Timeline for Is $\int_0^\infty{dx\over x^{x^{x^x}}}<\int_0^\infty{dx\over x^{x^{x^{x^{x^x}}}}}<\int_0^\infty{dx\over x^{x^{x^{x^{x^{x^{x^x}}}}}}}<\cdots$ true?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jan 13, 2023 at 4:23 | answer | added | Eric Naslund | timeline score: 17 | |
| S Apr 12, 2022 at 16:06 | history | bounty ended | CommunityBot | ||
| S Apr 12, 2022 at 16:06 | history | notice removed | CommunityBot | ||
| S Apr 4, 2022 at 14:51 | history | bounty started | TheSimpliFire | ||
| S Apr 4, 2022 at 14:51 | history | notice added | TheSimpliFire | Canonical answer required | |
| S Mar 8, 2022 at 17:03 | history | bounty ended | CommunityBot | ||
| S Mar 8, 2022 at 17:03 | history | notice removed | CommunityBot | ||
| Mar 6, 2022 at 11:51 | history | edited | TheSimpliFire | CC BY-SA 4.0 |
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| Mar 1, 2022 at 16:13 | comment | added | Holo | oh right, I forgot how tetretion works for x<1. a transformation between the m_k to the M_k looks possible, but I'm not sure what it would be | |
| Feb 28, 2022 at 15:42 | comment | added | TheSimpliFire | @Holo $m_k$ does not tend to infinity (its growth rate is below linear). We have $1/f_k(0)=1$ rather than $+\infty$ (as is the case for odd tetrations) and $\max1/f_k(x)=e$. In fact, $m_{k+1}-m_k\to0$ also, so it is difficult to compare them at first sight. | |
| S Feb 28, 2022 at 15:23 | history | bounty started | TheSimpliFire | ||
| S Feb 28, 2022 at 15:23 | history | notice added | TheSimpliFire | Canonical answer required | |
| Feb 26, 2022 at 16:41 | history | edited | TheSimpliFire | CC BY-SA 4.0 |
added 16 characters in body
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| Feb 26, 2022 at 16:23 | comment | added | Holo | > which can be interpreted as the area gained over (0,1) is greater than that lost over (1,∞) $$$$ Well, if we say $M_k=\int_1^\infty{dx\over f_k(x)}$ and $m_k=\int_0^1{dx\over f_k(x)}$, we have $0<M_k \to 0$ and $m_k\to\infty$, moreover $m_k$ grows rate is above linear, so after a certain point $M_{k}-M_{k+1}<m_{k+1}-m_k$, which is exactly "$\int_0^\infty{dx\over f_k(x)}$ is eventually monotonic increasing". It just looks like $k=1$ is that "certain point" | |
| Feb 26, 2022 at 14:40 | history | asked | TheSimpliFire | CC BY-SA 4.0 |