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**zetafunc.****Guest**

So far I did Q1, Q2, Q3, and am about to finish Q5 when I start under mock exam conditions again.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,548

That is very good!

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**zetafunc.****Guest**

Thanks -- do you agree that those are the easiest?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,548

I hope not, I have not looked at the others yet.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**zetafunc.****Guest**

You are writing solutions to them?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,548

Nope, I am wondering how to do them.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,858

Hi zf.

Do you have a link to the exam paper?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**zetafunc.****Guest**

maths helper . co . uk / oxb . htm (remove spaces)

Please don't post solutions or spoilers to any (yet) as I want to finish them under exam conditions.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,548

That is a better link , I can download it.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

Yeah it's the best link to use, although some of the links stopped working recently...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,548

I am looking at 1.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

I thought you might like that question, it is similar to some GF problems we did...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,548

That is what I am doing with it. It is a partition problem actually. We can look at it when you are done, if you like.

There are a few others that are mean. I am going shopping see you when I get back and good luck.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

Sure, that would be great.

Okay, see you later.

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,858

You are doing STEP II, year 2012?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**zetafunc.****Guest**

Yes, I just finished that one, haven't checked my answers yet though. Thought it was quite a nice paper, lots of accessible questions not requiring that much work. Hopefully the mark scheme will not be harsh.

**zetafunc.****Guest**

Just done marking... lots and lots of silly errors but definitely a solid 1 or possibly a low S if I got a generous examiner. Happy times! I am just glad I did not embarrass myself.

I did Q1-5 and 13. The cyclic quadrilateral one looked like it would take a long time to manipulate algebraically so I skipped that and did a shorter question so I could spend the last 5 minutes or so checking over.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,548

That is very good.

I was looking at Q12 also. Seemed like a nice problem.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

The probability one?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,548

Yes, I have not looked at it because I just got back home. It was like 110 or something over there.

Did you want to go over the coefficient problem or you are comfortable with the solution you have?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

I am interested in seeing how you did it -- given that they asked for the co-efficient of x^k for (relatively) small k, I just used arithmetic.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,548

Hope I can remember it offhand. Correct me if I make some silly mistake.

We are interested in the coefficient of x^24 of

is that correct?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

I think so. And the answer was 15?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,548

If we think of this as a partition problem then that has the polynomial representation of.

This is the generating polynomial for the number of ways you can form a sum using sixes and threes. We can now enumerate the possibilities by filling in the 3 polynomials.

You can have

(4 sixes)(0 threes) 1 way

(0 threes)(4 sixes) 1 way

(3 sixes)(1 six)(0 threes) 1 way

(1 six)(3 sixes)(0 threes) 1 way

(2 sixes)(2 sixes)(0 threes) 1 way

(2 sixes)(1 sixes)(2 threes) 1 way

(1 sixes)(2 sixes)(2 threes) 1 way

(3 sixes)(0 sixes)(2 threes) 1 way

(0 sixes)(3 sixes)(2 threes) 1 way

(1 sixes)(1 sixes)(4 threes) 1 way

(2 sixes)(0 sixes)(4 threes) 1 way

(0 sixes)(2 sixes)(4 threes) 1 way

(1 sixes)(0 sixes)(6 threes) 1 way

(0 sixes)(1 sixes)(6 threes) 1 way

(0 sixes)(0 sixes)(8 threes) 1 way

total ways 15.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

Oh I see, so you treated (1 - x^6)^(-2) as the square of a geometric series. That looks pretty neat. The binomial expansion of it has the triangular numbers as co-efficients (and then I used the same method as you, pairing up the exponents).

You did the same thing for the last part?