Can someone help me with what this means?

So essentially, you choose a number between 0-1 and use the answer key to then substitute that number into: 3.2 x ANS x (1 - ANS)

And you keep pressing =
Eventually you alternate between 0.513 and 0.799
But if you used algebra and tried to make the input the same as the output (ie x = 3.2x (1-x)) you get the answer as being 0.6875, which is a ‘fixed point’.

Could someone explain what is meant by a fixed point please?

Also, what does this actually mean? What does it mean for there to be disparity between what the algebra says the answer should be and what you get experimentally?

It’s supposed to be linked to the chaos theory if that helps. Thanks in advance!

Its a "tent map", so what have you covered/what is the full question? Not checked your numbers, but I presume the fixed point from algebra is unstable so any small deviation will cause the iterations to diverge from it, not converge back to it, and keying the answer into your calculator would mean that numerical precision would introduce an error which would (locally) grow expoentially.

A fixed point should mean that when you put a number in, you get the same number out. So if a pencil could balance perfectly, the upright angle would be a fixed point. However, if its unstable (like a balanced pencil) small errors will soon grow (exponentially).

It wasn’t a question as such but rather a taster of uni level maths so I’m unsure about it. I was told it’s not iteration but something similar.

I was wondering what it meant for the answer given by algebra to differ from the one reached experimentally?

Numerical precision is important when you have (locally) unstable systems.

A physical (simple) example of the above would be a pencil standing on end so its in an unstable equilibrium. A bit either side are two springs so that when the pencil hits them it bounces without losing energy. So a small disturbance would cause the pencil to topple and oscillate between the two springs. In theory, the pencil would stand upright. In practice, small (numerical) errors / disturbances will cause a very different behaviour.

Great, thanks for the explanation!

Great, thanks for the explanation!

And with a bit more thought, if you put the twice iterated function (a quartic) on there so
3.2*3.2*x*(1-x)*(1 - 3.2*x*(1-x))
you get
https://www.desmos.com/calculator/ypwxrvlxiw
So it has 3 fixed points of interest - the original one and the two ones which form a stable period 2 orbit of the original map. The tangents' gradients are < 1 which means theyre stable unlike the original fixed point where the gradient is >1 which means its unstable.

Obv 0 is an unstable fixed point as well

Thanks so much, it’s great to read even if it sort of flies well above my head (just started yr12) but I really appreciate it!!

Thought youd done it at a level when you referred to chaos. The fixed point idea
x = f(x)
should have been introduced at gcse so what an iterative sequence (may) converge to
https://thirdspacelearning.com/gcse-maths/algebra/iteration-maths/
and if a sequence has period of length two so it goes (tends to) a .-> b -> a -> b .-> ... then the chained function f(f(x)) must have two fixed points a and b corresponding to solutions of
x = f(f(x))

For both cases the line y=x represents the left hand side and the two curves are f(x) and f(f(x)) and when they intersect they give the solutions. Whether it converges or not is a bit beyond gcse, but its not that hard. When the tangent at the fixed point has |gradient| < 1 or > 1, it should be easy to trace a few iterations on the graph and note it locally converges to the fixed point for the former and locally diverges from the fixed point for the latter. Thats enough.

Some other bits of information.

Maps of the form $\lambda x(1-x)$ are called Logistic maps. (In this case $\lambda = 3.2$ obviously). The wiki link has a fair bit of discussion about connections with Chaos theory.

Also, if you have an interation $x_{n+1}=f(x_n)$ where f is a differentiable function, then:

If f(p)=p is a fixed point, it will be stable (i.e. if you start close to p you get closer as you iterate) if f'(p)<1 and unstable if f'(p) > 1.

I did iteration but wasn’t told exactly what it means/does. Do you have any suggestions as to where I could look into this in more depth?