Question

What is the Maclaurin series for? : `sqrt(1-x)`

Answer 1

`f(x) = 1 - 1/2x - 1/8x^2 - 1/16x^3 - 5/128x^4 + ...`

Explanation:

Let:

`f(x) = sqrt(1-x)`

We seek a Taylor Series, as no pivot is supplied, it is assumed that an expansion about the pivot `x=0` is required. This is known as a Maclaurin series and is given by
`f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n + ...`

Although we could use this method, it is actually quicker, in this case, to use a Binomial Series 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 + ...`

And so for the given function, we have:

`f(x) = (1-x)^(1/2)`

`\ \ \ \ \ \ \ = 1 + (1/2)(-x) + (1/2(-1/2))/(2!)(-x)^2 + (1/2(-1/2)(-3/2))/(3!)(-x)^3 + (1/2(-1/2)(-3/2)(-5/2))/(4!)(-x)^4 + ...`

`\ \ \ \ \ \ \ = 1 - 1/2x - (1/4)/(2)x^2 - (3/8)/(6)x^3 - (15/16)/(24)x^4 - ...`

`\ \ \ \ \ \ \ = 1 - 1/2x - 1/8x^2 - 1/16x^3 - 5/128x^4 + ...`