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**zetafunc.****Guest**

I will have to learn to adapt.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,691

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 agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zetafunc.****Guest**

Until then, it is a waiting game...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,691

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 agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 91,691

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 agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 91,691

Hi;

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

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

A paper?

Okay, see you later.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,691

Only thing it did not have the constant term.

**In mathematics, you don't understand things. You just get used to them.**

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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: 91,691

You can play spot the pattern.

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,691

Supposing you needed the coefficient of x^5 in

**In mathematics, you don't understand things. You just get used to them.**

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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: 91,691

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:

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

How do you know that?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,691

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?

**In mathematics, you don't understand things. You just get used to them.**

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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: 91,691

You can put that into a more succinct combinatoric notation.

**In mathematics, you don't understand things. You just get used to them.**

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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: 91,691

You can phrase wholly in terms of the power n:

**In mathematics, you don't understand things. You just get used to them.**

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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: 91,691

Yes, because of symmetry:

**In mathematics, you don't understand things. You just get used to them.**

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