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March 29, 2024, 09:52:26 am

Author Topic: Maths Extension 2  (Read 9336 times)  Share 

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AhmadTalal

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Maths Extension 2
« on: January 09, 2020, 08:57:57 pm »
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Hi, I don't quite understand how to solve this question. :'(

RuiAce

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Re: Maths Extension 2
« Reply #1 on: January 09, 2020, 09:06:07 pm »
+3
To 'iterate' means to apply the function over and over.

That means, in general, \( z_{n+1} = f(z_n)\). So you'd start with \( z_1 = f(z_0)\).

(Note: There's definitely a typo in the question. It should say \(f(z) = z^2+c\), instead of some \((z)\) thing by itself.)

If you have any further trouble, you should outline where those are.

AhmadTalal

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Re: Maths Extension 2
« Reply #2 on: January 13, 2020, 09:33:51 pm »
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Hi tried to work it out, is this how i need to prove that part is a prisoner set (a)? I substituted, the value of z0 into the z^2 part of, then i substituted the value of Z1 into the 3rd equation, is the way to go about it or am i on the wrong track. If so, is my explanation good enough, thanks in advance  :)!

RuiAce

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Re: Maths Extension 2
« Reply #3 on: January 13, 2020, 09:57:20 pm »
+2
Although your method seemed to head the right way, for some reason you used \(c = 0\). Part a) required you to use that \(c=i\).

So the first iteration is:
\begin{align*}
z_1 &= f(z_0)\\
&= f(0+0i)\\
&=(0+0i)^2 + i\\
&= 0+1i.
\end{align*}

AhmadTalal

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Re: Maths Extension 2
« Reply #4 on: January 15, 2020, 09:20:48 pm »
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Hi, thank you once again for the guidance :). I can't seem to find an understandable definition of a prisoner set or escape set.

For q4a, i found two answers to be recurring. I am not sure if z0 is a prisoner set as the recurrences start happening from z2, if it is a prisoner set then how do i prove it?
Regards

RuiAce

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Re: Maths Extension 2
« Reply #5 on: January 16, 2020, 09:29:10 pm »
+1
I wouldn't know what "prisoner set" means either in this context.

Please provide the source of the question in full. (The font looks like something from the NESA documents. If it is, feel free to just provide the URL or instructions on how to get there. If not, you may need to link/upload an attachment.)


RuiAce

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Re: Maths Extension 2
« Reply #7 on: January 18, 2020, 09:29:55 pm »
+2
My bad for the delayed reply.

So I'm not exactly sure why they insist on definitions of "prisoner" and "escape" sets here. They are indeed terms that frequently appear in fractal geometry, but they're not conventional terms that everyone's supposed to know how to use.

Loosely speaking,
- Escape occurs if \( \lim_{n\to\infty} |z_n| =\infty \). That is, if you keep iterating the function \(f(z) =z^2+c\) for a given \(c\),  over and over (starting with \(z_0 = 0\) in this context), the modulus of the numbers you get will always grow bigger and bigger. The escape set is the set of all complex numbers \(c\) that have this property.
- The prisoner set refers to the set of all complex numbers \(c\) that fail this property. This essentially occurs when if you keep iterating with the function \(f(z)=z^2+c\), the modulus of the numbers does not grow bigger and bigger.

As a consequence, if you find that the modulus of the values of \(z_0\), \(z_1\), \(z_2\), \(z_3\) start growing closer and closer to 0, or start oscillating somehow, they should be in the prisoner set.

(Note that for your example, where \(c=i\), the values themselves literally start oscillating. So you'd expect the modulus to do something similar as well.)
« Last Edit: January 18, 2020, 09:31:35 pm by RuiAce »

AhmadTalal

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Re: Maths Extension 2
« Reply #8 on: January 18, 2020, 09:52:10 pm »
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WOW!, thank you so much that made it so much more clearer for me. For once i was actually on the right track ;D.

AhmadTalal

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Re: Maths Extension 2
« Reply #9 on: January 20, 2020, 09:04:52 pm »
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Sorry to bother you about the same question again. I think I worked it out, sort of. In q 4A, the modulus is oscillating in the attachment below, but the oscillations do not start at z0, but at z2 and the question asks if z0 is in the prisoner set of f(z), therefore, would z0 not be included in the prisoner set, or would it?
Regards and thanks in advance :)

AhmadTalal

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Re: Maths Extension 2
« Reply #10 on: January 27, 2020, 12:39:54 am »
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Hi plz reply
thanks :)

RuiAce

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Re: Maths Extension 2
« Reply #11 on: January 29, 2020, 11:41:38 pm »
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Sorry. Although I've been busy, I genuinely missed this post.

We're not concerned if \(z_0\) is in the prisoner set, because \(z_0\) is always equal to \(0\). We're considered if \(c\) is in the prisoner set.

So long as oscillations happen for your value of c, that c is in the prisoner set. You should do some more research on the Mandelbrot set in regards to this. (Strong emphasis on that \(z_0\) is always 0, and it's c that we're always changing.)

esteban

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Re: Maths Extension 2
« Reply #12 on: January 30, 2020, 06:34:37 am »
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I don't think what you said is correct RuiAce, namely I DO think the question is about whether or not the point z_0=0 is in the prisoner set.

It is true that c is a parameter you are varying, but for each c you get a corresponding prisoner set and escape set. The question is about whether or not 0 is in the prisoner set for two particular values of c. I believe the answer is no in both cases (for the first one, the sequence is periodic, for the second it escapes to infinity.)

esteban

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Re: Maths Extension 2
« Reply #13 on: January 30, 2020, 06:57:06 am »
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To prove the claimed escape to infinity in the latter case it suffices to iterate the function f a few times (I think 4ish) until you reach a z that satisfies: |z|^2-|c| > |z|.

Such an inequality already tells you 0 is not a prisoner for your given c. To get an escape statement is not much harder (put a constant like k=1.00001 in front of the RHS of the preceding inequality. Once you satisfy this equality, you are guaranteed |f(z)| > k|z|, and exponential growth is ensured).

RuiAce

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Re: Maths Extension 2
« Reply #14 on: January 31, 2020, 03:12:40 am »
+2
My fault for not double checking the word document and screenshot.

In saying that, the concern I have here is that this sample assignment was set around Mandelbrot sets. The criteria you're providing appears to resemble the foundation in Julia sets.

Most of what I found regarding the terms "prisoner set" and "escape set" did seem to relate to Julia sets. But although Julia sets are closely related to the Mandelbrot set, I feel that it's still it's distinct enough and falls outside the main purpose of this assignment.

I also don't see why periodic behaviour would imply not being in the prisoner set here.  I could not find reason to believe that something cannot be in the prisoner set, if it does not escape to infinity. The little information I could find on these terms seemed to suggest that they were mutually exclusive and exhaustive categories.


Edit: I seem to find one website suggesting that there's supposedly something in the middle?
- http://www.shodor.org/interactivate/discussions/PrisonersAndEscapees/
But these pages that just require \(|z_n|\) to be a bounded sequence <=> being in the prisoner set seem to make more sense to me.
- http://www.combinatorialmath.ca/G&G/FractalOverview.html
- https://www.mcgoodwin.net/julia/juliajewels.html
- http://www.csun.edu/~acl23054/aclewis.html
« Last Edit: January 31, 2020, 03:15:57 am by RuiAce »