We shall establish the correctness of the degree elevation algorithm. Again, it requires simple, but tedious, calculations. Suppose the given Bézier curve C(u) of degree n is defined by control points P0, P1 , ..., Pn, and the degree elevated Bézier curve D(u) of degree n+1 is defined by control points Q0 = P0, Q1, Q2, ..., Qn, Qn+1 = Pn. We shall find the n unknown control points Qi's.
Our strategy is very simple. Since C(u) and D(u) are the same curve, their derivatives at any u should be equal. Hence, by comparing the first, second, third, ..., and n-th derivatives at 0, we should be able to determine what Qi's are. To simplify our calculation, we shall use mathematical indiction. More precisely, we find Q1 by comparing the first derivatives at 0. Then, we assume that Q1, Q2, ..., Qk-1 are computed and use these known results to compute Qk. Iterating this process n times, we will compute all new control points.
The first derivative of C(u) is
Thus, the derivative at u = 0 is n(P1 - P0). Similarly, the derivative of D(u) at u = 0 is (n+1)(Q1-Q0) = (n+1)(Q1-P0). Because these two derivatives are equal, we have
Solving for Q1, we have our first result:
For the induction step, we shall assume that the computation of Q0 = P0, Q1, ..., Qk-1 are correct and show that Qk is also correct. This is done by comparing the k-th derivatives of the original curve C(u) and the degree elevated curve D(u).
Just like in the base case, we actually do not need the complete curves. Let the k-th derivative of C(u) and D(u) be C[k](u) and D[k](u), respectively. The k-th derivative vectors at u = 0 are C[k](0) and D[k](0). Because C(u) and D(u) are the same curve, their k-th derivatives at u = 0 should be identical. More precisely, we should have C[k](0) = D[k](0). From this relation and the known Q0 = P0, Q1, ..., Qk-1, we shall compute Qk.
In the discussion of higher derivatives, we know that the k-th derivative function of curve C(u) is
where Dki is
and the combinatorial coefficient C(k,j) is
The computation of the Dki's is an exercise at the end of this unit. Click here if you wish to see the details. Therefore, C[k](0) is computed as follows:
Similarly, we can compute D[k](0). Note that curve D(u) is of degree n+1. This is the reason that the multiplier is (n+1)n(n-1)...(n-k+2).
Setting C[k](0) equal to D[k](0) and canceling some common factors yields the following:
In fact, we can do more. Since we know (1) Qk is unknown and Q0 = P0, we can take these two terms out of the summation of D[k](0). Now, the summation runs from 1 to k-1. To make the summation in C[k](0) compatible, we also take the terms of Pk and P0 out of the summation. This gives the following:
Rearranging the terms yields the following:
By the induction hypothesis, for j in the range of 0 and k-1, Qj is known and is computed as follows:
Plug Qj into the term in the summation and we have:
Plug this result back into the expression of (n+1)Qk and split the summation into two:
After taking Pk-1 out of the first summation, the first summation becomes:
Then, we take the P0 term out of the second summation:
However, in this expression, the index j runs from 2 to k-1. To make it compatible with that of the first summation, we can use a new index h = j - 1. After this substitution, the second summation becomes the following:
Note that the corresponding terms in these two summations are equal with different signs (i.e., (-1)k-j+1=(-1)×(-1)k-j in the first summation and (-1)k-h in the second). Hence, adding these two summations together will cancel out all terms in the summation and yield the following:
Putting this back into the expression of (n+1)Qk yields
Dividing the expression by n+1 gives us the desired result:
Therefore, the degree elevation computation for k also holds, and the degree elevation algorithm is correct.