• 2022-06-03
    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: ${x_{k + 1}} = {1 \over {\sqrt {{x_k} - 1} }}$
  • A

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    • 0

      ${X_1},{X_2},...,{X_n}$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: ${(\frac{2}{n}\sum\limits_{k = 1}^n {x_k} )^{\frac{1}{2}}}$

    • 1

      求函数$y = {{1 + \root 3 \of {{x^2}} - \sqrt {2x} } \over {\sqrt x }}$的导数$y' = $( ) A: $ {1 \over 2}{x^{ - {3 \over 2}}} + {1 \over 6}{x^{ - {5 \over 6}}}$ B: $ - {1 \over 2}{x^{ - {3 \over 2}}} + {1 \over 6}{x^{ - {5 \over 6}}}$ C: ${1 \over 2}{x^{ - {3 \over 2}}} - {1 \over 6}{x^{ - {5 \over 6}}}$ D: ${1 \over 3}{x^{ - {3 \over 2}}} - {1 \over 6}{x^{ - {5 \over 6}}}$

    • 2

      以下变换$\cal{A}$是线性变换的有( )。 A: $R^{3}$上变换:$\cal{A}(x_{1},x_{2},x_{3})=(x_{1},x_{3},x_{2}+1)$ B: $R^{3}$上变换:$\cal{A}(x_{1},x_{2},x_{3})=(\mid x_{1}\mid ,x_{3},x_{2})$ C: $R[x]$上变换:$\cal{A}(f(x))=f(x+3)$ D: $R[x]$上变换:$\cal{A}(f(x))=f(x+1)-f(x)$

    • 3

      【多选题】设新息序列ε(k)=y(k)-y^(k|k-1),则针对随机向量x有以下关系式 A. proj(x|y(1),y(2),……,y(k))=proj(x|ε(1),ε(2),……,ε(k)) B. C. 设A为常数矩阵,则proj(Ax|y(1),y(2),……,y(k))=Aproj(x|y(1),y(2),……,y(k)) D. 若E(x)=0,则proj(x|ε(1),ε(2),……,ε(k))=proj(x|ε(1)+proj(x|ε(2))+……+proj(x|ε(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,……‌(答案符号输入均为英文符号)‌‌