作者doom8199 (～口卡口卡 修～)
看板trans_math
標題Re: 求極值
時間Sun May 1 02:12:03 2011
※ 引述《bighte95016 (笨明)》之銘言：
: 求 |sinx+cosx+tanx+cotx+secx+cscx| 的最小值??
: 請教微積分高手幫忙求解0.0
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令 z = tan(x/2) = tan(ψ)
則:
f(x) = │ sin(x) + cos(x) + tan(x) + cot(x) + sec(x) + csc(x)│
2z + 1-z^2 1 + z 1
= │ ────── + ─── + ── │
1 + z^2 1 - z z
cosψ + sinψ
= │ sin(2ψ) + cos(2ψ) + ─────── + cotψ│
cosψ - sinψ
(denote)
= g(ψ)
let g'(ψ) = 0
2 1
=> 2[cos(2ψ) - sin(2ψ)] + ───────── - ───── = 0
(cosψ - sinψ)^2 (sinψ)^2
( denote c = cosψ & s = sinψ in order to simplify the eq. )
2 1
=> 2(c-s)^2 - 4s^2 + ──── - ── = 0
(c-s)^2 s^2
2(c-s)^4 + 2 4s^4 + 1
=> ────── = ────
(c-s)^2 s^2
4 2 2 2 4 2
=> 2(c-s) *s + 2s = 4(c-s) *s + (c-s)
2 2 2 2
=> [2(c-s) *s - 1][(c-s) - 2s ] = 0
=> sin(x) + cos(x) = 1 ± √2 or tan(x)=1
when sin(x) + cos(x) = 1 ± √2
=> sin(x)*cos(x) = 1 ± √2
1 + sin(x) + cos(x)
and f(x) = ｜ sin(x) + cos(x) + ────────── ｜
sin(x)*cos(x)
= 2√2 ± 1
when tan(x)=1
=> f(x) = 3√2 ± 2
所以 min{f(x)} = 2√2 - 1 , 發生於 sin(x) + cos(x) = 1 - √2
----- 分隔線 -----
剛剛有想到另外一做法
令 m = sin(x) + cos(x)
n = sin(x)*cos(x)
則題目可以改寫成:
f(x) = h(m,n) = │ m + n + (m/n)│ constrain with m^2 = 1 + 2n
可以用 Lagrange multiplier 去解它
這個做法就留給原 po 練習看看
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◆ From: 140.113.211.136
※ 編輯: doom8199 來自: 140.113.211.136 (05/01 02:47)
1F：推 bighte95016:感謝高手解答!!140.117.211.116 05/01 10:27
2F：→ suker:x=-π/4 ; =0? 118.169.82.149 05/01 19:24
3F：→ a016258: =2*sqrt(2) 吧? 114.45.91.242 05/01 20:09
4F：→ suker:我打錯了 XD 118.169.82.149 05/01 20:15
5F：推 znmkhxrw:怎麼直接對絕對值函數│f(x)│微分阿?? 1.169.129.100 05/01 20:40
6F：→ znmkhxrw:在 g(ψ)=│‧‧│ 那邊 1.169.129.100 05/01 20:40
7F：→ doom8199:把 │f│看成 √(f^2) , 則140.113.211.136 05/01 21:29
8F：→ doom8199:(|f|)' = 2ff'/[2√(f^2)]140.113.211.136 05/01 21:31
9F：→ doom8199: = f'*(f/|f|)140.113.211.136 05/01 21:32