How do you use the binomial series to expand the function f(x)=(1-x)^(2/3)f(x)=(1−x)23 ?

1 Answer
dansmath
Sep 27, 2014

This is a natural extension of raising a binomial to a whole number power:
(a+b)^n=sum_(k=0)^n (n/k)a^(n-k)b^k(a+b)n=n∑k=0(nk)an−kbk (sorry about the inner bar)
Where (n/k)=(n!)/(k!(n-k)!)=(n(n-1)…(n-k+1))/(k!)(nk)=n!k!(n−k)!=n(n−1)…(n−k+1)k!

So we can apply this to any exponent r even if r is an arbitrary real number.

(a+b)^r=sum_(k=0)^oo (r(r-1)…(r-k+1))/(k!) a^(r-k)b^k(a+b)r=∞∑k=0r(r−1)…(r−k+1)k!ar−kbk

Now put your info in, with r=2/3, a=1, b=-x:r=23,a=1,b=−x:

(1-x)^(2/3)=sum_(k=0)^oo (2/3(2/3-1)…(2/3-k+1))/(k!) 1^(2/3-k)(-x)^k(1−x)23=∞∑k=023(23−1)…(23−k+1)k!123−k(−x)k

=(1)^(2/3)+(2/3)/1(1)^(-1/3)(-x)^1+((2/3)(2/3-1))/2(1)^(-4/3)(-x)^2+…=(1)23+231(1)−13(−x)1+(23)(23−1)2(1)−43(−x)2+…

=1-2/3x-1/9x^2+…=1−23x−19x2+…

There's the start of the series; I dare you to compute the next two terms. Take the \dansmath challenge/!

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