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

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

α мәнін табыңыз (complex solution)

$$\alpha =\frac{\ln(\frac{9\sigma _{x}^{2}}{4}-\frac{1}{2})}{\ln(2)+\pi i}+\frac{2\pi n_{1}i}{\ln(2)+\pi i}$$

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

$$\sigma _{x}\neq -\frac{\sqrt{2}}{3}\text{ and }\sigma _{x}\neq \frac{\sqrt{2}}{3}$$

σ_x мәнін табыңыз (complex solution)

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

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

α мәнін табыңыз

$$\alpha =\log_{2}\left(\frac{9\sigma _{x}^{2}}{4}-\frac{1}{2}\right)$$

$$Numerator(\log_{2}\left(\frac{9\sigma _{x}^{2}}{4}-\frac{1}{2}\right))\text{bmod}2=0\text{ and }Denominator(\log_{2}\left(\frac{9\sigma _{x}^{2}}{4}-\frac{1}{2}\right))\text{bmod}2=1\text{ and }|\sigma _{x}|>\frac{\sqrt{2}}{3}$$

σ_x мәнін табыңыз

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

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