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Jun 25 at 13:14 comment added ZENG When $y \leq x^{\alpha}$, we use taylor-lagrange formula for $f(x+y) - f(x)$ like what's done in the paper. When $y > x^{\alpha}$, we could use $f(x+y) - f(x) \leq \log(1+y) < y^{\beta}$. So the theorem remains correct, there is no problem.
Jun 24 at 15:04 vote accept ZENG
Jun 23 at 18:10 history edited YCor CC BY-SA 4.0
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Jun 23 at 16:33 answer added John Dawkins timeline score: 1
S Jun 23 at 14:34 review First questions
Jun 23 at 15:17
S Jun 23 at 14:34 history asked ZENG CC BY-SA 4.0