Assume I have a chain of real numbers, s.th.
$x\_0 < y\_0 < x\_1<y\_1<x\_2<\dots <x\_n<y\_n$.


I'm trying to explicitely solve the expression


$$ \sum\_{i=0}^n \frac{\prod\_{j=0}^n(x\_j-y\_i)}{\prod\_{j=0, j\neq i}^{n}(y\_j-y\_i)}$$


Calculation until $n=3$ it seems the expression is equal to $\sum\_{j=0}^n x\_j - \sum\_{j=0}^{n}y\_j$.


As it is such a symmetric solution and even the above equation is somehow symmetric, I'm wondering if my assumption is right, that both expressions are the same and I'm asking myself what's the trick to show this equality.


The equation would be already true, if we had the following identity:


$$\sum\_{i=0}^n \frac{\prod\_{j=0}^{n-1}(x\_j-y\_i)}{\prod\_{j=0, j\neq i}^{n}(y\_j-y\_i)}=1$$


Maybe someone knows this expressions?



equality of expressions for sum
Consider polynomial $f(t)=\prod\_{i=0}^{n-1}(t-x\_i)$, interpolate it in points $y\_0,\dots,y\_n$:
$$
f(t)=\sum\_{i=0}^n f(y\_i)\frac{\prod\_{j\ne i}(t-y\_j)}{\prod\_{j\ne i}(y\_i-y\_j)}.
$$
Now compare coefficients of $t^{n}$.


