- #1
rmiller70015
- 110
- 1
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.
Last edited: