# How do I construct a Taylor series for f(x)=1/sqrt(x) centered at x=4?

Mar 5, 2015

The Taylor series expansion of a function $f \left(x\right)$ around a point, say $a$, is given by,

f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/(2!) + f'''(a)(x-a)^3/(3!) + ...

Applying this to our function,

f(x) = f(4) + f'(4)(x-4) + f''(4)(x-4)^2/(2!) + f'''(4)(x-4)^3/(3!) + ...

Therefore,

f(x) = 1/4^(1/2) +(-1/2)(1/4)^(3/2)(x-4)+(-1/2)(-3/2)(1/4)^(5/2)(x-4)^2/(2!)+(-1/2)(-3/2)(-5/2)(1/4)^(7/2)(x-4)^3/(3!)+...

The most reduced form of the equation would be,

$f \left(x\right) = \frac{1}{2} + \left(- \frac{1}{16}\right) \left(x - 4\right) + \left(\frac{3}{256}\right) {\left(x - 4\right)}^{2} + \left(- \frac{5}{2048}\right) {\left(x - 4\right)}^{3} + \ldots$