题目
Problem
g ( x ) = 1 4 − x 2 \mathrm{g}(x)=\frac{1}{\sqrt{4-x^2}} g ( x ) = 4 − x 2 1 (a) Find, in ascending powers of x x x g ( x ) \mathrm{g}(x) g ( x )
(5)
(b) State the range of values of x x x
(1)
(c) Use the expansion from part (a) to find a fully simplified rational approximation for 3 \sqrt{3} 3
Show your working and make your method clear.
(2)
题目中文翻译
g ( x ) = 1 4 − x 2 \mathrm{g}(x)=\frac{1}{\sqrt{4-x^2}} g ( x ) = 4 − x 2 1 (a) 按 x x x g ( x ) \mathrm{g}(x) g ( x )
(5)
(b) 写出该展开式成立时 x x x
(1)
(c) 利用 (a) 中的展开式，求 3 \sqrt{3} 3
请写出过程，并清楚说明你的方法。
(2)
解答
(a)
解法一
思路
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先从括号中提出 4 4 4 1 2 ( 1 − x 2 4 ) − 1 / 2 \frac12(1-\frac{x^2}{4})^{-1/2} 2 1 ( 1 − 4 x 2 ) − 1/2 − 1 2 -\frac12 − 2 1 x 2 , x 4 , x 6 x^2,x^4,x^6 x 2 , x 4 , x 6
答题过程
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First rewrite
g ( x ) = 1 2 ( 1 − x 2 4 ) − 1 2 . \mathrm{g}(x)
=\frac12\bigg(1-\frac{x^2}{4}\bigg)^{-\frac12}. g ( x ) = 2 1 ( 1 − 4 x 2 ) − 2 1 . Let n = − 1 2 n=-\frac12 n = − 2 1
b 1 = n = − 1 2 , b 2 = n ( n − 1 ) 2 ! = 3 8 , b 3 = n ( n − 1 ) ( n − 2 ) 3 ! = − 5 16 . \begin{align*}
b_1=&\,n=-\frac12,\\[4mm]
b_2=&\,\frac{n(n-1)}{2!}=\frac38,\\[4mm]
b_3=&\,\frac{n(n-1)(n-2)}{3!}\\[4mm]
=&\,-\frac5{16}.
\end{align*} b 1 = b 2 = b 3 = = n = − 2 1 , 2 ! n ( n − 1 ) = 8 3 , 3 ! n ( n − 1 ) ( n − 2 ) − 16 5 . Hence
( 1 + u ) − 1 / 2 = 1 − 1 2 u + 3 8 u 2 − 5 16 u 3 + ⋯ . \begin{align*}
&\,(1+u)^{-1/2}\\[4mm]
&\,=1-\frac12u+\frac38u^2\\[4mm]
&\,\hspace{2pt}-\frac5{16}u^3+\cdots.
\end{align*} ( 1 + u ) − 1/2 = 1 − 2 1 u + 8 3 u 2 − 16 5 u 3 + ⋯ . Substitute u = − x 2 4 u=-\frac{x^2}{4} u = − 4 x 2 1 2 \frac12 2 1
T 2 = 1 2 ( − 1 2 ) ( − x 2 4 ) = x 2 16 , T 3 = 1 2 ⋅ 3 8 ( − x 2 4 ) 2 = 3 x 4 256 , T 4 = 1 2 ( − 5 16 ) ( − x 2 4 ) 3 = 5 x 6 2048 . \begin{align*}
T_2=&\,\frac12\bigg(-\frac12\bigg)
\bigg(-\frac{x^2}{4}\bigg)\\[4mm]
=&\,\frac{x^2}{16},\\[4mm]
T_3=&\,\frac12\cdot\frac38
\bigg(-\frac{x^2}{4}\bigg)^2\\[4mm]
=&\,\frac{3x^4}{256},\\[4mm]
T_4=&\,\frac12\bigg(-\frac5{16}\bigg)
\bigg(-\frac{x^2}{4}\bigg)^3\\[4mm]
=&\,\frac{5x^6}{2048}.
\end{align*} T 2 = = T 3 = = T 4 = = 2 1 ( − 2 1 ) ( − 4 x 2 ) 16 x 2 , 2 1 ⋅ 8 3 ( − 4 x 2 ) 2 256 3 x 4 , 2 1 ( − 16 5 ) ( − 4 x 2 ) 3 2048 5 x 6 . Therefore,
g ( x ) = 1 2 + x 2 16 + 3 x 4 256 + 5 x 6 2048 + ⋯ . \begin{align*}
&\,\mathrm{g}(x)=\frac12+\frac{x^2}{16}\\[4mm]
&\,\hspace{2pt}+\frac{3x^4}{256}
+\frac{5x^6}{2048}+\cdots.
