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

1 Answer
Mar 5, 2015

The Taylor series expansion of a function #f(x)# 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(x) = 1/2 + (-1/16)(x-4)+(3/256)(x-4)^2+(-5/2048)(x-4)^3+...#