Iterative Equations in Wolfram Alpha

Question

I am trying to solve an iterated equation, but it is quite messy and I am having trouble. Here are my equations:

$$ \mu_{t+1} = \frac{h_t \mu_t + h_\epsilon Z_t}{h_t + h_\epsilon} $$

and

$$h_{t+1} = \frac{(h_t + h_\epsilon) h_\delta}{(h_t + h_\epsilon + h_\delta)} $$

$h_0$, $m_0$, $h_\epsilon$, $h_\delta$ and $Z_t$ are all given. I would like to solve for $m_t$ in terms of the givens.

Is it possible to solve this equation in Wolfram Alpha? If not, is there a different software I could use to solve it? If both those fail, does anyone have ideas as to how I might solve it by hand? I've been struggling with it for some time, but it gets very messy very quickly!

Answer 1

I think this will be useless...
The second equation does not involve $\mu_t$ at all, so we solve only for $h_t$. We must find the $t$ power of a certain $2 \times 2$ matrix.
Result (from Maple): $$ h_t = \frac{A_t h_0+ B_t}{C_t h_0 + D_t} $$ Where $$ A_t = \left( \sqrt {h_\epsilon\, \left( 4\,h_\delta+h_\epsilon \right) }h_0+h_\epsilon\, \left( h_0-2\,h_\delta \right) \right) \left( { \frac {2\,h_\delta+h_\epsilon-\sqrt {h_\epsilon\, \left( 4\,h_\delta+h_\epsilon \right) }}{h_\delta}} \right) ^{t}\\ B_t = \left( \sqrt {h_\epsilon\, \left( 4\,h_\delta+h_\epsilon \right) }h_0-h_\epsilon\, \left( h_0-2\,h_\delta \right) \right) \left( {\frac {2\,h_\delta+h_\epsilon+ \sqrt {h_\epsilon\, \left( 4\,h_\delta+h_\epsilon \right) }}{h_\delta}} \right) ^{t} \\ C_t = \left( \sqrt {h_\epsilon\, \left( 4\,h_\delta+h_\epsilon \right) }-2\,h_0-h_\epsilon \right) \left( {\frac {2\,h_\delta+h_\epsilon-\sqrt {h_\epsilon\, \left( 4\,h_\delta+h_\epsilon \right) }}{h_\delta}} \right) ^{ t} \\D_t= \left( \sqrt {h_\epsilon\, \left( 4\,h_\delta+h_\epsilon \right)}+2h_0+h_\epsilon \right) \left( {\frac {2\,h_\delta+h_\epsilon+\sqrt {h_\epsilon\, \left( 4\,h_\delta+h_\epsilon \right) }}{h_\delta}} \right) ^{t} $$

Answer 2

Your second equation is of the form :

$$h_{n+1}=\dfrac{ah_n+b}{ch_n+d}\tag{1}$$

Therefore there is an explicit solution (see below in (2)) of $h_n$ as a fonction of $h_0$ and $n$.

Having this expression, it remains to plug it into the first equation.

Explanation for the explicit solution :

There is a formula one can find as "rational difference equation" in (https://en.wikipedia.org/wiki/Iterated_function) :

$$h_n=\frac{a}{c}+\frac {bc-ad}{c}\left[\dfrac{A\alpha^{n-1}-B\beta^{n-1}}{A\alpha^{n}-B\beta^{n}}\right]\tag{2}$$

with

$$\alpha =\dfrac {a+d+{\sqrt {(a-d)^{2}+4bc}}}{2} \ \text{and} \ \beta =\dfrac {a+d-{\sqrt {(a-d)^{2}+4bc}}}{2}$$

$$A=c \; h_0-a+\alpha \ \ \ \text{and} \ \ \ B=c \; h_0-a+\beta$$

(I thought yesterday that there is a sign error in the numerator ; but this is not the case).

Remarks :

1) Important : the formulas assume that $\Delta=(a-d)^2+4bc \geq 0$ which is the case for you ($\Delta=h_{\delta}^2>0$). Usually, sequence $h_n$ converges to a fixed value which is one of the roots of quadratic "fixed point" equation $H=(aH+b)/(cH+d).$

2) one can give a nice linear algebra explanation for (2) where $\alpha$ and $\beta$ upcome as eigenvalues of matrix $\begin{pmatrix} a&b\\c&d \end{pmatrix}$.