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Converting an explicit series to a recursive form

  1. Aug 12, 2014 #1
    Let [itex]e_i [/itex] be a unit vector with one 1 in the [itex]i[/itex]-th element. Is the following expression has a recursive presentation?

    $$y_N = \sum_{k=0}^N {\frac{{{X^k} e_i}}{\|{{{X^k} e_i}\|}_2}} $$

    where [itex]X[/itex] is a [itex]n \times n[/itex] square matrix, and [itex]{\| \cdot \|}_2[/itex] is a vector norm defined as [itex]{\|z\|}_2 = \sqrt{|z_1|^2+|z_2|^2+...+|z_n|^2}[/itex].

    ---

    EDIT: I know that if [itex]y_N = \sum_{k=0}^N {{X^k} e_i} [/itex], it is easy to obtain the following recursive formula:

    $$y_{k+1} = X y_{k} + e_i, \quad (k=0,1,2,...) \textrm{ with } \ \ \ y_0=e_i$$

    However, after we add a normalized factor, is there a similar recursive expression? Thanks.
     
  2. jcsd
  3. Aug 13, 2014 #2

    HallsofIvy

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    Isn't it just
    [tex]y_{k+1}= \frac{Xy_k+e_i}{|Xy_k+ e_i|}[/tex]

    (And I don't see why you wouldn't just write that as
    [tex]y_{k+1}= \frac{Xy_k+ y_0}{|Xy_k+ y_0|}[/tex])
     
  4. Aug 13, 2014 #3
    Hi, it seems your formula is not right because when I checked [itex]y_0,y_1,y_2[/itex], I find the following:

    $$\begin{align}
    & {{y}_{1}}=\frac{X{{e}_{i}}+{{e}_{i}}}{\left\| X{{e}_{i}}+{{e}_{i}} \right\|} \\
    & {{y}_{2}}=\frac{X{{y}_{1}}+{{e}_{i}}}{\left\| X{{y}_{1}}+{{e}_{i}} \right\|} \\
    & =\frac{X\frac{X{{e}_{i}}+{{e}_{i}}}{\left\| X{{e}_{i}}+{{e}_{i}} \right\|}+{{e}_{i}}}{\left\| X\frac{X{{e}_{i}}+{{e}_{i}}}{\left\| X{{e}_{i}}+{{e}_{i}} \right\|}+{{e}_{i}} \right\|} \\
    & =\frac{\frac{{{X}^{2}}{{e}_{i}}+X{{e}_{i}}}{\left\| X{{e}_{i}}+{{e}_{i}} \right\|}+{{e}_{i}}}{\left\| \frac{{{X}^{2}}{{e}_{i}}+X{{e}_{i}}}{\left\| X{{e}_{i}}+{{e}_{i}} \right\|}+{{e}_{i}} \right\|} \\
    \end{align}$$
     
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