How do you find the first terms of the taylor series #f(x)=1/sqrt(x)# centered at #x=4#?

Answer 1
If #f(x) = 1/(sqrt(x)) = x^(-1/2)# then #f'(x) = - 1/2x^(-3/2) = - 1/2 (1/sqrt(x))^3# #f''(x) = 3/4x^(-5/2) - 1/2 (1/sqrt(x))^5# #f'''(x) = -(15)/8x^(-7/2) - 1/2 (1/sqrt(x))^7# and so on
The Taylor series is given by #sum_(n=0)^oo = (f^('n)(a))/(n!) * (x - a)^n#
so the first #4# terms of the Taylor series of #f(x) = 1/(sqrt(x))# centered at #a=4# would be (after noting that #sqrt(4) = 2#)
#1//2 + (- 1/2 (1/2)^3)(x-4) + ((3/4 (1/2)^5)(x-4)^2)/2 + ((- 7/2 (1/2)^7)(x-4)^3)/(3!)#

From there on, it's just arithmetic.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the first terms of the Taylor series for ( f(x) = \frac{1}{\sqrt{x}} ) centered at ( x = 4 ), we first find the derivatives of ( f(x) ) and evaluate them at ( x = 4 ). Then, we express the Taylor series using these derivatives.

  1. Find the derivatives of ( f(x) ): [ f(x) = x^{-\frac{1}{2}} ] [ f'(x) = -\frac{1}{2}x^{-\frac{3}{2}} ] [ f''(x) = \frac{3}{4}x^{-\frac{5}{2}} ] [ f'''(x) = -\frac{15}{8}x^{-\frac{7}{2}} ]

  2. Evaluate the derivatives at ( x = 4 ): [ f(4) = \frac{1}{2} ] [ f'(4) = -\frac{1}{16} ] [ f''(4) = \frac{3}{128} ] [ f'''(4) = -\frac{15}{1024} ]

  3. Express the Taylor series: [ f(x) = f(4) + f'(4)(x-4) + \frac{f''(4)}{2!}(x-4)^2 + \frac{f'''(4)}{3!}(x-4)^3 + \ldots ]

Substitute the evaluated derivatives: [ f(x) = \frac{1}{2} - \frac{1}{16}(x-4) + \frac{3}{128 \cdot 2!}(x-4)^2 - \frac{15}{1024 \cdot 3!}(x-4)^3 + \ldots ]

These are the first terms of the Taylor series for ( f(x) = \frac{1}{\sqrt{x}} ) centered at ( x = 4 ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To find the first terms of the Taylor series ( f(x) = \frac{1}{\sqrt{x}} ) centered at ( x = 4 ), we first need to find the derivatives of ( f(x) ) at ( x = 4 ). Then, we evaluate these derivatives at ( x = 4 ) to find the coefficients of the Taylor series.

  1. Find the derivatives of ( f(x) ): [ f(x) = x^{-\frac{1}{2}} ] [ f'(x) = -\frac{1}{2}x^{-\frac{3}{2}} ] [ f''(x) = \frac{3}{4}x^{-\frac{5}{2}} ] [ f'''(x) = -\frac{15}{8}x^{-\frac{7}{2}} ]

  2. Evaluate the derivatives at ( x = 4 ): [ f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2} ] [ f'(4) = -\frac{1}{2}(4)^{-\frac{3}{2}} = -\frac{1}{4} ] [ f''(4) = \frac{3}{4}(4)^{-\frac{5}{2}} = \frac{3}{32} ] [ f'''(4) = -\frac{15}{8}(4)^{-\frac{7}{2}} = -\frac{15}{128} ]

  3. Write down the Taylor series using these derivatives: [ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots ]

[ f(x) = \frac{1}{2} - \frac{1}{4}(x-4) + \frac{3}{32}(x-4)^2 - \frac{15}{128}(x-4)^3 + \cdots ]

Therefore, the first terms of the Taylor series for ( f(x) = \frac{1}{\sqrt{x}} ) centered at ( x = 4 ) are:

[ \frac{1}{2} - \frac{1}{4}(x-4) + \frac{3}{32}(x-4)^2 - \frac{15}{128}(x-4)^3 ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

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.

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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7