Optimization of parameter for recursive Cauchy sequence I have the following recursive sequence I'm analyzing:
$$V_0 = 50, V_1 = (1-10k)V_0,$$
$$V_{n+1} = (1-10k)V_n - 5kV_{n-1}$$
where $k > 0$ is a parameter that I'm investigating by running simulations in R.
(This is from the "docking problem" in Meerschaert's "Mathematical Modeling").
$\langle V \rangle$ isn't Cauchy for all $k$. The sequence seems to diverge if $k > 0.18$ or so.
My analysis is to examine how fast the sequence converges as $k$ varies. I did this by simulating the sequence and picking the first $n$ for which $|V_n| < 0.1$
Here's a plot of said $n$ as a function of $k$. The horizontal axis are various $k$'s, and the vertical axis is the first $n$ for which $|V_n| < 0.1$ with that $k$ as the parameter.


My question is: What is this function? Why does it behave the way it does?
How do I find local minima of this function other than by simulation?
And what is the exact $k$ at which the sequence $\langle V \rangle$ stops being Cauchy?

Do you remember:
Given
$${x^2} + p \cdot x + q = 0$$
and
$$D = {p^2} - 4q$$
we have the roots
$$x = \left\{ {\begin{array}{*{20}{c}}
  {\frac{{ - p - \sqrt D }}{2}} \\ 
  {\frac{{ - p + \sqrt D }}{2}} 
\end{array}} \right.$$
As an application, we consider Fibbonacci-numbers:
$$\begin{gathered}
  {f_0} = 1,{f_1} = 1 \hfill \\
  {f_n} = {f_{n - 1}} + {f_{n - 2}} \hfill \\ 
\end{gathered} $$
and
$${f_n} = {f_{n - 1}} + {f_{n - 2}} \Leftrightarrow {f_n} - {f_{n - 1}} - {f_{n - 2}} = 0$$
We build qudratic
$${x^2} = x + 1 \Leftrightarrow {x^2} - x - 1 = 0$$
with
$$p =  - 1,q =  - 1,D = {p^2} - 4 \cdot q = 1 + 4 = 5$$
and get roots
$$x = \left\{ {\begin{array}{*{20}{c}}
  {\frac{{1 - \sqrt D }}{2}} \\ 
  {\frac{{1 + \sqrt D }}{2}} 
\end{array}} \right.$$
With these roots we can write:
$${f_n} = \frac{1}{{\sqrt D }} \cdot \left( {{{\left( {\frac{{1 + \sqrt D }}{2}} \right)}^{n + 1}} - {{\left( {\frac{{1 - \sqrt D }}{2}} \right)}^{n + 1}}} \right) = \frac{1}{{\sqrt 5 }} \cdot \left( {{{\left( {\frac{{1 + \sqrt 5 }}{2}} \right)}^{n + 1}} - {{\left( {\frac{{1 - \sqrt 5 }}{2}} \right)}^{n + 1}}} \right)$$
We resolved the reccurence!
We are now going to apply this to your problem. Steps are exactly the same.
Given:
$$\begin{gathered}
  A(k) = 1 - 10 \cdot k,{V_0} = P \hfill \\
  {V_1}(k) = A(k) \cdot {V_0} \hfill \\
  {V_n}(k) = A(k) \cdot {V_{n - 1}}(k) - 5 \cdot k \cdot {V_{n - 2}}(k) \hfill \\ 
\end{gathered} $$
Then
$${V_n}(k) - A(k) \cdot {V_{n - 1}}(k) + 5 \cdot k \cdot {V_{n - 2}}(k) = 0$$
with quadratic
$${x^2} - A(k) \cdot x + 5 \cdot k = 0$$
Here
$$p =  - A(k),q = 5 \cdot k,D(k) = A{(k)^2} - 20 \cdot k$$
With the roots
$$x = \left\{ {\begin{array}{*{20}{c}}
  {\frac{{A(k) - \sqrt D }}{2}} \\ 
  {\frac{{A(k) + \sqrt D }}{2}} 
\end{array}} \right.$$
Now we set, like before:
$${p_k}(n) = \frac{1}{{\sqrt {D(k)} }}\left( {{{\left( {\frac{{A(k) + \sqrt {D(k)} }}{2}} \right)}^{n + 1}} - {{\left( {\frac{{A(k) - \sqrt {D(k)} }}{2}} \right)}^{n + 1}}} \right)$$
and we have got:
$${V_n}(k) = {p_k}(n) \cdot {V_0}$$
We resolved the reccurence. So far.
What to with this? Perfect answer is given by @Alex Ravsky!
