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It is based on [[Lagrange polynomial|Lagrange interpolation]] and techniques of [[generating function]]s.
Consider <math>S(x)\Lambda(x),
:<math>S(x)\Lambda(x)=\sum_{j=0}^{\infty}\sum_{i=0}^j s_{j-i+1}\lambda_i x^j.</math>
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\end{align}</math>
We want to compute unknowns <math>e_j,</math> and we could simplify the context by removing the <math>\left(x\alpha^{i_j}\right)^{d-1}</math> terms. This leads to the error evaluator polynomial
:<math>\Omega(x) \equiv S(x) \Lambda(x) \bmod{x^{d-1}}.</math>
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As [[formal derivative]]
:<math>\Lambda'(x)=\lambda_0\sum_{j=1}^v \alpha^{i_j}\prod_{\ell\in\{1,\cdots,v\}\setminus\{j\}} \left (\alpha^{i_\ell}x-1 \right ),</math>▼
▲:<math>\Lambda'(x)=\lambda_0\sum_{j=1}^v \alpha^{i_j}\prod_{\ell\in\{1,\cdots,v\}\setminus\{j\}} \left (\alpha^{i_\ell}x-1 \right ),</math>
we get again only one summand in
:<math>\Lambda'(\alpha^{-i_k})=\lambda_0\alpha^{i_k}\prod_{\ell\in\{1,\cdots,v\}\setminus\{k\}} \left (\alpha^{i_\ell}\alpha^{-i_k}-1 \right ).</math>
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This formula is advantageous when one computes the formal derivative of <math>\Lambda</math> form
:<math>\Lambda(x)=\sum_{i=1}^v\lambda_ix^i</math>
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where
:<math>i\cdot x := \sum_{k=1}^i x.</math>
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