Prove that a collection of certain polynomials sum to unity

Question

Consider a (finite) collection of $n$ distinct constants $\{f_1,\dots,f_n\}$ and define the polynomial in $x$ $$P_i(x):=\prod_{j\in\{1,\dots,n\}}^{j\neq i}\frac{(x-f_j)}{(f_i-f_j)}~,$$ for $1\leq i \leq n$. I wish to prove the property that the sum of all such polynomials is unity, $$\sum_{i=1}^{n}P_i(x)=1~.$$ The brute force method of finding a common denominator and naively summing doesn't seem to be very practical, so how should I approach this?

Note that this problem actually appears within the context of orthogonal projectors. In particular, the $P_i(x)$ is really a projector to the eigenspace $V_i$ associated with an eigenvalue $f_i$ of an operator $x$ defined to act on some $n$ dimensional vector space $V$. However, I believe this property should hold regardless of this context, and I do not want to use this information in the proof.

Answer 1

The polynomials $P_i(x)$ are called the Lagrange polynomials associated to the nodes $f_i$. They are commonly used to compute interpolating polynomials. In particular, since $P_i(f_j) = \delta_{ij}$, the interpolating polynomial of some function $g(x)$ in the nodes $f_1, \cdots, f_n$ is given by $$ p(x)= \sum_{i=1}^n g(f_i) P_i(x). $$

If you consider the constant function $g(x) = 1$, the result follows.

Answer 2

It is not the cleanest way to do it, but it is functional. Define $T(x)=\sum_{i=1}^n P_i(x)$, now using log-derivation, find $T'(x)$, and you can first check some small cases, I mean, $n=2,3$, so you can see that $T'(x)=0$. Then, the function $T(x)$ is constant. It is not difficult to show that $T(0)=1$. Then, $T(x)=1$.