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.
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Definition of polynomial.
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:<math>p(x) = \sum_{i=0}^n a_i x^i </math>
 
:<math>p(x) = \sum_{i=0}^n a_i x^i </math>
  
:<math>(x - 1)^n = \sum_{i=0}^n {n \choose i } x^{n-i} (-1)^i </math>
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Binomial theorem.
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:<math>(a + b)^n = \sum_{i=0}^n {n \choose i} a^{n-1} b^i </math>
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Special case.
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:<math>(x - 1)^n = \sum_{i=0}^n {n \choose i} x^{n-i} (-1)^i </math>
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Binomial coefficient simmetry.
  
 
:<math>{n \choose k} = {n \choose n-k} </math>
 
:<math>{n \choose k} = {n \choose n-k} </math>
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Hence:
  
 
:<math>(x - 1)^n = \sum_{i=0}^n {n \choose i} x^i (-1)^i </math>
 
:<math>(x - 1)^n = \sum_{i=0}^n {n \choose i} x^i (-1)^i </math>
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= \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>
  
:<math>p(x-1) = \sum_{i=0}^n a_i (x-1)^i </math>
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zZzZ...
 
 
:<math>t = x-1</math>
 
  
:<math>p(t) = \sum_{i=0}^n a_i t^i </math>
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Revision as of 15:43, 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-1} 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 (a_i \sum_{k=0}^n {n \choose k} x^k (-1)^k).

zZzZ...