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

Question

Christopher Agresta · Accepted Answer

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

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 + ...#

Scarlett Allison · Answer

undefined The Maclaurin series for $ \sqrt{1-x} $ is:

$$ \sqrt{1-x} = 1 - \frac{x}{2} + \frac{x^2}{8} - \frac{x^3}{16} + \frac{5x^4}{128} - \cdots $$

This series can be used to approximate the value of $ \sqrt{1-x} $ for values of $ x $ close to 0.