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# A Question

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
3 years, 9 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 $$p(10) = \tfrac{1}{5}$$. · 3 years, 9 months ago

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. Staff · 3 years, 9 months ago

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 $$x$$. · 3 years, 9 months ago

Try remainder theorem, although it will take many steps. · 3 years, 9 months ago

This will do it : http://www.mualphatheta.org/problemcorner/MathematicalLog/Issues/0402/MathematicalLogProblemistSpring02.aspx · 3 years, 9 months ago