The Bernstein polynomials are defined like this: $b_k(m,x)= {{m}\choose{k}} x^k(1-x)^{m-k}$, if $k<m$
I want to prove that $\sum\limits_{j=k}^m b_j(m,x) = m {{m-1}\choose{k-1} }\int\limits_{0}^x t^{k-1} (1-t)^{m-k} dt$. (The second equality in 1.7 here)
What I tried so far:
1) Going from left to right, and integrate using a) parts, and b) rewriting $(1-t)^{m-k}$ using Newton's binomial theorem. Both times I got somewhere that didn't look like what I was trying to prove. (Later I will upload my work, or at least where I got to).
2) Reeplace, in the left, $b_k(m,x)$ with $(1-x) b_k(m-1,x)+x b_{k-1}(m-1,x)$ (it's an equality that's easy to prove). Again, I didn't get anywhere.
3) Crying a little bit.
4) Integrate the expression in the right with Wolfram Alpha. It told me the result is the hypergeometric function (multiplied by something), but it's weird because when I integrate it by myself I get something much simpler.
5) ...I think they are wrong. Note that, at least, there's a typo in the paper, because they wrote $\sum\limits_{j=k}^m b_k(m,x) = m {{m-1}\choose{k-1} }\int\limits_{0}^x t^{k-1} (1-t)^{m-k} dt$ ($b_k$ instead of $b_j$)
Any ideas? Thanks!
Well, I solved it, with help from a friend. What I had tried (integrate using parts) was right; I must have had an error and didn't realise it.