MHB Polynomial of Degree 98: Value of $p(100)$

[juantheron]

If $p(x)$ be a polynomial of degree $98$ such that $\displaystyle p(x) = \frac{1}{x}$ for $x=1,2,3,...,98$

Then value of $p(100)=$

 

[CaptainBlack]

If $p(x)$ be a polynomial of degree $98$ such that $\displaystyle p(x) = \frac{1}{x}$ for $x=1,2,3,...,98$

Then value of $p(100)=$

Is there no other condition?

The polynomial \(p(x)\) being of degree 98 has 99 degrees of freedom, there are 98 constraints which leaves a degree of freedom. Now if we knew that it was a monic polynomial we could answer:

\(p(100)=99!\)

(that is assuming I have the algebra right, the method is related to Opalg's approach)

CB

 

[Opalg]

I wonder whether this problem should be:

If $p(x)$ be a polynomial of degree $98$ such that $\displaystyle p(x) = \frac{1}{x}$ for $x=1,2,3,...,\color{red}{99}$

Then value of $p(100)=$

In that case, let $q(x) = xp(x)-1$. Then $q(x)$ is a polynomial of degree 99, and $q(x)=0$ for $x=1,2,3,\ldots,99$. By the factor theorem, $q(x) = k(x-1)(x-2)\cdots(x-99)$ for some constant $k$. Also, $q(0)=-1$, and therefore $k=1/99!$. It follows that $q(100) = 1$, from which $p(100) = 1/50$.

 

[juantheron]

Sorry caption and opalg i have missed that