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I'm reading some notes here.

At the bottom of page 12, there is the equality $$ \log\prod_{i=1}^m\frac{1}{1-x_it}=\sum_{i=1}^m\sum_{j\geq 0}(x_it)^j. $$

I know the formal identity $$ \frac{1}{1-x_it}=\sum_{j\geq 0}(x_it)^j $$ but where did the $\log$ go on the right hand side? I thought a $\log$ or a product is the sum of the $\log$ of the terms.

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    It should go between the two sums. – 2017-01-07
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    @SquirtleSquad So it's a typo, yes? – 2017-01-07
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    Yes, I believe so. – 2017-01-07

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I suppose that this starts from $$\log\prod_{i=1}^m\frac{1}{1-x_it}=-\sum_{i=1}^m\log(1-x_it)$$ followed by Taylor expansion $$\log(1-a)=-\sum_ {j=1}^\infty \frac {a^j}j\implies \log(1-x_it)=-\sum_ {j=1}^\infty \frac {(x_it)^j}j$$ So, $$\log\prod_{i=1}^m\frac{1}{1-x_it}=-\sum_{i=1}^m\log(1-x_it)=\sum_{i=1}^m\sum_ {j=1}^\infty \frac {(x_it)^j}j=\sum_{i=1}^m\sum_ {j=1}^\infty \frac {(x_i)^j}j t^j$$