Partial fractions for any function?

Question

I recently discovered a property of polynomials that have all roots distinct. That is, if $a_i \neq a_j$ for $i \neq j$, and if $$f(x) = \prod_{i=1}^{n} (x-a_i)$$ Then, $$\sum_{i=1}^{n} \frac {f(x)}{(x-a_i)f'(a_i)} = 1$$ Which can be proved by noting that the LHS is an $n-1$ degree polynomial and if it equals the RHS for $n$ distinct $x$, it must be true always. But it's obviously true for the $n$ roots of $f(x)$ (in the limit as $x$ tends to $a_i$)

This was useful because if I had the reciprocal of such a polynomial, I could easily break it into partial fractions like: $$\boxed{\frac {1}{f(x)} = \frac{1/ f'(a_1)}{x-a_1} +\frac{1/f'(a_2)}{x-a_2} + ... + \frac{1/f'(a_n)}{x-a_n}}$$

Now, my question is: Is this property satisfied by ANY function (apart from polynomials) that doesn't have repeated roots? (Continuous, differentiable) For example, consider $f(x) = \tan x$ Then, $f(x)$ has roots at $x=k \pi$ for every integer $k$ And it doesn't have repeated roots, i.e., if $f(t)=0$ then $f'(t) \neq 0$

So, if we want to verify the equation in the box, it would be: $$\cot x = \sum_{n= -\infty}^{\infty} \frac {1}{x-n \pi}$$ Which is correct because it is the same as the well known $$\pi \cot \pi x - \frac {1}{x} = \sum_{n=1}^{\infty} \frac {2x}{x^2 - n^2}$$

So, would ANY continuous, differentiable function with no repeated roots satisfy this? If not, then what issues do I need to care about other than convergence?

Answer 1

If you look at the function in the complex plane, what you wrote is equivalent to a Laurent Series. You can prove that if a function has isolated poles and no essential singularity then you can write your function around a pole in a Laurent series $\Sigma_{n=-m}^{\infty} {a_n}(z-z_0)^n$. Thus you can extract out the function as $\Sigma_{poles={z_i}}\frac{g_i(z)}{z-z_i}$ where $g_i$ are meromorphic. See also https://en.wikipedia.org/wiki/Mittag-Leffler%27s_theorem.