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はじめに
3次splineを勉強していたらB-splineも見つけたので寄り道してみた。Wikipediaの記事で紹介されている性質の証明を与える。
定義
$\dotsb < t_{-1} < t_0 < t_1 < t_2 < \dotsb$に対して次の漸化式で定められる関数$B_{i,n}$を$n$次のB-spline関数と呼ぶ。
\[ B_{i,0}(t) \coloneqq \begin{cases} 1 & (t_i \leq t < t_{i+1}) \\ 0 & (\text{otherwise}) \end{cases} \] \[ B_{i,n+1}(t) \coloneqq \omega_{i,n}(t)B_{i,n}(t) + (1-\omega_{i+1,n}(t))B_{i+1,n}(t),\quad \omega_{i,n}(t) \coloneqq \frac{t – t_i}{t_{i+n+1} – t_i} \]
この定義から、$B_{i,n}$の台は$[t_i,t_{i+n+1})$であることが判る。
性質
和
次式が成り立つ。
\[\forall t \in [t_j,t_{j+1}), \sum_{i=j-n}^j B_{i,n}(t) = 1\]
$\textit{Proof}$
$n = 0$のときは明らかに成り立つ。$n = m \in \integers$のときに成り立つと仮定して$n = m+1$のときに成り立つことを示す。
\begin{align*}
\sum_{i=j-m-1}^j B_{i,m+1}(t) &= \sum_{i=j-m-1}^j \left[ \omega_{i,m}(t)B_{i,m}(t) + (1-\omega_{i+1,m}(t))B_{i+1,m}(t) \right] \\
&= \omega_{j-m-1,m}(t)B_{j-m-1,m}(t) + \sum_{i=j-m-1}^{j-1} B_{i+1,m}(t) + (1-\omega_{j+1,m}(t))B_{j+1,m}(t)
\end{align*}
$B_{i,n}$の台が$[t_i,t_{i+n+1})$であることから、上式の$B_{j-m-1,m}(t)$と$B_{j+1,m}(t)$は$t \in [t_j,t_{j+1})$に対して0となるので
\[ \sum_{i=j-m-1}^j B_{i,m+1}(t) = \sum_{i=j-m-1}^{j-1} B_{i+1,m}(t) = \sum_{k=j-m}^j B_{k,m}(t) = 0 \]
$\square$
微分
次式が成り立つ。
\[\derivLong{B_{i,k}(t)}{t}{} = k\left(\frac{B_{i,k-1}(t)}{t_{i+k}-t_i} – \frac{B_{i+1,k-1}(t)}{t_{i+1+k} – t_{i+1}}\right)\]
$\textit{Proof}$
$k=1$のときに成り立つことは容易に解る。$k=l \in \naturalNumbers$のときに成り立つと仮定して$k=l+1$のときに成り立つことを示す。
\begin{align*}
\derivLong{B_{i,l+1}(t)}{t}{} &= \derivLong{\left(\frac{t-t_i}{t_{i+l+1}-t_i}B_{i,l}(t) + \frac{t_{i+l+2}-t}{t_{i+l+2}-t_{i+1}}B_{i+1,l}(t)\right)}{t}{} \\
&= \frac{B_{i,l}(t)}{t_{i+l+1}-t_i} + \frac{t-t_i}{t_{i+l+1}-t_i}\derivLong{B_{i,l}(t)}{t}{} – \frac{B_{i+1,l}(t)}{t_{i+l+2} – t_{i+1}} + \frac{t_{i+l+2}-t}{t_{i+l+2} – t_{i+1}}\derivLong{B_{i+1,l}(t)}{t}{} \\
&= \frac{B_{i,l}(t)}{t_{i+l+1}-t_i} – \frac{B_{i+1,l}(t)}{t_{i+l+2} – t_{i+1}} + \underbrace{\frac{t-t_i}{t_{i+l+1}-t_i}l\left(\frac{B_{i,l-1}(t)}{t_{i+l}-t_i} – \frac{B_{i+1,l-1}(t)}{t_{i+1+l} – t_{i+1}}\right)}_{(1)} \\
&\phantom{=} + \underbrace{\frac{t_{i+l+2}-t}{t_{i+l+2} – t_{i+1}}l\left(\frac{B_{i+1,l-1}(t)}{t_{i+1+l}-t_{i+1}} – \frac{B_{i+2,l-1}(t)}{t_{i+2+l} – t_{i+2}}\right)}_{(2)}
\end{align*}
ここで
\begin{align*}
(1) &= \frac{l}{t_{i+l+1}-t_i}\left[\frac{t-t_i}{t_{i+l}-t_i}B_{i,l-1}(t) + \frac{t_{i+1+l} – t}{t_{i+1+l} – t_{i+1}}B_{i+1,l-1}(t) + \frac{t_i – t_{i+1+l}}{t_{i+1+l} – t_{i+1}}B_{i+1,l-1}(t)\right] \\
&= \frac{l}{t_{i+l+1}-t_i}B_{i,l}(t) – \frac{l}{t_{i+1+l} – t_{i+1}}B_{i+1,l-1}(t)
\end{align*}
また
\begin{align*}
(2) &= -\frac{l}{t_{i+l+2} – t_{i+1}}\left[\frac{t – t_{i+1}}{t_{i+1+l}-t_{i+1}}B_{i+1,l-1}(t) + \frac{t_{i+l+2}-t}{t_{i+2+l} – t_{i+2}}B_{i+2,l-1}(t) + \frac{t_{i+1} – t_{i+l+2}}{t_{i+1+l}-t_{i+1}}B_{i+1,l-1}(t)\right] \\
&= -\frac{l}{t_{i+l+2}-t_{i+1}}B_{i+1,l}(t) + \frac{l}{t_{i+1+l} – t_{i+1}}B_{i+1,l-1}(t)
\end{align*}
以上より
\[ \derivLong{B_{i,l+1}(t)}{t}{} = (l+1)\left[\frac{B_{i,l}(t)}{t_{i+l+1}-t_i} – \frac{B_{i+1,l}(t)}{t_{i+l+2} – t_{i+1}}\right] \]
$\square$
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