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**zetafunc.****Guest**

I will have to learn to adapt.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

Once they begin to see you in one of the 9 or 10 forms that they can not do without, then you will do better.

**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**

Until then, it is a waiting game...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

I suggest doing your math and forgetting about them. Not completely but putting them in the last and smallest compartment of your mind.

**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**

I find it difficult to forget about them...

I finally found a STEP question that was easily done with generating functions, getting the co-efficient was just a bit tedious.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

The tough part of any generating function approach is getting the coefficient.

**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**

Yes, unfortunately no calc/CAS is allowed for STEP. Luckily all I had to do was get the coefficient of x[sup]24[/sup] in (1 + x + x[sup]2[/sup] + x[sup]3[/sup] + x[sup]4[/sup] + x[sup]5[/sup] + x[sup]6[/sup])[sup]5[/sup]. If it was an EGF that would have been more tedious.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

Hi;

I was working on a paper for just that sort of problem. I will back soon from doing a chore.

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**

A paper?

Okay, see you later.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

Only thing it did not have the constant term.

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**

Sorry, I had to go to school, I am back now.

So you have a good way of getting the co-efficient?

My method was

and then just get the co-efficient of x^24 from that but then the arithmetic can become a bit long...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

You can play spot the pattern.

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**

Okay............

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

Supposing you needed the coefficient of x^5 in

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**

By hand, I think I would just go back to sleep if someone asked me to do that

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

That could be difficult even by computer. That is when math becomes an art and ceases to be a science.

Anyway this is the answer:

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**

How do you know that?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

Sometimes in combinatorics we have a big problem. We can often solve it by looking at smaller problems that have the same flavor and then spotting the pattern. For instance it is not too difficult to expand these:

What do you notice about the coefficient of x^5 in each case?

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 co-efficients are constant for the first sequence, (n+1) for the second sequence, and the triangular numbers for the third.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

You can put that into a more succinct combinatoric notation.

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**

(n+1) choose (k-1), where k is the power of the big bracket?

**zetafunc.****Guest**

Hmm maybe (n + k - 1) choose (k - 1).

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

You can phrase wholly in terms of the power n:

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 in my example I could have just done

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

Yes, because of symmetry:

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.

**Online**