# what are they asking? logstic iterative

• Apr 6th 2011, 03:12 PM
jwxie
what are they asking? logstic iterative
The problem is in the attachment.
Attachment 21386

I have to do this in Matlab and I did. The previous problem gave me the logistic iterative equation

y(k+1) = r(1-y(k))y(k)

i need help with understanding the two questions posted. What are they asking?

Can someone clarify the questions for me, please? Thank you!

ps: I don't know where to post, but I guess the form of this iterative equation is difference equation, which is at least college level. So excuse me for posting here.
• Apr 6th 2011, 04:40 PM
TKHunny
Iterative is just iterative. Please look up the word. You get the next value from the current value (or some combination of the current value and some previous values). You obtained the curent value from the previus value. (or some combination of previous values.)

Not everyone opens attachments, even if they are innocent-looking "png"s. Learn to communicate through written media.
• Apr 6th 2011, 04:47 PM
jwxie
Hi Hunny thanks for the reply

Quote:

Find the iterates of the logistic equation for the logistic equation for the following values of r: { 3.1, 3.236068, 3.3, 3.498561699, 3.566667, and 3.569946}, assuming the following three initial conditions:
y(1) = 0.2, y(1) = 0.5, and y(1) = 0.7
In particular, specify for each case:

a. The period of the orbit for large N, and the values of each of the iterates.
b. Whether the orbit is superstable (i.e, the periodicity is present for all values of N).
I understand what iterative means, and as I stated I did the Matlab (which implies I did the iteration).

I am troubled by question a and b. I have no idea what they mean by period of the orbit, and thus I cannot answer b either.
Now that I have written out the problem, can anyone please guide me through question a and b?

How would I go about answering them?

Thank you very much.
• Apr 6th 2011, 05:14 PM
TKHunny
The "Period of the Orbit" is a technical term usually defined in terms of the iterative function. If the function is $f(y)$, then $f^{n}(y)$, is the function iterated n times. So the Period of the Orbit is the least value of m for which we have $f^{n}(y) = f^{n+m}(y)$, or in other words, how many iterations does it take before we quit getting something new and we are just repeating ourselves.

For your first example, r = 3.1 and y(1) = 0.2, we see that after only 3 or 4 iterations, we're esentially just jumping back and forth between two values. It's not exact, but it looks like m = 2, at least vaguely.

Trying what looks like the worst one to me, r = 3.498561699 and y(1) = 0.7. It looks like a nice m = 4. Around iteration 30, it becomes quite stable.
• Apr 7th 2011, 08:23 AM
jwxie
Hi, Thank you very much!

As you said the orbit is the least mth iteration before we see the repeation.

So if I hit

0.8746 0.31 0.8746
I would take the mth iteration that generates the first 0.8746 to be the orbit?
• Apr 9th 2011, 06:23 AM
TKHunny
Maybe. Four decimal places does not equivalence make. I'd try a few orbits before I believed it.