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Let P be a quadratic B-spline with three control points $ p_{0},p_{1},p_{2}, $ and knot vector $\tau = (0,0,0,1,1,1) $. How to prove explicitly that P is a Bezier curve?

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using the definition of B-spline (basis) functions of degree d, that if you let the underlying knot sequence be 0,…0, 1…1 — where the knots 0 and 1 each have multiplicity d + 1, then the B-spline basis functions are exactly the Bernstein polynomials.

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Yes, your approach is correct. Just use the usual (recursive) definition of b-spline basis functions. You will find that the b-spline basis functions of degree $1$ are $1-t$ and $t$. From these, we can compute the basis functions of degree 2. You will find that these are $(1-t)^2$, $2t(1-t)$, and $t^2$, which are the Bernstein polynomials of degree 2. Continue from there.

Another approach ...
You know that any b-spline curve is just a sequence of polynomial segments (i.e. Bezier curves), strung together end-to-end. There is one polynomial segment corresponding to each non-trivial knot span $[t_i, t_{i+1}]$, where $t_i < t_{i+1}$. In our case, there is only one non-trivial knot span, namely $[0,1]$, so the entire b-spline curve consists of just one polynomial segment. In other words, the b-spline curve is a Bezier curve.

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