What is the Maclaurin series for? : #sqrt(1-x)#
1 Answer
# 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
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 + ...#
