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

Question

Ezra Adams · Accepted Answer

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+...#

Christopher Agresta · Answer

undefined To construct the Taylor series for $ f(x) = \frac{1}{\sqrt{x}} $ centered at $ x = 4 $, we first need to find the derivatives of $ f(x) $ at $ x = 4 $ to determine the coefficients of the series. The nth derivative of $ f(x) $ is given by:

$$ f^{(n)}(x) = (-1)^n \cdot \frac{(2n-3)!!}{2^n} \cdot \frac{1}{x^{(3/2)+n}} $$

where $ !! $ represents the double factorial. Evaluating these derivatives at $ x = 4 $, we get:

$$ f(4) = \frac{1}{2} $$
$$ f'(4) = -\frac{1}{8} $$
$$ f''(4) = \frac{3}{128} $$
$$ f'''(4) = -\frac{15}{2048} $$

The Taylor series for $ f(x) $ centered at $ x = 4 $ is then:

$$ f(x) = f(4) + f'(4)(x-4) + \frac{f''(4)}{2!}(x-4)^2 + \frac{f'''(4)}{3!}(x-4)^3 + \cdots $$

Substituting the values of the derivatives at $ x = 4 $ into the series, we get:

$$ f(x) = \frac{1}{2} - \frac{1}{8}(x-4) + \frac{3}{128}\frac{(x-4)^2}{2!} - \frac{15}{2048}\frac{(x-4)^3}{3!} + \cdots $$

Simplifying, we have:

$$ f(x) = \frac{1}{2} - \frac{1}{8}(x-4) + \frac{3}{256}(x-4)^2 - \frac{5}{8192}(x-4)^3 + \cdots $$