De Boor遞推算法

De Boor算法

設$u\in\left[u_j,u_{j+1}\right)$，$V_{i,0}=V_i$，對於$i=j-p,\cdots,j$

令
$V_{i,k}=\dfrac{u_{i+p+1-k}-u}{u_{i+p+1-k}-u_i}V_{i-1,k-1}+\dfrac{u-u_i}{u_{i+p+1-k}-u_i}V_{i,k-1},\quad k=1,\cdots,p,\quad i=j-p+k,\cdots,j$
其中$V_i$爲控制點，$p$爲B樣條的冪次，$P(u)$爲B樣條曲線，則
$P(u)=V_{j,p}.$

De Boor遞推算法求B樣條曲線上的點

設$u\in\left[u_j,u_{j+1}\right)$，則
$\begin{aligned} P\left(u\right)=& \sum\limits_{i=0}^{n}N_{i,p}(u)V_i =\sum\limits_{i=j-p}^{j}N_{i,p}(u)V_i \\ =& \sum\limits_{i=j-p}^{j}\left(\dfrac{u-u_i}{u_{i+p}-u_i}N_{i,p-1}(u)+\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u)\right)V_i \\ =& \sum\limits_{i=j-p}^{j}\dfrac{u-u_i}{u_{i+p}-u_i}N_{i,p-1}(u)V_i +\sum\limits_{i=j-p}^{j}\dfrac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u)V_i \\ =& \sum\limits_{i=j-p+1}^{j}\dfrac{u-u_i}{u_{i+p}-u_i}N_{i,p-1}(u)V_i +\sum\limits_{i=j-p+1}^{j}\dfrac{u_{i+p}-u}{u_{i+p}-u_i}N_{i,p-1}(u)V_{i-1} \\ = &\sum\limits_{i=j-p+1}^{j}\left(\dfrac{u-u_i}{u_{i+p}-u_i}V_i+\dfrac{u_{i+p}-u}{u_{i+p}-u_i}V_{i-1}\right)N_{i,p-1}(u). \end{aligned}$
令
$\begin{aligned} V_{i,k}=\begin{cases} V_i,\quad k=0 \\ \dfrac{u_{i+p+1-k}-u}{u_{i+p+1-k}-u_i}V_{i-1,k-1}+\dfrac{u-u_i}{u_{i+p+1-k}-u_i}V_{i,k-1},\quad k=1,\cdots,p \end{cases} \end{aligned}$
則
$P\left(u\right)=\sum\limits_{i=j-p+k}^{j}N_{i,p-k}(u)V_{i,k}=V_{j,p}$

