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How the following relation can be proved: If $$l\left(x\right)=\left(x-x_{0}\right)\left(x-x_{1}\right)...\left(x-x_{k}\right)$$ if a polynomial,then

$$l_{j}\left(x\right)=\frac{l\left(x\right)}{l^{'}\left(x_{j}\right)\left(x-x_{j}\right)}$$ where $$l_{j}\left(x\right)=\prod_{m=0}^{k}\frac{x-x_{m}}{x_{j}-x_{m}}$$ when $m$ and $j$ are not equal. I tried to see thr formula in some example, but still I'm not able to proof that.

Absurd
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    Note that $l'(x)=\sum_{n=0}^{k}\prod_{0\leq m\leq k, m\neq n}(x-x_m)$. Then $l'(x_j)=\prod_{0\leq m\leq k, m\neq j}(x_j-x_m)$, since all other terms in the sum have the factor $x-x_j$ and will vanish once you evaluate at $x=x_j$. The $l(x)/(x-x_j)$ gives you the numerator $\prod_{0\leq m\leq k, m\neq j}(x-x_m)$. – conditionalMethod Dec 06 '19 at 19:16
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    Do a small example: $l(x)=(x-x_1)(x-x_2)(x-x_3)$ apply the Leibniz's product rule $l'(x)=(x-x_2)(x-x_3)+(x-x_1)(x-x_3)+(x-x_1)(x-x_2)$. See, how it is a sum of products just like $l(x)$ but missing each time one of the factors. – conditionalMethod Dec 06 '19 at 19:20
  • @conditionmethod,thnks,I got my answer. – Absurd Dec 06 '19 at 19:31

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