Skip to main content
5 events
when toggle format what by license comment
Feb 8, 2012 at 5:21 comment added zapkm @Ralph, You are right,... I cnt see reason for $\frac{F(g^{-1}(v))}{G(g^{-1}(v))}$ to be independent of $v$.
Feb 8, 2012 at 5:19 history edited zapkm CC BY-SA 3.0
added 98 characters in body; added 23 characters in body
Feb 7, 2012 at 13:42 comment added Ralph Why is $\frac{F(g^{-1}(v))}{G(g^{-1}(v))}$ independent from $v$, i.e. constant ?
Feb 7, 2012 at 12:29 history edited KConrad CC BY-SA 3.0
added 2 characters in body
Feb 7, 2012 at 4:20 history answered zapkm CC BY-SA 3.0