De Boor遞推算法求B樣條曲線的一階導矢

設$u\in\left[u_j,u_{j+1}\right)$，則
$\begin{aligned} P^{'}\left(u\right)=& \sum\limits_{i=0}^{n}N^{'}_{i,p}(u)V_i=\sum\limits_{i=j-p}^{j}N^{'}_{i,p}(u)V_i \\ =&\sum\limits_{i=j-p}^{j}\left(\dfrac{p}{u_{i+p}-u_i}N_{i,p-1}(u)-\dfrac{p}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u)\right)V_i \\ =&\sum\limits_{i=j-p}^{j}\dfrac{p}{u_{i+p}-u_i}N_{i,p-1}(u)V_i-\sum\limits_{i=j-p}^{j}\dfrac{p}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u)V_i \\ =&\sum\limits_{i=j-p+1}^{j}\dfrac{p}{u_{i+p}-u_i}N_{i,p-1}(u)V_i-\sum\limits_{i=j-p+1}^{j}\dfrac{p}{u_{i+p}-u_i}N_{i,p-1}(u)V_{i-1} \\ =&p\sum\limits_{i=j-p+1}^{j}\dfrac{V_i-V_{i-1}}{u_{i+p}-u_i}N_{i,p-1}(u). \end{aligned}$
記
$\begin{aligned} \Delta=& \sum\limits_{i=j-p+1}^{j}\dfrac{V_i-V_{i-1}}{u_{i+p}-u_i}N_{i,p-1}(u) \\ =& \sum\limits_{i=j-p+1}^{j}\dfrac{V_i-V_{i-1}}{u_{i+p}-u_i}\left(\dfrac{u-u_i}{u_{i+p-1}-u_i}N_{i,p-2}(u)+ \dfrac{u_{i+p}-u}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)\right) \\ =& \sum\limits_{i=j-p+1}^{j}\dfrac{V_i-V_{i-1}}{u_{i+p}-u_i}\dfrac{u-u_i}{u_{i+p-1}-u_i}N_{i,p-2}(u) +\sum\limits_{i=j-p+1}^{j}\dfrac{V_i-V_{i-1}}{u_{i+p}-u_i}\dfrac{u_{i+p}-u}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u) \\ =& \sum\limits_{i=j-p+2}^{j}\dfrac{V_i-V_{i-1}}{u_{i+p}-u_i}\dfrac{u-u_i}{u_{i+p-1}-u_i}N_{i,p-2}(u) +\sum\limits_{i=j-p+2}^{j}\dfrac{V_{i-1}-V_{i-2}}{u_{i+p-1}-u_{i-1}}\dfrac{u_{i+p-1}-u}{u_{i+p-1}-u_{i}}N_{i,p-2}(u) \\ =& \sum\limits_{i=j-p+2}^{j}\left(\dfrac{V_i-V_{i-1}}{u_{i+p}-u_i}\dfrac{u-u_i}{u_{i+p-1}-u_i}+ \dfrac{V_{i-1}-V_{i-2}}{u_{i+p-1}-u_{i-1}}\dfrac{u_{i+p-1}-u}{u_{i+p-1}-u_i}\right)N_{i,p-2}(u). \end{aligned}$
因此
$\begin{aligned} P^{'}\left(u\right)=& p\sum\limits_{i=j-p+1}^{j}\dfrac{V_i-V_{i-1}}{u_{i+p}-u_i}N_{i,p-1}(u) \\ =& p\sum\limits_{i=j-p+2}^{j}\left(\dfrac{V_i-V_{i-1}}{u_{i+p}-u_i}\dfrac{u-u_i}{u_{i+p-1}-u_i}+ \dfrac{V_{i-1}-V_{i-2}}{u_{i+p-1}-u_{i-1}}\dfrac{u_{i+p-1}-u}{u_{i+p-1}-u_i}\right)N_{i,p-2}(u). \end{aligned}$
令
