Path between fixed points in a logistic map

[Decimal]

Hello,

I have a question about period doubling and fixed points in the logistic map. Let's say I have a basic logistic map, $f(x) = 4\lambda x(1-x)$. Let me then compare 1,2 and 4 iterations of this map on fixed points. I assume that $\lambda$ is large enough such that two period doublings have occurred, and a 4-cycle exists.

Now if $\bar{x}$ represents a fixed point for $f^4(x)$ (4 iterations of the map), then we have by definition, $$f(\bar{x}) = \bar{x}.$$ Obviously from this it follows that applying the function $f^2(\bar{x})$ twice should also take one back to point $\bar{x}$, however this is where my confusion comes in. We are in a 4 cycle, so 4 different fixed points relative to $f^4$ exist. Let's call the fixed points closest to $\bar{x}$ (nearest neighbour) $\bar{x}'$.

My lecture notes state that, $f^2(\bar{x}) = \bar{x}'$. In other words, applying the map twice to a fixed point always takes one to the nearest neighbour. It is stated that this can also be extended to the general n-cycle case, where one would have, $\bar{x}' = f^{2^{n-1}}(\bar{x})$.

I do not understand why these properties are true. There are 4 different fixed points, and I don't see why the path through these different points has to follow this form, since as I understand it all these points are practically equivalent. Why if I apply $f^2$ can't I go to a different point further away?

I suppose there is some connection between the nearest neighbours $\bar{x}$ and $\bar{x}'$, but I am just not seeing it. Any help would be greatly appreciated!

 

[Mark44]

Hello,

I have a question about period doubling and fixed points in the logistic map. Let's say I have a basic logistic map, $f(x) = 4\lambda x(1-x)$. Let me then compare 1,2 and 4 iterations of this map on fixed points. I assume that $\lambda$ is large enough such that two period doublings have occurred, and a 4-cycle exists.

Now if $\bar{x}$ represents a fixed point for $f^4(x)$ (4 iterations of the map), then we have by definition, $$f(\bar{x}) = \bar{x}.$$

Wouldn't $\bar x$ be a fixed point for $f^4$? If $\bar x$ is a fixed point for $f^4$, then it should be the case that $f^4(\bar x) = \bar x$.

Obviously from this it follows that applying the function $f^2(\bar{x})$ twice should also take one back to point $\bar{x}$, however this is where my confusion comes in. We are in a 4 cycle, so 4 different fixed points relative to $f^4$ exist. Let's call the fixed points closest to $\bar{x}$ (nearest neighbour) $\bar{x}'$.

My lecture notes state that, $f^2(\bar{x}) = \bar{x}'$. In other words, applying the map twice to a fixed point always takes one to the nearest neighbour. It is stated that this can also be extended to the general n-cycle case, where one would have, $\bar{x}' = f^{2^{n-1}}(\bar{x})$.

I do not understand why these properties are true. There are 4 different fixed points, and I don't see why the path through these different points has to follow this form, since as I understand it all these points are practically equivalent. Why if I apply $f^2$ can't I go to a different point further away?

I suppose there is some connection between the nearest neighbours $\bar{x}$ and $\bar{x}'$, but I am just not seeing it. Any help would be greatly appreciated!