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

Applying this to our function,

Therefore,

The most reduced form of the equation would be,

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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 ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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