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> |
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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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+ | |||
⚫ | |||
+ | |||
+ | Binomial coefficient simmetry. |
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:<math>{n \choose k} = {n \choose n-k} </math> |
:<math>{n \choose k} = {n \choose n-k} </math> |
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+ | |||
+ | Hence: |
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:<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> |
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+ | zZzZ... |
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− | |||
− | :<math>t = x-1</math> |
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+ | ---- |
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− | :<math>p(t) = \sum_{i=0}^n a_i t^i </math> |
Revision as of 15:43, 14 September 2006
Show that if is a polynomial of degree , then is a polynomial of the same degree.
Definition of polynomial.
Binomial theorem.
Special case.
Binomial coefficient simmetry.
Hence:
zZzZ...