分からない問題はここに書いてね２５７
>230
(1) fは奇函数で周期2πをもつから、-0≦x≦π で考える。
f '(x) = cos(x)/{2-cos(x)}^n - n{sin(x)}^2 /{2-cos(x)}^(n+1)
= {(n-1)cos(x)^2 +2cos(x)-n}/{2-cos(x)}^(n+1) = g_n(cos(x))/{2-cos(x)}^(n+1),
ここに, g_n(t) = (n-1)t^2 +2t -n, その根で|t|≦1 をみたすものが A_n.
n=1 のとき cos(A_1) = 1/2 より A_1 = π/3,
n>1 のとき cos(A_n) = {-1 +√(n^2 -n+1)}/(n-1) → 1 (n→∞),
0 ≦ A_n < (4/π)sin({A_n}/2) = (4/π)√{[1-cos(A_n)]/2} → 0 (n→∞).
(2)
√(n^2 -n+1) = √{(n -1/2)^2 +(3/4)} = {n -(1/2)}{1+O(1/n^2)} = n -(1/2) +O(1/n).
sin({A_n}/2) = √{[1-cos(A_n)]/2} = 1/√{2[n+√(n^2 -n+1)]} = 1/{2√[n-(1/4)+O(1/n)]} = 1/(2√n) + O(1/{n√n}),
A_n = 2arcsin( 1/{2√n} + O(1/{n√n}) ) = 1/(√n) + O(1/{n√n}),
(1) fは奇函数で周期2πをもつから、-0≦x≦π で考える。
f '(x) = cos(x)/{2-cos(x)}^n - n{sin(x)}^2 /{2-cos(x)}^(n+1)
= {(n-1)cos(x)^2 +2cos(x)-n}/{2-cos(x)}^(n+1) = g_n(cos(x))/{2-cos(x)}^(n+1),
ここに, g_n(t) = (n-1)t^2 +2t -n, その根で|t|≦1 をみたすものが A_n.
n=1 のとき cos(A_1) = 1/2 より A_1 = π/3,
n>1 のとき cos(A_n) = {-1 +√(n^2 -n+1)}/(n-1) → 1 (n→∞),
0 ≦ A_n < (4/π)sin({A_n}/2) = (4/π)√{[1-cos(A_n)]/2} → 0 (n→∞).
(2)
√(n^2 -n+1) = √{(n -1/2)^2 +(3/4)} = {n -(1/2)}{1+O(1/n^2)} = n -(1/2) +O(1/n).
sin({A_n}/2) = √{[1-cos(A_n)]/2} = 1/√{2[n+√(n^2 -n+1)]} = 1/{2√[n-(1/4)+O(1/n)]} = 1/(2√n) + O(1/{n√n}),
A_n = 2arcsin( 1/{2√n} + O(1/{n√n}) ) = 1/(√n) + O(1/{n√n}),
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