MS3011: Week 12

March 27, 2014 in MS 3011

I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.com and I will add you to the mailing list.

Homework

The closing date is 11 April at 2 pm. Work handed up late will be assigned a mark of zero. Note what the homework says”

Answer the following using geometric arguments.

Week 12

We finished on Monday by talking a little bit more about taking roots of complex numbers.

Week 13 Review Week (21-25 April)

I will book a two hour slot and we will first go through the layout of the paper and then I will answer any questions that ye might have.

Math.Stack Exchange

If you find yourself stuck and for some reason feel unable to ask me the question you could do worse than go to the excellent site math.stackexchange.com. If you are nice and polite, and show due deference to these principles you will find that your questions are answered promptly.

19 comments

Comments feed for this article

April 19, 2014 at 12:15 pm

Student

Hi J.P.,

If you get a chance can you quickly run through the Summer 2012 paper Q. 4 parts e and f please. It is about eventually periodic points.

Thanks!

April 19, 2014 at 12:28 pm

J.P. McCarthy

We looked at these in a tutorial.

For (e) look at the first few iterates of $z_0$ :

$f(z_0)=\left(e^{2\pi i/2^q} \right)^2=e^{2\left(2\pi i/2^q\right)}=e^{2\pi i/2^{q-1}}$

$f^2(f(z_0))=\left(e^{2\pi i/2^{q-1}}\right)^2=e^{2\left({2\pi i/2^{q-1}}\right)}=e^{2\pi i/2^{q-2}}$ .

Inductively, we see that

$f^q(z_0)=e^{2\pi i/2^{q-q}}=e^{2\pi i/2^0}=e^{2\pi i}=e^{0 i}=1,$

which is a fixed point of $f$ ; i.e. $z_0$ is eventually fixed because some iterate of it is fixed.

For (f), we disagree with the statement. Note that $e^{2\sqrt{2} \pi i}$ is on the unit circle, and $f$ maps from the unit circle to the unit circle, so that the only way that $e^{2\sqrt{2} \pi i}$ could be eventually fixed is if some iterate is mapped to $z=1$ — the only fixed point of $f$ on the unit circle,

Suppose for the sake of contradiction that $f^N\left(e^{2\sqrt{2}\pi i}\right)=1$ . Now the argument of $1$ is $0, 2\pi, 4\pi,\dots$ , or indeed any negative of these… the argument is any multiple of $2\pi$ so that we can write $1=e^{2k \pi i}$ , for $k\in\mathbb{Z}$ .

Now we are assuming $f^N\left(e^{2\sqrt{2} \pi i}\right)=e^{2k\pi i}$ for some $N\in\mathbb{N}$ . That is

$e^{2\sqrt{2}\pi i2^N}=e^{2k\pi i}$

for some $k\in\mathbb{Z}$ . That is

$2\sqrt{2}\pi 2^N=2k\pi$

$\Rightarrow \sqrt{2}=\frac{k}{2^N}$ ,

which contradicts the fact that $\sqrt{2}$ is not a fraction. Hence there is no $N\in\mathbb{N}$ such that $f^N(e^{2\sqrt{2}\pi i})=1$ , that is $e^{2\sqrt{2}\pi i}$ is NOT eventually fixed.

Regards,
J.P.

PS: I clarified earlier in the term that a question like part (f) will NOT be on your paper.

April 29, 2014 at 3:53 pm

David culhane

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

April 19, 2014 at 12:34 pm

Student

Hi J.P.,

Having a bit of a mental block on Q. 56. found one fixed point $1/2 i$ but more by instinct than understanding. You’re probably on hols now but if you could cover it briefly in the review session i’d appreciate it.

Regards.

