How to solve $n$ from $c \leq 1.618^{n+1} -(-0.618)^{n+1}$

Question

I need to solve the bound for $n$ from this inequality:

$$c \leq 1.618^{n+1} -(-0.618)^{n+1},$$

where $c$ is some known constant value. How can I solve this? At first I was going to take the logarithm, but the difference of the two exponentials trouble me...

Any hints? :) Thnx for any help !

Answer 1

To solve $$c=\phi^n-(-\phi)^{-n}$$ If $n$ is even, then $$c=\phi^n-\phi^{-n}\\(\phi^n)^2-c(\phi^n)-1=0$$ and you can solve a quadratic for $\phi^n$ as a function of $c$. Similar if $n$ is odd.

Answer 2

Hints:

First note that $$\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618$$ and $$-\frac1\phi = \frac{1 - \sqrt{5}}{2} \approx -0.618$$ both are the roots of $x^2 - x - 1 = 0$.

Note that $$ \lim_{n \to \infty}(-0.618)^{n+1} = 0 $$ as $|-0.618| < 1$.

Now note the $n$ element in the Fibonachi sequence is $$ F_n = \frac{1}{\sqrt5} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right) $$

Answer 3

Here is the answer I got by using the hints given to me:

First I select $c = \frac{\sqrt{5}}{0.05}$, so my equation becomes:

$$\frac{\sqrt{5}}{0.05} =1.618^{n+1} - (-0.618)^{n+1}$$

I set $\phi = 1.618$ and $\displaystyle -\frac{1}{\phi} = -0.618$ and I get

$$\frac{\sqrt{5}}{0.05} = \phi^{n+1} - (-\phi)^{-(n+1)}.$$

Now I consider two cases: $n$ is odd or even. I consider the case $n$ is odd and I get:

$$\frac{\sqrt{5}}{0.05} = \phi^{n+1} - \phi^{-(n+1)},$$

and from this I get:

$$(\phi^{n+1})^2 - \frac{\sqrt{5}}{0.05}\phi^{n+1}-1 = 0$$

So I get:

$$\phi^{n+1} = \frac{\frac{\sqrt{5}}{0.05}\pm \sqrt{\frac{5}{0.05^2}+4}}{2}$$

$$\phi^{n+1} \approx 44.7437, -0.0223$$

from here I solve:

$$n = \frac{\ln(44.7437)}{\ln(\phi)}-1 = \frac{\ln(44.7437)}{\ln(1.618)}-1 \approx 6.89 $$

The other possibility evaluates into a complex number so I discard it, because I need a real valued answer. I do similarly for the $n$ is even case.