Exploring Oscillatory Behavior of Sinusoidal Functions

[rmiller70015]

Homework Statement

This isn't really part of my homework, my homework was to draw a pretty graph, but I am curious about some behavior.
I was given a picture of a sinusoidal function. I found it was $2sin(\frac{\pi}{3}t-\frac{\pi}{6}) + 6$. Then I used trig identities to get $\sqrt{3}sin(\frac{\pi}{3}t) - cos(\frac{\pi}{3}t) +6$ and set it equal to $t$ to get $\sqrt{3}sin(\frac{\pi}{3}t) - cos(\frac{\pi}{3}t) - t =-6$. I then plotted $\sqrt{3}sin(\frac{\pi}{3}t) - cos(\frac{\pi}{3}t) - t = 0$ and $y = -6$ and found the intercepts to get my equilibrium points of 4.392, 7, and 8.

This is where I am having issues. The web diagram's behavior around the first equilibrium is sensitive to initial conditions. Graphically I found that when t = 4, 5, or 6, the diagram will oscillate around the fixed point indefinitely. I'm having trouble explaining why this is.

Homework Equations

The Attempt at a Solution

I've tried starting off by saying the oscillation condition is $t_n = t_{n+2}$ and $t_{n+1} = t_{n+3}$.

I think that the updating formula is $t_{n+1} = (1-n)t_n + (1-n)t_n[2sin(\frac{\pi}{3}t - \frac{\pi}{6})+6]$

Then,
$t_{n+1} = (1-n)t_n[7+2sin(\frac{\pi}{3}t - \frac{\pi}{6})] = t_{n+3} = (1-(n+2))t_{n+2}[7+2sin(\frac{\pi}{3}t-\frac{\pi}{6})]$
Division gives:
$\frac{t_n}{t_{n+2}} = \frac{-(1+n)t_{n+2}[7+2sin(\frac{\pi}{3}t - \frac{\pi}{6})]}{(1-n)t_n[7 + 2sin(\frac{\pi}{3}t - \frac{\pi}{6})]}$
But this ends up telling me that 1 = -1, so, I'm not sure what to do from here or if this was the correct way to do things. I need to find some expression that allows the oscillation conditions to be met.

Edit: Upon closer inspection it appears that this oscillatory behavior occurs at the inflection points and half way between the inflection points.

 

[FactChecker]

I can't understand. If you fix t=4, then what is "oscillating"? And what initial conditions are you referring to?