April 19, 2014 at 12:49 pm

J.P. McCarthy

We want to find a point $z$ that is sent to itself by $Q$ so we are looking for a $z$ such that

$Q(z)=z$

$\Rightarrow z^2+0.5i+0.25=z.$

The best we can say about this is that it is a quadratic in $z$ and because the complex numbers are a ‘field’ the $-b\cdots$ formula still applies (although we have a little bit of work to do to interpret ‘ $\sqrt{}$ ‘). Note that we do NOT have the conjugate root theorem (aka $-\frac12 i$ is not necessarily a fixed point because the coefficients are not real). Let us set it up

$z^2-z+(0.25+0.5i)=0$

$\Rightarrow z_{\pm}=\frac{1\pm \sqrt{1-4(1)(0.25+0.5i)}}{2}$

$=\frac{1\pm \sqrt{1-1-2i}}{2}$

$=\frac{1\pm \sqrt{-2i}}{2}$ .

Now we need to find ‘ $\sqrt{-2i}$ ‘ — or moreover the roots of $w^2=-2i$ . We can do this in our head following the homework. The magnitude of $-2i$ is $2$ so our root will have a magnitude of $\sqrt{2}$ . The argument of $-2i$ is $3\pi/2$ so the principal square root as a magnitude of half of this $3\pi/4$ which is basically $\pi/4$ in the second quadrant where cosine is negative and sine is positive so we have

$'\sqrt{-2i}'=\sqrt{2}\left(\cos (3\pi/4)+i\sin(3\pi/4)\right)$

$=\sqrt{2}\left(-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)$

$=-1+i$ .

Now if $(-1+i)^2=-2i$ then we also have $\left(-(-1+i)\right)^2=-2i$ so that $1-i$ is the other square root (and of course there are only two).

So we have

$z_{+}=\frac{1+(-1+i)}{2}=\frac12 i$ , and

$z_{-}=\frac{1-(-1+i)}{2}=\frac{2-i}{2}=1-\frac12 i$ .

Now to test their nature — attracting, repelling or indifferent — we look at $|Q'(z_{\pm})|$ . We have $Q'(z)=2z$ so we calculate

$Q'(z_+)=2(i/2)=i$ , which has a magnitude of one so that $z_+$ is indifferent.

$Q'(z_-)=2(1-i/2)=2-i$ which clearly lies outside the unit circle so has a magnitude of greater than one so it repelling. Another way to see this is to calculate $|2-i|=\sqrt{4+1}=\sqrt{5}>\sqrt{4}=2>1$ using Pythagoras.

Regards,
J.P.

April 19, 2014 at 12:57 pm

Student

Hi J.P.,

Can you start me off for answering Q. 41 in the exercises? Do I plot the graphs for different values of $\mu<4$ ?

Thanks!

April 19, 2014 at 1:02 pm

J.P. McCarthy

Unfortunately that would NOT be sufficient. $\mu$ can take on the continuum of values between $0\leq \mu\leq 4$ so plotting a finite number of those graphs is not enough.

For part (a) you would calculate, for a general $\mu\in [0,4]$ and $x\in(0,1/2]$ (why only $x\in(0,1/2]$ ?), $Q_\mu(x-1/2)$ and $Q_{\mu}(x+1/2)$ . If they are equal then you are done.
Similarly calculate, for a general $x\in(0,1/2]$ , $T(x-1/2)$ and $T(x+1/2)$ and show that they are equal.

For part (b), show that $Q_{\mu}(x)$ is increasing for $x1/2$ . We did this last year… how do you show that a function is increasing/decreasing. Do the same for $T(x)$ and you are done.

Regards,
J.P.

April 22, 2014 at 2:24 pm

Student

Hi J.P.,

I was wondering have you a solution for Autumn 2012 Q. 2… or even a rough guideline, just so i know if I am on the right track.

Thanks.

April 22, 2014 at 2:27 pm

J.P. McCarthy

Hi,

From what you sent on, your Q.2 (a) (ii) is fine but for part (i) you need something like this:

“Suppose that $x\approx 0$ so that $x^2\ll x$ . Then the population $x$ grows according to $Q_{\mu}(x)=\mu x(1-x)=\mu(x-x^2)$ , but $x\gg x^2$ so that $x-x^2\approx x$ so we have $Q_{\mu}(x)\approx \mu x$ , i.e. approximately geometric”.

