Taylor
Curves
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 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 colors of the arrows represent \(
C'(0)\) , \( C''(0)\)
, \( C^{(3)}(0)\)
, \( C^{(4)}(0)\)
, \( C^{(5)}(0)\)
, \( C^{(6)}(0)\).
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
\(\begin{bmatrix} t_0^4 & 4t_0^3 & 12t_0^2 & 24t_0 &
24\end{bmatrix}\).