How do you find a power series converging to #f(x)=sqrt(1+2x)# and determine the radius of convergence?
1 Answer
Dec 31, 2017
# sqrt(1-x^2) = 1 + x - 1/2x^2 + 1/2x^3 - 5/8x^4 + ...#
# | x | lt 2 #
Explanation:
Let:
# f(x) = sqrt(1+2x) #
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+2x)^(1/2) #
# \ \ \ \ \ \ \ = 1 + (1/2)(2x) + (1/2(-1/2))/(2!)(2x)^2 + (1/2(-1/2)(-3/2))/(3!)(2x)^3 + (1/2(-1/2)(-3/2)(-5/2))/(4!)(2x)^4 + ...#
# \ \ \ \ \ \ \ = 1 + 1/2(2x) - (1/4)/(2)(4x^2) + (3/8)/(6)(8x^3) - (15/16)/(24)(16x^4) + ...#
# \ \ \ \ \ \ \ = 1 + x - 1/2x^2 + 3/6x^3 - 15/24x^4 + ...#
# \ \ \ \ \ \ \ = 1 + x - 1/2x^2 + 1/2x^3 - 5/8x^4 + ...#
The series is valid provided
# | 2x | lt 1 => 1/2| x | lt 1 => |x| lt 2#
