How do you find a power series converging to f(x)=sqrt(1-x^2) and determine the radius of convergence?
1 Answer
Dec 30, 2017
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
