Difference between revisions of "User talk:PaoloMartini"

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:<math>p(x-1) = \sum_{i=0}^n a_i (x - 1)^i
 
:<math>p(x-1) = \sum_{i=0}^n a_i (x - 1)^i
 
= \sum_{i=0}^n (a_i \sum_{k=0}^n {n \choose k} x^k (-1)^k). </math>
 
= \sum_{i=0}^n (a_i \sum_{k=0}^n {n \choose k} x^k (-1)^k). </math>
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:<math>p(x-1) = \sum_{i=0}^n a_i (x-1)^i </math>
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:<math>t = x-1</math>
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:<math>p(t) = \sum_{i=0}^n a_i t^i </math>

Revision as of 15:33, 14 September 2006