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> |
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| − | Binomial theorem. |
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| − | :<math>(a + b)^n = \sum_{i=0}^n {n \choose i} a^{n-i} 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. |
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| − | :<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> |
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| − | :<math>p(x-1) = \sum_{i=0}^n a_i (x - 1)^i |
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| − | = \sum_{i=0}^n \left[ a_i \left( \sum_{k=0}^n {n \choose k} x^k (-1)^k \right) \right] |
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| − | = \sum_{i=0}^n \sum_{k=0}^i a_i {n \choose k} x^k (-1)^k |
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| − | = \sum_{i=0}^n \sum_{k=0}^i a_k {n \choose i} x^i (-1)^i.</math> |
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| − | QED. |
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| − | ---- |
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