\end{align*} g ( x ) = 2 1 + 16 x 2 + 256 3 x 4 + 2048 5 x 6 + ⋯ .
解法二
思路
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也可以直接展开 ( 4 − x 2 ) − 1 / 2 (4-x^2)^{-1/2} ( 4 − x 2 ) − 1/2 ( a + u ) n (a+u)^n ( a + u ) n a = 4 a=4 a = 4 u = − x 2 u=-x^2 u = − x 2 n = − 1 2 n=-\frac12 n = − 2 1 u , u 2 , u 3 u,u^2,u^3 u , u 2 , u 3 u = − x 2 u=-x^2 u = − x 2
答题过程
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Use the direct binomial expansion
( a + u ) n = C 0 + C 1 u + C 2 u 2 + C 3 u 3 + ⋯ , \begin{align*}
&\,(a+u)^n=C_0+C_1u\\[4mm]
&\,\hspace{2pt}+C_2u^2+C_3u^3+\cdots,
\end{align*} ( a + u ) n = C 0 + C 1 u + C 2 u 2 + C 3 u 3 + ⋯ , where
C 0 = a n , C 1 = n a n − 1 , C 2 = n ( n − 1 ) 2 ! a n − 2 , C 3 = n ( n − 1 ) ( n − 2 ) 3 ! × a n − 3 . \begin{align*}
C_0=&\,a^n,\\[4mm]
C_1=&\,na^{n-1},\\[4mm]
C_2=&\,\frac{n(n-1)}{2!}a^{n-2},\\[4mm]
C_3=&\,\frac{n(n-1)(n-2)}{3!}\\[4mm]
&\,\hspace{2pt}\times a^{n-3}.
\end{align*} C 0 = C 1 = C 2 = C 3 = a n , n a n − 1 , 2 ! n ( n − 1 ) a n − 2 , 3 ! n ( n − 1 ) ( n − 2 ) × a n − 3 . For a = 4 a=4 a = 4 n = − 1 2 n=-\frac12 n = − 2 1
C 0 = 4 − 1 / 2 = 1 2 , C 1 = − 1 2 ⋅ 4 − 3 / 2 = − 1 16 , C 2 = 3 8 ⋅ 4 − 5 / 2 = 3 256 , C 3 = − 5 16 ⋅ 4 − 7 / 2 = − 5 2048 . \begin{align*}
C_0=&\,4^{-1/2}=\frac12,\\[4mm]
C_1=&\,-\frac12\cdot4^{-3/2}\\[4mm]
=&\,-\frac1{16},\\[4mm]
C_2=&\,\frac38\cdot4^{-5/2}\\[4mm]
=&\,\frac3{256},\\[4mm]
C_3=&\,-\frac5{16}\cdot4^{-7/2}\\[4mm]
=&\,-\frac5{2048}.
\end{align*} C 0 = C 1 = = C 2 = = C 3 = = 4 − 1/2 = 2 1 , − 2 1 ⋅ 4 − 3/2 − 16 1 , 8 3 ⋅ 4 − 5/2 256 3 , − 16 5 ⋅ 4 − 7/2 − 2048 5 . Substituting u = − x 2 u=-x^2 u = − x 2
g ( x ) = 1 2 + x 2 16 + 3 x 4 256 + 5 x 6 2048 + ⋯ . \begin{align*}
&\,\mathrm{g}(x)=\frac12+\frac{x^2}{16}\\[4mm]
&\,\hspace{2pt}+\frac{3x^4}{256}
+\frac{5x^6}{2048}+\cdots.
\end{align*} g ( x ) = 2 1 + 16 x 2 + 256 3 x 4 + 2048 5 x 6 + ⋯ .
