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Given $$\alpha\{v(a + 1) - v(1) \} = \beta v(a)$$ Deduce $$v(a) = \frac{1-(\beta/\alpha)^m}{1-(\beta/\alpha)} v(1),$$ and derive $$v(a) = \frac{1-(\beta/\alpha)^a}{1-(\beta/\alpha)^{m+n}}.$$

I am new to difference equations, but I am following wiki.

$$v(a+1) = \frac{\beta}{\alpha}v(a) + v(1),$$ which has steady state value $$y^*= \frac{v(1)}{1-\frac{\beta}{\alpha}}.$$

Moving on from here I find the solution to the homogeneous problem as

$$x(a) = \bigg(\frac{\beta}{\alpha}\bigg)^av(1).$$

This is not correct. What am I doing wrong? Even if I find the first part I can not see how to progress to the second part ( the derivation part).

HMPtwo
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2 Answers2

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There seems to be some information missing regarding $m,n$. But we can make some progress without its knowledge. We start with the recurrence relation \begin{align*} \alpha\left(\nu(a+1)-\nu(1)\right)=\beta \nu(a)\qquad\qquad a\geq 1, \alpha,\beta \ne 0 \end{align*} We can write the recurrence relation as \begin{align*} \color{blue}{\nu(a+1)-\frac{\beta}{\alpha}\nu(a)=\nu(1)\qquad\qquad a\geq 1}\tag{1} \end{align*}

We use (1) to write down a chain of recurrence equations, each equation multiplied with a factor so that we can apply telescoping conveniently. We obtain \begin{align*} \color{blue}{\nu(a)}-\frac{\beta}{\alpha}\nu(a-1)&=\nu(1)\\ \frac{\beta}{\alpha}\,\nu(a-1)-\left(\frac{\beta}{\alpha}\right)^2\nu(a-2)&=\frac{\beta}{\alpha}\nu(1)\\ \left(\frac{\beta}{\alpha}\right)^2\nu(a-2)-\left(\frac{\beta}{\alpha}\right)^3\nu(a-3)&=\left(\frac{\beta}{\alpha}\right)^2\nu(1)\\ &\ \ \vdots\\ \left(\frac{\beta}{\alpha}\right)^{a-2}\nu(2)\color{blue}{-\left(\frac{\beta}{\alpha}\right)^{a-1}\nu(1)}&=\left(\frac{\beta}{\alpha}\right)^{a-2}\nu(1)\\ \end{align*} Summing up these lines all but the blue marked terms at the left side cancel and we obtain using the finite geometric series formula \begin{align*} \nu(a)-\left(\frac{\beta}{\alpha}\right)^{a-1}\nu(1)&=\nu(1)\sum_{k=0}^{a-2}\left(\frac{\beta}{\alpha}\right)^{k}\\ \color{blue}{\nu(a)}&=\nu(1)\sum_{k=0}^{a-1}\left(\frac{\beta}{\alpha}\right)^{k}\\ &\color{blue}{=\nu(1)\frac{1-\left(\frac{\beta}{\alpha}\right)^{a}}{1-\frac{\beta}{\alpha}}}\tag{2} \end{align*}

Setting $ \nu(1)=\frac{1-\frac{\beta}{\alpha}}{1-\left(\frac{\beta}{\alpha}\right)^{m+n}} $ we get from (2) OP's final expression \begin{align*} \color{blue}{\nu(a)=\frac{1-\left(\frac{\beta}{\alpha}\right)^{a}}{1-\left(\frac{\beta}{\alpha}\right)^{m+n}}} \end{align*}

Markus Scheuer
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  • Thanks for your solution. Could you have a look over mine to verify it please? – HMPtwo Oct 15 '22 at 18:11
  • @HMPtwo: You're welcome. If you want to prove the problem using induction you need to clearly state the base step, induction hypothesis, induction step and show the validity of the induction step. – Markus Scheuer Oct 15 '22 at 18:25
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There is no need to invoke difference equations. To obtain $$v(a) = \frac{1-(\beta/\alpha)^a}{1-(\beta/\alpha)} v(1),$$ we just use induction and $$v(a + 1) = \frac{\beta}{\alpha} v(a) + v(1).$$ Then we use the fact that $v(m+n) = 1$ (This was missing from my original post) to find $v(1)$ as $$\frac{1-\beta /\alpha}{1-(\beta /\alpha)^{m+n}}.$$

Plugging this in the equation for $v(a)$ at the top of this reply and we get the desired result.

HMPtwo
  • 433