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bobbym
Yesterday 03:18:51

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.

zetafunc.
Yesterday 03:13:20

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
Yesterday 02:57:27

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?

zetafunc.
Yesterday 02:53:42

Oh...

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

bobbym
Yesterday 02:51:49

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

zetafunc.
Yesterday 02:44:59

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
Yesterday 02:42:54

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

zetafunc.
Yesterday 02:42:15

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

bobbym
Yesterday 02:21:28

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.

zetafunc.
Yesterday 02:12:11

Okay, thank you.

bobbym
Yesterday 02:08:12

Hi;

Hold on am looking at it.

zetafunc.
Yesterday 01:50:49

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
Yesterday 00:21:54

Yes, but it is hard to barbecue them.

phanthanhtom
Yesterday 00:10:10

Looks like we are discussing cooking. I love fried eggs.

bobbym
Yesterday 00:06:08

Very good! I cook the same but have not been able to barbecue for many years.