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VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: cipherpol on April 06, 2011, 09:52:43 pm

Title: some math questions
Post by: cipherpol on April 06, 2011, 09:52:43 pm: Gonna need help understanding some of these proof questions :)

If there is a statement P(n) about natural numbers n, out of the 8 truth tables ([TTT], [TTF], [TFT], etc.) what values can the truth table [P(k), P(k+1), P(k)⇒P(k+1)] take?

Can someone give an example about one of the possibilities? Not sure how to go about writing this.

Thanks :)

EDIT: done
Title: Re: some math questions
Post by: Mao on April 06, 2011, 10:20:32 pm: If I'm understanding this correctly, P(n) is an arbitrary function of n that returns T or F. k+1 and k are both arbitrary points on the domain of P. That is, I can rewrite the truth table as [P(m), P(n), P(m) implies P(n)] where m, n are both natural numbers.

Then the possible combinations are:
[TTT], [TFF], [FTT], [FFT]
Title: Re: some math questions
Post by: cipherpol on April 06, 2011, 11:09:29 pm: Ah, I should have mentioned this was to do with induction. I remember my lecturer going on about how a false statement can imply anything, so I'm wondering whether it would be [FTT] instead of [FTF] in your post.
Title: Re: some math questions
Post by: Mao on April 07, 2011, 12:15:49 am: Quote from: cipherpol on April 06, 2011, 11:09:29 pm
Ah, I should have mentioned this was to do with induction. I remember my lecturer going on about how a false statement can imply anything, so I'm wondering whether it would be [FTT] instead of [FTF] in your post.

Quite right, I wasn't thinking at the time. You are absolutely right.

(the source of my knowledge of truth tables goes as far as minecraft allows)
Title: Re: some math questions
Post by: cipherpol on April 07, 2011, 06:49:50 pm: Find the 80th number in the bijection of the positive rational numbers with the natural numbers

1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, 1/7, 3/5, ...

lol, is there a better way than drawing out the table to find the 80th number? :P
Title: Re: some math questions
Post by: /0 on April 07, 2011, 08:21:37 pm: The easiest way is probably to just draw out the table and in the manner of the sieve of eratosthenes, cross out rational numbers $a/b$ with $gcd(a,b) \neq 1$ . I mean, it's a practical way to reach 80, but might not be so useful for higher numbers.

Another way which might work... in the bijection, note that you only have rational numbers $a/b$ such that $gcd(a,b)=1$ . So you could count all the pairs of numbers with $gcd(a,b)=1$ until you reach the 80th.

A useful tool you might want to use is Euler's totient function. Let $\phi(n)$ be the number of integers less than $n$ which are relatively prime to $n$ . Then

$\phi(n) = n\sum_{i}\left(1-\frac{1}{p_i}\right)$ where $p_i$ are prime factors of $n$ . Proof can be found on wikipedia.

Actually, the diagonal counting method in your question makes it really annoying. If we used a counting method like in my picture, it would be much easier to find the 80th number :/
Title: Re: some math questions
Post by: cipherpol on April 07, 2011, 09:25:53 pm: Thanks /0 =]

Not sure if we've even been through the material in the next question... :S

Identify the interesting points for the function f(z)=z+1/z.
Title: Re: some math questions
Post by: /0 on April 07, 2011, 11:04:23 pm: Let $z=x+iy$ , then $z+\frac{1}{z}=\frac{x}{x^2+y^2}+x+(y-\frac{y}{x^2+y^2})i$

We can view this as a parametric equation:

$(u,v) = \left(\frac{x}{x^2+y^2}+x,y-\frac{y}{x^2+y^2}\right)$

After playing around with the graph in Mathematica, I realised that as you vary $y$ and keep $x$ fixed, a point will do a loop-the-loop around one of the 'interesting points'.

If you do this for different $x$ , you find that the size of the loop changes. It is reasonable to assume that for a particular $x$ , the loop is so small that it becomes a cusp.

Then to find this cusp, we must simultaneously solve $\frac{\partial u}{\partial y}=0$ , $\frac{\partial v}{\partial y}=0$ .

Here's the notebook. If you don't have mathematica you can download a free mathematica player.
Title: Re: some math questions
Post by: Mao on April 07, 2011, 11:12:31 pm: +1 for mathematica.