Timeline for Symmetric power series over $\mathbb{F}_2$
Current License: CC BY-SA 3.0
6 events
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| Aug 2, 2016 at 14:22 | comment | added | darij grinberg | (3) At the very end, you are using the fact that if $p$ is any power series in $\mathbb{F}_2\left[\left[x_i \mid i \in J\right]\right]$, then $p + \sum\limits_{i \in J} x_i \dfrac{\partial}{\partial x_i} p$ is a power series whose all monomials have even degree. This is easy to check by linearity. | |
| Aug 2, 2016 at 14:20 | comment | added | darij grinberg | (2) You are using the fact that $\left(g\left(x\right)-1\right) g\left(x\right) = x$. This follows from the computation $\left(g\left(x\right)\right)^2 = \left(1+x+x^2+x^4+\cdots\right)^2 = 1+x^2+x^4+x^8+\cdots = g\left(x\right) - x = g\left(x\right) + x$. | |
| Aug 2, 2016 at 14:19 | comment | added | darij grinberg | Wonderful proof! Let me just spell out a few of the steps in more detail. (1) You are using the fact that $g\left(x_{i_0}\right) = 1 + \sum\limits_{i\in J}\left(g\left(x_i\right)-1\right)$. This follows from the observation that the power series $h\left(x\right) := g\left(x\right) - 1$ is additive (i.e., it satisfies $h\left(x+y\right) = h\left(x\right) + h\left(y\right)$) and thus satisfies $h\left(\sum\limits_{i \in J} x_i\right) = \sum\limits_{i \in J} h\left(x_i\right)$. Of course, the additivity of $h\left(x\right)$ follows from the additivity of $x^{2^i}$ for each $i > 0$. | |
| Aug 2, 2016 at 13:18 | vote | accept | Jens Reinhold | ||
| Aug 2, 2016 at 13:11 | history | edited | darij grinberg | CC BY-SA 3.0 |
deleted 1 character in body
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| Aug 2, 2016 at 13:03 | history | answered | user95545 | CC BY-SA 3.0 |