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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: 91,458

Hi;

Hold on am looking at it.

**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**

Okay, thank you.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,458

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 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**

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: 91,458

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 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**

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: 91,458

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

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

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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: 91,458

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?

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

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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: 91,458

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.

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

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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: 91,458

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.

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

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**zetafunc.****Guest**

What is PIE?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,458

Principle of Inclusion and Exclusion

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

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**zetafunc.****Guest**

And why is the method ugly?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,458

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

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

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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: 91,458

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

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

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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: 91,458

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

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

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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: 91,458

It is probably not worth the trouble to even keep it.

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

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**zetafunc.****Guest**

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