Polynomials based on Taylor expansion that
form a basis for Taylor curves,, see expression (5.16) in *Blending techniques in
Curve and Surface constructions*

A Taylor curve is expressed by the
following formula: \( C(t) = \displaystyle\sum_{i=0}^{d} C^{(i)}(t_0)\
\frac{(t-t_0)^i}{i!} \), where \(d\) is the polynomial degree,
and where \(c^{(0)}(t_0) = C(t_0)\), \(c^{(1)}(t_0) =
C'(t_0)\) and so on, and where \( \frac{(t-t_0)^i}{i!}, \
i=0,1,...,d \) are basis functions based on Taylor expansion. In the
figure above, a set of basis functions of degree \(d\) is plotted. Note
that the parameter domain is determined in the menu at the top right.

In the figure below, we have a point (grey pentagon) with associated 1st derivatives (red vector), 2nd derivatives (yellow vector), ... up to \(d\)th derivatives. The green curve is the result of the Taylor expansion over the parameter interval [parameter start, parameter end]. In the figure is \( t_0 = 0\).

In the figure below, we have a point (grey pentagon) with associated 1st derivatives (red vector), 2nd derivatives (yellow vector), ... up to \(d\)th derivatives. The green curve is the result of the Taylor expansion over the parameter interval [parameter start, parameter end]. In the figure is \( t_0 = 0\).

- In the menu at the top right, you can determine the polynomial degree \(d\) of the curve at the top.
- In the next two fields in the menu, you determine the parameter domain of the curve. However, the positon is locked to the parameter value \(t_0=0\).
- In the figure below, you can change the arrows, which represent derivatives of different orders. You can change them by pressing the left mouse button when the cursor is over an arrowhead. The tip of the arrow moves as long as you hold the button down. When you release the button, the arrow (the corresponding derivative) is locked and the curve is updated.

The plot is programmed according to chapter 5.5 in the book.

A Taylor curve is a curve defined by a Taylor expansion, starting from a point and a set of derivatives of successive order.

To create a Taylor curve, we need a point with an associated parameter value and a degree that indicates how many vectors representing derivatives of successive order we need.

If we look at the Taylor basis of degree 3, we get in matrix form: \( \begin{bmatrix} 1 & 0 & 0 & 0 \\ -t_0 & 1 & 0 & 0 \\ \frac{t_0^2}{2} & -t_0 & \frac{1}{2} & 0\\ \frac{-t_0^3}{6} & \frac{t_0^2}{2} & \frac{-t_0}{2} & \frac{1}{6} \end{bmatrix} \begin{bmatrix} 1 \\ t\\ t^2 \\ t^3 \end{bmatrix} \). This matrix is then the basis change matrix from a monomial degree 3 basis to the Taylor basis of degree 3.

Hvis vi inverterer matrisen får vi : \( \begin{bmatrix} 1 \\ t\\ t^2 \\ t^3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ t_0 & 1 & 0 & 0 \\ t_0^2 & 2t_0 & 2 & 0\\ t_0^3 & 3t_0^2 & 6t_0 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ t-t_0\\ \frac{(t-t_0)^2}{2} \\ \frac{(t-t_0)^3}{6} \end{bmatrix}\). This matrix is then the basis change matrix from a monomial degree 3 basis to the Taylor basis of degree 3.

If \(t_0=0 \) we get a simple diaginal matrix. If the degree is greater than 3, just add new rows following the same pattern. At degree 4, the first matrix gets a new row \(\begin{bmatrix} \frac{t_0^4}{24} & \frac{-t_0^3}{6} & \frac{t_0^2}{4} & \frac{-t_0}{6} & \frac{1}{24}\end{bmatrix}\) and the second matrix a new row \(\begin{bmatrix} t_0^4 & 4t_0^3 & 12t_0^2 & 24t_0 & 24\end{bmatrix}\).