Binomial Series

Key Questions

  • Binomial Series

    #(1+x)^{alpha}=sum_{n=0}^infty((alpha),(n))x^n#,

    where #((alpha),(n))={alpha(alpha-1)(alpha-2)cdot cdots cdot(alpha-n+1)}/{n!}#.

    Let us look at this example below.

    #1/{sqrt{1+x}}#

    by rewriting a bit,

    #=(1+x)^{-1/2}#

    by Binomial Series,

    #=sum_{n=0}^infty((-1/2),(n))x^n#

    by writing out the binomial coefficients,

    #=sum_{n=0}^infty{(-1/2)(-3/2)(-5/2)cdots(-{2n-1}/2)}/{n!}x^n#

    by simplifying the coefficients a bit,

    #=sum_{n=0}^infty(-1)^n{1cdot3cdot5cdot cdots cdot(2n-1)}/{2^n n!}x^n#

    I hope that this was helpful.

  • Pascal's triangle gives the binomial coefficients.

    Pascal's Triangle

    1
    1 1
    1 2 1
    1 3 3 1
    1 4 6 4 1
    ...

    Binomial Coefficients

    #((0),(0))#

    #((1),(0))# #((1),(1))#

    #((2),(0))# #((2),(1))# #((2),(2))#

    #((3),(0))# #((3),(1))# #((3),(2))# #((3),(3))#

    #((4),(0))# #((4),(1))# #((4),(2))# #((4),(3))# #((4),(4))#
    ...


    I hope that this was helpful.

  • To understand Ismail's answer, it is worth recalling some notations:

    #((n),(k))=(n!)/((n-k)!k!)#, where #n,k in NN#

    #n! =n.(n-1)...2.1#

Questions