$\begin{aligned} Q_{i,k}=\begin{cases} \dfrac{V_i-V_{i-1}}{u_{i+p}-u_i},\quad k=1 \\ \dfrac{u_{i+p+1-k}-u}{u_{i+p+1-k}-u_i}Q_{i-1,k-1} +\dfrac{u-u_i}{u_{i+p+1-k}-u_i}Q_{i,k-1},\quad k=2,\cdots,p \end{cases} \end{aligned}$
下面用數學歸納法證明
$Q_{i,k}=\dfrac{V_{i,k-1}-V_{i-1,k-1}}{u_{i+p+1-k}-u_i},\quad k=1,\cdots,p.$
當$k=1$時，$Q_{i,k}=\dfrac{V_{i,k-1}-V_{i-1,k-1}}{u_{i+p+1-k}-u_i}$，於是有
$\begin{aligned} Q_{i,k+1}=& \dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i}Q_{i-1,k}+ \dfrac{u-u_i}{u_{i+p-k}-u_i}Q_{i,k} \\ =& \dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i}\dfrac{V_{i-1,k-1}-V_{i-2,k-1}}{u_{i+p-k}-u_i} +\dfrac{u-u_i}{u_{i+p-k}-u_i}\dfrac{V_{i,k-1}-V_{i-1,k-1}}{u_{i+p+1-k}-u_i} \\ =& \dfrac{1}{u_{i+p-k}-u_i}\left[\dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i}\left(V_{i-1,k-1}-V_{i-2,k-1}\right) +\dfrac{u-u_i}{u_{i+p+1-k}-u_i}\left(V_{i,k-1}-V_{i-1,k-1}\right)\right]. \end{aligned}$
又
$\begin{aligned} V_{i,k}-V_{i-1,k}=& \dfrac{u_{i+p+1-k}-u}{u_{i+p+1-k}-u_i}V_{i-1,k-1} +\dfrac{u-u_i}{u_{i+p+1-k}-u_i}V_{i,k-1} \\ &-\dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i}V_{i-2,k-1} -\dfrac{u-u_i}{u_{i+p-k}-u_i}V_{i-1,k-1} \\ =& \left(V_{i-1,k-1} +\dfrac{u_i-u}{u_{i+p+1-k}-u_i}V_{i-1,k-1}\right) -\dfrac{u-u_i}{u_{i+p-k}-u_i}V_{i-1,k-1} \\ &+\dfrac{u-u_i}{u_{i+p+1-k}-u_i}V_{i,k-1} -\dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i}V_{i-2,k-1} \\ =& \left(V_{i-1,k-1} -\dfrac{u-u_i}{u_{i+p-k}-u_i}V_{i-1,k-1}\right) +\dfrac{u_i-u}{u_{i+p+1-k}-u_i}V_{i-1,k-1} \\ &+\dfrac{u-u_i}{u_{i+p+1-k}-u_i}V_{i,k-1} -\dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i}V_{i-2,k-1} \\ =& \dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i}V_{i-1,k-1} -\dfrac{u-u_i}{u_{i+p+1-k}-u_i}V_{i-1,k-1} \\ &+\dfrac{u-u_i}{u_{i+p+1-k}-u_i}V_{i,k-1} -\dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i}V_{i-2,k-1} \\ =& \dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i}\left(V_{i-1,k-1}-V_{i-2,k-1}\right) +\dfrac{u-u_i}{u_{i+p+1-k}-u_i}\left(V_{i,k-1}-V_{i-1,k-1}\right). \end{aligned}$
因此，可得
$Q_{i,k+1}=\dfrac{V_{i,k}-V_{i-1,k}}{u_{i+p-k}-u_i}.$
證畢。

