题目
Problem
(a) State the sum of the series 1 + w + w 2 + w 3 + … + w n − 1 1+w+w^2+w^3+\ldots+w^{n-1} 1 + w + w 2 + w 3 + … + w n − 1 w ≠ 1 w\ne1 w = 1
(1)
(b) Show that ( 1 + i tan θ ) k = sec k θ ( cos k θ + i sin k θ ) (1+\mathrm{i}\tan\theta)^k=\sec^k\theta(\cos k\theta+\mathrm{i}\sin k\theta) ( 1 + i tan θ ) k = sec k θ ( cos k θ + i sin k θ ) θ \theta θ 1 2 π \frac{1}{2}\pi 2 1 π
(2)
(c) By considering ∑ k = 0 n − 1 ( 1 + i tan θ ) k \displaystyle\sum_{k=0}^{n-1}(1+\mathrm{i}\tan\theta)^k k = 0 ∑ n − 1 ( 1 + i tan θ ) k
∑ k = 0 n − 1 sec k θ sin k θ = cot θ ( 1 − sec n θ cos n θ ) , \sum_{k=0}^{n-1}\sec^k\theta\sin k\theta=\cot\theta(1-\sec^n\theta\cos n\theta), k = 0 ∑ n − 1 sec k θ sin k θ = cot θ ( 1 − sec n θ cos n θ ) , provided θ \theta θ 1 2 π \frac{1}{2}\pi 2 1 π
(5)
(d) Hence find ∑ k = 0 6 m − 1 2 k sin ( 1 3 k π ) \displaystyle\sum_{k=0}^{6m-1}2^k\sin\big(\frac{1}{3}k\pi\big) k = 0 ∑ 6 m − 1 2 k sin ( 3 1 k π ) m m m
(2)
题目中文翻译
(a) 写出级数 1 + w + w 2 + w 3 + … + w n − 1 1+w+w^2+w^3+\ldots+w^{n-1} 1 + w + w 2 + w 3 + … + w n − 1 w ≠ 1 w\ne1 w = 1
(1)
(b) 证明 ( 1 + i tan θ ) k = sec k θ ( cos k θ + i sin k θ ) (1+\mathrm{i}\tan\theta)^k=\sec^k\theta(\cos k\theta+\mathrm{i}\sin k\theta) ( 1 + i tan θ ) k = sec k θ ( cos k θ + i sin k θ ) θ \theta θ 1 2 π \frac{1}{2}\pi 2 1 π
(2)
(c) 通过考虑 ∑ k = 0 n − 1 ( 1 + i tan θ ) k \displaystyle\sum_{k=0}^{n-1}(1+\mathrm{i}\tan\theta)^k k = 0 ∑ n − 1 ( 1 + i tan θ ) k
∑ k = 0 n − 1 sec k θ sin k θ = cot θ ( 1 − sec n θ cos n θ ) , \sum_{k=0}^{n-1}\sec^k\theta\sin k\theta=\cot\theta(1-\sec^n\theta\cos n\theta), k = 0 ∑ n − 1 sec k θ sin k θ = cot θ ( 1 − sec n θ cos n θ ) , 其中 θ \theta θ 1 2 π \frac{1}{2}\pi 2 1 π
(5)
(d) 由此求 ∑ k = 0 6 m − 1 2 k sin ( 1 3 k π ) \displaystyle\sum_{k=0}^{6m-1}2^k\sin\big(\frac{1}{3}k\pi\big) k = 0 ∑ 6 m − 1 2 k sin ( 3 1 k π ) m m m
(2)
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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用首项为 1 1 1 w w w
答题过程
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For w ≠ 1 w\ne1 w = 1 ( w n − 1 ) / ( w − 1 ) \boxed{(w^n-1)/(w-1)} ( w n − 1 ) / ( w − 1 )
(b)
解法一
思路
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先把 1 + i tan θ 1+\mathrm{i}\tan\theta 1 + i tan θ sec θ \sec\theta sec θ
答题过程
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Since 1 + i tan θ = sec θ ( cos θ + i sin θ ) 1+\mathrm{i}\tan\theta=\sec\theta(\cos\theta+\mathrm{i}\sin\theta) 1 + i tan θ = sec θ ( cos θ + i sin θ )
( 1 + i tan θ ) k = sec k θ ( cos k θ + i sin k θ ) . (1+\mathrm{i}\tan\theta)^k=\sec^k\theta(\cos k\theta+\mathrm{i}\sin k\theta). ( 1 + i tan θ ) k = sec k θ ( cos k θ + i sin k θ ) .
