Re: [轉錄][問題] 救救我呀

[Frobenius (i^(-i)= e^(π/2))]

※ 引述《Frobenius (i^(-i)= e^(π/2))》之銘言：
: ※ 引述《Jasy (Transferers)》之銘言：
:      -1           ( s + 1 )
:    Ｌ  { ─────────────── }
:           ( s^2 + 1 )( s^2 + 4s + 13 )
:      -1   1     1       1     s      -1         3
:  = Ｌ  { ─ ──── + ─ ──── + ─ ─────────
:          10  s^2 + 1   20  s^2 + 1   15 [ (s+2)^2 + 3^2 ]
:     -1        (s+2)
:   + ─ ───────── }
:     20 [ (s+2)^2 + 3^2 ]
:  = (1/10)sin(t) + (1/20)cos(t) - (1/15)e^(-2t)sin(3t) - (1/20)e^(-2t)cos(3t)

        1            s                 3                     (s+2)
設 A ──── + B ──── + C ───────── + D ─────────
     s^2 + 1      s^2 + 1      [ (s+2)^2 + 3^2 ]      [ (s+2)^2 + 3^2 ]

(至於為什麼要這樣設，因為這樣比較好看出反轉換)

( A + B s ) ( s^2 + 4s + 13 ) + ( 3 C + D(s+2) ) ( s^2 + 1 ) = s + 1 = 1 + s

令 s = i，(A + B i) ( -1 + 4i + 13 ) = (12A - 4B) + i(4A + 12B) = 1 + i

12A - 4B = 1       36A - 12B = 3
             =>
4A + 12B = 1    +)  4A + 12B = 1
                ────────
                   40A       = 4 => A = (1/10) => 12B = (6/10) => B = (1/20)

s^2 + 4s + 13 = 0 => s = -2 ± 3i

令 s = -2 + 3i，1 + s = 1 - 2 + 3i = -1 + 3i，s + 2 = 3i

 s^2 = -5 - 12i，s^2 + 1 = -4 -12i，

( 3 C + 3 D i ) ( -4 -12i ) = (-12 C + 36 D) + i(-36 C - 12 D) = -1 + 3i

-12C + 36D = -1       -12C + 36D = -1
                =>
-36C - 12D =  3    -) -12C -  4D =  1
                   ─────────
                             40D = -2 => D = (-1/20)

                      => -12C = (6/5) => C = (-1/10)

A = (1/10)，B = (1/20)，C = (-1/10)，D = (-1/20)

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