Timeline for What properties characterize the function $L(x) = x+\exp(x) \log(x)$?
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Oct 14, 2019 at 5:47 | comment | added | user6671 | @user142929 thanks for your comment | |
Oct 12, 2019 at 20:35 | comment | added | user142929 | I think (from the last paragraph of the article due to Lagarias) that maybe it is interesting consider $l(x)=e^x\log x$. For $L(x)$ or $l(x)$, maybe can be interesting to state Möbius inversion, and related asymptotic formulas, section Generalizations of the Wikipedia Möbius inversion formula as starting point. I refer also (section 3) of Manuel Benito and Luis M. Navas and Juan Luis Varona, Möbius inversion from the point of view of arithmetical semigroup flows, Biblioteca de la Revista Matemática Iberoamericana, Proceedings of the Segundas Jornadas de Teoría de Números (Madrid, 2007). | |
May 24, 2019 at 18:07 | history | edited | user6671 | CC BY-SA 4.0 |
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May 19, 2019 at 18:03 | history | edited | user6671 |
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May 19, 2019 at 4:23 | history | edited | user6671 | CC BY-SA 4.0 |
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May 18, 2019 at 20:21 | history | edited | user6671 | CC BY-SA 4.0 |
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May 18, 2019 at 20:19 | comment | added | user6671 | @NathanielJohnston: Ok, I will update the question, with further properties. | |
May 18, 2019 at 20:13 | comment | added | Nathaniel Johnston | @orgesleka -- It definitely can't be true if you allow arbitrary complex eigenvalues. Try the diagonal matrix $A$ with diagonal entries $1$ and $i$. Then $L(|A|) = 1$ but $|L(A)| \approx 2.27$. | |
May 18, 2019 at 16:43 | comment | added | user6671 | @NathanielJohnston: Some of the matrices I am considering do not have real eigenvalues, but still $L(|A|) = |L(A)|$. | |
May 18, 2019 at 16:17 | comment | added | user6671 | @NathanielJonston: Ok, thanks for your comment. Would you mind writing your comment in a bit more detail as an answer? (If it is not asked too much than maybe with reference?) | |
May 18, 2019 at 16:14 | comment | added | Nathaniel Johnston | Oh I see. You're right, that property holds then simply by usual functional calculus. $|A|$ is the largest eigenvalue of $A$, whereas $|L(A)|$ is the largest eigenvalue of $L(A)$. But the eigenvalues of $L(A)$ are just $L$ applied to the eigenvalues of $A$. Since $L$ is monotone, the largest eigenvalue of $L(A)$ is just $L(|A|)$. This works as long as $A$ has positive real eigenvalues (but if that's not the case then even defining $L(A)$ in the first place is going to be somewhat icky, so I hope it's OK). | |
May 18, 2019 at 15:52 | comment | added | user6671 | @NathanialJohnston: Spectral norm. Can you give your examples? The matrices I am looking at seem to have this property. | |
May 18, 2019 at 15:42 | comment | added | Nathaniel Johnston | Can you clarify your third edit? What is the Euclidean norm of a matrix? Frobenius/Hilbert-Schmidt norm? Or operator/spectral norm? Or something else? I've tried a few guesses, but numerics quickly disprove $L(|A|) = |L(A)|$ for every guess that I've made for what the norm should be. | |
May 18, 2019 at 14:38 | history | edited | user6671 | CC BY-SA 4.0 |
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May 10, 2019 at 18:29 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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May 8, 2019 at 7:49 | history | edited | user6671 | CC BY-SA 4.0 |
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May 8, 2019 at 7:37 | history | edited | user6671 | CC BY-SA 4.0 |
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May 6, 2019 at 10:24 | history | edited | user6671 | CC BY-SA 4.0 |
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Apr 30, 2019 at 13:38 | comment | added | user6671 | @MattF. You are right. The uniqueness of your characterization is given. | |
Apr 30, 2019 at 12:42 | comment | added | user44143 | WolframAlpha confirms the uniqueness of the characterization I proposed: wolframalpha.com/input/…. The alternative suggestion has $L(1)=e$. | |
Apr 30, 2019 at 11:28 | comment | added | user6671 | @MattF. I think you forgot the condition $L'(1) = e+1$. Thanks for your help. | |
Apr 30, 2019 at 11:14 | comment | added | user44143 | One way to characterize it is $x L'' - (2 x - 1) L' + (x - 1) L =x^2 - 3 x + 1$, $L(1)=1$, $L''(1)=e$ | |
Apr 30, 2019 at 10:48 | history | asked | user6671 | CC BY-SA 4.0 |