Fergal,

MS3011 banter: I’ve seen it all now.

Regards.

]]>who is this loser? julia and mandlebrot sets arent even on the exam man

]]>To find an eventually periodic point (which is not itself period) you need to find points that are sent to fixed points which are not themselves fixed:

Solve for the fixed points ().

Solve to find points sent to fixed points.

The fixed points will always be solutions to this equation but the solutions which aren’t themselves are not periodic but are eventually periodic (namely eventually period-1 or fixed).

Proofs of non-existence can be tough. The reason is there is a possibly infinite amount of cases to check. If you believe that there are no eventually periodic points then your best bet is to use a proof by contradiction.

Example

Prove that for some fixed has no non-zero periodic points if and with .

Proof: Suppose on the contrary that such a non-zero periodic point exists, say of period .

Then . Now so we have

as .

Now and must equal one hence have an argument of the form for some :

which is a contradiction as .

Hence no such periodic point exists.

Now what is a period- point of the doubling mapping? Well a solution to

.

Let have binary representation . All the time when in talking binary the . chops the first digits off.

Hence

.

Hence recurring.

Eventually periodic points are points for which some iterate is periodic:

with periodic.

Now, with respect to the Doubling Mapping (whenever we are talking about the binary representation it will be in relation to the Doubling Mapping), eventually periodic points are just that

with periodic.

That is

a recurring binary representation. Hence the eventually periodic points have a finite number of arbitrary digits but then begin to recur. They have the form:

for some and .

I gave my best explanation of density of periodic points and a dense orbit in the review lecture but I’ll give a more technical explanation here.

A subset of a set is said to be dense in set if for for any point , there exists a point arbitrarily close to ; i.e.

and , such that

.

So to show that a dynamical system (a set together with a function ) has a dense orbit you must find a seed such that the orbit of ;

is dense in . That means given any , you must find an such that can be made arbitrarily close to for some .

To show that the periodic points are dense you must, for any , find a period point such that is arbitrarily close to .

We don’t need to convert fractions into binary but http://cs.furman.edu/digitaldomain/more/ch6/dec_frac_to_bin.htm .

I should have allowed ternary but I have said it will be the Logistic Map or the Doubling Map. But ternary is very similar to binary and if you can do binary ternary should be no problem at all.

Regards,

J.P.

Also, With regard to chaos theory, one needs to prove sensitivity on initial conditions, that it has a dense orbit and that it has a dense set of periodic points. The sensitivity part if fine, but I am not grasping the other two. Can you please expand it out clearly for me to understand?

How do you convert fractions into binary form?

Will ternary form be coming up on the test? I kind of understand how to expand something into binary form, but the notes have slightly confused me as regards ternary form. Can you please explain?

]]>Plain English? Don’t know what you are talking about!!

A dynamical system consists of a set and a(n) (iterator) function which maps elements of the set to other elements of the set.

We have to start somewhere and the seed is an element of the set which we usually denote by and is considered the initial state of the system.

The set of iterates of under is the orbit of :

.

Example

Let and be given by

.

Suppose we begin at . Thence:

Then the orbit of is given by:

.

Regards,

J.P.

Conor,

Chill out.

Regarding the Newton’s method question from last year I sent this to another student:

“Question 3 will be an unseen question. If you have a keen understanding of the terms dynamical system, iterator function, orbit and fixed point you will be rewarded. This question is to reward those who understand the material.

Past papers are useless for question 3 and as I have dubbed the question “unseen” it is very unlikely that Newton’s Method will be seen in question 3 (considering that I did the question in a lecture). It does not belong anywhere else in the paper neither.”

Julia sets and Mandlebrot sets are not examinable. We stopped at p.6 of the complex numbers section of the notes and instead worked on the board.

Regards,

J.P.

I’m one of the students doing your MS3011 test tomorrow and there’s been one quesiton that I can’t quite get. I don’t know how many times that I thought I was on the right track and then realised that i was miles off…

Its Q. 4 of the past papers, Summer 2011… All about Julia’s set and the madlebrot set… Help is URGENTLY needed J.P…. I’m lost… ]]>