Converting an explicit series to a recursive form

[regina_lee]

Let $e_i$ be a unit vector with one 1 in the $i$-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 $X$ is a $n \times n$ square matrix, and ${\| \cdot \|}_2$ is a vector norm defined as ${\|z\|}_2 = \sqrt{|z_1|^2+|z_2|^2+...+|z_n|^2}$.

---

EDIT: I know that if $y_N = \sum_{k=0}^N {{X^k} e_i}$, 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.

 

[HallsofIvy]

Isn't it just
$$y_{k+1}= \frac{Xy_k+e_i}{|Xy_k+ e_i|}$$

(And I don't see why you wouldn't just write that as
$$y_{k+1}= \frac{Xy_k+ y_0}{|Xy_k+ y_0|}$$)

 

[regina_lee]

Let $e_i$ be a unit vector with one 1 in the $i$-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 $X$ is a $n \times n$ square matrix, and ${\| \cdot \|}_2$ is a vector norm defined as ${\|z\|}_2 = \sqrt{|z_1|^2+|z_2|^2+...+|z_n|^2}$.

---

EDIT: I know that if $y_N = \sum_{k=0}^N {{X^k} e_i}$, 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.

Hi, it seems your formula is not right because when I checked $y_0,y_1,y_2$, I find the following:

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