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 $
A
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
内容
- 0
阅读以下代码，x=[0 1 0 2 0 3 0 4];for k=1:8if x(k)==0 x(k)=k;elsex(k)=2*k+1;endend回答：x(2)=______ ，x(5)=______
- 1
有语句：k＝x＜y(y＜z1：0)：0；，以下选项中，与此语句功能相同的是（ ）。 A: if(x＜y || y＜z)k＝1；else k＝0； B: if(x＜y)k＝0；else if(y＜z)k＝1； C: if(x＜y)if(y＜z)k＝1；else k＝0； D: if(x＜y＆＆y＜z)k＝1；else k＝0；
- 2
（1）已知函数f(x)x+2(x>=2)若f(f(f(k)))=25/4,求k
- 3
若x(t)傅立叶变换为X(f),则x(kt)的傅立叶变换为(1/k)X(f/k)
- 4
以下三个中___可以是分布律： （1）P{X=k}=1/2×(1/3)^k, k=0,1,2,…… （2）P{X=k}=(1/2)^k, k=1,2,3,…… （3）P{X=k}=1/[k(k+1)], k=1,2,3,……