Re: [分析] 台大102高微
※ 引述《kerwinhui (kezza)》之銘言：
: : Let g(x,y) be a function satisfying
: : -1 < g(x,y) < 1 for x in R , y > 1, and
: : ln[(1+g(x,y))/(1-g(x,y))] + (2y)arctan[yg(x,y)] = 2(1+y^2)x
: : a) Show that g(x,y) is increasing in x (用反證，已解決)
: : b) Find lim(y->oo) g(x,y)
: : c) 證明 g(x,y) 在其定義域上可微
: : d) 求 lim(y->oo) g_x(x,y)
d) 我想到一個作法，不知道可不可以
g_x(x,y) = [1-g^2(x,y)][1+(y^2)(g^2(x,y))]
又 x>0 => g(x,y) -> 1 as y -> oo
x<0 => g(x,y) -> -1 as y -> oo
x=0 => g(x,y) -> 0 as y -> oo
1) x=0 時原方程式為 ln[(1+g(0,y))/(1-g(0,y))] + (2y)arctan[yg(0,y)] = 0
讓 y->oo => 0 + (oo) * arctan[lim(y->oo) yg(0,y)] = 0
故 lim(y->oo) yg(0,y) 必為 0
=> g_x(0,y) = [1-g^2(0,y)][1+(y^2)(g^2(0,y))] -> (1-0)(1+0^2) = 1
as y -> oo
2) x>0 時原方程式兩邊同除以 (1+g)(1-g) 再讓 y -> oo
則等號左邊為 ln[(1+g)/(1-g)]/[(1+g)/(1-g)] -> 0 as y -> oo
(同 lim(y->oo) (lny)/y = 0)
等號右邊整理後
2x(1-g) + 2x(y^2)(1-g) - 2y(1-g)arctan(yg)
= lim(y->oo) ------------------------------------------- (*)
1 + g
2x(1-g) + 2y(1-g)[xy - arctan(yg)]
= lim(y->oo) -----------------------------------
1 + g
0 + [lim(y->oo) 2y(1-g)]*[oo - pi/2]
= ------------------------------------- = 0
2
故 lim(y->oo) y(1-g) 必為 0，代入 (*)
0 + 2x[lim(y->oo) (y^2)(1-g)] - [0 * (pi/2)]
=> --------------------------------------------- = 0
2
故 lim(y->oo) (y^2)(1-g) 必為 0，代回原式
=> g_x(x,y) = [1-g^2(x,y)][1+(y^2)(g^2(x,y))]
= (1+g){(1-g)[1+(y^2)(g^2)]}
= (1+g){(1-g) + [(1-g)y^2]g^2}
-> 2 * (0 + 0 * 1^2) = 0 as y -> oo #
x< 0 時 作法類似
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