Bonus: Come up with a general statement that extends to any arithmetical function that has a sufficiently "nice" asymptotic approximation.
Here is one instructive way to do it (probably not the fastest way):
First choose more natural coordinates, u_n=x_n-1, v_n=y_n-2 centred at the fixed point.
Our recurrence then transforms to:
The matrix in question is symmetric, which implies that the eigenvalues are real and the eigenspaces are orthogonal. (An easy computation shows that there are indeed two distinct eigenvalues.)
A consequence of this fact is that these matrix satisfies
(Exercise).
The maximal (maximal means maximal in absolute value btw) eigenvalue of the n-th matrix is
and so convergence of (u_n,v_n) to 0 is proven by showing that the product of these maximal eigenvalues tends to zero, which I am sure you are capable of.
(*) Note that this method establishes convergence for ANY choice of initial point (x_0,y_0). Note also that this is quite related to my most recently posted problem in the linear algebra marathon, which addresses sequences like this but for a fixed multiplication matrix (which is not specified and need not be symmetric).
Last edited by seanieg89; 30 Nov 2016 at 1:28 PM.
Bonus: Come up with a general statement that extends to any arithmetical function that has a sufficiently "nice" asymptotic approximation.
Last edited by Paradoxica; 6 May 2017 at 10:30 PM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
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