Polynomial Intepolation

Algebra Level 5

If P ( x ) P(x) is a polynomial function of degree 98 such that P ( K ) = 1 K P(K)=\dfrac{1}{K} for K = 1 , 2 , 3 , 4 , , 99 K=1,2,3,4,\cdots,99 . Calculate P ( 100 ) P(100)

100 ! + 1 100!+1 1 50 \dfrac{1}{50} None of the other choices 1 100 ! \dfrac{1}{100!} 1 100 \dfrac{1}{100}

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1 solution

Pranjal Jain
Apr 11, 2015

Consider a polynomial Q ( x ) Q(x) of degree 99 99 such that Q ( x ) = x P ( x ) 1 Q(x)=xP(x)-1 . Clearly, x = 1 , 2 , 3 , , 99 x=1,2,3,\cdots,99 are the roots of Q ( x ) Q(x) .

Thus, Q ( x ) = x P ( x ) 1 = λ ( x 1 ) ( x 2 ) ( x 99 ) Q(x)=xP(x)-1=\lambda(x-1)(x-2)\cdots(x-99) .

Substituting x = 0 x=0 , 1 = λ 99 ! λ = 1 99 ! -1=-\lambda 99!\Rightarrow \lambda=\dfrac{1}{99!}

Substituting x = 100 x=100 , 100 P ( 100 ) 1 = 1 99 ! × 99 ! = 1 P ( 100 ) = 2 100 = 1 50 100P(100)-1=\dfrac{1}{99!}\times 99!=1\Rightarrow P(100)=\dfrac{2}{100}=\dfrac{1}{50}

Given in our Reso material? Seems like I made a tiny error by assuming Q ( x ) = P ( x ) 1 x Q(x)=P(x)-\frac{1}{x} as it won't remain a polynomial. Nice solution!

Arpan Banerjee - 6 years, 2 months ago

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Yeah, we are given material for advanced these days and they have lots of nice questions.

Pranjal Jain - 6 years, 2 months ago

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We here in Mumbai are following the Kota schedule as well, albeit with a few changes.

Arpan Banerjee - 6 years, 2 months ago

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

Karan Shekhawat - 6 years, 2 months ago

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.)

James Wilson - 3 years, 9 months ago

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P(x)= λ ( x 1 ) ( x 2 ) ( x 99 ) + 1 x \dfrac{\lambda(x-1)(x-2)\cdots(x-99)+1}{x} it still has term (1-99!)/x. Is P(x) a polynomial?

Nikita Sharma - 3 years, 9 months ago

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Pranjal forced it to be a polynomial by taking λ = 1 99 ! \lambda =\frac{1}{99!} . 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).

James Wilson - 3 years, 9 months ago

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@James Wilson Thank you James.

Nikita Sharma - 3 years, 9 months ago

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@Nikita Sharma No problem.

James Wilson - 3 years, 9 months ago

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