1. In MS3011, given a function , it is points that we call periodic rather than the function. We call a period--point if

.

For example fixed points are period-one points for they satisfy

A point is an eventually periodic point if some iterate of is a periodic point. For example, if two has the following orbit under some :

,

then we say that two is eventually periodic. Note that two itself is NOT periodic.

If you are to get a question like in Q.24 or Q.25 you will be asked to find an eventually fixed point that is not fixed. What you need to do in this case is find all of the fixed points of by solving . Then take one fixed point, say , and find all of the points that are sent to by solving

.

As it is fixed, itself is eventually fixed — but it is fixed also and we want to find an eventually fixed point that is NOT fixed.

Hence you will hope to solve this equation and find that one of the solutions is not fixed and this will be your answer.

2. This is bookwork from the lectures and is in the notes too.

3. It depends. If you put your complex number at around 20 on the unit circle then will have angle 40 and will have angle 60 so all these points are in the first quadrant. Obviously the situation will be different if you place around 80.

To plot the points… look up how to plot roots of unity.

Regards,

J.P.

1. How do I go about finding an eventually periodic point of a function which is not periodic?

These come up on part (c) and part (d) of Q.24 and Q.25 respectively.

2. Would it be possible for you to give me the solutions for autumn 2013 Q.3 part (d) and (e)?

I’m having trouble understanding the whole concept of density in relation to this topic.

3. Finally just a small thing in relation to complex numbers. When plotting points on an Argand diagram, for Q. 4 (a) of the summer paper what quadrants are and in and how does this change when our expression changes to ?

And for part (d) I’m left with four points. How do I plot these? Just a bit confused

Any help at all on the above would be greatly appreciated.

]]>Firstly if is a fixed point, then is also period-two, three, four, five, etc (if a point is period- then it is also period- for any .).

Secondly, if is prime-period-two, this means that is period-two and this is the lowest period; i.e. is prime-period-two means that it is period-two but not period-one. Similarly a point is prime-period-three if it is period-three but not period-two or period-one.

Therefore , while it is period-two, is not prime-period-two because it is also period-one: it is prime-period-one. This means that you can’t use the fixed points because they are not prime-period-two.

In fact the result does not hold when is a fixed point. For suppose that is fixed, that is period-one. Therefore is period- for any . In particular, is period-eight, however it is also period-nine.

Here are three perfectly good solutions to the problem.

1. Once you find the (complex) prime-period-two points, let one of them be . Now because is period-two it is also period-eight. Suppose for the sake of contradiction that is also period-nine. That is

,

but so we this implies

,

i.e. is a fixed, period-one point. But this contradicts the fact that is prime-period-two so we cannot have that is period-nine.

2. A slicker proof uses the fact that a point is period- if and only if is a multiple of the prime period. Now eight is a multiple of the prime-period, two, but nine is not. Therefore is period-eight, but not period-nine.

3. Another proof goes as follows. Denote one of the prime-period-two points by and the other by . We have that all of the points in a periodic orbit of period- are also period- so if w consider the orbit of :

,

then must be period-two. Clearly cannot be a fixed point therefore it must be prime-period-two and the only other prime-period-two point is so we have .

Therefore the orbit of is:

so clearly is period-eight but not period-nine.

Regards,

J.P.

Sorry for the last minute question, I just want to clear something up. If is a fixed point (i.e. period one), and is a prime period two point. Is also prime period two or just period two.

This question is coming from the 2013 paper Q.1 (a) (iii). The prime period two points are complex whereas the fixed points are much easier to use. Can I use a fixed point to represent a prime period two point because I feel like I can’t.

Regards.

]]>Don’t worry about part (d), I won’t ask that in your exam (although I address it here https://jpmccarthymaths.com/2013/03/27/ms3011-week-12/#comment-666).

The rest is bookwork, covered in the notes.

Q. 43 (a) is covered in great detail on P. 3 of the pdf notes. However a more compact answer would be as follows.

“Let be the the proportion of a maximum population. Suppose that we want for small populations that the growth rate is approximately geometric.

Also we want extinction to ensue whenever the maximum population is reached; i.e. . The model achieves this.

For as required.

Also when , we have so that, , as required.”

Q. 43 (b) and (c) are covered on the bottom of p.26/ top of p.27.

Regards,

J.P.

Is it possible to get the solution please to Q. 43 in the exercise set, particularly part (d)? It concerns the logistical mapping.

Kind regards.

]]>Hi JP, just wondering where the -1 comes from at the end of the last line.

thanks

*This answer refers to Q.62 of the exercises.*

From what you sent me Q.1 (a) is pretty good but explicitly show that if has argument then has argument , i.e. by annotating the graph.

Q.1 (b) is pretty sweet and (c) is fine.

Q. 1 (d)… your picture is incorrect. First of all is a fixed point and that is found at the origin. Now is looking for the third roots of unity. The first is and the rest are found by chopping the unit circle up into three, i.e. . So we have one at , another at and another at .

Q. 1 (e) is pretty good but maybe write that so we have , which is fixed. Also you actually have and in fact it is .

Regarding Q. 1 (f), please see https://jpmccarthymaths.com/2014/03/27/ms3011-week-12-2/#comment-1525

Regards,

]]>I need some help with Q. 62.

Regards.

]]>Two issues before we start.

You don’t evaluate an equation such as you solve it. Secondly please don’t use decimal rounding in a maths exam unless you want an approximate solution. The solution of , by taking the log of both sides, is . Now but not equal to it so leave it as .

I suspect we dealt with the half rather than just dropped it. O.K. we have and we want to find the fixed points so we solve :

or .

Now to classify these we look at . Now is the product of and so we apply the product rule:

is attracting.

is repelling.

Regards,

J.P.