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Let p(x) be a polynomial of degree 8, such that p(k)=\(\frac{1}{k}\) for k=1,2,3,4,5,6,7,8,9. What is the value of p(10)?

Note by Snehdeep Arora
4 years, 5 months ago

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The mualphatheta link has (almost) this question set as a problem - the answer link did not work for me. Note that \(xp(x)-1\) is a degree \(9\) polynomial with \(1,2,3,4,5,6,7,8,9\) as zeros, and hence \[ xp(x) - 1 \; = \; A(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) \] Putting \(x=0\) we see that \(-1 \,=\, -A\times9!\) and so \(A = \tfrac{1}{9!}\). Then \(10p(10)-1 \,=\, A\times9! = 1\), and hence \(p(10) = \tfrac{1}{5}\).

Mark Hennings - 4 years, 5 months ago

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As a side note, you should verify that \[ \frac{ 1 + \frac{1}{9!} \prod_{i=1}^9 (x-i) } { x } \]

is indeed a polynomial. This follows because the constant term in the numerator cancels out, so it is a multiple of \(x\).

It is possible (especially in scenarios where the exact roots are uncertain), for the function that you define to end up being a rational function, instead of a polynomial.

Calvin Lin Staff - 4 years, 4 months ago

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I already did this when I evaluated \(A\), because I chose the value of \(A\) to be such that the value of \[ 1 + A\prod_{j=1}^9(x-j) \] when \(x=0\) was \(0\), which means that this polynomial has no constant term and hence can be safely divided by \(x\).

Mark Hennings - 4 years, 4 months ago

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Try remainder theorem, although it will take many steps.

Siddharth Kumar - 4 years, 5 months ago

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This will do it : http://www.mualphatheta.org/problemcorner/MathematicalLog/Issues/0402/MathematicalLogProblemistSpring02.aspx

Vikram Waradpande - 4 years, 5 months ago

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you can use newton-ramphson techhnique, newton forward or backward formula . it will be easy .

Ashok Kumar - 4 years, 5 months ago

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