Logistic Sequence?

Question

i have a problem to solve for my math class that's related to the logistic equation. I've been unable to come up with a solution:

can you find a number Y(1) such that
Y(1)=Y(3)=Y(5)=...=Y(2n+1)=...
with Y(n+1)=4Y(n)(1-Y(n))?

thanks for the help!

Answer 1

You have

y2 = 4y1(1 - y1)
y3 = 4y2(1 - y2)
= 4[4y1(1 - y1)](1 - [4y1(1 - y1)])
= 16y1(1 - 5y1 + 8(y1)² - 4(y1)³)

SInce you want y3 = y1, we must have

y1 = 16y1(1 - 5y1 + 8(y1)² - 4(y1)³)

So y1 = 0 or, dividing both sides of the above by y1,

1 = 16(1 - 5y1 + 8(y1)² - 4(y1)³)

or

64(y1)³ - 128(y1)² + 80y1 - 15 = 0

This equation has three solutions: y1 = 3/4, y1 = (5 + √5)/8, and y1 = (5 - √5)/8

If y1 = 3/4, then y2 = 3/4 = y3 = y4 = ...

If y1 = (5 + √5)/8, then y2 = (5 - √5)/8, y3 = (5 + √5)/8, y4 = (5 - √5)/8, etc.

If y1 = (5 - √5)/8, then y2 = (5 + √5)/8, y3 = (5 - √5)/8, y4 = (5 + √5)/8, etc.

All of these are numerically unstable. That is, if you alter one of the solutions for y1 by a slight amount (or can't represent it exactly in your calculator or computer), y3, y5, etc. will move away from that value of y1 and you will get chaos.