A: 1
B: 1/(πn)sin n
C: 2/(πn)sin n
D: 1/(2πn)sin n
举一反三
- 函数$f(x)=\arcsin(\sin x)$的傅里叶级数展开式为 A: $x$ B: $$\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^n\sin(2n+1)x}{(2n+1)^2}$$ C: $$\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n\sin(2n+1)x}{(2n+1)^2}$$ D: $$\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sin(2n+1)x}{(2n+1)^2}$$
- 函数\(f(x) = x^2,\; x \in [-\pi,\pi]\)的Fourier级数为 A: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) B: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\) C: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) D: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\)
- 若n∈Z,在①sin(nπ+π3),②sin(2nπ±π3),③sin[nπ+(−1)nπ3)],④cos[2nπ+(−1)nπ6]中,与sinπ3相等的是( )
- 下面数列{xn}是单调递增的为()。 A: (1+1/n)(1/n) B: (-1)n+2n C: 1/n D: sin(1/n)
- sin1+sin(1/根号2)+……+sin(1/根号n)收敛吗
内容
- 0
5. 下列数列中,极限为$1$的是 A: $\frac{n}{{{a}^{n}}}\ \ (a\gt 1)$ B: ${{a}^{\frac{1}{n}}}\ \ (a\gt 1)$ C: $\frac{\sin {{n}^{2}}}{n}$ D: $\frac{n\sqrt{n+1}}{\sqrt{n}(2n-1)}$
- 1
Which one of the following sequences has a finite limit? A: $\ln(n),\;n=1,2,\cdots$ B: $\ln(\sin(n)),\;n=1,2,\cdots$ C: $\sqrt{n^2-1}-n^{1/3},\;n=1,2,\cdots$ D: $ \sin\frac{1}{n},\;n=1,2,\cdots$
- 2
求方程\(x = \cos x\)根的牛顿迭代公式是 。 A: \({x_{n + 1}} = {x_n} - { { {x_n} - \cos {x_n}} \over {1 + \sin {x_n}}},n = 0,1,2 \cdots \) B: \({x_{n + 1}} = {x_n} + { { {x_n} - \cos {x_n}} \over {1 + \sin {x_n}}},n = 0,1,2 \cdots \) C: \({x_{n + 1}} = {x_n} - { { {x_n} - \sin {x_n}} \over {1 + \sin {x_n}}},n = 0,1,2 \cdots \) D: \({x_{n + 1}} = {x_n} - { { {x_n} - \cos {x_n}} \over {1 + \cos{x_n}}},n = 0,1,2 \cdots \)
- 3
lim2^n*sin(x/2^n)
- 4
下列数列中,无界但不是无穷大的是 A: $\frac{n}{\ln n}$ B: ${{(-1)}^{n}}{{n}^{2}}+n$ C: $n\sin \frac{n\text{ }\!\!\pi\!\!\text{ }}{2}$ D: $\frac{{{\text{e}}^{n}}}{n!}$