题目
Problem
(a) By considering the binomial expansion of ( z + 1 z ) 7 \left(z + \frac{1}{z}\right)^7 ( z + z 1 ) 7 z = cos θ + i sin θ z = \cos\theta + \mathrm{i}\sin\theta z = cos θ + i sin θ
cos 7 θ = a cos 7 θ + b cos 5 θ + c cos 3 θ + d cos θ , \cos^7\theta = a\cos 7\theta + b\cos 5\theta + c\cos 3\theta + d\cos\theta, cos 7 θ = a cos 7 θ + b cos 5 θ + c cos 3 θ + d cos θ , where a a a b b b c c c d d d
[5]
Let
I n = ∫ 0 1 4 π cos n θ d θ . I_n = \int_0^{\frac{1}{4}\pi} \cos^n\theta\,\mathrm{d}\theta. I n = ∫ 0 4 1 π cos n θ d θ . (b) Show that
n I n = 2 − 1 2 n + ( n − 1 ) I n − 2 . nI_n = 2^{-\frac{1}{2}n} + (n - 1)I_{n - 2}. n I n = 2 − 2 1 n + ( n − 1 ) I n − 2 . [4]
(c) Using the results given in parts (a) and (b), find the exact value of I 9 I_9 I 9
[5]
题目中文翻译
(a) 考虑二项式展开 ( z + 1 z ) 7 \left(z + \frac{1}{z}\right)^7 ( z + z 1 ) 7 z = cos θ + i sin θ z = \cos\theta + \mathrm{i}\sin\theta z = cos θ + i sin θ
cos 7 θ = a cos 7 θ + b cos 5 θ + c cos 3 θ + d cos θ \cos^7\theta = a\cos 7\theta + b\cos 5\theta + c\cos 3\theta + d\cos\theta cos 7 θ = a cos 7 θ + b cos 5 θ + c cos 3 θ + d cos θ 其中 a a a b b b c c c d d d
(5)
令
I n = ∫ 0 1 4 π cos n θ d θ I_n = \int_0^{\frac{1}{4}\pi} \cos^n\theta\,\mathrm{d}\theta I n = ∫ 0 4 1 π cos n θ d θ (b) 证明
n I n = 2 − 1 2 n + ( n − 1 ) I n − 2 nI_n = 2^{-\frac{1}{2}n} + (n - 1)I_{n - 2} n I n = 2 − 2 1 n + ( n − 1 ) I n − 2 (4)
(c) 使用 (a) 和 (b) 部分给出的结果，求 I 9 I_9 I 9
(5)
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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用 z + z − 1 = 2 cos θ z+z^{-1}=2\cos\theta z + z − 1 = 2 cos θ
答题过程
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Since z + z − 1 = 2 cos θ z+z^{-1}=2\cos\theta z + z − 1 = 2 cos θ
( z + z − 1 ) 7 = z 7 + 7 z 5 + 21 z 3 + 35 z + 35 z − 1 + 21 z − 3 + 7 z − 5 + z − 7 . \begin{align*}
(z+z^{-1})^7=&\,z^7+7z^5+21z^3+35z\\[4mm]
&\,\hspace{2pt}+35z^{-1}+21z^{-3}+7z^{-5}+z^{-7}.
\end{align*} ( z + z − 1 ) 7 = z 7 + 7 z 5 + 21 z 3 + 35 z + 35 z − 1 + 21 z − 3 + 7 z − 5 + z − 7 . Grouping matching powers gives
( 2 cos θ ) 7 = ( z 7 + z − 7 ) + 7 ( z 5 + z − 5 ) + 21 ( z 3 + z − 3 ) + 35 ( z + z − 1 ) . \begin{align*}
(2\cos\theta)^7=&\,(z^7+z^{-7})+7(z^5+z^{-5})\\[4mm]
&\,\hspace{2pt}+21(z^3+z^{-3})+35(z+z^{-1}).
