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 \left[ a_i \left( \sum_{k=0}^n {n \choose k} x^k (-1)^k \right) \right] </math>
  
 
zZzZ...
 
zZzZ...
  
 
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Revision as of 15:52, 14 September 2006

Show that if p(x) is a polynomial of degree n, then p(x - 1) is a polynomial of the same degree.

Definition of polynomial.

p(x) = \sum_{i=0}^n a_i x^i

Binomial theorem.

(a + b)^n = \sum_{i=0}^n {n \choose i} a^{n-i} b^i

Special case.

(x - 1)^n = \sum_{i=0}^n {n \choose i} x^{n-i} (-1)^i

Binomial coefficient simmetry.

{n \choose k} = {n \choose n-k}

Hence:

(x - 1)^n = \sum_{i=0}^n {n \choose i} x^i (-1)^i
p(x-1) = \sum_{i=0}^n a_i (x - 1)^i
= \sum_{i=0}^n \left[ a_i \left( \sum_{k=0}^n {n \choose k} x^k (-1)^k \right) \right]

zZzZ...