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**zetafunc.****Guest**

Sort of an emergency because STEP II is tomorrow morning...

In post #9524, we claimed that, if we have this:

then the co-efficient of x^n is

if n is less than or equal to the middle co-efficient (if it isn't, just use the opposite value of n, using the symmetry of the binomial expansion).

But trying this out on other examples, this doesn't seem to work and at best seems to be an approximation, getting weaker for higher powers. Do you know the correct closed form for the co-efficient of x^n in such an expansion?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

Hi;

Hold on am looking at it.

**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**

Okay, thank you.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

I think I have it, I do not know how we got confused back then . Let's try again.

with k is the power, n is x^n. This should work fine.

**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**

Yeah, that's what I wrote at the top of the page -- but it doesn't always seem to work...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

Which one does it not work for? Do you have an example?

**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**

Suppose we want the co-efficient of x^7 in:

(1 + x + x^2 + x^3 + x^4)^10

By the formula above, we get (10 + 7 - 1) C (10 - 1) = 16C9 = 11440...

...but WolframAlpha is saying it is 10890.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

That is because that one is too short. To be on the safe side it should be at least:

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...

So how would you get the co-efficient of x^7 in (1 + x + x^2 + x^3 + x^4)^10?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

That is a different problem then the one we were working on back around #9500.

For one thing these are computational problems and as such are best done with computers. It will be tedious to the extreme to do by hand and down right wrong! Sort of like asking the student 17254367465112 x 98102673846512.

There may be a trick or two but you can not expect the same trick to work every time. Do you expect a problem like that on the test?

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 understand, but this sort of thing will probably come up in either STEP I, II or III. It's usually the first question and they want you to do it 'systematically' -- which takes 15-25 minutes. But if I use a shortcut, it could take 2 minutes! In an exam where time is of the essence, those extra minutes could be a whole grade.

If I were posed the problem outside of an exam, I would just use a computer, however.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

What ideas do you have to get it?

To show how difficult these medium range problems are, the answer for the coefficient of x^7 for any power n is:

If I were posed the problem outside of an exam, I would just use a computer,

It is also not quite as simple as that either. What if you needed x^7's coefficient and the power was 1 000 000 000. Wolfram can not get that.

As far as I know this is an open field of research. This is my field, computation and mathematics joining forces and getting answers that neither can get alone.

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 was not aware it was an open field of research. Did you say you were writing a paper on it? How many different formulae have they found?

Just did STEP II.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

Yes, I am partly through a very ugly method to do a slightly harder one then you proposed. Unfortunately earlier today I came across a method that uses PIE but I do not follow it. So this may not be any mystery to anyone but me.

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**

What is PIE?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

Principle of Inclusion and Exclusion

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**

And why is the method ugly?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

It is long and arduous. There is no discernible pattern which to me is a sure sign of the "wrong method!"

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**

Hmm... I am perplexed. I always assumed that something like this would be trivial for a computer to solve, but if it's an open problem, that is interesting. I suppose it is a bit like the Tower of Hanoi problem with the number of discs and pegs both variable; seems like it would be trivial to find experimentally but it turns out it is not at all.

There was a girl doing a maths exam with me today -- just the two of us (she's a year older than me). She was from our maths class a year ago (with F and H in it). She started telling me about how a guy in our class started making moves towards her, putting his arm around her, etc. but when she told her boyfriend, the BF started threatening him and he was too scared to go to school. What a shame, given that she finds me attractive and last I remember didn't want to be in a relationship with her BF anyway.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

Hmmm, the bone bender Brodsky type. Is she very flirtatious?

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**

Not really... she is just good-looking, no surprise guys are after her.

I don't know if I'm justified in saying she finds me attractive. Last year, upon hearing that F rejected me, she said "Who would reject you?" and her talking to me suggests she enjoys my company. I won't pursue her though, for pretty obvious reasons. Her BF is known for getting into fights and hurting people. I think he went to jail once.

That would be interesting if she did find me attractive, though. Then all of the girls who have admitted that they like me will have been Oriental.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

Looks like as long as she is with "The Bone Bender" you should steer clear of her.

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**

She will be gone by Monday anyway, that is the last exam we have together. I have her e-mail though.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

It is probably not worth the trouble to even keep 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.

**Online**

**zetafunc.****Guest**

Yep. Her BF probably has her e-mail address and PIN number too.