Regards,
J.P.

April 26, 2014 at 10:28 am

Student

Hi J.P.,

I’m getting stuck with the August 2013 paper Q.1 part c. It is Q. 32 in the exercises. You did answer this in the tutorial, however you dropped the half and now I am not sure how to continue. I evaluated $f(x)=x$ and got $x=0$ or $e^x = 2$ . By taking the $\ln$ of both sides I got $x=0.693$ .

Is this the correct what to answer the question?

Thanks.

April 26, 2014 at 10:49 am

J.P. McCarthy

Two issues before we start.

You don’t evaluate an equation such as $f(x)=x$ you solve it. Secondly please don’t use decimal rounding in a maths exam unless you want an approximate solution. The solution of $e^x=2$ , by taking the log of both sides, is $x=\ln 2$ . Now $\ln 2\approx 0.693$ but not equal to it so leave it as $\ln 2$ .

I suspect we dealt with the half rather than just dropped it. O.K. we have $f(x)=xe^x/2$ and we want to find the fixed points so we solve $f(x)=x$ :

$xe^x/2=x\Rightarrow xe^x=2x\Rightarrow xe^x-2x=0$

$\Rightarrow x(e^x-2)=0\Rightarrow x=0$ or $e^x-2=0\Rightarrow x=\ln 2$ .

Now to classify these we look at $f'(x)$ . Now $f(x)$ is the product of $x/2$ and $e^x$ so we apply the product rule:

$f'(x)=(x/2)e^x+e^x(1/2)$

$f'(0)=0+e^0(1/2)=1/2\Rightarrow x=0$ is attracting.

$f'(\ln 2)=(\ln 2/2)e^{\ln 2}+e^{\ln 2}(1/2)$
$=(\ln 2/2)2+2(1/2)=\ln 2+1>1\Rightarrow x=\ln 2$ is repelling.

Regards,
J.P.

April 29, 2014 at 12:32 pm

Student

J.P.,

I need some help with Q. 62.

Regards.

April 29, 2014 at 12:37 pm

J.P. McCarthy

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 $z$ has argument $\theta$ then $z^2$ has argument $2\theta$ , 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 $z=0$ is a fixed point and that is found at the origin. Now $z^3=1$ is looking for the third roots of unity. The first is $z=1$ and the rest are found by chopping the unit circle up into three, i.e. $2\pi/3\equiv 120^\circ$ . So we have one at $z=1$ , another at $e^{2\pi i/3}$ and another at $e^{4\pi i/3}$ .

Q. 1 (e) is pretty good but maybe write that $4\pi\equiv 0$ so we have $e^{4\pi i}=e^{0i}=1$ , which is fixed. Also you actually have $f^{q+1}(z_0)=e^{4\pi i}$ and in fact it is $f^q(z_0)=e^{2\pi i}=e^{0i}=1$ .

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

Regards,

April 29, 2014 at 7:37 pm

Student

Dear 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.

April 29, 2014 at 7:43 pm

J.P. McCarthy

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 $x$ be the the proportion of a maximum population. Suppose that we want for small populations $x \approx 0$ that the growth rate is approximately geometric.

Also we want extinction to ensue whenever the maximum population is reached; i.e. $x_n=1 \Rightarrow x_{n+1}=0$ . The model $Q_\mu(x)=\mu x(1-x)$ achieves this.

For $x=1,\, Q_\mu(1)=\mu 1(1-1)=0$ as required.

Also when $x_n \approx 0$ , we have $x_n\gg x_n^2$ so that, $x_{n+1}=Q_{\mu}(x_n)=\mu(x_n-x_n^2)\approx \mu x_n$ , as required.”

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

Regards,
J.P.

April 30, 2014 at 8:06 am

Student

Hi J.P.,

Sorry for the last minute question, I just want to clear something up. If $x=1$ is a fixed point (i.e. period one), and $x=2$ is a prime period two point. Is $x=1$ 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.

