Question

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

Answer 1

`(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 - ....`

Answer 2

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)`