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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?

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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 $t^{n}$.

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  • $\begingroup$ That's a very interesting approach. So to clearify it a bit more, I'm considering the datapoints $(y_0, f(y_0)), \dots ,(y_n, f(y_n)))$ and use the interpolation polynomial in Lagrange form. And by comparing the coefficients of $t^n$ the second identity is proven. Thank you very much for your quick help. $\endgroup$ – Olorin Jan 3 '16 at 19:03
  • $\begingroup$ Yes, exactly. You may use for the first equality as well: interpolating polynomial $f(t)=\prod_{i=0}^n(t-x_i)$ of degree $n+1$ in points $y_0,\dots,y_n$ gives $f(t)$ modulo $\prod (t-y_i)$, which in our situation is just $f(t)-\prod (t-y_i)=(\sum y_i-\sum x_i) t^{n}+\dots$ $\endgroup$ – Fedor Petrov Jan 3 '16 at 19:08
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OR $$ \sum_{i=0}^{n}\frac{\prod_{j=0}^{n-1}(x_{j}-y_{i})}{\prod_{j=0, \, j\neq i}^{n}(y_{j}-y_{i})} = \frac{1}{2\pi i} \int_{BigDisk}\frac{\prod_{i=0}^{n-1}(z-x_{j})}{\prod_{i=0}^{n}(z-y_{j})}dz=-\mathrm{Res}(\infty)=1 $$

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  • $\begingroup$ Thank you very much for this answer, too. Using the Residue theorem it is also very easy to show the first identity. $\endgroup$ – Olorin Jan 4 '16 at 11:59
  • $\begingroup$ Yes, exactly. The sum of all residues (including at $\infty$) of a rational function is zero. Now we just need to choose the right function $f(z)=\frac{ \prod_{j=0}^{n}(z-x_{j}) } { \prod_{j=0}^{n}(z-y_{j}) }$ $\endgroup$ – Paata Ivanishvili Jan 4 '16 at 16:18

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