2
$\begingroup$

Let $x_1<x_2<\dots<x_n$ be $n$ real numbers. I'm trying to prove the following polynomial identity (I found this identity in an undergrad book): $$ P(Y):=1+Y+Y^2+\cdots+Y^{n-1}= \sum_{k=1}^n \prod_{\underset{j\neq k}{1\le j \le n}} \frac{Yx_k-x_j}{x_k-x_j}$$

Since it is very similar to the Lagrange basis, I started to write that $$P(Y)=\sum_{k=1}^n P(x_k)\prod_{\underset{j\neq k}{1\le j \le n}} \frac{Y-x_j}{x_k-x_j} $$ but apparently I didn't find a way to transform this expression into the first one. Since the roots of $P$ are the $n$-th roots of unity (except $1$), I've considered also to prove that the RHS of the first equality as the same roots (but due to the fact that $x_1<x_2<\dots<x_n$ are random numbers it didn't seem like a good idea).

Do you have any hint for me?

$\endgroup$
1
$\begingroup$

First, we should show that the right-hand side is independant of the values $x_1,\ldots,x_n$.

By reducing everything to the same denominator, the right-hand side is of the form $Q(Y,x_1,\ldots,x_n)/\prod_{i<j} (x_i-x_j)$, where $Q$ is of degree $n(n-1)/2$ in the $x_i$ (it has the same degree as the denominator). We want to show that $Q$ is actually of the form $R(Y)\prod (x_i-x_j)$. To do that we show that $Q$ is divisible by $(x_1-x_2)$.

By looking at the first two terms of the right-hand side, they were originally of the form $(Yx_1-x_2)f(x_1,Y,x_3,\ldots,x_n)/(x_1-x_2)$ and $(Yx_2-x_1) f(x_2,Y,x_3,\ldots,x_n)/(x_2-x_1)$ where $f$ is some rational fraction. After putting them on the denominator $\prod_{i<j} (x_i-x_j)$, the numerators become of the form $(Yx_1-x_2)g(x_1,x_2,Y,x_3,\ldots,x_n)$ and $-(Yx_2-x_1)g(x_2,x_1,Y,x_3,\ldots,x_n)$ where $g$ is some polynomial. When you evaluate this sum when $x_1 = x_2$ you obtain $0$, which shows that the whole thing is a multiple of $(x_1-x_2)$.

As for the remaining $n-2$ terms, $(x_1-x_2)$ is a factor introduced when putting them on the denominator $\prod_{i<j} (x_i-x_j)$, so their numerators are all multiples of $x_1-x_2$.

Hence $(x_1-x_2)$ divides $Q(Y,x_1,\ldots,x_n)$. By a symmetry argument, so does $(x_i-x_j)$ for any pair $i<j$. Since the degree of $Q$ in the $x_i$ is $n(n-1)/2$, it follows that $Q$ is of the form $R(Y)\prod (x_i-x_j)$, and the right-hand side simplifies to $R(Y)$ where $R$ is a degree $n-1$ polynomial.

Now, to identify $R$, we can choose whatever $x_i$ we want. In particular, take the complex numbers $x_i = \exp(2i\pi/n)$.

Now, using the well-known fact that $\prod_{i \neq n} (X-x_i) = 1+X+\ldots+X^{n-1}$, the right-hand side becomes :
$$\sum_k \prod_{j \neq k} \frac {Yx_k-x_j}{x_k-x_j} = \sum_k \prod_{j \neq k} \frac {Y-x_j/x_k}{1-x_j/x_k} = \sum_k \prod_{j \neq n} \frac {Y-x_j}{1-x_j} = n \prod_{j \neq n} \frac {Y-x_j}{1-x_j} = \\ n \frac{1+Y+\ldots+Y^{n-1}}{1+1+\ldots+1} = 1+Y+\ldots+Y^{n-1} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.