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Polynomial Curves

Consider an nnth degree polynomial curve R\boldsymbol{R} parametrized by tt:

R(t)=∑i=0nAiti,\boldsymbol{R}(t) = \sum_{i=0}^n \boldsymbol{A_i} t^i,

with the initial conditions:

R(tj)=Pj,\boldsymbol{R}(t_j) = \boldsymbol{P_j},

where ti∈[0,1]t_i \in [0, 1].

The initial conditions imply:
∑i=0nAitji=Pj, \sum_{i=0}^n \boldsymbol{A_i} t_j^i = \boldsymbol{P_j},

this may be represented as matrix multiplication:

[tj0…tjn][A0⋮An]=Pj, \begin{bmatrix}t_j^0 & \dots & t_j^n\end{bmatrix} \begin{bmatrix} \boldsymbol{A_0} \\ \vdots \\ \boldsymbol{A_n} \end{bmatrix} = \boldsymbol{P_j},

or:

[t00…t0n⋮⋱⋮tn0…tnn][A0⋮An]=[P0⋮Pn]. \begin{bmatrix} t_0^0 & \dots & t_0^n \\ \vdots & \ddots & \vdots \\ t_n^0 & \dots & t_n^n \end{bmatrix} \begin{bmatrix} \boldsymbol{A_0} \\ \vdots \\ \boldsymbol{A_n} \end{bmatrix} = \begin{bmatrix} \boldsymbol{P_0} \\ \vdots \\ \boldsymbol{P_n} \end{bmatrix}.

Solving for AiA_i:

[A0⋮An]=[t00…t0n⋮⋱⋮tn0…tnn]−1[P0⋮Pn]. \begin{bmatrix} \boldsymbol{A_0} \\ \vdots \\ \boldsymbol{A_n} \end{bmatrix} = \begin{bmatrix} t_0^0 & \dots & t_0^n \\ \vdots & \ddots & \vdots \\ t_n^0 & \dots & t_n^n \end{bmatrix}^{-1} \begin{bmatrix} \boldsymbol{P_0} \\ \vdots \\ \boldsymbol{P_n} \end{bmatrix}.
x(t)=t,y(t)=t. \begin{align*} x(t) &= t, \\ y(t) &= t. \end{align*}