利用上面的結果，可以得到
$P^{'}\left(u\right)=p\sum\limits_{i=j-p+k}^{j}N_{i,p-k}(u)Q_{i,k}=pQ_{j,p}=p\dfrac{V_{j,p-1}-V_{j-1,p-1}}{u_{j+1}-u_j}.$

De Boor遞推算法求B樣條曲線的二階導矢

設$u\in\left[u_j,u_{j+1}\right)$，則
$\begin{aligned} P^{''}\left(u\right)=& \left(P\left(u\right)\right)^{'} =\left(\sum\limits_{i=j-p}^{j}N^{'}_{i,p}(u)V_i\right)^{'} \\ =& \left(\sum\limits_{i=j-p}^{j}\dfrac{pV_i}{u_{i+p}-u_i}N_{i,p-1}(u) -\sum\limits_{i=j-p}^{j}\dfrac{pV_i}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u)\right)^{'} \\ =& \sum\limits_{i=j-p}^{j}\dfrac{pV_i}{u_{i+p}-u_i}N^{'}_{i,p-1}(u) -\sum\limits_{i=j-p}^{j}\dfrac{pV_i}{u_{i+p+1}-u_{i+1}}N^{'}_{i+1,p-1}(u) \\ =& \sum\limits_{i=j-p}^{j}\dfrac{pV_i}{u_{i+p}-u_i}\left(\dfrac{p-1}{u_{i+p-1}-u_i}N_{i,p-2}(u) -\dfrac{p-1}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)\right) \\ &-\sum\limits_{i=j-p}^{j}\dfrac{pV_i}{u_{i+p+1}-u_{i+1}}\left(\dfrac{p-1}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u) -\dfrac{p-1}{u_{i+p+1}-u_{i+2}}N_{i+2,p-2}(u)\right) \\ =& p\left(p-1\right)\sum\limits_{i=j-p}^{j}\left[\dfrac{V_i}{u_{i+p}-u_i}\dfrac{1}{u_{i+p-1}-u_i}N_{i,p-2}(u) -\dfrac{V_i}{u_{i+p}-u_i}\dfrac{1}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u)\right. \\ &\left.-\dfrac{V_i}{u_{i+p+1}-u_{i+1}}\dfrac{1}{u_{i+p}-u_{i+1}}N_{i+1,p-2}(u) +\dfrac{V_i}{u_{i+p+1}-u_{i+1}}\dfrac{1}{u_{i+p+1}-u_{i+2}}N_{i+2,p-2}(u)\right]. \end{aligned}$
於是
$\begin{aligned} \dfrac{P^{''}\left(u\right)}{p\left(p-1\right)}=& \sum\limits_{i=j-p}^{j}\dfrac{V_i}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ &-\sum\limits_{i=j-p}^{j}\dfrac{V_i}{\left(u_{i+p}-u_i\right)\left(u_{i+p}-u_{i+1}\right)}N_{i+1,p-2}(u) \\ &-\sum\limits_{i=j-p}^{j}\dfrac{V_i}{\left(u_{i+p+1}-u_{i+1}\right)\left(u_{i+p}-u_{i+1}\right)}N_{i+1,p-2}(u) \\ &+\sum\limits_{i=j-p}^{j}\dfrac{V_i}{\left(u_{i+p+1}-u_{i+1}\right)\left(u_{i+p+1}-u_{i+2}\right)}N_{i+2,p-2}(u) \\ =& \sum\limits_{i=j-p+2}^{j}\dfrac{V_i}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ &-\sum\limits_{i=j-p+1}^{j+1}\dfrac{V_{i-1}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ &-\sum\limits_{i=j-p+1}^{j+1}\dfrac{V_{i-1}}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ &+\sum\limits_{i=j-p+2}^{j+2}\dfrac{V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ =& \sum\limits_{i=j-p+2}^{j}\dfrac{V_i}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ &-\sum\limits_{i=j-p+2}^{j}\dfrac{V_{i-1}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ &-\sum\limits_{i=j-p+2}^{j}\dfrac{V_{i-1}}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ &+\sum\limits_{i=j-p+2}^{j}\dfrac{V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ =& \sum\limits_{i=j-p+2}^{j}N_{i,p-2}(u)\left[\dfrac{V_i-V_{i-1}}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)} -\dfrac{V_{i-1}-V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)}\right]. \end{aligned}$
記
$\begin{aligned} \Delta_1=& \sum\limits_{i=j-p+2}^{j}\dfrac{V_i-V_{i-1}}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ =&\sum\limits_{i=j-p+2}^{j}\dfrac{V_i-V_{i-1}}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)}\left(\dfrac{u-u_i}{u_{i+p-2}-u_i}N_{i,p-3}(u) +\dfrac{u_{i+p-1}-u}{u_{i+p-1}-u_{i+1}}N_{i+1,p-3}(u)\right) \\ =&\sum\limits_{i=j-p+3}^{j}\dfrac{V_i-V_{i-1}}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)}\dfrac{u-u_i}{u_{i+p-2}-u_i}N_{i,p-3}(u) \\ &+\sum\limits_{i=j-p+3}^{j}\dfrac{V_{i-1}-V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-2}-u_{i-1}\right)}\dfrac{u_{i+p-2}-u}{u_{i+p-2}-u_i}N_{i,p-3}(u). \end{aligned}$

