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Suppose $x_{0},\ldots,x_{n}$ are distinct nodes and m are in natural numbers. prove exist polynomial $P_{n}$ of degree n, satisfying $P_{n}^{(k)}$=$y_{i}^{(k)}$ that k=$0,\ldots,m$ and i=$0,\ldots,n$. To achieve this goal we must assume $P_{n}$=$a_{0}+a_{1}x+\ldots+a_{n}x^{n}$ and prove the system that is result of $P_{n}^{(k)}$=$y_{i}^{(k)}$ have a solution.To achieve this goal we must prove that Determinants of coefficient matrix is nonzero. How we reached this goal?

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a) If you post closely related questions, please link them up so that people don't waste time on one without knowing what's been done on the other one. b) This question doesn't make sense to me. If I understand correctly, you're setting $(m+1)(n+1)$ conditions on a polynomial of degree at most $n$? That can't work. Also, what are the "determinants" of the coefficient matrix? Is the plural intentional? The matrix isn't square, so it doesn't have a determinant. Do you mean minors? –  joriki Oct 19 '11 at 9:47
Also, you haven't interacted with my answer at the other question. Did it answer your question? If so, the idea is to accept it so the question doesn't remain marked as unanswered. If not, please point out what's missing so that it can be improved or someone else can come up with a better answer. Also, I suggest that you consider letting someone with a higher proficiency in English proofread your questions. Part of what makes them difficult to understand may be the high density of grammatical errors. –  joriki Oct 19 '11 at 9:49

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