Re: [微積] ln(1+x)的不等式

看板Math作者transt (transt)時間9年前 (2016/07/11 15:58)推噓2(2推 0噓 3→)

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我覺得這應該不是出題者的解法... Let F(t) = t - t^2/2 + t^3/3 * 1/(1+t) + c, where c is a constant thus F(0) = c. Let G(t) = ln(1+t) Let H(t) = F'(t) = 1/(1+t) - t^3/3 * 1/(1+t)^2, G'(t) = 1/(1+t) obviously, H(t) < G'(t) ∀ t > 0, H(0) = G'(0) = 0 x x x therefore, ∫ H(t) dt < ∫ G'(t) dt, and ∫ H(t) dt = F(x) - F(0) 0 0 0 希望有高手能用原PO的脈絡做出來，或是有別的比較漂亮的解法 -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 1.163.240.195 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1468223892.A.AE7.html

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