Question

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

Answer 1

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+...`