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

Question

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^n1/(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?)

Cameron Alexander · Accepted Answer

#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))#

Cameron Alexander · Answer

undefined 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:

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

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

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