求解 sigma_x^2=(-2-0)^alpha*4/9+(0*0)^2*3/9+(1*0)^2*2/9=

$$\sigma _ { x } ^ { 2 } = ( - 2 - 0 ) ^ { \alpha } \times \frac { 4 } { 9 } + ( 0 \times 0 ) ^ { 2 } \times \frac { 3 } { 9 } + ( 1 \times 0 ) ^ { 2 } \times \frac { 2 } { 9 } =$$

求解 α 的值 (复数求解)

$$\alpha =\frac{2\left(\ln(\sigma _{x})+\ln(\frac{3}{2})\right)}{\ln(2)+\pi i}+\frac{2\pi n_{1}i}{\ln(2)+\pi i}$$

$$n_{1}\in \mathrm{Z}$$

$$\sigma _{x}\neq 0$$

求解 σ_x 的值 (复数求解)

$$\sigma _{x}=-\frac{2\left(-2\right)^{\frac{\alpha }{2}}}{3}$$

$$\sigma _{x}=\frac{2\left(-2\right)^{\frac{\alpha }{2}}}{3}$$

求解 α 的值

$$\alpha =\frac{\ln(\sigma _{x}^{2})+\ln(\frac{9}{4})}{\ln(2)}$$

$$\sigma _{x}\neq 0\text{ and }Numerator(\frac{\ln(\sigma _{x}^{2})+2\ln(3)}{\ln(2)}-2)\text{bmod}2=0\text{ and }Denominator(\frac{\ln(\sigma _{x}^{2})+2\ln(3)}{\ln(2)})\text{bmod}2=1$$

求解 σ_x 的值

$$\sigma _{x}=\frac{2\sqrt{\left(-2\right)^{\alpha }}}{3}$$

$$\sigma _{x}=-\frac{2\sqrt{\left(-2\right)^{\alpha }}}{3}\text{, }Denominator(\alpha )\text{bmod}2=1\text{ and }\left(-2\right)^{\alpha }\geq 0$$