题目
Problem
(a) Find the Maclaurin series for sin − 1 x \sin^{-1}x sin − 1 x up to and including the term in x 3 x^3 x 3 .
(5)
(b) Deduce an approximation to ∫ 0 1 5 1 1 − u 2 d u \displaystyle\int_0^{\frac{1}{5}}\frac{1}{\sqrt{1-u^2}}\,\mathrm{d}u ∫ 0 5 1 1 − u 2 1 d u , giving your answer as a fraction.
(1)
题目中文翻译
(a) 求 sin − 1 x \sin^{-1}x sin − 1 x 的麦克劳林级数,展开至并包括 x 3 x^3 x 3 项。
(5)
(b) 由此推导出 ∫ 0 1 5 1 1 − u 2 d u \displaystyle\int_0^{\frac{1}{5}}\frac{1}{\sqrt{1-u^2}}\,\mathrm{d}u ∫ 0 5 1 1 − u 2 1 d u 的近似值,将答案写成分数。
(1)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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从 arcsin x \arcsin x arcsin x 的导数出发,用乘积法则求到三阶。代入 x = 0 x=0 x = 0 后,一次项和三次项留下。
答题过程
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Let f ( x ) = arcsin x f(x)=\arcsin x f ( x ) = arcsin x . Then
f ′ ( x ) = ( 1 − x 2 ) − 1 / 2 , f ′ ′ ( x ) = x ( 1 − x 2 ) − 3 / 2 , f ′ ′ ′ ( x ) = ( 1 − x 2 ) − 3 / 2 + 3 x 2 ( 1 − x 2 ) − 5 / 2 . \begin{align*}
f'(x)=&\,(1-x^2)^{-1/2},\\
f''(x)=&\,x(1-x^2)^{-3/2},\\
f'''(x)=&\,(1-x^2)^{-3/2}+3x^2(1-x^2)^{-5/2}.
\end{align*} f ′ ( x ) = f ′′ ( x ) = f ′′′ ( x ) = ( 1 − x 2 ) − 1/2 , x ( 1 − x 2 ) − 3/2 , ( 1 − x 2 ) − 3/2 + 3 x 2 ( 1 − x 2 ) − 5/2 .
At x = 0 x=0 x = 0 , f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 , f ′ ( 0 ) = 1 f'(0)=1 f ′ ( 0 ) = 1 , f ′ ′ ( 0 ) = 0 f''(0)=0 f ′′ ( 0 ) = 0 and f ′ ′ ′ ( 0 ) = 1 f'''(0)=1 f ′′′ ( 0 ) = 1 . Using
f ( x ) = f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) 2 ! x 2 + f ( 3 ) ( 0 ) 3 ! x 3 + ⋯ , f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\cdots, f ( x ) = f ( 0 ) + f ′ ( 0 ) x + 2 ! f ′′ ( 0 ) x 2 + 3 ! f ( 3 ) ( 0 ) x 3 + ⋯ ,
the terms through degree three are x + x 3 6 \boxed{x+\frac{x^3}{6}} x + 6 x 3 .
解法二
思路
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也可令 y = arcsin x y=\arcsin x y = arcsin x ,用 sin y = x \sin y=x sin y = x 隐式求导。这样各阶导数都可先用 y y y 表示,再在 y = 0 y=0 y = 0 处取值。
答题过程
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From sin y = x \sin y=x sin y = x , implicit differentiation gives y ′ = sec y y'=\sec y y ′ = sec y . Using the chain rule twice gives
y ′ ′ = ( sec y tan y ) y ′ = sec 2 y tan y , y ′ ′ ′ = ( 2 sec 2 y tan 2 y + sec 4 y ) y ′ = 2 sec 3 y tan 2 y + sec 5 y . \begin{align*}
y''=&\,(\sec y\tan y)y'=\sec^2y\tan y,\\
y'''=&\,\big(2\sec^2y\tan^2y+\sec^4y\big)y'\\
=&\,2\sec^3y\tan^2y+\sec^5y.
\end{align*} y ′′ = y ′′′ = = ( sec y tan y ) y ′ = sec 2 y tan y , ( 2 sec 2 y tan 2 y + sec 4 y ) y ′ 2 sec 3 y tan 2 y + sec 5 y .
At x = 0 x=0 x = 0 , y = 0 y=0 y = 0 , so these derivatives have values 1 , 0 , 1 1,0,1 1 , 0 , 1 respectively. The expansion through the term in x 3 x^3 x 3 is x + x 3 / 6 x+x^3/6 x + x 3 /6 .
(b)
解法一
思路
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被积函数是 arcsin u \arcsin u arcsin u 的导数。用 (a) 在上限 u = 1 / 5 u=1/5 u = 1/5 处近似,减去零点的值。
答题过程
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Since arcsin 0 = 0 \arcsin0=0 arcsin 0 = 0 , the integral equals arcsin ( 1 / 5 ) \arcsin(1/5) arcsin ( 1/5 ) . Using (a), its approximation is
1 5 + 1 6 ( 1 5 ) 3 = 151 750 . \frac15+\frac16\bigg(\frac15\bigg)^3=\boxed{\frac{151}{750}}. 5 1 + 6 1 ( 5 1 ) 3 = 750 151 .