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Can anyone help me to break this product into the series based on ${x_j}$ ?

$$ \prod_{i=1}^{K}(1-x_i)$$

I want to break it to some function as below:

$$ \sum_j\Psi(x_j) $$

I saw something like this long time ago in an article, but I cant remember it right now! I appreciate any suggestion for starting point! or any hint.

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    $\begingroup$ I guess you mean $x^i$ in the product? $\endgroup$ – gammatester Jul 17 '15 at 12:09
  • $\begingroup$ In your product there are $K$ numbers $x_1, \dots x_K$ but no $x$! $\endgroup$ – gammatester Jul 17 '15 at 12:14
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    $\begingroup$ have you tried to derive a formula for $K=1,K=2$ etc. and find a pattern? $\endgroup$ – Surb Jul 17 '15 at 12:15
  • $\begingroup$ @ Surb, yes it does not make sense ! maybe i made mistake $\endgroup$ – Cardinal Jul 17 '15 at 12:16
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    $\begingroup$ Your edit will be hard to reach, since there are mixed terms appearing in your product. E.g. $K=2 \implies (1-x_1)(1-x_2)=1-x_1-x_2+{\color{red}{x_1x_2}}$ $\endgroup$ – Surb Jul 17 '15 at 12:21
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Would it be something along the lines of considering the logarithm of the product, whence \begin{eqnarray} \log \prod_{i=1}^{K}(1-x_{i}) &=& \sum_{i=1}^{K}\log(1-x_{i}) \end{eqnarray}

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  • $\begingroup$ and if i break the right hand side log we have log(PROD), haven't I ? $\endgroup$ – Cardinal Jul 17 '15 at 12:14
  • $\begingroup$ I'm not sure what you mean by break but I imagine it would be difficult to find a function $\Psi$ since asymptotically you could consider the Taylor expansions of $\log (1-x_{i})$ etc $\endgroup$ – Autolatry Jul 17 '15 at 12:41
  • $\begingroup$ you solution gives the logarithm of a product. ok ? but i want a new representation of a products to sums $\endgroup$ – Cardinal Jul 17 '15 at 12:45

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