$\begin{aligned} \Delta_2=& \sum\limits_{i=j-p+2}^{j}\dfrac{V_{i-1}-V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)}N_{i,p-2}(u) \\ =&\sum\limits_{i=j-p+2}^{j}\dfrac{V_{i-1}-V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)} \left(\dfrac{u-u_i}{u_{i+p-2}-u_i}N_{i,p-3}(u) +\dfrac{u_{i+p-1}-u}{u_{i+p-1}-u_{i+1}}N_{i+1,p-3}(u)\right) \\ =&\sum\limits_{i=j-p+3}^{j}\dfrac{V_{i-1}-V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)}\dfrac{u-u_i}{u_{i+p-2}-u_i}N_{i,p-3}(u) \\ &+\sum\limits_{i=j-p+3}^{j}\dfrac{V_{i-2}-V_{i-3}}{\left(u_{i+p-2}-u_{i-2}\right)\left(u_{i+p-2}-u_{i-1}\right)}\dfrac{u_{i+p-2}-u}{u_{i+p-2}-u_i}N_{i,p-3}(u). \end{aligned}$

因此，得到
$\begin{aligned} \dfrac{P^{''}\left(u\right)}{p\left(p-1\right)}=& \sum\limits_{i=j-p+2}^{j}N_{i,p-2}(u)\left[\dfrac{V_i-V_{i-1}}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)} -\dfrac{V_{i-1}-V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)}\right] \\ =&\sum\limits_{i=j-p+3}^{j}N_{i,p-3}(u) \cdot \\ &\left\{\left[\dfrac{V_i-V_{i-1}}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)} -\dfrac{V_{i-1}-V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)}\right]\dfrac{u-u_i}{u_{i+p-2}-u_i}\right. \\ &\left. +\left[\dfrac{V_{i-1}-V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-2}-u_{i-1}\right)} -\dfrac{V_{i-2}-V_{i-3}}{\left(u_{i+p-2}-u_{i-2}\right)\left(u_{i+p-2}-u_{i-1}\right)}\right]\dfrac{u_{i+p-2}-u}{u_{i+p-2}-u_i}\right\}. \end{aligned}$
令
$\begin{aligned} Q_{i,k}=\begin{cases} \dfrac{V_i-V_{i-1}}{\left(u_{i+p}-u_i\right)\left(u_{i+p-1}-u_i\right)} -\dfrac{V_{i-1}-V_{i-2}}{\left(u_{i+p-1}-u_{i-1}\right)\left(u_{i+p-1}-u_i\right)},\quad k=2 \\ \dfrac{u_{i+p+1-k}-u}{u_{i+p+1-k}-u_i}Q_{i-1,k-1} +\dfrac{u-u_i}{u_{i+p+1-k}-u_i}Q_{i,k-1},\quad k=3,\cdots,p \end{cases} \end{aligned}$
下面用數學歸納法證明
$Q_{i,k}=\dfrac{V_{i,k-2}-V_{i-1,k-2}}{\left(u_{i+p+2-k}-u_i\right)\left(u_{i+p+1-k}-u_i\right)} -\dfrac{V_{i-1,k-2}-V_{i-2,k-2}}{\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p+1-k}-u_i\right)},\quad k=2,\cdots,p$
當$k=2$時，
$Q_{i,k}=\dfrac{V_{i,k-2}-V_{i-1,k-2}}{\left(u_{i+p+2-k}-u_i\right)\left(u_{i+p+1-k}-u_i\right)} -\dfrac{V_{i-1,k-2}-V_{i-2,k-2}}{\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p+1-k}-u_i\right)}.$
於是有
$\begin{aligned} Q_{i,k+1}=& \dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i}Q_{i-1,k} +\dfrac{u-u_i}{u_{i+p-k}-u_i}Q_{i,k} \\ =&\dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_i} \left[\dfrac{V_{i-1,k-2}-V_{i-2,k-2}}{\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p-k}-u_{i-1}\right)} -\dfrac{V_{i-2,k-2}-V_{i-3,k-2}}{\left(u_{i+p-k}-u_{i-2}\right)\left(u_{i+p-k}-u_{i-1}\right)}\right] \\ &+\dfrac{u-u_i}{u_{i+p-k}-u_i}\left[\dfrac{V_{i,k-2}-V_{i-1,k-2}}{\left(u_{i+p+2-k}-u_i\right)\left(u_{i+p+1-k}-u_i\right)} -\dfrac{V_{i-1,k-2}-V_{i-2,k-2}}{\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p+1-k}-u_i\right)}\right] \\ =& \dfrac{u_{i+p-k}-u}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p-k}-u_{i-1}\right)}\left(V_{i-1,k-2}-V_{i-2,k-2}\right) \\ &-\dfrac{u_{i+p-k}-u}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p-k}-u_{i-2}\right)\left(u_{i+p-k}-u_{i-1}\right)}\left(V_{i-2,k-2}-V_{i-3,k-2}\right) \\ &+\dfrac{u-u_i}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+2-k}-u_i\right)\left(u_{i+p+1-k}-u_i\right)}\left(V_{i,k-2}-V_{i-1,k-2}\right) \\ &-\dfrac{u-u_i}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p+1-k}-u_i\right)}\left(V_{i-1,k-2}-V_{i-2,k-2}\right). \end{aligned}$
記
$\begin{aligned} \Delta=& \dfrac{V_{i,k-1}-V_{i-1,k-1}}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)} -\dfrac{V_{i-1,k-1}-V_{i-2,k-1}}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)} \\ =& \dfrac{V_{i,k-1}}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)} -\dfrac{V_{i-1,k-1}}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)} \\ &-\dfrac{V_{i-1,k-1}}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)} +\dfrac{V_{i-2,k-1}}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)} \\ =&\dfrac{1}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)} \left[\dfrac{u_{i+p+2-k}-u}{u_{i+p+2-k}-u_i}V_{i-1,k-2} +\dfrac{u-u_i}{u_{i+p+2-k}-u_i}V_{i,k-2}\right] \\ &-\dfrac{1}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)} \left[\dfrac{u_{i+p+1-k}-u}{u_{i+p+1-k}-u_{i-1}}V_{i-2,k-2} +\dfrac{u-u_{i-1}}{u_{i+p+1-k}-u_{i-1}}V_{i-1,k-2}\right] \\ &-\dfrac{1}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)} \left[\dfrac{u_{i+p+1-k}-u}{u_{i+p+1-k}-u_{i-1}}V_{i-2,k-2} +\dfrac{u-u_{i-1}}{u_{i+p+1-k}-u_{i-1}}V_{i-1,k-2}\right] \\ &+\dfrac{1}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)} \left[\dfrac{u_{i+p-k}-u}{u_{i+p-k}-u_{i-2}}V_{i-3,k-2} +\dfrac{u-u_{i-2}}{u_{i+p-k}-u_{i-2}}V_{i-2,k-2}\right] \\ =& \dfrac{u_{i+p+2-k}-u}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+2-k}-u_i\right)}V_{i-1,k-2} \\ &+\dfrac{u-u_i}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+2-k}-u_i\right)}V_{i,k-2} \\ &-\dfrac{u_{i+p+1-k}-u}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)}V_{i-2,k-2} \\ &-\dfrac{u-u_{i-1}}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)}V_{i-1,k-2} \\ &-\dfrac{u_{i+p+1-k}-u}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)}V_{i-2,k-2} \\ &-\dfrac{u-u_{i-1}}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)}V_{i-1,k-2} \\ &+\dfrac{u_{i+p-k}-u}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p-k}-u_{i-2}\right)}V_{i-3,k-2} \\ &+\dfrac{u-u_{i-2}}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p-k}-u_{i-2}\right)}V_{i-2,k-2}. \end{aligned}$
令$Q_{i,k+1}-\Delta=\alpha_1V_{i,k-2}+\alpha_2V_{i-1,k-2}+\alpha_3V_{i-2,k-2}+\alpha_4V_{i-3,k-2}$，則有
$\begin{aligned} \alpha_1=& \dfrac{u-u_i}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+2-k}-u_i\right)\left(u_{i+p+1-k}-u_i\right)} \\ &-\dfrac{u-u_i}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+2-k}-u_i\right)} \\ =&0, \end{aligned}$

