牛顿迭代法的迭代格式以下正确的是：
A: ${x_{k + 1}} = {x_k} - {{f({x_k})} \over {f'({x_k})}},k = 0,1, \cdots $
B: ${x_{k + 1}} = {x_k} - {{f'({x_k})} \over {f({x_k})}},k = 0,1, \cdots $
C: ${x_{k + 1}} = {x_k} - {{f'({x_{k + 1}})} \over {f({x_k})}},k = 0,1, \cdots $
D: ${x_{k + 1}} = {x_k} - {{f({x_{k + 1}})} \over {f'({x_k})}},k = 0,1, \cdots $
A: ${x_{k + 1}} = {x_k} - {{f({x_k})} \over {f'({x_k})}},k = 0,1, \cdots $
B: ${x_{k + 1}} = {x_k} - {{f'({x_k})} \over {f({x_k})}},k = 0,1, \cdots $
C: ${x_{k + 1}} = {x_k} - {{f'({x_{k + 1}})} \over {f({x_k})}},k = 0,1, \cdots $
D: ${x_{k + 1}} = {x_k} - {{f({x_{k + 1}})} \over {f'({x_k})}},k = 0,1, \cdots $
举一反三
- 为求方程${x^3} - {x^2} - 1 = 0$在${x_0} = 1.5$附近的一个根，以下迭代格式收敛的是： A: ${x_{k + 1}} = 1 + {1 \over {x_k^2}}$ B: ${x_{k + 1}} = 1 - {1 \over {x_k^2}}$ C: ${x_{k + 1}} = \root 3 \of {x_k^2 - 1} $ D: ${x_{k + 1}} = {1 \over {\sqrt {{x_k} - 1} }}$
- 已知序列x[n]={1,1},n=0,1，则x[n]的2点DFT为（） A: X[k]={1,1},k=0,1 B: X[k]={1,0},k=0,1 C: X[k]={0,1},k=0,1 D: X[k]={-1,-1},k=0,1
- ${X_1},{X_2},...,{X_n}$是来自均匀分布X~U(-a,a)的样本，用矩估计法估计参数a为() A: ${(\frac{3}{n}\sum\limits_{k = 1}^n {x_k^2} )^{\frac{1}{2}}}$ B: ${(\frac{2}{n}\sum\limits_{k = 1}^n {x_k^2} )^{\frac{1}{2}}}$ C: ${(\frac{3}{n}\sum\limits_{k = 1}^n {x_k} )^{\frac{1}{2}}}$ D: ${(\frac{2}{n}\sum\limits_{k = 1}^n {x_k} )^{\frac{1}{2}}}$
- x(k 2) 2x(k 1) x(k) = u(k)，x(0)=0，x(1)=0，u(k)=k (k=0,1,2,…)，符合描述的选项为（）。_
- f(x)是分段函数f(x)=xsin(1/x)+2,x≠0并且f(x)=k,x=0如果f(x)在x=0处连续,则k=多少