Douglas B. Staple
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 Mar 9 awarded Yearling Sep 30 awarded Explainer May 27 awarded Nice Answer May 16 comment “Dependence” or “dependencies” of a function on multiple variables. It seems like 'dependencies' or even 'dependences' is a reasonable choice when one wants to highlight distinct functional dependence on one variable versus another. However, 'dependence' certainly sounds more natural, and is in standard usage, while 'dependencies' does not and is not. May 16 comment “Dependence” or “dependencies” of a function on multiple variables. I guess I really am referring to different types of functional dependence. For example, suppose $f$ varies exponentially on $x$, linearly on $y$, and logarithmically on $z$. May 16 comment “Dependence” or “dependencies” of a function on multiple variables. @ClementC. it came up in physics paper I wrote discussing the "dependencies" of a physical quantity on experimental variables. In that context it sounds a lot more reasonable. I'm trying to decide if I should tell the editor to change it to "dependence". May 16 asked “Dependence” or “dependencies” of a function on multiple variables. May 8 comment Strange notation for a decimal expansion of a transcendental number Also, very funny and interesting comment you made about $\zeta(3)$! It hadn't occurred to me that it might not be transcendental. (This only came up in this post, not in the paper!) May 8 comment Strange notation for a decimal expansion of a transcendental number I've written up my long list of comments with line-numbers and sent them to my coauthors. I'm going to push for a second round of page proofs to try to iron the typos out. May 7 accepted Strange notation for a decimal expansion of a transcendental number May 7 comment Strange notation for a decimal expansion of a transcendental number Thanks, I think you are right. There are some other strange changes that appear to be done by someone with no background in math. May 7 asked Strange notation for a decimal expansion of a transcendental number Mar 9 awarded Yearling Aug 21 awarded Revival Jul 15 awarded Nice Answer Jun 24 comment $y - y_0 = \frac{d^2}{dx^2}\left[ \ln\left( \frac{y}{y_0} \right) \right]$, solve for $y$. @O.L. Ah, yes, I see it now. Jun 24 comment $y - y_0 = \frac{d^2}{dx^2}\left[ \ln\left( \frac{y}{y_0} \right) \right]$, solve for $y$. @O.L. Well, we can't have $y_0 = 0$ because of the $1/y_0$ in the log. I guess what you wrote had $y_0=0$ substituted on the LHS, and $y_0=1$ on the RHS. Tricked me too. Still interesting, though. Jun 24 comment $y - y_0 = \frac{d^2}{dx^2}\left[ \ln\left( \frac{y}{y_0} \right) \right]$, solve for $y$. Thanks to GEdgar and O.L., we have an implicit solution right away. Unfortunately, inverting that solution doesn't look like much fun! Now the search for explicit solutions begins... Jun 24 comment $y - y_0 = \frac{d^2}{dx^2}\left[ \ln\left( \frac{y}{y_0} \right) \right]$, solve for $y$. This would give something for $y$ in terms of $y_0$ only, which would imply that $y(x)$ is a constant function of $x$. My suspicion is that the trivial solution $y=y_0$ is the only solution to follow from this equation. Jun 24 answered $y - y_0 = \frac{d^2}{dx^2}\left[ \ln\left( \frac{y}{y_0} \right) \right]$, solve for $y$.