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Jun 26 at 0:51 comment added GH from MO This question was asked and answered before: mathoverflow.net/q/81732
Jun 25 at 22:26 history became hot network question
Jun 25 at 15:09 vote accept loup blanc
Jun 25 at 14:47 history edited loup blanc CC BY-SA 4.0
added 28 characters in body
Jun 25 at 14:41 history edited loup blanc CC BY-SA 4.0
added 28 characters in body
Jun 25 at 14:36 history edited LSpice CC BY-SA 4.0
$\textbf{Question}.$ -> **Question.**
Jun 25 at 14:36 answer added Will Sawin timeline score: 15
Jun 25 at 14:35 comment added YCor You certainly need to assume that $f$ is not constant.
Jun 25 at 14:34 comment added YCor @AlekseiKulikov I know, I erased it before you replied. Sorry.
Jun 25 at 14:33 comment added Aleksei Kulikov @YCor Jochen does not say that you can replace $b$ by $0$, but that you can replace $b$ by $1$.
Jun 25 at 14:32 history edited loup blanc CC BY-SA 4.0
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Jun 25 at 14:32 comment added Jochen Wengenroth ($b$ is unnecessary since you can replace $f-a$ by $(f-a)/b$, hence $b=1$ is enough)
Jun 25 at 14:30 comment added YCor ($a$ is unnecessary since you can replace $f$ with $f-a$)
Jun 25 at 14:29 comment added YCor What is the question?
Jun 25 at 14:23 history asked loup blanc CC BY-SA 4.0