A [[Polynomial]] [[Interpolation]] method between points (similar to [[Lagrange Polynomial|lagrange polynomials]]) using [[Divided Difference|Divided Differences]]. $\huge P(x) = \sum_{i=1}^{n} f[x_{1}, \dots, x_{i}] \prod_{j=1}^{i-1}(x-x_{j}) $ Note that because a polynomial connecting these points is unique, this is simply a different representation of the [[Lagrange Polynomial]]. Given a [[Set|set]] of values $\set{x_{1},x_{2},\dots,x_{n}}$ and $y_{1},y_{2},\dots, y_{n}$ such that $f(x_{i})=y_{i}$. Our [[Polynomial]] for [[Interpolation]] using this method would be the following: $ \begin{align} P(x) &= f[x_{1}] + f[x_{1},x_{2}](x-x_{1}) + \cdots + f[x_{1},\dots,x_{n}](x-x_{j}) \cdots\pa{x-x_{n-1}}\\ &= \sum_{i=1}^{n} f[x_{1}, \dots, x_{i}] \prod_{j=1}^{i-1}(x-x_{j}) \end{align}$ ### Constraining Slopes In order to not only constraint the location of certain [[Point|points]] for $P$, but also the [[Derivative|derivatives]] at one (or more) of those points, a trick is to add a new constraint point that is very close to the point whose derivative you want to constrain. If you position this new point such that the slope of the line between them is the target derivative, you can constrain the slope. In order to be exact, the offset of this new point $\Delta x$ must be 0. In this case, when we follow our divided difference charts and encounter a division by $0$, we instead take the limit as $\Delta x\to 0$, which is the target derivative. Given a set of points we want to interpolate $\set{p_{1},p_{2},\dots,p_{n}}$, and a constraint that $P'(x_k) = m_{k}$, we create an interpolating polynomial with the points $\set{p_{1},p_{2},\dots,p_{n}, (x_{k}+\Delta x,y_{k}+m_{k}\Delta x)}$ as $\Delta x \to 0$. This can be generalized to constraining higher order $m$-th derivatives by having $n+1$ points. $\large \left\{ \dots, (x_{k},y_{k}) , \dots ,\pa{ x_{k} + \d x, y_{k} + \frac{1}{d!} \deriv{^{m} P}{x^{m}}(x_{k}), } \right\} $