Question

How do you find a power series converging to `f(x)=sqrt(1-x^2)` and determine the radius of convergence?

Answer 1

`sqrt(1-x^2) = 1 - 1/2x^2 - 1/8x^4 - 1/16x^6 - 5/128x^8 + ...`

`| x | lt 1`

Explanation:

Let:

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

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 + ...`

Which converges for:

`| x | lt 1`

And so for the given function, we have:

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

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

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

`\ \ \ \ \ \ \ = 1 - 1/2x^2 - 1/8x^4 - 1/16x^6 - 5/128x^8 + ...`

The series is valid provided

`| -x^2 | lt 1 => | x | lt 1`