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Proof of the binomial theorem

  1. Mar 30, 2005 #1
    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...
  2. jcsd
  3. Mar 30, 2005 #2
  4. Mar 30, 2005 #3
    Oh, thanks.

    Edit: But are you sure there isn't a easier one. I find this rather complex.
    Last edited: Mar 30, 2005
  5. Mar 30, 2005 #4
    I've here an outline for an, according to me,simpler one.

    [tex](a+1)^n = (a+1)...(a+1) ~(n times)[/tex]
    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:
    [tex]a_1x^n + a_2x^{n-1} + ... + a_nx + a_{n-1}x^0[/tex]
    I'm not so sure about the combination interpretation needed next. Maybe someone can help me with this.
  6. Mar 30, 2005 #5
    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...
  7. Mar 30, 2005 #6
    I dunno... I think it's funny to figure something out all by myself... :) But indeed I don't get any fare without help.
  8. Mar 30, 2005 #7


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    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??
  9. Mar 30, 2005 #8
    Nicely stated, mathwork. It was exactly what I had in mind. Thanks!
  10. Mar 30, 2005 #9

    matt grime

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    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

    [tex]\prod_{n}( 1- q^n)^{-1} = \sum_n P(n)q^{n}[/tex]

    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.
  11. Mar 31, 2005 #10
    Sorry matt grime. I cannot see how the apply of the aguments of mathwork works.
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