$\begin{aligned} \alpha_2=& \dfrac{u_{i+p-k}-u}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p-k}-u_{i-1}\right)} \\ &-\dfrac{u-u_i}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+2-k}-u_i\right)\left(u_{i+p+1-k}-u_i\right)} \\ &-\dfrac{u-u_i}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p+1-k}-u_i\right)} \\ &-\dfrac{u_{i+p+2-k}-u}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+2-k}-u_i\right)} \\ &+\dfrac{u-u_{i-1}}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)} \\ &+\dfrac{u-u_{i-1}}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)} \\ =&-\dfrac{1}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)} \\ &+\dfrac{1}{\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)} \\ &-\dfrac{u-u_i}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p+1-k}-u_i\right)} \\ &+\dfrac{u-u_{i-1}+u_{i+p+1-k}-u_{i+p+1-k}}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)} \\ =&\dfrac{u-u_{i+p+1-k}-u+u_i}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)} \\ &+\dfrac{1}{\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)} \\ =&0, \end{aligned}$

$\begin{aligned} \alpha_3=& -\dfrac{u_{i+p-k}-u}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p-k}-u_{i-1}\right)} \\ &-\dfrac{u_{i+p-k}-u}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p-k}-u_{i-2}\right)\left(u_{i+p-k}-u_{i-1}\right)} \\ &+\dfrac{u-u_i}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p+1-k}-u_i\right)} \\ &+\dfrac{u_{i+p+1-k}-u}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)} \\ &+\dfrac{u_{i+p+1-k}-u}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)} \\ &-\dfrac{u-u_{i-2}}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p-k}-u_{i-2}\right)} \\ =&-\dfrac{1}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p-k}-u_{i-1}\right)} \\ &+\dfrac{1}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)} \\ &-\dfrac{u_{i+p-k}-u}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)\left(u_{i+p-k}-u_{i-1}\right)} \\ &+\dfrac{u_{i+p+1-k}-u-u_{i-1}+u_{i-1}}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)} \\ =&\dfrac{u_{i-1}-u-u_{i+p-k}+u}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)} \\ &+\dfrac{1}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p+1-k}-u_{i-1}\right)} \\ =&0, \end{aligned}$

$\begin{aligned} \alpha_4=& \dfrac{u_{i+p-k}-u}{\left(u_{i+p-k}-u_i\right)\left(u_{i+p-k}-u_{i-2}\right)\left(u_{i+p-k}-u_{i-1}\right)} \\ &-\dfrac{u_{i+p-k}-u}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)\left(u_{i+p-k}-u_{i-2}\right)} \\ =&0. \end{aligned}$

從而，得到
$\begin{aligned} Q_{i,k+1}=\Delta=\dfrac{V_{i,k-1}-V_{i-1,k-1}}{\left(u_{i+p+1-k}-u_i\right)\left(u_{i+p-k}-u_i\right)} -\dfrac{V_{i-1,k-1}-V_{i-2,k-1}}{\left(u_{i+p-k}-u_{i-1}\right)\left(u_{i+p-k}-u_i\right)}. \end{aligned}$
證畢。

利用上面的結果，可以得到
$\begin{aligned} P^{''}\left(u\right)=& p\left(p-1\right)\sum\limits_{i=j-p+k}^{j}N_{i,p-k}(u)Q_{i,k}=p\left(p-1\right)Q_{j,p} \\ =& p\left(p-1\right)\left[\dfrac{V_{j,p-2}-V_{j-1,p-2}}{\left(u_{j+2}-u_j\right)\left(u_{j+1}-u_j\right)} -\dfrac{V_{j-1,p-2}-V_{j-2,p-2}}{\left(u_{j+1}-u_{j-1}\right)\left(u_{j+1}-u_j\right)}\right]. \end{aligned}$

