How can the binomial theorem be proven?

[danne89]

Hi! I haven't found any good proofs of the binomial theroem. But I've discovered how to go from (a+1)= bla bla to (a+b) = bla bla. So if anyone could told me how to prove (a+1) = bla bla...

 

[Muzza]

http://planetmath.org/?op=getobj&from=objects&id=338
http://planetmath.org/?op=getobj&from=objects&id=6785

 

[danne89]

Oh, thanks.

Edit: But are you sure there isn't a easier one. I find this rather complex.

 

[danne89]

I've here an outline for an, according to me,simpler one.

$$(a+1)^n = (a+1)...(a+1) ~(n times)$$
According to the Polynomial Factor Theorem has the polynom on the right hand side the zero -1 with multiplicity n. And then, using the Fundamental Theorem of Algebra, the polynom must be of degree n. Thus the mission is to find the coefficent a of:
$$a_1x^n + a_2x^{n-1} + ... + a_nx + a_{n-1}x^0$$
I'm not so sure about the combination interpretation needed next. Maybe someone can help me with this.

 

[Muzza]

I've here an outline for an, according to me,simpler one.

But how would you know if it is simpler if you haven't completed the most crucial step? You also rely on the fundamental theorem of algebra, a very non-trivial result...

 

[danne89]

I dunno... I think it's funny to figure something out all by myself... :) But indeed I don't get any fare without help.

 

[mathwonk]

look your approach is fine... just look at the product (a+1)(a+1)...(a+1).


every term in it looks like a certain number of a's tiems a certain number of 1's, and all you want to do is figure out how many terms have no a's. how many have one a, how many have 2 a's and so on.


obviously there is only one term with no a's, since you have to choose 1 from every factor, and similarly only one term with n a's since you choose a from evert factor.

ok how many terms have just one a? well it depends which factor you choose the a from, and there are n of them, so it is n. so you get 1 + na + ...


and how many ways give 2 a's? well of the n factors you have to choose two of them to take a's from, so that's "n choose 2" ways.

so we ghet 12 + na + "n choose 2" a^2 +... see??

 

[danne89]

Nicely stated, mathwork. It was exactly what I had in mind. Thanks!

 

[matt grime]

And if you think about it that way then you're on the verge of learning about formal power series and such elegant and powerful things they are...

Prove that

$$\prod_{n}( 1- q^n)^{-1} = \sum_n P(n)q^{n}$$

Where P(n) is the number of ways of writing n as a sum of positive integers (ordernot important). Note P(0)=1 by convention.

Do this using mathwonks argument: write (1-q^m)^{-1} as a power series and work out how to get the coefficient of q^n in the RHS by multiplying these power series together. (we don't care about convergence, that is why they are formal -they do not represent any function and need not converge if we put in any value for q.

 

[danne89]

Sorry matt grime. I cannot see how the apply of the aguments of mathwork works.