업데이트:

베지어 곡선의 explicit form 증명

베지어 곡선의 explicit form을 증명한다.


Lemma

(n−1i−1)+(n−1i)=(ni)\binom{n-1}{i-1} + \binom{n-1}{i} = \binom{n}{i}

(ni)\binom{n}{i}는 binomial coefficient이다.

Proof

(n−1i−1)+(n−1i)=(n−1)!(i−1)!(n−i)!+(n−1)!i!(n−i−1)!=(n−1)!(n−i)+ii!(n−i)!=n!i!(n−i)!=(ni)□\begin{aligned} \binom{n-1}{i-1} + \binom{n-1}{i} &= \frac{(n-1)!}{(i-1)!(n-i)!}+\frac{(n-1)!}{i!(n-i-1)!} \\\\ &=(n-1)!\frac{(n-i)+i}{i!(n-i)!} \\\\ &=\frac{n!}{i!(n-i)!} \\\\ &=\binom{n}{i} \quad \square \end{aligned}

Explicit form of Bézier curve

다음 두 식은 동치이다.

BP0(0)(t)=P0, t∈[0,1]BP0,P1,...,Pn(n)(t)=(1−t)BP0,...,Pn−1(t)+tBP1,...,Pn(t), t∈[0,1]ifn>0−(1)\begin{aligned} B_{P_0}^{(0)}(t) &= P_0,\ t\in[0,1] \\\\ B_{P_0,P_1,...,P_n}^{(n)}(t) &= (1-t)B_{P_0,...,P_{n-1}}(t)+tB_{P_1,...,P_n}(t),\ t\in[0,1] \qquad if \quad n > 0 \qquad -(1) \end{aligned} BP0,...,Pn(n)(t)=∑i=0n(ni)(1−t)n−itiPi−(2)B_{P_0,...,P_n}^{(n)}(t) = \sum^n_{i=0} \binom{n}{i}(1-t)^{n-i}t^iP_i \qquad -(2)

Proof

간단하게 BP0,...,Pn(t)B_{P_0,...,P_n}(t)을 B0...n(t)B_{0...n}(t)이라고 쓰자.

By induction.

i) When n = 0: trivial

ii) When n > 0:

B0...n(n)(t)=(1−t)B0...(n−1)+tB1...n=[(1−t)∑i=0n−1(n−1i)(1−t)n−1−itiPi]+[t∑i=0n−1(n−1i)(1−t)n−1−itiPi+1]=∑i=0n−1(n−1i)(1−t)n−itiPi+∑i=1n(n−1i−1)(1−t)n−itiPi=(n−10)(1−t)nP0+∑i=1n−1[(n−1i)+(n−1i−1)](1−t)n−itiPi+(n−1n−1)tnPn=(1−t)nP0+∑i=1n−1(ni)(1−t)n−itiPi+tnPnby above lemma=∑i=0n(ni)(1−t)n−itiPi□\begin{aligned} B_{0...n}^{(n)}(t) &= (1-t)B_{0...(n-1)}+tB_{1...n} \\\\ &= [(1-t)\sum^{n-1}_{i=0} \binom{n-1}{i} (1-t)^{n-1-i}t^iP_i]+[t\sum^{n-1}_{i=0} \binom{n-1}{i} (1-t)^{n-1-i}t^iP_{i+1}] \\\\ &= \sum^{n-1}_{i=0} \binom{n-1}{i} (1-t)^{n-i}t^iP_i+\sum^{n}_{i=1} \binom{n-1}{i-1} (1-t)^{n-i}t^{i}P_{i} \\\\ &= \binom{n-1}{0}(1-t)^nP_0+\sum^{n-1}_{i=1} \left[ \binom{n-1}{i}+\binom{n-1}{i-1} \right] (1-t)^{n-i}t^iP_i+\binom{n-1}{n-1}t^nP_n \\\\ &= (1-t)^nP_0+\sum^{n-1}_{i=1}\binom{n}{i}(1-t)^{n-i}t^iP_i+t^nP_n \qquad by\ above\ lemma\\\\ &= \sum^{n}_{i=0}\binom{n}{i}(1-t)^{n-i}t^iP_i \qquad \square \end{aligned}