for fx=(1/x) centered at c=1
Assume that $\displaystyle f(x)$ can be expressed as a polynomial.
Since it is centred at $\displaystyle c = 1$, it will be of the form
$\displaystyle x^{-1} = a + b(x - 1) + c(x - 1)^2 + d(x - 1)^3 + e(x - 1)^4 + \dots$.
At $\displaystyle x = 1$ we find $\displaystyle a = 1$.
Take the derivative of both sides
$\displaystyle -x^{-2} = b + 2c(x - 1) + 3d(x - 1)^2 + 4e(x - 1)^3 + 5f(x - 1)^4 + \dots$
At $\displaystyle x = 1$ we find $\displaystyle b = -1$.
Take the derivative of both sides
$\displaystyle 2x^{-3} = 2c + 6d(x - 1) + 12e(x - 1)^2 + 20f(x - 1)^3 + \dots$
At $\displaystyle x = 1$ we find $\displaystyle c = 1$.
Take the derivative of both sides
$\displaystyle -6x^{-4} = 6d + 24e(x - 1) + 60f(x - 1)^2 + \dots$
At $\displaystyle x = 1$ we find $\displaystyle d = -1$.
Take the derivative of both sides
$\displaystyle 24x^{-5} = 24e + 120f(x - 1) + \dots$
At $\displaystyle x = 1$ we find $\displaystyle e = 1$.
Take the derivative of both sides
$\displaystyle -120x^{-6} = 120f + \dots$.
At $\displaystyle x = 1$ we find $\displaystyle f = -1$.
Looking at our polynomial we have
$\displaystyle \frac{1}{x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots$
$\displaystyle \frac{1}{x} = \sum_{n = 0}^{\infty}{(-1)^n(x-1)^n}$.