Difference between revisions of "User talk:PaoloMartini"

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Show that if <math>p(x)</math> is a polynomial of degree <math>n</math>, then <math>p(x - 1)</math> is a polynomial of the same degree.
 
  
Definition of polynomial.
 
 
:<math>p(x) = \sum_{i=0}^n a_i x^i </math>
 
 
Binomial theorem.
 
 
:<math>(a + b)^n = \sum_{i=0}^n {n \choose i} a^{n-i} b^i </math>
 
 
Special case.
 
 
:<math>(x - 1)^n = \sum_{i=0}^n {n \choose i} x^{n-i} (-1)^i </math>
 
 
Binomial coefficient simmetry.
 
 
:<math>{n \choose k} = {n \choose n-k} </math>
 
 
Hence:
 
 
:<math>(x - 1)^n = \sum_{i=0}^n {n \choose i} x^i (-1)^i </math>
 
 
:<math>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]
 
= \sum_{i=0}^n \sum_{k=0}^i a_i {n \choose k} x^k (-1)^k
 
= \sum_{i=0}^n \sum_{k=0}^i a_k {n \choose i} x^i (-1)^i.</math>
 
 
QED.
 
 
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Revision as of 19:36, 14 September 2006