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

Aug 23, 2017

$f \left(x\right) = 1 - \frac{1}{2} x - \frac{1}{8} {x}^{2} - \frac{1}{16} {x}^{3} - \frac{5}{128} {x}^{4} + \ldots$

Explanation:

Let:

$f \left(x\right) = \sqrt{1 - x}$

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:

$f \left(x\right) = {\left(1 - x\right)}^{\frac{1}{2}}$

 \ \ \ \ \ \ \ = 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 + ...

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 1 - \frac{1}{2} x - \frac{\frac{1}{4}}{2} {x}^{2} - \frac{\frac{3}{8}}{6} {x}^{3} - \frac{\frac{15}{16}}{24} {x}^{4} - \ldots$

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