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author | Jaron Kent-Dobias <jaron@kent-dobias.com> | 2017-06-14 14:10:53 -0400 |
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committer | Jaron Kent-Dobias <jaron@kent-dobias.com> | 2017-06-14 14:10:53 -0400 |
commit | 1b990cf214d7e2ebf003db5f7c36565b36391633 (patch) | |
tree | 706eb2bb4a373ed4f8c4717aed9aeccda80c2711 /essential-ising.tex | |
parent | 3650d1101e092048cd4f29e444385e4a44509d63 (diff) | |
download | paper-1b990cf214d7e2ebf003db5f7c36565b36391633.tar.gz paper-1b990cf214d7e2ebf003db5f7c36565b36391633.tar.bz2 paper-1b990cf214d7e2ebf003db5f7c36565b36391633.zip |
changed the way data is processed for the plots
Diffstat (limited to 'essential-ising.tex')
-rw-r--r-- | essential-ising.tex | 14 |
1 files changed, 7 insertions, 7 deletions
diff --git a/essential-ising.tex b/essential-ising.tex index 1eadf68..4238355 100644 --- a/essential-ising.tex +++ b/essential-ising.tex @@ -185,19 +185,19 @@ explicitly given our scaling ansatz, yielding \def\eqthreedeeone{ \mathcal F^{\text{\textsc{3d}}}(X)&= \frac{AB^{1/3}}{12\pi X^2}e^{-1/(BX)^2} - \bigg[\Gamma(\tfrac16)E_{7/6}((BX)^{-2}) + \bigg[\Gamma(\tfrac16)E_{7/6}(-(BX)^{-2}) } \def\eqthreedeetwo{ - -4BX\Gamma(\tfrac23)E_{5/3}((BX)^{-2})\bigg] + -4BX\Gamma(\tfrac23)E_{5/3}(-(BX)^{-2})\bigg] } \def\eqfourdeeone{ \mathcal F^{\text{\textsc{4d}}}(X)&= - \frac{A}{9\pi X^2}e^{1/(BX)^3} - \Big[3\Gamma(0,(BX)^{-3}) + -\frac{A}{9\pi X^2}e^{1/(BX)^3} + \Big[3\ei(-(BX)^{-3}) } \def\eqfourdeetwo{ - -3\Gamma(\tfrac23)\Gamma(\tfrac13,(BX)^{-3}) - -\Gamma(\tfrac13)\Gamma(-\tfrac13,(BX)^{-3})\Big] + +3\Gamma(\tfrac23)\Gamma(\tfrac13,(BX)^{-3}) + +\Gamma(\tfrac13)\Gamma(-\tfrac13,(BX)^{-3})\Big] } \ifreprint \begin{align} @@ -240,7 +240,7 @@ $C=C_{0-}/T_c$, with $C_{0-}=0.025\,536\,971\,9$ then be extracted by integration and comparison with known exact results at zero field, yielding \[ - \mathcal M^{\text{\textsc{2d}}}(X)=\frac{D}{BX}(BX-1)e^{1/BX}\ei(-1/BX)-D+\mathcal M(0) + \mathcal M^{\text{\textsc{2d}}}(X)=\mathcal M(0)+D-\frac{D}{BX}(BX-1)e^{1/BX}\ei(-1/BX) \label{eq:mag_scaling} \] with $\mathcal M(0)=\big(2(\sqrt2-1)\big)^{1/4}\big((4+3\sqrt2)\sinh^{-1}1\big)^{1/8}$ |