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!

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!

Original post by subbhy

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!

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).

(edited 8 months ago)

Original post by mqb2766

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).

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?

Original post by subbhy

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?

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.

Original post by mqb2766

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.

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!

Original post by subbhy

Great, thanks for the explanation!

Starting from 0.6, you get for a few iterations

https://www.desmos.com/calculator/ip2mafjgxr

Keep going if you want ...

Obviously the fixed point is when the line y=x intersects the parabola. Note the tent map ref in the first reply is really for a "/\" type function rather than a "n" parabola, which has a bit different dynamics.

(edited 8 months ago)

Original post by subbhy

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

Original post by mqb2766

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

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!!

Original post by subbhy

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.

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.

Original post by mqb2766

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.

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.

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?

Original post by DFranklin

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.

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.

Thanks!

Original post by subbhy

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?

The short answer is that it depends on what the iteration is referring to as it crops up in many different parts of maths. In the context of the OP (chaos, dynamical systems) you could start with the logistic map (the OP)

https://plus.maths.org/content/maths-minute-logistic-map

https://geoffboeing.com/2015/03/chaos-theory-logistic-map/

https://www.google.co.uk/books/edition/Chaos/LDvhBwAAQBAJ?hl=en (chapter 1 in particular)

and as above code up/visualise what happens as the 3.2 parameter varies from 0 to 4 and see if you can understand it (where do fixed points occur, how many are there, are the stable, ...). Just google for other introductions.

There are many (popular/good) maths books about how this leads to chaos

https://www.shortform.com/best-books/genre/best-chaos-theory-books-of-all-time

and it would be good for you just to browse a few and see which you find interesting. Stewarts is old but good, Gleick is newer, ...

"iteration" in terms of discrete dynamics really starts with a simple geometric sequence so a linear recurrence relationship

x_{n} = r*x_{n-1)

which obviously has a closed form "exponential" solution

x_{n} = ar^n

The initial value x_0=a=0 is the only fixed point as 0->0, as its the only solution to (assuming r!=1 which gives a trivial relationship)

x = rx

and any perturbation from this will grow exponentially (r>1) as the closed form solution shows. So if |r|>1, the fixed point at 0 is unstable whereas if |r|<1 then the fixed point at 0 is stable as the sequence asymptotically tends to 0 for an initial perturbation.

Youve just looked at a nonlinear (quadratic) version fo the basic recurrence relation so

x_{n} = r*x_{n-1} (1 - x_{n-1})

and immediately, for different values of r, there is no closed form solution and some values of r (around 4) lead to chaotic (complex, unpredictable dynamics). You could expand it as

x_{n} = r*x_{n-1} - r*x_{n-1}^2

so for small values of x where the squared term on the right will be very small, this is just the previous geometric/exponential growth. However, as x gets larger, the negative squared term kicks in and would dominate. The period 2, ... stable solutions represent the interacting dynamics between the growth term rx and the decay term -rx^2 and this depends on the value of r.

A simple introduction with a few examples to discrete dynamics is

https://mathinsight.org/discrete_dynamical_system_introduction

though as above there are many others.

(edited 8 months ago)

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