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I have an iterated function (a 2D dynamical map) that I'm trying to find the derivative for. Here is an example:
$x_{t+1} = \rho - x_{t}^2 + \gamma y_{t}$
$y_{t+1} = x_{t}$

where the initial value of $y$ and $x$ are user defined, and $\rho , \lambda$ are small positive values.
I'm doing this to try to find fixed points, but my attempts with partial derivatives haven't seemed to work.

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    What kind of derivative are you trying to find? i.e. derivative of what wrt. what? – 2017-02-04

1 Answers 1

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You are considering the dynamical system $$ (x,y)\mapsto (\rho-x^2+\gamma y,x). $$

To find the fixed points, simply set $(x,y)$ equal to $(\rho-x^2+\gamma y,x)$. Then $x=y$ and $x$ is a root of the quadratic equation $$ x^2+(1-\gamma)x-\rho=0. $$ By the quadratic formula, the fixed points are therefore $$ \left(\frac{\gamma-1\pm\sqrt{(\gamma-1)^2+4\rho}}{2},\frac{\gamma-1\pm\sqrt{(\gamma-1)^2+4\rho}}{2}\right). $$ You mentioned that $\gamma$ and $\rho$ are small constants. We can approximate the square root using a Taylor expansion to get that $\sqrt{(\gamma-1)^2+4\rho}\approx 1-\gamma+2\rho$, to first order. Thus, the two roots are roughly $$ (\rho,\rho)\text{ and }(\gamma-1-\rho,\gamma-1-\rho). $$

Old response, prior to clarification: Since the only variable is $t$, I am not sure which partial derivatives you are referring to. In any case, by the chain rule $$ x'(t+1)=-2x(t)x'(t)+\gamma \ y'(t),\qquad y'(t+1)=x'(t). $$

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    I fixed the formatting to be clearer. I think it is more accurate to view $x_t$ as the x value at time t, than view x as a function. But maybe I'm on the wrong track, I'll try your solution. – 2017-02-04