Question:If $\displaystyle x_{1}, ... , x_{n}$ are distinct numbers, find a polynomial $\displaystyle f_{i}$ of degree $\displaystyle n-1$ which is $\displaystyle 1$ at $\displaystyle x_{i}$ and 0 at $\displaystyle x_{j}$ for $\displaystyle j \ne i$. Hint: the product of all $\displaystyle (x-x_{i})$ for $\displaystyle j \ne i$ is zero at $\displaystyle x_{j}$ if $\displaystyle j \ne i$. .

Confusion:The part of the question that I don't understand is the 'which is $\displaystyle 1$ at $\displaystyle x_{i}$ and 0 at $\displaystyle x_{j}$ for $\displaystyle j \ne i$.' What are $\displaystyle x_{i}$ and $\displaystyle x_{j}$? Does it mean that for any two numbers we choose from the sequence $\displaystyle x_{1}, ..., x_{n}$, we get $\displaystyle f_{i}$ as 0, for one, and 1, for the other? And how would you go about finding this polynomial?