If P ( x ) is a polynomial function of degree 98 such that P ( K ) = K 1 for K = 1 , 2 , 3 , 4 , ⋯ , 9 9 . Calculate P ( 1 0 0 )
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Given in our Reso material? Seems like I made a tiny error by assuming Q ( x ) = P ( x ) − x 1 as it won't remain a polynomial. Nice solution!
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Yeah, we are given material for advanced these days and they have lots of nice questions.
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We here in Mumbai are following the Kota schedule as well, albeit with a few changes.
oopes ... I forget to devide by two ... and report as .. 1/100 ! ... anyway nice question ... It is request that please post more such question from ur advance material for physics also if possible... so that it helps our community helps also .. Thanks
Nice problem! I did it the long way and I got my answer to be sum from n=1 to 100 of (-1)^(n+1)(100)C(n)/100. But I messed up and didn't realize the index of the sum was only suppose to go to 99. So I got 1/100 as the answer instead of 1/50. (The sum has a lot of cancellation between terms on opposite ends of Pascal's triangle.)
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P(x)= x λ ( x − 1 ) ( x − 2 ) ⋯ ( x − 9 9 ) + 1 it still has term (1-99!)/x. Is P(x) a polynomial?
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Pranjal forced it to be a polynomial by taking λ = 9 9 ! 1 . I used a Lagrange interpolation, so in my case it was easy to see it was a polynomial. In Pranjal's case, the +1 at the end cancels with the constant term of Q(x).
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@James Wilson – Thank you James.
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Consider a polynomial Q ( x ) of degree 9 9 such that Q ( x ) = x P ( x ) − 1 . Clearly, x = 1 , 2 , 3 , ⋯ , 9 9 are the roots of Q ( x ) .
Thus, Q ( x ) = x P ( x ) − 1 = λ ( x − 1 ) ( x − 2 ) ⋯ ( x − 9 9 ) .
Substituting x = 0 , − 1 = − λ 9 9 ! ⇒ λ = 9 9 ! 1
Substituting x = 1 0 0 , 1 0 0 P ( 1 0 0 ) − 1 = 9 9 ! 1 × 9 9 ! = 1 ⇒ P ( 1 0 0 ) = 1 0 0 2 = 5 0 1