Find the Taylor expansion #\color(red)\bb\text(formula)#... for #f(x)=1/x^2# given #a=4#?

Please check my work (it has to be done using the colored parts, sorry):

#f'=(-2)x^-3#
#f''=(-2)(-3)x^-4#
#f'''=(-2)(-3)(-4)x^-5#

#f'(4)=(-2)*4^-3#
#f''(4)=(-2)(-3)*4^-4#
#f'''(4)=(-2)(-3)(-4)*4^-5#

#\color(green)(f^n(4))=(-1)^n(n-1)!4^-n=\color(olive)(((-1)^n(n-1)!)/4^n)#

#\color(red)(C_n=f^n(a)*1/(n!))=((-1)^n(n-1)!)/4^n*1/(n!)=((-1)^n(n-1))/4^n#
#\rarr\color(red)(f(x)=\sum_(n=0)^\inftyC_n(x-a)^n)=\sum_(n=0)^\infty((-1)^n(n-1))/4^n*(x-2)^n#

(Can I simplify this further?)

Answer 1

#1/x^2=sum_(n=0)^oo(-1)^n((x-4)^n(n+1))/(4^(n+2))#

Let's determine the pattern for the #nth# derivative evaluated at #4#. We adopt the convention that the #0th# derivative is just the function itself.
#f(x)=x^-2, f(4)=4^-2#
#f'(x)=-2x^-3, f'(4)=-2(4)^-3#
#f''(x)=(-2)(-3)x^-4, f''(4)=(-2)(-3)4^-4#
#f'''(x)=(-2)(-3)(-4)x^-5, f'''(4)=(-2)(-3)(-4)4^-5#
We can see that we start out with a positive term and then alternate between negative and positive terms, so, we have an instance of #(-1)^n.#
The first exponent on the #4# is #-2#, so, since we're starting at #n=0,# we're going to have a #4^-(n+2)=1/4^(n+2).#
Finally, for the factorial, the 0th derivative has a coefficient of #1#, or #1!#, the first derivative has a coefficient of #-2,# or #-(2!),# the second derivative has a coefficient of #(-2)(-3),# or #3!#.
Knowing that the negatives are handled by the #(-1)^n,# we can represent the factorials by #(n+1)!#

Thus,

#f^((n))(4)=(-1)^n((n+1)!)/4^(n+2)#
Now, using the general form of a Taylor series about #a,#
#sum_(n=0)^oof^((n))(a)(x-a)^n/(n!),# we get
#1/x^2=sum_(n=0)^oo(-1)^n((x-4)^n(n+1)!)/(n!4^(n+2))#
We can simplify the factorials, as #(n+1)! = (n+1)(n!)#
#1/x^2=sum_(n=0)^oo(-1)^n((x-4)^n(n+1)cancel(n!))/(cancel(n!)4^(n+2))#
#1/x^2=sum_(n=0)^oo(-1)^n((x-4)^n(n+1))/(4^(n+2))#
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

The Taylor expansion formula for a function ( f(x) ) centered at ( a ) is:

[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n ]

For ( f(x) = \frac{1}{x^2} ) centered at ( a = 4 ), we need to find the derivatives of ( f(x) ) and evaluate them at ( x = 4 ). Then, we substitute these values into the Taylor series formula.

First, let's find the derivatives of ( f(x) ):

[ f(x) = \frac{1}{x^2} ] [ f'(x) = -\frac{2}{x^3} ] [ f''(x) = \frac{6}{x^4} ] [ f'''(x) = -\frac{24}{x^5} ]

Now, let's evaluate these derivatives at ( x = 4 ):

[ f(4) = \frac{1}{4^2} = \frac{1}{16} ] [ f'(4) = -\frac{2}{4^3} = -\frac{1}{8} ] [ f''(4) = \frac{6}{4^4} = \frac{3}{32} ] [ f'''(4) = -\frac{24}{4^5} = -\frac{3}{16} ]

Now, we substitute these values into the Taylor series formula:

[ f(x) = f(4) + f'(4)(x - 4) + \frac{f''(4)}{2!}(x - 4)^2 + \frac{f'''(4)}{3!}(x - 4)^3 + \ldots ]

[ = \frac{1}{16} - \frac{1}{8}(x - 4) + \frac{3}{64}(x - 4)^2 - \frac{1}{16}(x - 4)^3 + \ldots ]

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