# How can the equation of a Bézier curve be transformed from a Bézier basis function to a bivariate function?

Several nights ago, I was researching the problem of identifying self-intersections in arbitrary curves, particularly Bézier curves. (The reason being is that I want to write a program that inserts the little gaps typically seen in 2D knot diagrams.) I came across this paper, which seems to have what I want. However, I'm running into a logic gap that is hindering my application of their work to my program. Below, I've reproduced the relevant bits.

Here, they have a subtraction of the same Bézier basis function ($C(x)$) with two independent variables.

\begin{align} C(u)-C(v) &= \sum^{n}_{i=0}{a_i u^i} - \sum^{n}_{i=0}{a_i v^i} \\ &= \sum^{n}_{i=0}{a_i (u^i - v^i)} \end{align}

Then they let

$$I(u,v) = C(u)-C(v)$$

From there, they want to find $\hat{I}$ such that

$$(u-v)\hat{I}(u,v) = I(u,v)$$

Then they somehow make the jump to "in Bézier basis function terms, this reduces to":

$$(u-v)\sum^{m-1}_{i=0}\sum^{n-1}_{j=0} p_{ij}\theta_{i}^{m-1}(u)\theta_{j}^{n-1}(v) \;\; = \;\; \sum^{m}_{k=0}\sum^{n}_{l=0} q_{kl}\theta_{k}^{m}(u)\theta_{l}^{n}(v)$$ Where $\theta_{k}^{m}$ is the $k$th Bézier basis function of order $m$.

...how in the world did they get from a subtraction of two univariate Bézier basis functions to a bivariate function that has a product of two Bézier basis functions?

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