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

2 Answers
Jan 29, 2018

Answer:

#(1+x^3)^(-1/2) = 1 - x^3/2 + (3/8)x^6 - (5/16)x^9 + (35/128)x^12 - ....#

Explanation:

It is in the form (a+b)^n where
#a = 1, b = x^3, n = -(1/2).

The difference here is that the sum on the right doesn't have just n+1 terms, but infinitely many. This is called an infinite series.

In the binomial expansion formula for #(1+x)^n = 1 +nx+ (n(n-1))/(2!)x^2 + ...#

#1+(-1/2)(x)^3 + ((-1/2)(-3/2) (x)^6) /(2!) + ((-1/2)(-3/2)(-5/2)x^9)/(3!) + ((-1/2)(-3/2)(-5/2)(-7/2)x^12)/(4!) + ....#

#(1+x^3)^(-1/2) = 1 - x^3/2 + (3/8)x^6 - (5/16)x^9 + (35/128)x^12 - ....#

Jan 29, 2018

Answer:

The answer is #=1-1/2x^3+3/8x^6-5/16x^9+o(x^9)#

Explanation:

The binomial series is

#(a+b)^n=a^n+((n),(1))a^(n-1)b^+((n),(2))a^(n-2)b^2+((n),(3))a^(n-3)b^3+.#

#=a^n+na^(n-1)b+(n(n-1))/(1xx2)a^(n-2)b^2+(n(n-1)(n-2))/(1xx2xx3)a^(n-3)b^3+...#

Here,

#a=1#

#b=x^3#

#n=-1/2#

Therefore,

#(1+x^3)^(-1/2)=1+(-1/2)x^3+((-1/2)(-3/2))/(1*2)x^6+((-1/2)(-3/2)(-5/2))/(1*2*3)x^9+...#

#=1-1/2x^3+3/8x^6-5/16x^9+o(x^9)#