Question

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))`

Explanation:

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))`