题目
Problem
(a) State the sum of the series 1+z+z2+⋯+zn−1, for z=1.
[1]
(b) By letting z=cosθ+isinθ, where cosθ=1, show that
\begin{aligned}
1+\cos\theta+\cos 2\theta+\cdots+\cos(n-1)\theta
=&\,\frac{1}{2}\left(1-\cos n\theta\\
&\,+\frac{\sin n\theta\sin\theta}{1-\cos\theta}\right).
\end{aligned}
[7]
The diagram shows the curve with equation y=cosx for 0≤x≤1, together with a set of n rectangles of width n1.
(c) By considering the sum of the areas of these rectangles, show that
∫01cosxdx<2n1(1−cos1+1−cos(n1)sin1sin(n1)).
[4]
(d) Use a similar method to find, in terms of n, a lower bound for ∫01cosxdx.
[3]
题目中文翻译
(a) 写出级数 1+z+z2+⋯+zn−1 的和,其中 z=1。
(b) 令 z=cosθ+isinθ,其中 cosθ=1,证明
\begin{aligned}
1+\cos\theta+\cos 2\theta+\cdots+\cos(n-1)\theta
={}&\,\frac{1}{2}\left(1-\cos n\theta\\
&\,+\frac{\sin n\theta\sin\theta}{1-\cos\theta}\right)
\end{aligned}
图中显示曲线 y=cosx,其中 0≤x≤1,并显示了一组宽度为 n1 的 n 个矩形。
(c) 考虑这些矩形面积之和,证明
∫01cosxdx<2n1(1−cos1+1−cos(n1)sin1sin(n1))
(d) 使用类似方法,求出 ∫01cosxdx 的一个关于 n 的下界。
解答