i was reading a book about Bezier curves and Berstein polynomials and i have a big doubt.
Why $B_{-1}^n(t)=B_n^{n-1}(t)=0$ where $B$ are the Berstein polynomials?
Link of the book: http://math.aalto.fi/~ahniemi/hss2012/Notes06.pdf
Pag 4, Theorem 5.
My attempt:
By definition of Berstein polynomial i know that:
$$B_n^{n-1}(t)=\frac{(n-1)!}{n!(n-1+n)!}\cdot t^n(1-t)^{n-1+n}$$
But if i continue then i will get: $$\frac{(n-1)!}{n!(n-1+n)!}\cdot t^n(1-t)^{n-1+n}=\frac{1}{n(-1)!}\cdot t^n(1-t)^{-1}$$
This is wrong, because $(-1)!$ doesn't exists.
Can someone help me? Thanks.
The definition of Bernstein polynomials is $$ B_{n,k}(t)=\binom{n}{k}t^k(1-t)^{n-k}. $$ Therefore, for $k$ outside the range $\{0,1,\dots,n\}$, the Bernstein polymomial is just zero, since the binomial coefficient $\binom{n}{k}$ is defined to be zero when $k>n$ or when $k<0$. This is very typical convention.
In this case, you should not blindly apply the formula $$ \binom{n}{k}=\frac{n!}{k!(n-k)!} $$ because you end up with nonsense as you observed. Although, if you insist, you can interpret division by $(-1)!$ as division by "infinity" which gives you zero because the gamma function has poles at negative integers. This can be made rigorous within complex analysis.
One version of the formula that does work for arbitrary $n$ for any nonnegative integer $k$ is $$ \binom{n}{k}=\frac{(\alpha)_k}{k!}=\frac{\alpha(\alpha-1)\dotsm(\alpha-k+1)}{k!} $$ where $(\alpha)_k$ is called the Pochhammer symbol (falling factorial). You can extend to negative $k$ by use of $\binom{n}{k}=\binom{n}{n-k}$.