Question

What is the Maclaurin series for? : `1/root(3)(8-x)`

Answer 1

`1/root(3)(8-x) = 1 + 1/48x + 1/576 x^2 + 7/41472x^3 + ...`

`|x| lt 8`

Explanation:

We can derive a MacLaurin Series by using the Binomial Expansion: The binomial series tell us that:

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

We can write:

`1/root(3)(8-x) = 1/(8-x)^(1/3)`
`" " = 1/((8)(1-x/8))^(1/3)`
`" " = 1/2(1-x/8)^(-1/3)`

And so for the given function, we can replace "`x`" by `-x/8` and substitute `n=-1/3` in the binomial series:

`1/root(3)(8-x) = 1/2{1 + (-1/3)(-x/8) + ((-1/3)(-4/3))/(2!)(-x/8)^2 + ((-1/3)(-4/3)(-7/3))/(3!)(-x/8)^3 + ...`

`" " = 1/2{1 + x/24 + (4/9)/(2) x^2/64 + (28/27)/(6)x^3/512 + ... }`

`" " = 1/2{1 + 1/24x + 1/288 x^2 + 7/20736x^3 + ... }`

`" " = 1 + 1/48x + 1/576 x^2 + 7/41472x^3 + ...`

The Radius of Convergence is given by:

`| -x/8 | lt 1 => |x| lt 8`