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May 11, 2023 at 12:13 comment added The Amplitwist There seems to be a bug due to which a link in the last paragraph is not being displayed, so I'm just reposting that link for convenience: Steve Huntsman's answer to "The matrix tree theorem for weighted graphs"
Jan 12, 2010 at 2:20 history edited Steve Huntsman CC BY-SA 2.5
fixed per VA's comment
Jan 11, 2010 at 19:07 comment added Steve Huntsman Fixed the 6th line, thanks. I probably used a different convention for the Bernoullis. And I think you are correct about the convergence issue. I was basically just pasting old notes and am not current on this. Sorry.
Jan 11, 2010 at 19:05 history edited Steve Huntsman CC BY-SA 2.5
Fixed minus sign per VA comment
Jan 9, 2010 at 17:10 comment added VA. The signs in your expansion of $Td$ are wrong; the question gives the right expansion. In particular, $Td(x)$ starts with $1=B_0$, not with $-1$. Further, in the 6th line it should be $e^{az}/(1-e^z)$.
Jan 6, 2010 at 22:58 comment added VA. I think for a general smooth function $f$ convergence is a far more delicate question than you indicate. Indeed, I think the equality does not hold for some very simple functions. But this formula with two $Td$ operators is an identity if $f$ is a hyperpolynomial.
Jan 4, 2010 at 17:24 comment added S. Carnahan You can put a backslash in front of the underscores to escape them.
Jan 4, 2010 at 16:49 comment added Steve Huntsman David--How did you fix the TeX?
Jan 4, 2010 at 16:27 history edited David E Speyer CC BY-SA 2.5
added 9 characters in body
Jan 4, 2010 at 15:27 history edited Steve Huntsman CC BY-SA 2.5
added 575 characters in body
Jan 4, 2010 at 14:57 history edited Steve Huntsman CC BY-SA 2.5
edited body
Jan 4, 2010 at 14:25 history answered Steve Huntsman CC BY-SA 2.5