Editing BCH code (section)

==== Explanation of Forney algorithm computation ====
It is based on [[Lagrange polynomial|Lagrange interpolation]] and techniques of [[generating function]]s.

Consider <math>S(x)\Lambda(x),</math> and for the sake of simplicity suppose <math>\lambda_k = 0</math> for <math>k > v,</math> and <math>s_k = 0</math> for <math>k > c + d - 2.</math> Then

:<math>S(x)\Lambda(x) = \sum_{j=0}^{\infty}\sum_{i=0}^j s_{j-i+1}\lambda_i x^j.</math>

:<math>\begin{align}
  S(x)\Lambda(x)
    &= S(x) \left \{ \lambda_0\prod_{\ell=1}^v \left (\alpha^{i_\ell}x-1 \right ) \right \} \\
    &= \left \{ \sum_{i=0}^{d-2}\sum_{j=1}^v e_j\alpha^{(c+i)\cdot i_j} x^i \right \} \left \{ \lambda_0\prod_{\ell=1}^v \left (\alpha^{i_\ell}x-1 \right ) \right \} \\
    &= \left \{ \sum_{j=1}^v e_j \alpha^{c i_j}\sum_{i=0}^{d-2} \left (\alpha^{i_j} \right )^i x^i \right \} \left \{ \lambda_0\prod_{\ell=1}^v \left (\alpha^{i_\ell}x-1 \right ) \right \} \\
    &= \left \{ \sum_{j=1}^v e_j \alpha^{c i_j} \frac{\left (x \alpha^{i_j} \right )^{d-1}-1}{x \alpha^{i_j}-1} \right \} \left \{ \lambda_0 \prod_{\ell=1}^v \left (\alpha^{i_\ell}x-1 \right ) \right \} \\
    &= \lambda_0 \sum_{j=1}^v e_j\alpha^{c i_j} \frac{ \left (x\alpha^{i_j} \right)^{d-1}-1}{x\alpha^{i_j}-1} \prod_{\ell=1}^v \left (\alpha^{i_\ell}x-1 \right ) \\
    &= \lambda_0  \sum_{j=1}^v e_j\alpha^{c i_j} \left ( \left (x\alpha^{i_j} \right)^{d-1}-1 \right ) \prod_{\ell\in\{1,\cdots,v\}\setminus\{j\}} \left (\alpha^{i_\ell}x-1 \right )
\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>

Thanks to <math>v\leqslant d-1</math> we have
:<math>\Omega(x) = -\lambda_0\sum_{j=1}^v e_j\alpha^{c i_j} \prod_{\ell\in\{1,\cdots,v\}\setminus\{j\}} \left(\alpha^{i_\ell}x - 1\right).</math>

Thanks to <math>\Lambda</math> (the Lagrange interpolation trick) the sum degenerates to only one summand for <math>x = \alpha^{-i_k}</math>

:<math>\Omega \left(\alpha^{-i_k}\right) = -\lambda_0 e_k\alpha^{c\cdot i_k}\prod_{\ell\in\{1,\cdots,v\}\setminus\{k\}} \left(\alpha^{i_\ell}\alpha^{-i_k} - 1\right).</math>

To get <math>e_k</math> we just should get rid of the product. We could compute the product directly from already computed roots <math>\alpha^{-i_j}</math> of <math>\Lambda,</math> but we could use simpler form.

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>

we get again only one summand in
:<math>\Lambda'\left(\alpha^{-i_k}\right) = \lambda_0\alpha^{i_k}\prod_{\ell\in\{1,\cdots,v\}\setminus\{k\}} \left(\alpha^{i_\ell}\alpha^{-i_k} - 1\right).</math>

So finally
:<math>e_k = -\frac{\alpha^{i_k}\Omega \left(\alpha^{-i_k}\right)}{\alpha^{c\cdot i_k}\Lambda' \left(\alpha^{-i_k}\right)}.</math>

This formula is advantageous when one computes the formal derivative of <math>\Lambda</math> form
:<math>\Lambda(x) = \sum_{i=1}^v \lambda_i x^i</math>

yielding:
:<math>\Lambda'(x) = \sum_{i=1}^v i \cdot \lambda_i x^{i-1},</math>

where
:<math>i\cdot x := \sum_{k=1}^i x.</math>