推 yusd24 :moving h 10/09 11:50
請問您的意思是?!
※ 編輯: lavender003 來自: 125.231.164.167 (10/09 15:45)
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.128.36.169
Suppose f is a twice-differentiable real function on (0,∞) and let
M0,M1,M2 be the least upper bounds of |f(x)|,|f'(x)|,|f''(x)|
respectively on (0,∞)。
prove that M1^2≦4M0*M2
(hint:Taylor's theorem shats that f'(x)=(1/2h*[f(x+2h)-f(x)] ) - h''(ξ) 。
so that |f'|≦hM2+M0/h )
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<pf>由三角不等式配合f'(x)=(1/2h*[f(x+2h)-f(x)] ) - h''(ξ)此方程式
我們可得到|f'|≦hM2+M0/h
又因M1為最小上界,所以M1≦hM2+M0/h
再來取雙邊平方:M1^2≦h^2*(M2^2)+2M0M2+(M0^2/h^2)
根據算術平均大於幾何平均,可得以下式子
[2*(h^2*(M2^2) * M0^2/h^2)^1/2] +2M0M2 ≦ h^2*(M2^2)+2M0M2+(M0^2/h^2)
=>4M0M2 ≦ h^2*(M2^2)+2M0M2+(M0^2/h^2)
目前證到這邊就卡關了,請問該如何推到M1^2≦4M0*M2?!
煩請解惑了,thx~!
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