Re: [積分] 中興的題目

看板trans_math作者LuisSantos (^______^)時間19年前 (2005/07/08 12:13)推噓8(8推 0噓 0→)

留言8則, 8人參與討論串2/2 (看更多)

※ 引述《damaka (九死一生之路)》之銘言： : 1. 求 ∫ x^5 e^(-x) dx : 2. tan^-1 (99) (0) : 題目是對arctan (0) 的99階微分 : 謝謝 1. ∫(x^5)*(e^(-x)) dx = (-e^(-x))*(x^5) - ∫(-e^(-x))*(5*x^4)dx = (-x^5)*(e^(-x)) + 5*∫(x^4)*(e^(-x)) dx = (-x^5)*(e^(-x)) + 5*[(-e^(-x))*(x^4) - ∫(-e^(-x))*(4*x^3) dx] = (-x^5 - 5*x^4)*(e^(-x)) + 20*∫(x^3)*(e^(-x)) dx = (-x^5 - 5*x^4)*(e^(-x)) + 20*[(-e^(-x))*(x^3) - ∫(-e^(-x))*(3*x^2) dx] = -(x^5 + 5*x^4 + 20*x^3)*(e^(-x)) + 60*∫(x^2)*(e^(-x)) dx = -(x^5 + 5*x^4 + 20*x^3)*(e^(-x)) + 60*[(-e^(-x))*(x^2) - ∫(-e^x)*(2x) dx] = -(x^5 + 5*x^4 + 20*x^3 + 60*x^2)*(e^(-x)) + 120*∫x*(e^(-x)) dx = -(x^5 + 5*x^4 + 20*x^3 + 60*x^2)*(e^(-x)) + 120*[(-e^(-x))*x - ∫(-e^(-x)) dx] = -(x^5 + 5*x^4 + 20*x^3 + 60*x^2 + 120x + 120)*e^(-x) + c -1 x 1 2. tan x = ∫ ------- dx 0 1 + t^2 x = ∫ (1 - t^2 + t^4 - t^6 + ... + ((-1)^n]*t^(2n)) + ...) dt 0 t^3 t^5 t^7 ((-1)^n)*(t^(2n+1)) |x = (t - --- + --- - --- + ... + ------------------- | 3 5 7 2n + 1 |0 x^3 x^5 x^7 ((-1)^n)*(x^(2n+1)) = x - --- + --- - --- + ... + ------------------- + ... |x|<1 3 5 7 2n + 1 1 1 1 (-1)^n x = 1 時, 1 - --- + --- - --- + ... + ------ + ... 收斂 3 5 7 2n + 1 -1 π ∞ (-1)^n π 且收歛於 tan 1 = --- , 即 Σ ------ = --- 4 n=0 2n + 1 4 1 1 1 (-1)^(n+1) x = -1 時, -1 + --- - --- + --- + ... + ---------- + ... 收斂 3 5 7 2n + 1 -1 π 且收斂於 tan (-1) = - --- 4 -1 ∞ ((-1)^n)*(x^(2n+1)) 所以 f(x) = tan x = Σ ------------------- , |x|≦1 n=0 2n + 1 (2n+1) f (0) (-1)^n (2n+1) ------------ = ------- 所以 f (0) = ((-1)^n)*(2n)! (2n + 1)! 2n + 1 (99) 2n + 1 = 99 , n = 49 所以 f (0) = ((-1)^49)*(98)! = -98! -1 所以 tan (0) 的99階微分為 -98! -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 61.66.173.21

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