2}|< a/6,
since
{|f_k|<δ} is included in the union of {|f_k-f|<δ/2} and {|f|<δ/2}
=> |{|f_k|<δ}|< a/3 as k > N
Now,
1 1
{∣— － —∣>ε}:= S_k is included in the union of {|f_k|<δ/2},{|f|<δ/2}
f_k f
and {|f_k-f|<(δ^2)ε/4}. Because |{|f_k-f|<4(δ^2)ε}|< a/3 for k > M.
|S_k|< a/6 +a/3+ a/3 < a when k > max{N,M}. We finish the proof.
和高微證什麼倒數函數連續是一樣的，寫一寫就出來了。
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