Harmonic entropy: Difference between revisions
→Equivalence of exp-UHE and UHE for a \leq 2: fix dropped exponent |
→Equivalence of exp-UHE and UHE for a \leq 2: latex typo |
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Note that you have to be careful if you choose to compute the analytically continued <math>a=2</math> numerically: this corresponds to the vertical slice of zeta function at <math>\Re(z) = 1</math>, where there is a pole. | Note that you have to be careful if you choose to compute the analytically continued <math>a=2</math> numerically: this corresponds to the vertical slice of zeta function at <math>\Re(z) = 1</math>, where there is a pole. | ||
==Equivalence of exp-UHE and UHE for <math>a \leq 2</math>== | ==Apparent Equivalence of exp-UHE and UHE for <math>a \leq 2</math>== | ||
Let's go back to our original convolution expression for finite-<math>N</math> UHE: | Let's go back to our original convolution expression for finite-<math>N</math> UHE: | ||
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<math>\displaystyle \tilde{U}(c) \sim \log \tilde{U}(c)</math> | <math>\displaystyle \tilde{U}(c) \sim \log \tilde{U}(c)</math> | ||
additionally, noting that <math>\log \tilde{U}(c) \sim \frac{1}{1-a} \log \tilde{U}(c) = \log \tilde{U}(c)^{\frac{1}{1-a}} \sim \tilde{U}(c)^{\frac{1}{1-a}} = \exp(\text{ | additionally, noting that <math>\log \tilde{U}(c) \sim \frac{1}{1-a} \log \tilde{U}(c) = \log \tilde{U}(c)^{\frac{1}{1-a}} \sim \tilde{U}(c)^{\frac{1}{1-a}} = \exp(\text{UHE}_a(c))</math>, we get | ||
<math>\displaystyle \text{UHE}_a(c) \sim \exp(\text{ | <math>\displaystyle \text{UHE}_a(c) \sim \exp(\text{UHE}_a(c))</math> | ||
as <math>N \to \infty</math>, for <math>a \leq 2</math>. | as <math>N \to \infty</math>, for <math>a \leq 2</math>. | ||