Let $p(x)$ be an eighth degree polynomial. Given that:

$p(1) = 1$, $p(2) = 1/2$, $p(3) = 1/3$, .... $p(9) = 1/9$

Find the value of $p(10)$. I tried to approach this by taking $h(x) = p(x) - 1/x$ but then $h(x)$ won't be a polynomial. I tried to manipulate the equation but I failed to find anything helpful. I have found a relation between the coefficients and some values; 9 equations and 9 variables but it would be too cumbersome to solve.

Some guidance would be appreciated. Thanks! (Sorry for the poor title; if anyone can improve it please do so)

  • 3
    $\begingroup$ Consider $g(x) = x\cdot p(x) - 1$. $\endgroup$ – Daniel Fischer Apr 18 '17 at 15:57
  • 1
    $\begingroup$ $g(x)$ is a polynomial with degree $9$ that vanishes at $x=1,2,\ldots,8,9$, hence $g(x)=K(x-1)\cdot\ldots\cdot(x-8)(x-9)$. Since $g(0)=-1$ we have $K=\frac{1}{9!}$ and $$ p(10) = \frac{g(10)+1}{10} = \frac{1+1}{10}=\frac{1}{5}.$$ $\endgroup$ – Jack D'Aurizio Apr 18 '17 at 16:24

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