Re: 求極值

[doom8199 (～口卡口卡 修～)]

※ 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
---

    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


--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.113.211.136
※ 編輯: doom8199        來自: 140.113.211.136      (05/01 02:47)
推 bighte95016:感謝高手解答!!
→ suker:x=-π/4 ;   =0?
→ a016258:          =2*sqrt(2) 吧?
→ suker:我打錯了 XD
推 znmkhxrw:怎麼直接對絕對值函數│f(x)│微分阿??
→ znmkhxrw:在 g(ψ)=│‧‧│ 那邊
→ doom8199:把 │f│看成 √(f^2) , 則
→ doom8199:(|f|)' = 2ff'/[2√(f^2)]
→ doom8199:       = f'*(f/|f|)