Question

将函数f(x)=2x2(0<x≤)展开成周期为2π的正弦级数.设正弦级数和函数为S(x),求

Answer 1

问：将函数f(x)=2x2(0

答：根据傅里叶级数的公式，要将函数$f(x)$展开成周期为$2\pi$的正弦级数，需要求出其系数$a_n$和$b_n$。公式如下： $$f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos nx + b_n \sin nx \right)$$ 其中， $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx \mathrm{d}x$$ $$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx \mathrm{d}x$$ 由于$f(x)$在$(0,2\pi)$上是$2x^2$，在$(-\pi,0)$上是$2(\pi+x)^2$，因此有： $$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \mathrm{d}x = \frac{1}{\pi} \left( \int_{0}^{\pi} 2x^2 \mathrm{d}x + \int_{-\pi}^{0} 2(\pi+x)^2"

答：因此有： $$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \mathrm{d}x = \frac{1}{\pi} \left( \int_{0}^{\pi} 2x^2 \mathrm{d}x + \int_{-\pi}^{0} 2(\pi+x)^2 \mathrm{d}x \right) = 2\pi^2$$ $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx \mathrm{d}x = \frac{4}{\pi} \int_{0}^{\pi} x^2 \cos nx \mathrm{d}x = \frac{4}{\pi n^2} ((-1)^n - 1)$$ $$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx \mathrm{d}x = 0$$ 因此，$f(x)$的正弦级数展开为： $$S(x) = \pi^2 + \sum_{n=1}^{\infty} \frac{4}{\pi n^2} ((-1)^n - 1) \cos nx$$

答：因此，$f(x)$的正弦级数展开为：$$S(x) = \pi^2 + \sum_{n=1}^{\infty} \frac{4}{\pi n^2} ((-1)^n - 1) \cos nx$$