# Pair of two-variable polynomial equations of high order

I have the following pair of equations to be solved for two variables $\rho$ and $D$ resulting from a certain Maximum Likelihood Estimation for a time series $X_n > 0$, $n=0, \ldots, N+1$ with $N \ge 1$:

$$\sum\limits_{n=0}^N \frac{1}{\rho + D X_n^2} = \sum\limits_{n=0}^N \frac{(X_{n+1}-X_n)^2}{(\rho + D X_n^2)^2}$$

$$\sum\limits_{n=0}^N \frac{X_n^2}{\rho + D X_n^2} = \sum\limits_{n=0}^N \frac{X_n^2(X_{n+1}-X_n)^2}{(\rho + D X_n^2)^2}$$

Specifically I'm interested in a positive solution, which should be unique (I think)...

The problem is apparently equivalent to solving a pair of polynomial equations of order $2N-1$, but I was hoping there might be a "trick" or a way to obtain an analytical solution due to its form? I've tried some changes of variables and integral transforms, but with no luck... I'm sure there's a smarter way to analyze the equations, which I will do, but I'm hoping the smartest way is to ask here ;)

One random observation is that $\partial_D(\text{first eq}) = \partial_\rho(\text{secondeq})$. I have absolutely no idea if this is useful.

Also, what would be the best scheme for solving equations such as this one numerically? I use mostly Python and the tools in here will probably be helpful, but I'd appreciate some further tips! Maybe an OLS scheme for the equations?

EDIT:I should probably mention that setting $D=0$ and forgetting about the second equation gives the solution $\rho = \frac{1} {N+1} \sum (X_{n+1}-X_n)^2$.

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Should the sum on the right end at n=N-1? Or is $X_{N+1}=X_0$? –  Michael Renardy Jul 23 '13 at 15:06
Oops.. fixed the series definition (although the boundary condition would make sense too). –  H. Arponen Jul 23 '13 at 16:55
Letting $N=0$, it seems that $\rho+DX_0^2=(X_1-X_0)^2$, which does not have a unique positive solution for $\rho$ and $D$. –  Suvrit Jul 23 '13 at 17:30
Oops 2... $N$ needs to be larger than zero (edited the question)! –  H. Arponen Jul 23 '13 at 18:12
$\rho=+\infty$, $D=0$ "solves" the system :-) –  Suvrit Jul 23 '13 at 19:11