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| Temat postu: A broken language extension and application of _16 |
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One -off promotion and application of language
Mathematics and Physics , Taizhou 225300 ) Abstract : With a generalized broken language, describes the general solution of a class of problems . Key words : a definite answer ; common ; method CLC number: 0293.4 Document code : A Article ID :1008 -8458 (2008) 02-0024 a O1 I believe that as long as the combined terms . Proof: Suppose not, () is m +1 heavy roots, is not, (), or multiple root , ( ) k- re- root, but k ≠ m +1. (1 ) if they are , () the re- root , from that, ( ) the root, can be set  = (a ) p (), where (-Ot),} px), Tony Ⅱ) = ( a ) px) + p (x), so (-Ot) () which Ot is f (x) the m re- root conflict. (2 ) If Ot is, ( ) k- re- root, but k ≠ m +1, set ,()=(- Ot) (),[link widoczny dla zalogowanych], where (-Ot),} px), beta 0, x) = k ( -Ot) a P ()+(- Ot) kp () = ( a Ot) [Px) + ( a ) p ()] so (I Ot) 'f factory ( ), but x-Ot) x) by the Ot is the factory (), m- repeated roots , we can get m = k -1 is : k = m +1, which is a contradiction , which , if not established. Concluded with this promotion , you can directly solve the following example: Example : If a is P ( ), a k multiple root , prove that a is g (x )=()+ plant (o)] a ) X. , (o) of the Roots of a k +3 . Proof: From conditions can be obtained (): £ L Wu +()()=() is the result of a () a k multiple root , it is a is ( ) k + 1 , a re- root. Because of a Yes ( ) is a root , so a is gx) k +2 a multiple root . Is because of a g'x) a root , so a is g (x) a k +3 heavy roots.
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