HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · May 8, 2013

Re: HSC 2013 4U Marathon

Registered User said:
I think it should be $0 < \theta < 2\pi$

Can I do this using Fourier or Taylor series since they use elementary functions and some integration?

yes that is also true.

Yeah sure if you want, just make sure you can prove everything that you need to, and justify things rigorously (such as radius of convergence etc.)
I would prefer a HSC solution though.

EDIT: Also Heroic, I don't like the step 'I'll assume R(x) is a constant' are you able to justify that it is a constant?

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$x = p + \sqrt{q} $ is a root, hence $P(x) = [x-(p-\sqrt{q})]Q(x)$

???

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

Suppose we are flipping a biased coin such that:

$\frac{P(T)}{P(H)} = q$

Where:

$P(T)$ is the probability of flipping tails.

$P(H)$ is the probability of flipping heads.

$q$ is a positive integer.

Find the probability of flipping $n$ heads before $m$ tails.

Registered User · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I quite like the original, mostly because a lot of the others are too advanced for me to understand right now haha.

I like the one using Fourier series but I still don't completely understand it haha

I want to make a 4U level question to compute $\zeta (4)$ , don't know if that is possible though.

Registered User · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Suppose we are flipping a biased coin such that:

$\frac{P(T)}{P(H)} = q$

Where:

$P(T)$ is the probability of flipping tails.

$P(H)$ is the probability of flipping heads.

$q$ is a positive integer.

Find the probability of flipping $n$ heads before $m$ tails.

Seen a question slightly similar to this in 2000 paper. Don't know if any one in the state managed to get it lol

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

Registered User said:
I like the one using Fourier series but I still don't completely understand it haha

I want to make a 4U level question to compute $\zeta (4)$ , don't know if that is possible though.

You can use the exact same method as the question I posted up I'm pretty sure. The only problem is the assumptions don't make the question that rigorous, and it would be tedious.

seanieg89 · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I quite like the original, mostly because a lot of the others are too advanced for me to understand right now haha.

Haha except the ideas needed to make that proof rigorous are deeper than the ideas in most of the other proofs.

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Haha except the ideas needed to make that proof rigorous are deeper than the ideas in most of the other proofs.

It's pretty ironic isn't it.

seanieg89 · May 8, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Haha except the ideas needed to make that proof rigorous are deeper than the ideas in most of the other proofs.

Armed with the knowledge that it does happen to work though it does make for a pretty efficient way of computing zeta(2n) inductively.

HeroicPandas · May 8, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
yes that is also true.

Yeah sure if you want, just make sure you can prove everything that you need to, and justify things rigorously (such as radius of convergence etc.)
I would prefer a HSC solution though.

EDIT: Also Heroic, I don't like the step 'I'll assume R(x) is a constant' are you able to justify that it is a constant?

no... i'm afraid not

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Armed with the knowledge that it does happen to work though it does make for a pretty efficient way of computing zeta(2n) inductively.

Just equate co-efficients of 2n right?

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

not 100% sure about this actually...

seanieg89 · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
It's pretty ironic isn't it.

Haha a bit. Euler had a knack for doing some whacky things on intuition that turned out to be things you could make rigorous. Its a pretty bad habit to get into unless you have the intuition of Euler

. (He never claimed that the random manipulations comprised a proof I don't think, he just used such magic to convince himself of the plausible truth of statements before he rigorously proved them.)

On the whole the standard of rigour is higher these days as well.

Registered User · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
You can use the exact same method as the question I posted up I'm pretty sure. The only problem is the assumptions don't make the question that rigorous, and it would be tedious.

hmm I will try to do Apéry's constant then

PS: this can't be done, right? Since it doesn't converge to a nice value.

seanieg89 · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Just equate co-efficients of 2n right?

Its a bit more complicated than that but I remember that it can be done.

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Its a bit more complicated than that but I remember that it can be done.

Don't you end up with:

$x-\frac{x^3}{3!}+\frac{x^5}{5!} - \frac{x^7}{7!}+...$ is identical to:

$(1-\frac{x}{a})(1-\frac{x}{b})(1-\frac{x}{c})...$

where a,b,c.. are roots of sinx

And equate the co-efficient of 2n for whatever integer value of n you choose?

Registered User · May 8, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove$ $2^n \cos \left (\frac{n \pi}{2} \right ) = \sum_{k=0}^{n} (-1)^k \binom{2n}{2k}$

$\sum_{k=0}^n\binom{2n}{2k}x^{2k}$

$=\frac12\left(\sum_{k=0}^{2n}\binom{2n}{k}x^k+\sum_{k=0}^{2n}\binom{2n}{k}(-x)^k\right)$

$=\frac{(1+x)^{2n}+(1-x)^{2n}}{2}$

$Plug in$ $x=i$ $to get$

$\sum_{k=0}^n(-1)^k\binom{2n}{2k}$

$=\frac{(1+i)^{2n}+(1-i)^{2n}}{2}$

$=\frac{(2i)^n+(-2i)^n}{2}$

$=2^n\frac{e^{in\pi/2}+e^{-in\pi/2}}{2}$

$=2^n\cos(n\pi/2)$

seanieg89 · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Don't you end up with:

$x-\frac{x^3}{3!}+\frac{x^5}{5!} - \frac{x^7}{7!}+...$ is identical to:

$(1-\frac{x}{a})(1-\frac{x}{b})(1-\frac{x}{c})...$

where a,b,c.. are roots of sinx

And equate the co-efficient of 2n for whatever integer value of n you choose?

I don't think so but post your working for zeta(4) if you believe so.

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

From the infinite series of sine:

$\frac{sinx}{x} = 1-\frac{x^2}{3!} + \frac{x^4}{4!} - ...$

Also:

$\frac{sinx}{x} = (1-\frac{x}{\pi})(1-\frac{x}{-\pi})(1-\frac{x}{2\pi})(1-\frac{x}{-2\pi})...$

By difference of two squares:

$\frac{sinx}{x} = (1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})...$

So we can say that:

$1-\frac{x^2}{3!} + \frac{x^4}{5!} - ...$

is identical to

$(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})...$

hence they have equal co-efficients. Equating co-efficients of x^4 gives:

$\frac{1}{5!} = ...$

ok I see the problem.

Sy123 · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Suppose we are flipping a biased coin such that:

$\frac{P(T)}{P(H)} = q$

Where:

$P(T)$ is the probability of flipping tails.

$P(H)$ is the probability of flipping heads.

$q$ is a positive integer.

Find the probability of flipping $n$ heads before $m$ tails.

The problem simply boils down to computing:

$P = \sum_{k=0}^{m} \binom{n+k}{n} (P(T))^n (P(H))^k$

$\frac{1-P(H)}{P(H)} = q \ \ P(H) = \frac{1}{q+1} \ \ P(T) = \frac{q}{q+1}$

Is it possible to compute

$\sum_{k=0}^{m} \binom{n+k}{k} x^k$

? Its easy if x is 1 but....

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