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This Week

In lectures, we covered from section 1.4 to 2.1 and some of section 2.2

Tutorials start next week Thursday at 3 in Windle PDT. Email me if you have a timetable clash. Please indicate code of the module which is clashing. 

Problems

You need to do exercises – all of the following you should be able to attempt.

Exercise on page. 10.

From the Class

Nothing

Additional Notes

This is the question which I failed miserably to do in the Wednesday lecture.

August 2010 Question 2(c)(ii)

We want to find an iterator function for which the orbit of 3 is given by:

x_n=n^3+3.

That is a function f:\mathbb{R}\rightarrow\mathbb{R} such that:

x_0=(0)^3+3

x_1=f(3)=(1)^3+3=4

x_2=f(4)=(2)^3+3=11

x_3=f(11)=(3)^3+3=30

\vdots

x_{n+1}=f(x_{n})=f(n^3+3)=(n+1)^3+3.

The ordinary approach is to write

f(x_n)=x_{n+1}=(n+1)^3+3

\Rightarrow f(n^3+3)=(n+1)^3+3.

and get an expression/function G in terms of n^3+3 on the right-hand side; i.e.

f(n^3+3)=G(n^3+3)

and so

f(x)=G(x);

which’ll certainly generate the required orbit of 3. So we experiment:

f(n^3+3)=(n+1)^3+3,

=n^3+3n^2+3n+1+3,

(n^3+3)+(3n^2+3n+1).   (*)

Now

(n+1)^3=n^3+3n^2+3n+1,

\Rightarrow 3n^2+3n+1=(n+1)^3-n^3,

=(n+1)^3-((n^3+3)-3),

=(n+1)^3-(n^3+3)+3.

We now need to writes (n+1)^3 in terms of n^3+3. First :

n^3= (n^3+3)-3.

Now take the cube root to get n:

n= \sqrt[3]{(n^3+3)-3}.

Now add 1:

n+1= \sqrt[3]{(n^3+3)-3}+1,

and finally cube to get (n+1)^3 in terms of n^3+3:

(n+1)^3=\left(\sqrt[3]{(n^3+3)-3}+1\right)^3,

\Rightarrow 3n^2+3n+1=\left(\sqrt[3]{(n^3+3)-3}+1\right)^3-(n^3+3)+3.    (**)

So from (*) and we can write:

f(n^3+3)=(n^3+3)+\left(\sqrt[3]{(n^3+3)-3}+1\right)^3-(n^3+3)+3,

\Rightarrow f(x)=x+(\sqrt[3]{x-3}+1)^3-x+3,

=(\sqrt[3]{x-3}+1)^3+3.

This produces the orbit as required.

Alternative (although equivalent) Solution

When I did this first I kind of divined the answer by looking at the orbit of 3:

3,4,\dots,n^3+3,(n+1)^3+3,\dots,

and asked myself how to get from n^3+3 to (n+1)^3+3 and I kind of thought:

  • take away three to get n^3
  • take a cube root to get n
  • add one to get n+1
  • cube to get (n+1)^3
  • add three to get (n+1)^3+3

This is a composition of functions:

n^3+3\mapsto n^3\mapsto n\mapsto n+1\mapsto (n+1)^3\mapsto (n+1)^3.

We could also write this as

x\mapsto x-3\mapsto \sqrt[3]{x-3}\mapsto \sqrt[3]{x-3}+1

\mapsto (\sqrt[3]{x-3}+1)^3\mapsto (\sqrt[3]{x-3}+1)^3+3.

That is f(x)=(\sqrt[3]{x-3}+1)^3+3 as before \bullet

However, the ordinary situation is that you know the iterator function but would like the find an expression for x_n; i.e. a function g:\mathbb{R}\rightarrow \mathbb{R} such that x_n=g(n). It is unnatural to know that x_n=g(n) and want to find out an iterator function for such a dynamical system  — it might not even be possible. The reason it is unnatural is because if you know g(n) you can answer all the questions we might have about the dynamical system very easily: e.g. limiting behaviour is \lim_{n\rightarrow\infty}g(n), etc..

For this reason and the fact that I think this question is too hard this type of question will now not be seen in homework, tests or exams.

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