解法三
思路
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先用平方差分解 4 − x 2 = ( 2 − x ) ( 2 + x ) 4-x^2=(2-x)(2+x) 4 − x 2 = ( 2 − x ) ( 2 + x ) z = x 2 z=\frac{x}{2} z = 2 x ( 1 − z ) − 1 / 2 (1-z)^{-1/2} ( 1 − z ) − 1/2 ( 1 + z ) − 1 / 2 (1+z)^{-1/2} ( 1 + z ) − 1/2 1 2 \frac12 2 1 x 6 x^6 x 6 z 6 z^6 z 6
答题过程
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Factorise the difference of two squares:
4 − x 2 = ( 2 − x ) ( 2 + x ) . 4-x^2=(2-x)(2+x). 4 − x 2 = ( 2 − x ) ( 2 + x ) . Let z = x 2 z=\frac{x}{2} z = 2 x
g ( x ) = 1 2 ( 1 − z ) − 1 / 2 × ( 1 + z ) − 1 / 2 . \begin{align*}
&\,\mathrm{g}(x)=\frac12(1-z)^{-1/2}\\[4mm]
&\,\hspace{2pt}\times(1+z)^{-1/2}.
\end{align*} g ( x ) = 2 1 ( 1 − z ) − 1/2 × ( 1 + z ) − 1/2 . The binomial expansions, retaining terms through z 6 z^6 z 6
( 1 − z ) − 1 / 2 = 1 + 1 2 z + 3 8 z 2 + 5 16 z 3 + 35 128 z 4 + 63 256 z 5 + 231 1024 z 6 + ⋯ , \begin{align*}
&\,(1-z)^{-1/2}\\[4mm]
&\,=1+\frac12z+\frac38z^2\\[4mm]
&\,\hspace{2pt}+\frac5{16}z^3+\frac{35}{128}z^4\\[4mm]
&\,\hspace{4pt}+\frac{63}{256}z^5\\[4mm]
&\,\hspace{6pt}+\frac{231}{1024}z^6+\cdots,
\end{align*} ( 1 − z ) − 1/2 = 1 + 2 1 z + 8 3 z 2 + 16 5 z 3 + 128 35 z 4 + 256 63 z 5 + 1024 231 z 6 + ⋯ , and
( 1 + z ) − 1 / 2 = 1 − 1 2 z + 3 8 z 2 − 5 16 z 3 + 35 128 z 4 − 63 256 z 5 + 231 1024 z 6 + ⋯ . \begin{align*}
&\,(1+z)^{-1/2}\\[4mm]
&\,=1-\frac12z+\frac38z^2\\[4mm]
&\,\hspace{2pt}-\frac5{16}z^3+\frac{35}{128}z^4\\[4mm]
&\,\hspace{4pt}-\frac{63}{256}z^5\\[4mm]
&\,\hspace{6pt}+\frac{231}{1024}z^6+\cdots.
\end{align*} ( 1 + z ) − 1/2 = 1 − 2 1 z + 8 3 z 2 − 16 5 z 3 + 128 35 z 4 − 256 63 z 5 + 1024 231 z 6 + ⋯ . In their product, odd-power terms cancel in pairs: for an odd
total power, reversing the two factors gives the opposite sign.
For example, the coefficient of z 3 z^3 z 3
5 16 − 3 8 ⋅ 1 2 + 1 2 ⋅ 3 8 − 5 16 = 0. \begin{align*}
&\,\frac5{16}-\frac38\cdot\frac12\\[4mm]
&\,\hspace{2pt}+\frac12\cdot\frac38-\frac5{16}=0.
\end{align*} 16 5 − 8 3 ⋅ 2 1 + 2 1 ⋅ 8 3 − 16 5 = 0. Let h 2 , h 4 , h 6 h_2,h_4,h_6 h 2 , h 4 , h 6 z 2 , z 4 , z 6 z^2,z^4,z^6 z 2 , z 4 , z 6
h 2 = 2 ⋅ 3 8 − ( 1 2 ) 2 = 1 2 , \begin{align*}
h_2=&\,2\cdot\frac38-\bigg(\frac12\bigg)^2\\[4mm]
=&\,\frac12,
\end{align*} h 2 = = 2 ⋅ 8 3 − ( 2 1 ) 2 2 1 , h 4 = 2 ⋅ 35 128 − 2 ⋅ 1 2 ⋅ 5 16 + ( 3 8 ) 2 = 3 8 , \begin{align*}
&\,h_4=2\cdot\frac{35}{128}\\[4mm]
&\,\hspace{2pt}-2\cdot\frac12\cdot\frac5{16}\\[4mm]
&\,\hspace{4pt}+\bigg(\frac38\bigg)^2=\frac38,
\end{align*} h 4 = 2 ⋅ 128 35 − 2 ⋅ 2 1 ⋅ 16 5 + ( 8 3 ) 2 = 8 3 , and
h 6 = 2 ⋅ 231 1024 − 2 ⋅ 1 2 ⋅ 63 256 + 2 ⋅ 3 8 ⋅ 35 128 − ( 5 16 ) 2 = 5 16 . \begin{align*}
&\,h_6=2\cdot\frac{231}{1024}\\[4mm]
&\,\hspace{2pt}-2\cdot\frac12\cdot\frac{63}{256}\\[4mm]
&\,\hspace{4pt}+2\cdot\frac38\cdot\frac{35}{128}\\[4mm]
&\,\hspace{6pt}-\bigg(\frac5{16}\bigg)^2=\frac5{16}.
