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filip <filip~sladek~gmail~com> - 09. 11. 2009 - 01:14:07 z adsl-195-168-238-025.dynamic.nextra.skmato
napísal:
Ak to mam spravne tak ano...som si povedal ze P(x)=(x-1)^(2^k) * Q(x)
a ukazal ze koeficienty Q(x) su urcite cele cisla a ze koeficient pri
clene x^(2^k -1) je neparny(na to som vyuzil,ze 2^k nad m, kde
0<m<2^k je vzdy parne(je to vcelku pekne na dokazanie) a to ze
P(x) ma koeficienty iba +-1) ,cize urcite neni nula, takze P(x) bude
aspon stupna 2^k + 2^k - 1. |
Celkom sa mi toto riesenie paci. neznamena to nahodov, ze rovnake
tvrdenie plati aj ked su koeficienty polynomu P(x) lubovolne cele
cisla a navyse koeficient pri x^(2^k-1) je neparny?
cituj ma |
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mato - 08. 11. 2009 - 13:16:46 z dial-92-52-43-6-orange.orange.skfilip napísal:
Konecne nejaky pekny atypicky priklad na polynomy s celociselnymi
koeficientami. S otazkov odhadnite stupen polynomu som sa este nikde
nestretol a dost ma zaskocila. Dalo sa to aj inak ako cez komplexne
cisla? |
Ak to mam spravne tak ano...som si povedal ze P(x)=(x-1)^(2^k) * Q(x)
a ukazal ze koeficienty Q(x) su urcite cele cisla a ze koeficient pri
clene x^(2^k -1) je neparny(na to som vyuzil,ze 2^k nad m, kde
0<m<2^k je vzdy parne(je to vcelku pekne na dokazanie) a to ze
P(x) ma koeficienty iba +-1) ,cize urcite neni nula, takze P(x) bude
aspon stupna 2^k + 2^k - 1.
cituj ma |
|
filip <filip~sladek~gmail~com> - 08. 11. 2009 - 12:55:19 z adsl-195-168-245-128.dynamic.nextra.skKonecne nejaky pekny atypicky priklad na polynomy s celociselnymi
koeficientami. S otazkov odhadnite stupen polynomu som sa este nikde
nestretol a dost ma zaskocila. Dalo sa to aj inak ako cez komplexne
cisla? cituj ma |
|