Proving that Bernstein polynomials are basic B-splines I want to prove that if we use the knot vector $t_0=\dots=t_n=0, t_{n+1}=\dots =t_{2n+1}=1$ then $N^n_i=B^n_i$ on $[0,1)$. I have the following definitions:

*

*$N^0_i=1$ on $[t_i,t_{i+1})$, $0$ elsewhere.

*$N^r_i = \frac{t-t_i}{t_{r+i}-t_i} N^{r-1}_i + (1-\frac{t-t_{i+1}}{t_{r+i+1}-t_{i+1}})N^{r-1}_{i+1}$ for $0 \leq r,i \leq n$ (where the fractions is defined as $0$ if denominator vanishes).

*$B^n_i= \frac{n!}{(n-k)!k!}t^i(1-t)^{n-i}$ for $0 \leq i \leq n$.

I tried using a proof by induction, but I just could not finish it, because I always end up running into situations where I cannot apply my induction hypothesis, or I end up with a recursion that just does not seem to fit. Can you show me how to proof this, all textbooks I found online mention the fact, but do not prove it.
EDIT: I have now found a book which proves the statement by saying, choosing the knots this way lets the two recursion formulas coincide.
I do not understand how that works out.
 A: This answer uses following notations, which are slightly different from the OP:

*

*For $0 \leqslant i \leqslant 2n$,$$
N_{0, i}(t) = \begin{cases} 1; & t \in [t_i, t_{i + 1}) \\ 0; & t \in [0, 1) \setminus [t_i, t_{i + 1}) \end{cases};
$$

*For $1 \leqslant k \leqslant n$, $0 \leqslant i \leqslant 2n - k$,$$
N_{k, i}(t) = \frac{t - t_i}{t_{i + k} - t_i} N_{k - 1, i}(t) + \frac{t_{i + k + 1} - t}{t_{i + k + 1} - t_{i + 1}} N_{k - 1, i + 1}(t) \quad (t \in [0, 1)),
$$
where the fraction vanishes if its denominator vanishes;

*For $k \geqslant 0$, $i \in \mathbb{Z}$,$$
B_{k, i}(t) = c_{k, i} t^i (1 - t)^{k - i},
$$
where $\{c_{k, i}\}$ satisfies$$
c_{0, i} = \begin{cases} 1; & i = 0 \\ 0; & \text{otherwise} \end{cases}, \quad c_{k, i} = c_{k - 1, i - 1} + c_{k - 1, i} \quad (k \geqslant 1,\ i \in \mathbb{Z}).
$$
Note that $c_{k, i} = \dfrac{k!}{i!\, (k - i)!}$ for $0 \leqslant i \leqslant k$.
Now it will be proved by induction on $k$ that $N_{k, i} = B_{k, i + k - n}$ for $0 \leqslant k \leqslant n$, $0 \leqslant i \leqslant 2n - k$. Since $N_{0, n} = B_{0, n} = 1$ and $N_{0, i} = B_{0, i} = 0$ for $i \ne n$, the proposition holds for $k = 0$.
Assume that the proposition holds for $k - 1$. It suffices to prove that\begin{gather*}
B_{k, i + k - n}(t) = \frac{t - t_i}{t_{i + k} - t_i} B_{k - 1, i + k - n - 1}(t) + \frac{t_{i + k + 1} - t}{t_{i + k + 1} - t_{i + 1}} B_{k - 1, i + k - n}(t) \tag{1}
\end{gather*}
for $0 \leqslant i \leqslant 2n - k$.
Case 1: $0 \leqslant i \leqslant n - k - 1$ or $n + 1 \leqslant i \leqslant 2n - k$. Now $t_i = t_{i + 1} = t_{i + k} = t_{i + k + 1} = 0$, and $B_{k, i + k - n} = 0$, so (1) holds.
Case 2: $i = n - k$. Now $t_i = t_{i + 1} = t_{i + k} = 0$, $t_{i + k + 1} = 1$, and$$
\text{RHS of}\ (1) = 0 + \frac{1 - t}{1 - 0} \cdot (1 - t)^{k - 1} = (1 - t)^k = \text{LHS of}\ (1),
$$
so (1) holds.
Case 3: $n - k + 1 \leqslant i \leqslant n - 1$. Now $t_i = t_{i + 1} = 0$, $t_{i + k} = t_{i + k + 1} = 1$, and\begin{align*}
\text{RHS of}\ (1) &= \frac{t - 0}{1 - 0} c_{k - 1, i + k - n - 1} t^{i + k - n - 1} (1 - t)^{n - i} + \frac{1 - t}{1 - 0} c_{k - 1, i + k - n} t^{i + k - n} (1 - t)^{n - i - 1}\\
&= (c_{k - 1, i + k - n - 1} + c_{k - 1, i + k - n}) t^{i + k - n} (1 - t)^{n - i}\\
&= c_{k, i + k - n} t^k (1 - t)^{i + k - n} = \text{LHS of}\ (1),
\end{align*}
so (1) holds.
Case 4: $i = n$. Now $t_i = 0$, $t_{i + 1} = t_{i + k} = t_{i + k + 1} = 1$, and$$
\text{RHS of}\ (1) = \frac{t - 0}{1 - 0} \cdot t^{k - 1} + 0 = t^k = \text{LHS of}\ (1),
$$
so (1) holds. End of induction.
Finally, the proposition holds for $k = n$, i.e. $N_{n, i} = B_{n, i}$ for $0 \leqslant i \leqslant n$.
