Hello I want to show that the Bernstein polynomial $$B_{n,k}=\binom{n}{k}x^k(1-x)^{n-k}\,$$ is a basis. For linear independece I got a hint from my teacher to expand the binom $(1-x)^{n-k}$ This way I get: $$B_{n,k}=\binom{n}{k}x^k(1-x)^{n-k}\sum_{j=0}^{n-k}\binom{n-k}{j}(-1)^jx^j$$ And changing the index of summation gives: $$B_{n,k}=\sum_{j=k}^{n}\binom{n-k}{j-k}\binom{n}{k}(-1)^{j-k}x^{j-k+k}=\sum_{j=k}^{n}\binom{n}{j}\binom{j}{k}(-1)^{j-k}x^j$$ Now I have to show that $\alpha_i$ are $0$ in the relation $\sum_{i=0}^{n}\alpha_iB_{i,n}=0\,$ or$$\alpha_0\sum_{j=0}^n(-1)^j\binom{n}{j}\binom{j}{0}x^j+\alpha_1\sum_{j=1}^n(-1)^{j-1}\binom{n}{j}\binom{j}{1}x^j+...+\alpha_{n}\sum_{j=n}^n(-1)^{j-n}\binom{n}{j}\binom{j}{n}x^j=0$$ Now what can I do and how can I finish this problem? Thanks in advance! Login To add answer/comment