\end{align*} ( 2 cos θ ) 7 = ( z 7 + z − 7 ) + 7 ( z 5 + z − 5 ) + 21 ( z 3 + z − 3 ) + 35 ( z + z − 1 ) . By de Moivre’s theorem, z k + z − k = 2 cos ( k θ ) z^k+z^{-k}=2\cos(k\theta) z k + z − k = 2 cos ( k θ )
cos 7 θ = 1 64 cos 7 θ + 7 64 cos 5 θ + 21 64 cos 3 θ + 35 64 cos θ , \begin{align*}
\cos^7\theta=&\,\frac1{64}\cos7\theta+\frac7{64}\cos5\theta\\[4mm]
&\,\hspace{2pt}+\frac{21}{64}\cos3\theta+\frac{35}{64}\cos\theta,
\end{align*} cos 7 θ = 64 1 cos 7 θ + 64 7 cos 5 θ + 64 21 cos 3 θ + 64 35 cos θ , so ( a , b , c , d ) = ( 1 , 7 , 21 , 35 ) / 64 (a,b,c,d)=(1,7,21,35)/64 ( a , b , c , d ) = ( 1 , 7 , 21 , 35 ) /64
(b)
解法一
思路
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将被积函数拆成 cos n − 1 θ \cos^{n-1}\theta cos n − 1 θ cos θ \cos\theta cos θ sin 2 θ \sin^2\theta sin 2 θ 1 − cos 2 θ 1-\cos^2\theta 1 − cos 2 θ
答题过程
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Take u = cos n − 1 θ u=\cos^{n-1}\theta u = cos n − 1 θ d v = cos θ d θ \mathrm{d}v=\cos\theta\,\mathrm{d}\theta d v = cos θ d θ v = sin θ v=\sin\theta v = sin θ d u = − ( n − 1 ) cos n − 2 θ sin θ d θ \mathrm{d}u=-(n-1)\cos^{n-2}\theta\sin\theta\,\mathrm{d}\theta d u = − ( n − 1 ) cos n − 2 θ sin θ d θ ( 1 / 2 ) n − 1 ( 1 / 2 ) = 2 − n / 2 (1/\sqrt2)^{n-1}(1/\sqrt2)=2^{-n/2} ( 1/ 2 ) n − 1 ( 1/ 2 ) = 2 − n /2
I n = [ cos n − 1 θ sin θ ] 0 π / 4 + ( n − 1 ) ∫ 0 π / 4 cos n − 2 θ sin 2 θ d θ = 2 − n / 2 + ( n − 1 ) ( I n − 2 − I n ) . \begin{align*}
I_n=&\,[\cos^{n-1}\theta\sin\theta]_0^{\pi/4}\\[4mm]
&\,\hspace{2pt}+(n-1)\int_0^{\pi/4}\cos^{n-2}\theta\sin^2\theta\,\mathrm{d}\theta\\[4mm]
=&\,2^{-n/2}+(n-1)(I_{n-2}-I_n).
\end{align*} I n = = [ cos n − 1 θ sin θ ] 0 π /4 + ( n − 1 ) ∫ 0 π /4 cos n − 2 θ sin 2 θ d θ 2 − n /2 + ( n − 1 ) ( I n − 2 − I n ) . Rearranging gives n I n = 2 − n / 2 + ( n − 1 ) I n − 2 nI_n=2^{-n/2}+(n-1)I_{n-2} n I n = 2 − n /2 + ( n − 1 ) I n − 2
(c)
解法一
思路
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先将 (a) 的多倍角恒等式逐项积分求 I 7 I_7 I 7 sin ( 7 π / 4 ) \sin(7\pi/4) sin ( 7 π /4 ) sin ( 5 π / 4 ) \sin(5\pi/4) sin ( 5 π /4 ) I 9 I_9 I 9
答题过程
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Integrating the identity in (a) gives
I 7 = 1 64 [ sin 7 θ 7 + 7 sin 5 θ 5 + 7 sin 3 θ + 35 sin θ ] 0 π / 4 = 2 128 ( − 1 7 − 7 5 + 7 + 35 ) = 177 2 560 . \begin{align*}
I_7=&\,\frac1{64}\bigg[\frac{\sin7\theta}{7}+\frac{7\sin5\theta}{5}\\[4mm]
&\,\hspace{2pt}+7\sin3\theta+35\sin\theta\bigg]_0^{\pi/4}\\[4mm]
=&\,\frac{\sqrt2}{128}\bigg(-\frac17-\frac75+7+35\bigg)\\[4mm]
=&\,\frac{177\sqrt2}{560}.
\end{align*} I 7 = = = 64 1 [ 7 sin 7 θ + 5 7 sin 5 θ + 7 sin 3 θ + 35 sin θ ] 0 π /4 128 2 ( − 7 1 − 5 7 + 7 + 35 ) 560 177 2 . Taking n = 9 n=9 n = 9
I 9 = 1 9 ( 2 32 + 8 177 2 560 ) = 2867 2 10080 . I_9=\frac19\bigg(\frac{\sqrt2}{32}+8\frac{177\sqrt2}{560}\bigg)
=\boxed{\frac{2867\sqrt2}{10080}}. I 9 = 9 1 ( 32 2 + 8 560 177 2 ) = 10080 2867 2 . 