De Boor遞推算法求NURBS 曲線上的點

設$u\in\left[u_j,u_{j+1}\right)$，則
$P\left(u\right)=V_{j,p}.$
其中
$\begin{aligned} V_{i,k}=& \begin{cases} V_i,\quad k=0 \\ \left(1-\alpha_{i,k}\right)\dfrac{\omega_{i-1,k-1}}{\omega_{i,k}}V_{i-1,k-1} +\alpha_{i,k}\dfrac{\omega_{i,k-1}}{\omega_{i,k}}V_{i,k-1},\quad k=1,\cdots,p,\quad i=j-p+k,\cdots,j \end{cases} \\ \omega_{i,k}=& \begin{cases} \omega_i,\quad k=0 \\ \left(1-\alpha_{i,k}\right)\omega_{i-1,k-1} +\alpha_{i,k}\omega_{i,k-1},\quad k=1,\cdots,p,\quad i=j-p+k,\cdots,j \end{cases} \\ \alpha_{i,k}=& \dfrac{u-u_i}{u_{i+p+1-k}-u_i},\quad k=1,\cdots,p,\quad i=j-p+k,\cdots,j \end{aligned}$

De Boor遞推算法求NURBS 曲線的一階導矢

設$u\in\left[u_j,u_{j+1}\right)$，則
$P^{'}\left(u\right)=\dfrac{p}{u_{j+1}-u_j}\dfrac{\omega_{j-1,p-1}\omega_{j,p-1}}{\omega_{j,p}^{2}}\left(V_{j,p-1}-V_{j-1,p-1}\right).$
其中
$\begin{aligned} V_{i,k}=& \begin{cases} V_i,\quad k=0 \\ \left(1-\alpha_{i,k}\right)\dfrac{\omega_{i-1,k-1}}{\omega_{i,k}}V_{i-1,k-1} +\alpha_{i,k}\dfrac{\omega_{i,k-1}}{\omega_{i,k}}V_{i,k-1},\quad k=1,\cdots,p,\quad i=j-p+k,\cdots,j \end{cases} \\ \omega_{i,k}=& \begin{cases} \omega_i,\quad k=0 \\ \left(1-\alpha_{i,k}\right)\omega_{i-1,k-1} +\alpha_{i,k}\omega_{i,k-1},\quad k=1,\cdots,p,\quad i=j-p+k,\cdots,j \end{cases} \\ \alpha_{i,k}=& \dfrac{u-u_i}{u_{i+p+1-k}-u_i},\quad k=1,\cdots,p,\quad i=j-p+k,\cdots,j \end{aligned}$

萊布尼茨公式求NURBS曲線的高階導矢

令$P\left(u\right)=\dfrac{\sum\limits_{i=0}^{n}N_{i,p}(u)\omega_iV_i}{\sum\limits_{i=0}^{n}N_{i,p}(u)\omega_i} =\dfrac{A\left(u\right)}{W\left(u\right)}$，利用求積的高階導的萊布尼茨公式，得到
$P^{\left(k\right)}\left(u\right)=\dfrac{A^{\left(k\right)}\left(u\right) -\sum\limits_{i=1}^{k} C_{k}^{i} W^{\left(i\right)}\left(u\right) P^{\left(k-i\right)}\left(u\right)}{W\left(u\right)}.$
特別的，
$\begin{aligned} P^{'}\left(u\right)=&\dfrac{A^{'}\left(u\right)-W^{'}\left(u\right)P\left(u\right)}{W\left(u\right)} =\dfrac{A^{'}\left(u\right)W\left(u\right)-A\left(u\right)W^{'}\left(u\right)}{W^{2}\left(u\right)}, \\ P^{''}\left(u\right)=&\dfrac{A^{''}\left(u\right)-2W^{'}\left(u\right)P^{'}\left(u\right)-W^{''}\left(u\right)P\left(u\right)}{W\left(u\right)} \\ =&\dfrac{A^{''}\left(u\right)W\left(u\right)-2W^{'}\left(u\right)W\left(u\right)P^{'}\left(u\right) -W^{''}\left(u\right)A\left(u\right)}{W^{2}\left(u\right)}. \end{aligned}$