Question

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

Answer 1

`1/(1+x^3)^(1/2)=1-1/2x^3+3/8x^6-15/48x^9+105/384x^12+............`

Explanation:

Binomial theorem gives the expansion of `(1+x)^n` as

`(1+x)^n=1+nx+(n(n-1))/(2!)x^2+(n(n-1(n-2)))/(3!)x^3+(n(n-1)(n-2)(n-3))/(4!)x^4+....................`

Hence `1/(1+x^3)^(1/2)=(1+x^3)^(-1/2)`

= `1+(-1/2)x^3+((-1/2)(-1/2-1))/(2!)x^6+((-1/2)(-1/2-1)(-1/2-2))/(3!)x^9+((-1/2)(-1/2-1)(-1/2-2)(-1/2-3))/(4!)x^12+....................`

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

= `1-1/2x^3+3/8x^6-15/48x^9+105/384x^12+....................`