Proof of $E_\alpha(x^\alpha)E_\alpha(y^\alpha)=E_\alpha((x+y)^\alpha)$

The author obtains interesting results that seem to be groundbreaking. However they are based on the identity $E_\alpha(x^\alpha)E_\alpha(y^\alpha)=E_\alpha((x+y)^\alpha)$ (Equation(2.21)), which...

04 December 2012 9,343 5 View