does have an eventually periodic point… :

so the orbit is which is certainly eventually periodic… indeed right away. Could you modify very slightly to deal with this? You only need one function which does the trick.

For the second part, what is a prime number?

Regards,

J.P.

function? I picked f(x)=x^2?

My question is the second part of (b), just wondering could you help

me on starting it . I know we need some info on number theory to start

it but i can’t think of what it could be ?

The result that divides into applies (via the Factor Theorem) only when is a polynomial. This is not the case here and I have no idea how you divided an abstract into !

This result (iii) actually holds whenever has an odd number of elements and there is a clever argument which will give the result.

However, with only three elements in we could actually write out ALL of the functions . There are 27 of them. e.g.

, and is one of them.

You could then pick out all the period-2 ones and show that each have a fixed point. This would certainly suffice for an answer.

Obviously this is not ideal but is an option for you. If you do this you might see how the more general argument goes. Funnily enough, the general argument can use an implication of part (iv).

Even by playing around with some of these functions you might see how the general argument goes.

, and

is period-2 and has a fixed point.

Can you find a function that is period-2 and has no fixed point? Explain why not and there is your answer.

Regards,

J.P.

a fixed point thus prove it has a period-1 point i think? My

argument is that we are told all of the 3 elements and they are all

prime-period 2 points thus they will begin to repeat after 2 iterates.

In order to prove that a fn has a fixed point it is necessary to prove

there exists solns for f(x)=x. However we know there exists solns for

f2(x)=x because all the elements are prime period 2. Thus its true to

say that the solutions of f2(x)=x will contain fixed points if they

exist? So what i tried to do was divide f(x)-x into f2(x)=x since we

have to work in the general case. However i got a remainder of x^2-x??

Am i on the right track or completely off ? ]]>

Robert,

Sorry this notation isn’t clear you’re right. is the two-by-two identity:

.

Incidentally .

Regards,

J.P.

just about question 5, part j, did you mean that p^n = I^2? Just wondering if that was a typo, thanks. ]]>

Taking inspiration from parts (d) and (h) we can say that for some is indeed eventually fixed.

However part (i) wants us to find a function that is a prime-period- point of . That is we need a function such that if we differentiate times:

,

and furthermore none of

for (prime-period )

From parts (a) and (b) we see that is prime-period-one (or fixed), and is prime-period-two point. Hence these are answers for .

Compare now these functions with

… and

as mentioned in part 2, and note that and are primitive first and, respectively, second roots of unity (primitive meaning that is a solution of but not ).

Try differentiating three times and solve equal to …

Hopefully that might give you enough to see what to do.

Regards,

J.P.

Kind regards

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