Hello everyone,

English isn't my native language, so I hope I can make myself clear. I'm new on this forum, because I have a question. I put this topic in this subforum, but I am not sure if this is the right one. Sorry if I'm wrong.

Given is some vector $\displaystyle A(t)$ and $\displaystyle t \in \{0, 1, ...\}$ with following conditions:

$\displaystyle A_{j+1}(t+1) = A_j(t)p_j$ with $\displaystyle j \in \{0, ..., l - 1\}$

$\displaystyle A_0(t+1) = \sum^r_{i=0} A_i(t)f_i$

In this case: $\displaystyle 0 < r < l \in \mathbb{N}$. $\displaystyle r$ and $\displaystyle l$ are constants.

$\displaystyle f_i$ with $\displaystyle i \in \{0, ..., r\}$ and $\displaystyle p_j$ with $\displaystyle j \in \{0, ..., l - 1\}$ are constants too.

Let $\displaystyle B$ be a vector with above mentioned conditions ánd:

$\displaystyle \frac{B_{j+1}(t+1)}{B_j(t+1)} = \frac{B_{j+1}(t)}{B_j(t)}$

Define $\displaystyle \alpha_j = \alpha_j(t) = \frac{B_{j+1}(t+1)}{B_j(t+1)}$. Express $\displaystyle \alpha_j$ in $\displaystyle \alpha_0$, $\displaystyle p_j$ and $\displaystyle p_0$.

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I have a very difficult time doing this. This is how far I got:

$\displaystyle \alpha_0 = \alpha_0(t) = \frac{B_1(t+1)}{B_0(t+1)} = \frac{B_0(t)p_0}{\sum^r_{i=0} B_i(t)f_i}$.

So for $\displaystyle j \in \{1, ..., l - 1\}$:

$\displaystyle \alpha_j = \alpha_j(t) = \frac{B_{j+1}(t+1)}{B_j(t+1)} = \frac{B_j(t)p_j}{B_j(t+1)}$.

I think I have to substitute $\displaystyle \alpha_0$ somewhere in $\displaystyle \alpha_j$ in a way that only $\displaystyle \alpha_0$, $\displaystyle p_j$ and $\displaystyle p_0$ remain as constants/variables.

Later I found that:

$\displaystyle \alpha_0 = \alpha_0(t) = \frac{B_1(t+1)}{B_0(t+1)} = \frac{B_1(1)}{B_0(1)} = \frac{B_0(0)p_0}{\sum^r_{i=0} B_i(0)f_i} = \alpha_0(0)$

And for $\displaystyle j \in \{1, ..., l - 1\}$:

$\displaystyle \alpha_j = \alpha_j(t) = \frac{B_{j+1}(t+1)}{B_j(t+1)} = \frac{B_{j+1}(1)}{B_j(1)} = \frac{B_j(0)p_j}{B_j(1)}$.

But I don't know what I can do with that.

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Anyone who can help me with this?

Thanks in advance.