Difference between revisions of "User talk:PaoloMartini"

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

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