I am told to prove for any $n$, $p_n(0)=p_0$ and $p_n(1)=p_n$. Where $p_0,p_1,...,p_n$ are a set of points with $p_i=(x_i,y_i)$ in the plane and the n-th Bernstein polynomial is given by
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How about starting by writing the value for $p_n(0)$ and $p_n(1)$? – NoChance Nov 03 '19 at 18:41
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well for (0)=0 since 0 to any power would be zero and cancel the whole ting out and and (1)=0 since the (1-1)^i would give you zero. – Sherien Hassan Nov 03 '19 at 18:47
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So, you know the proof. What is missing? – NoChance Nov 03 '19 at 18:51
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Well the question did not set (0)=0 and (1)=0 is what I am confused about. and also what the point is of plugging in zero and 1, like what do these two numbers signify in terms of this problem. – Sherien Hassan Nov 03 '19 at 18:54
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1Keep in mind that $0^0 \neq 0$. – darij grinberg Nov 03 '19 at 19:03
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right! 0^0 would be undefined. But then how does any of this show that (0)=0 and (1)= for any n is what I cannot see. It is the setting it equal to 0 and that is confusing me. Would it suffice for this proof for me to show how each of these summations go to zero when 0 and 1 is plugged in for t? – Sherien Hassan Nov 03 '19 at 19:05
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To what evaluate $t^0$ and $(1-t)^0$ if one assumes that they are polynomials? – Lutz Lehmann Nov 03 '19 at 19:29
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Could you rephrase your question @Dr.LutzLehmann – Sherien Hassan Nov 03 '19 at 19:30
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In the sequence of monomials $t^k$ or $(1-t)^k$, what is the element for $k=0$? – Lutz Lehmann Nov 03 '19 at 21:30