Difference between revisions of "User talk:PaoloMartini"

Revision as of 20:14, 15 September 2006

$p(x)=\sum _{i=0}^{n}a_{i}x^{i},a_{n}\neq 0$

$p(x)=a_{n}x^{n}+\sum _{i=0}^{n-1}a_{i}x^{i}$

$p(x-1)=a_{n}(x-1)^{n}+\sum _{i=0}^{n-1}a_{i}(x-1)^{i}$

$(x-1)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}(-1)^{k}$

$p(x-1)=a_{n}\sum _{k=0}^{n}{n \choose k}x^{n-k}(-1)^{k}+\sum _{i=0}^{n-1}a_{i}(x-1)^{i}$

$p(x-1)=a_{n}x^{n}+a_{n}\sum _{k=1}^{n}{n \choose k}x^{n-k}(-1)^{k}+\sum _{i=0}^{n-1}a_{i}(x-1)^{i}$

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-<math>\sum_{i=0}^n a_i \sum_{j=0}^i {n \choose k} x^{n-j} (-1)^j</math>
+<math>p(x) = \sum_{i=0}^n a_i x^i, a_n \neq 0 </math>
-<math>\sum_{i=0}^n {n \choose k} x^{n-i} (-1)^i \sum_{j=i}^n a_j </math>
+<math>p(x) = a_n x^n + \sum_{i=0}^{n-1} a_i x^i </math>
+<math>p(x-1) = a_n (x-1)^n + \sum_{i=0}^{n-1} a_i (x-1)^i </math>
+<math>(x-1)^n = \sum_{k=0}^n {n \choose k} x^{n-k} (-1)^k </math>
+<math>p(x-1) = a_n \sum_{k=0}^n {n \choose k} x^{n-k} (-1)^k + \sum_{i=0}^{n-1} a_i (x-1)^i </math>
+<math>p(x-1) = a_n x^n + a_n \sum_{k=1}^n {n \choose k} x^{n-k} (-1)^k + \sum_{i=0}^{n-1} a_i (x-1)^i </math>