The discussion revolves around determining the value of a polynomial \( p(x) \) of degree 98 at \( x = 100 \), given that \( p(x) = \frac{1}{x} \) for \( x = 1, 2, \ldots, 98 \). Participants explore different conditions and approaches to solve the problem.
Participants express differing views on the conditions of the polynomial and the implications for calculating \( p(100) \). There is no consensus on the correct approach or final value of \( p(100) \), as multiple interpretations and methods are presented.
The discussion highlights the dependence on the specific conditions of the polynomial and the implications of whether it is monic or has additional constraints. The assumptions made by participants regarding the polynomial's properties remain unresolved.
jacks said: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 $p(100)=$
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 $p(100) = 1/50$.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 $p(100)=$