\end{align*} h 6 = 2 ⋅ 1024 231 − 2 ⋅ 2 1 ⋅ 256 63 + 2 ⋅ 8 3 ⋅ 128 35 − ( 16 5 ) 2 = 16 5 . Thus
g ( x ) = 1 2 ( 1 + 1 2 z 2 + 3 8 z 4 + 5 16 z 6 + ⋯ ) . \begin{align*}
&\,\mathrm{g}(x)=\frac12\bigg(1+\frac12z^2\\[4mm]
&\,\hspace{2pt}+\frac38z^4+\frac5{16}z^6+\cdots\bigg).
\end{align*} g ( x ) = 2 1 ( 1 + 2 1 z 2 + 8 3 z 4 + 16 5 z 6 + ⋯ ) . Finally, substituting z = x 2 z=\frac{x}{2} z = 2 x
g ( x ) = 1 2 + x 2 16 + 3 x 4 256 + 5 x 6 2048 + ⋯ . \begin{align*}
&\,\mathrm{g}(x)=\frac12+\frac{x^2}{16}\\[4mm]
&\,\hspace{2pt}+\frac{3x^4}{256}
+\frac{5x^6}{2048}+\cdots.
\end{align*} g ( x ) = 2 1 + 16 x 2 + 256 3 x 4 + 2048 5 x 6 + ⋯ .
(b)
解法一
思路
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二项式展开要求变量部分的绝对值小于 1 1 1 − x 2 4 -\frac{x^2}{4} − 4 x 2
答题过程
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The expansion is valid when
∣ − x 2 4 ∣ < 1. \bigg|-\frac{x^2}{4}\bigg|<1. − 4 x 2 < 1. Thus
x 2 < 4 , x^2<4, x 2 < 4 , so the required range is
− 2 < x < 2 . \boxed{-2<x<2}. − 2 < x < 2 .
(c)
解法一
思路
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取 x = 1 x=1 x = 1 g ( 1 ) = 1 3 \mathrm{g}(1)=\frac1{\sqrt3} g ( 1 ) = 3 1 3 = 3 g ( 1 ) \sqrt3=3\mathrm{g}(1) 3 = 3 g ( 1 ) x = 1 x=1 x = 1 3 3 3
答题过程
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At x = 1 x=1 x = 1
g ( 1 ) = 1 3 , \mathrm{g}(1)=\frac{1}{\sqrt3}, g ( 1 ) = 3 1 , and hence
3 = 3 g ( 1 ) . \sqrt3=3\mathrm{g}(1). 3 = 3 g ( 1 ) . Using the first four terms of the expansion,
g ( 1 ) ≈ 1 2 + 1 16 + 3 256 + 5 2048 = 1181 2048 . \begin{align*}
&\,\mathrm{g}(1)\approx\frac12+\frac1{16}\\[4mm]
&\,\hspace{2pt}+\frac3{256}+\frac5{2048}\\[4mm]
&\,=\frac{1181}{2048}.
\end{align*} g ( 1 ) ≈ 2 1 + 16 1 + 256 3 + 2048 5 = 2048 1181 . Therefore,
3 ≈ 3 ( 1181 2048 ) = 3543 2048 . \begin{align*}
\sqrt3
\approx&\,3\bigg(\frac{1181}{2048}\bigg)\\
=&\,\boxed{\frac{3543}{2048}}.
\end{align*} 3 ≈ = 3 ( 2048 1181 ) 2048 3543 .
也可以对同一代入结果直接取倒数：
3 = 1 g ( 1 ) ≈ 2048 1181 . \begin{align*}
\sqrt3=&\,\frac1{\mathrm{g}(1)}\\[4mm]
\approx&\,\boxed{\frac{2048}{1181}}.
\end{align*} 3 = ≈ g ( 1 ) 1 1181 2048 . 