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For $a_1,a_2,\cdots,a_n$, let $f_i(x)$ be $$f_i(x)=\frac{\prod_{j\neq i}(x-a_j)}{\prod_{j\neq i}(a_i-a_j)}.$$

For $b_1,b_2,\cdots,b_n$, let $g_i(x)$ be $$g_i(x)=\frac{\prod_{j\neq i}(x-b_j)}{\prod_{j\neq i}(b_i-b_j)}.$$

Can we prove such an equality: $$\sum_{i=1}^n f_j(-b_i)g_i(-a_j)=1?$$

I need this in the calculation of the detrminant of $$\left(\frac{1}{a_i+b_j}\right)_{1\leq i,j\leq n}$$ by induction on $n$.

XLDD
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  • That's not an inequality. – Kaster Jun 22 '13 at 01:09
  • This seems highly unlikely, to me (the equality, anyway). The first thing I would do is try a few simple random examples, and see if the formula is true in those cases. Either you'll get a counterexample, or you'll get some motivation to look harder for a proof. – bubba Jun 22 '13 at 01:13
  • Or, plug the whole mess into Mathematica and see what you get. It's just algebraic manipulation, after all. – bubba Jun 22 '13 at 01:14
  • With reference to your determinant question, check out Hilbert Matrix. In particular, the determinant is $ \frac{ \prod_{i\neq j} (a_i-a_j)(b_i-b_j)}{\prod (a_i+b_j)}$. – Calvin Lin Jun 22 '13 at 01:31
  • @CalvinLin No. the determinant is $\prod_{i<j}(a_i-a_j)\prod_{k<l}(b_k-b_l)/\prod (a_i+b_j)$. – XLDD Jun 22 '13 at 03:31

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It's true and no horrible algebra needed to prove it. Remember that the Lagrange polynomials are defined in this way so that if given the points $ (b_1, y_1), \dots, (b_n, y_n) $, the function: $$ L(x) = \sum_i y_i g_i(x) $$ will go through the points, ie. $ L(b_i) = y_i $ for all i. Now for your sum, let: $$ f(x) = \sum_{i=1}^n f_j(-b_i)g_i(x) $$ This is the polynomial that goes through the n points $ (b_1, f_j(-b_1)), \cdots, (b_n, f_j(-b_n)) $. There is one unique (n-1)-order polynomial that goes through n distinct points, so $f(x)$ must be exactly $ f_j(-x) $. The fact that $ f(-a_j) = f_j(a_j) = 1 $ is the usual normalization, easily checked.

user81327
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