(c)
解法一
思路
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对 w = 1 + i tan θ w=1+\mathrm{i}\tan\theta w = 1 + i tan θ − i cot θ -\mathrm{i}\cot\theta − i cot θ
答题过程
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Let w = 1 + i tan θ w=1+\mathrm{i}\tan\theta w = 1 + i tan θ θ \theta θ w ≠ 1 w\ne1 w = 1
∑ k = 0 n − 1 w k = w n − 1 w − 1 = w n − 1 i tan θ = − i cot θ ( w n − 1 ) . \begin{align*}
\sum_{k=0}^{n-1}w^k
=&\,\frac{w^n-1}{w-1}\\
=&\,\frac{w^n-1}{\mathrm{i}\tan\theta}\\
=&\,-\mathrm{i}\cot\theta\,(w^n-1).
\end{align*} k = 0 ∑ n − 1 w k = = = w − 1 w n − 1 i tan θ w n − 1 − i cot θ ( w n − 1 ) . By (b), the imaginary part of w k w^k w k sec k θ sin k θ \sec^k\theta\sin k\theta sec k θ sin k θ
− i cot θ [ sec n θ ( cos n θ + i sin n θ ) − 1 ] = cot θ sec n θ sin n θ + i cot θ ( 1 − sec n θ cos n θ ) . \begin{align*}
&\,-\mathrm{i}\cot\theta\big[\sec^n\theta(\cos n\theta+\mathrm{i}\sin n\theta)-1\big]\\
=&\,\cot\theta\sec^n\theta\sin n\theta\\
&\,\hspace{2pt}+\mathrm{i}\cot\theta\big(1-\sec^n\theta\cos n\theta\big).
\end{align*} = − i cot θ [ sec n θ ( cos n θ + i sin n θ ) − 1 ] cot θ sec n θ sin n θ + i cot θ ( 1 − sec n θ cos n θ ) . Equating imaginary parts gives
∑ k = 0 n − 1 sec k θ sin k θ = cot θ ( 1 − sec n θ cos n θ ) . \sum_{k=0}^{n-1}\sec^k\theta\sin k\theta
=\cot\theta(1-\sec^n\theta\cos n\theta). k = 0 ∑ n − 1 sec k θ sin k θ = cot θ ( 1 − sec n θ cos n θ ) .
(d)
解法一
思路
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在 (c) 中取 θ = π / 3 \theta=\pi/3 θ = π /3 n = 6 m n=6m n = 6 m sec θ = 2 \sec\theta=2 sec θ = 2 cos ( n θ ) = 1 \cos(n\theta)=1 cos ( n θ ) = 1
答题过程
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With θ = π / 3 \theta=\pi/3 θ = π /3 n = 6 m n=6m n = 6 m cot θ = 1 / 3 \cot\theta=1/\sqrt3 cot θ = 1/ 3 cos ( n θ ) = cos ( 2 m π ) = 1 \cos(n\theta)=\cos(2m\pi)=1 cos ( n θ ) = cos ( 2 mπ ) = 1
∑ k = 0 6 m − 1 2 k sin ( k π / 3 ) = 1 − 2 6 m 3 . \sum_{k=0}^{6m-1}2^k\sin(k\pi/3)
=\boxed{\frac{1-2^{6m}}{\sqrt3}}. k = 0 ∑ 6 m − 1 2 k sin ( k π /3 ) = 3 1 − 2 6 m . 