5.4 critical fit NG β=5.85 ρ=3

we fit mPCAC and M2PS with the formule

{mPCAC=P[0]+P[2](ηηcr)+P[4]μ+P[6](m0mcr)+P[9]μ(ηηcr)M2PS=P[1]+P[3](ηηcr)+P[5]μ+P[7](m0mcr)+P[8](ηηcr)2.

The coefficient P[0] and P[1] represent the value of mPCAC and M2PS at ηcr and μ=0. We are assuming that we are simulating at mcr

χ2/d.o.f.=1.81664P[0]=0.0139226±(0.0035)P[1]=0.0171355±(0.0075)P[2]=0.158279±(0.014)P[3]=0.329532±(0.064)P[4]=1.25664±(0.12)P[5]=2.88677±(0.042)P[6]=1.06658±(0.065)P[7]=0.509116±(0.11)P[8]=1.5204±(0.096)P[9]=6.63147±(0.42) {C=(10.002740.001280.002830.002130.0002370.0008290.0009820.0009050.0008030.0027410.0003810.007120.006110.001830.0009390.002520.001997.32e050.001280.00038110.0005010.004050.00180.005180.004830.004930.007660.002830.007120.00050110.04960.006540.007160.03130.02990.005920.002130.006110.004050.049610.0270.008110.008910.001470.04480.0002370.001830.00180.006540.02710.006940.001750.01040.003280.0008290.0009390.005180.007160.008110.0069410.01860.007270.01530.0009820.002520.004830.03130.008910.001750.018610.08980.01120.0009050.001990.004930.02990.001470.01040.007270.089810.02460.0008037.32e050.007660.005920.04480.003280.01530.01120.02461)}

−2.4−2.2−2−1.8−1.6−0.25−0.2−0.15−0.1−0.0500.050.10.150.2
(0.012,-0.048)(0.012,-0.045)(0.0224,-0.048)(0.0224,-0.045)(0.0224,-0.04)(0.04,-0.045)(0.04,-0.04)(ribbon,-0.045)(ribbon,-0.04)(ribbon,-0.048)$\eta$$m_{pcac}$$(\mu_3,\color{red}{m}_{\color{cyan}0})$
−2.4−2.2−2−1.8−1.60.10.20.30.40.5
(0.012,-0.048)(0.012,-0.045)(0.0224,-0.048)(0.0224,-0.045)(0.0224,-0.04)(0.04,-0.045)(0.04,-0.04)(ribbon,-0.045)(ribbon,-0.04)(ribbon,-0.048)$\eta$$M_{PS}^2$$(\mu_3,\color{red}{m}_{\color{cyan}0})$