April 30, 2014 at 8:23 am

J.P. McCarthy

Firstly if $x=1$ is a fixed point, then $x=1$ is also period-two, three, four, five, etc (if a point is period- $q$ then it is also period- $nq$ for any $n\in\mathbb{N}$ .).

Secondly, if $x=2$ is prime-period-two, this means that $x=2$ is period-two and this is the lowest period; i.e. $x=2$ 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 $x=1$ , 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 $\alpha$ is a fixed point. For suppose that $\alpha$ is fixed, that is period-one. Therefore $\alpha$ is period- $n\times 1=n$ for any $n\in\mathbb{N}$ . In particular, $\alpha$ 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 $\alpha$ . Now because $\alpha$ is period-two it is also period-eight. Suppose for the sake of contradiction that $\alpha$ is also period-nine. That is

$f^9(\alpha)=\alpha\Rightarrow f(f^8(\alpha))=\alpha$ ,

but $f^8(\alpha)=\alpha$ so we this implies

$f(\alpha)=\alpha$ ,

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

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

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

$\text{orb}(\alpha)=\{\alpha,f(\alpha),\alpha,f(\alpha),\dots\}$ ,

then $f(\alpha)$ must be period-two. Clearly $f(\alpha)$ cannot be a fixed point therefore it must be prime-period-two and the only other prime-period-two point is $\beta$ so we have $f(\alpha)=\beta$ .

Therefore the orbit of $\alpha$ is:

$\text{orb}(\alpha)=\{\alpha,\beta,\alpha,\beta,\alpha,\beta,\alpha,\beta,\alpha,\beta,\dots\}$

so clearly $\alpha$ is period-eight but not period-nine.

Regards,
J.P.

August 8, 2014 at 3:23 pm

Student

Just want to clear up a few things with you that I’m bothered about before the test on Monday.

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 $z$ and $z^2$ in and how does this change when our expression changes to $z^3$ ?

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.

August 8, 2014 at 3:38 pm

J.P. McCarthy

1. In MS3011, given a function $f:S\rightarrow S$ , it is points $x_0\in S$ that we call periodic rather than the function. We call $x_0\in S$ a period- $n$ -point if

$f(f(\cdots f(x_0)))=f^n(x_0)=x_0$ .

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

$f(x_f)=x_f$

A point $x_1$ is an eventually periodic point if some iterate of $x_1$ is a periodic point. For example, if two has the following orbit under some $f$ :

$\text{orb}(2)=\{2,3,4,6,7,4,6,7,4,6,7,\dots\}$ ,

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 $f$ by solving $f(x)=x$ . Then take one fixed point, say $x_2$ , and find all of the points that are sent to $x_2$ by solving

$f(x)=x_2$ .

As it is fixed, $x_2$ 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 $z$ at around 20 $^\circ$ on the unit circle then $z^2$ will have angle 40 $^\circ$ and $z^3$ will have angle 60 $^\circ$ so all these points are in the first quadrant. Obviously the situation will be different if you place $z$ around 80 $^\circ$ .

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

Regards,
J.P.

	J.P. McCarthy on MATH6040: Spring 2022, Week…
	J.P. McCarthy on MATH6040: Spring 2022, Week…
	J.P. McCarthy on MATH7019: Winter 2020, Week…
	J.P. McCarthy on MATH7019: Winter 2020, Week…
	A Sufficient Conditi… on Almost All Trees have Quantum…

MS3011: Week 12

Homework

Week 12

Week 13 Review Week (21-25 April)

Math.Stack Exchange

Recent Comments

Categories

J.P. McCarthy Maths

19 comments

Leave a Reply Cancel reply

MS3011: Week 12

Homework

Week 12

Week 13 Review Week (21-25 April)

Math.Stack Exchange

Share this:

Like this:

Related

Recent Comments

Categories

J.P. McCarthy Maths

19 comments

Leave a Reply Cancel reply