Quiz 7 answer 1. 求 x sin−1(x)在 x = 0 的泰勒展式.
< Sol >
先找出 sin−1(x) 的泰勒展式(p.127),
再利用p.118習題2.7求 x sin−1(x)的泰勒展式.
由p.118習題2.5和p.126二項式公式的推廣,得到 (sin−1(x))′ = 1
√1− x2 = (1 + (−x2))−12
= 1 + C−
1 2
1 (−x2) + C−
1 2
2 (−x2)2+· · · + Cn−12(−x2)n+· · ·
= 1 + 1
2x2+3
8x4+· · · + (−1)nC−
1
n2x2n+· · · (⋆) (p.118 習題2.5) ⇒
sin−1(x) = sin−1(0) +
∫ x 0
(⋆) dx
= 0 + x +1
6x3+ 3
40x5+· · · + (−1)n 2n + 1C−
1
n2x2n+1+· · · (p.118 習題2.7) ⇒
x sin−1(x) = x2+ 1
6x4+ 3
40x6+· · · + (−1)n 2n + 1C−
1
n2x2n+2+· · · 即為所求.
(註)
(−1)nC−
1
n2 = (−1)n(−12)(−12 − 1) · · · (−12 − n + 1) n!
= (−1)n(−12)(−32)· · · (−2n2−1) n!
= 1∗ 3 ∗ · · · ∗ (2n − 1)
2nn! (分式上下同乘偶數)
= 1
2nn! ∗ 1∗ 2 ∗ · · · ∗ 2n 2∗ 4 ∗ · · · ∗ 2n
= 1
2nn! ∗ (2n)!
2n(1∗ 2 ∗ · · · ∗ n) = (2n)!
22n(n!)2
1
2. 求
xlim→∞
∫x 2 ln sds
x2ln x .
< Sol >
利用p.132的Fact: l′H ˆopital 法則可用於 ( ∞∞ 型)
xlim→∞
∫x 2 ln sds x2ln x (∞
∞)
L′H
= lim
x→∞
ln x
2x ln x + x (∞
∞)
L′H
= lim
x→∞
1 x
2 ln x + 2 + 1
= lim
x→∞
1
2x ln x + 3x = 0 . (註)
由p.78微積分基本定理得知 (∫x
2 ln sds)′ = ln x . 若要套用p.79習題2.2: dxd(∫g(x)
h(x) f (s)ds) = f (g(x))g′(x)− f(h(x))h′(x) 則 dxd(∫x
2 ln sds) = (ln x)(x)′− (ln 2)(2)′
= (ln x)(1)− (ln 2)(0) = ln x . 常數的微分為0,是常被疏忽的地方.
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