Why is the Padé approximant typically written in this form? $$R(x) = \frac{\sum_{j = 0}^m a_{j} x^j}{1+\sum_{k=1}^n b_k x^k}$$
I've started computing these to approximate my coefficients for a regression and others have been asking me how the Padé approximant works and why it is written how it is. I don't get why we add one to the sum in the denominator instead of just taking care of it as part of the summation in the denominator.
Numerical Recipes in C, where I got this function, doesn't really explain how it works or why. Wikipedia doesn't really cover this topic. Both just state the theorem and go on. I've requested those books through my university library but they are in the mail. 
Is it just a historical holdover?
Thank you for your time.
 A: Well the most obvious reason is that if we had $b_0$ replace $1$ it would be a useless degree of freedom (at least over a field).  This is basically it. One area where rational functions arise is in signal processing and in the (frequency/Z domain) transfer functions of infinite impulse response filters, i.e. filters defined by a linear recurrence.
Let $x$ and $y$ be sequences and $y = R(Z^{-1})x$ where $Z^{-1}$ is a delay operator $(Z^{-1}x)_n = x_{n-1}$. We have $y + \sum_{k=1}^N b_k Z^{-k} y = \sum_{j=0}^M a_j Z^{-j} x$ which is $$y_n = \sum_{j=0}^M a_j x_{n-j} - \sum_{k=1}^N b_k y_{n-k}$$ This is a causal linear recurrence equation.  That is, it calculates the next output value $y_n$ in terms of previous and current input values $x_{n-j}$ and previous output values $y_{n-k}$. There's little reason to have the equation solve for $b_0y_n$; we'd just need to divide through anyway to get the current output. It's simpler and cheaper to just include it in the polynomial coefficients.
To make it closer to Padé approximants, just replace the delay operators with inverse differential operators.
A: Instead of computing $f (x) $ by its Taylor expansion
$$f (x)=\sum_{i=0}^{n+m}c_ix^i+o (x^{n+m}) $$
we use a rational approximation with the same precision.
$$f (x)\approx \frac {P_m (x)}{Q_n (x)} $$
with $Q_n (0)=1$ and
$$P_m (x)-f (x)Q_(x)=o (x^{n+m}) .$$
