first question here so I hope it's appropriate.
I'm looking at the following equation:
And I'm struggling to figure it out. In the sum, there's a symbol that looks like a Kronecker delta (but that the text calls 'the kronecker symbol', which I know is different). If this is the Kronecker delta as the symbol suggests, doesn't this sum only contain one term (m=0)? As for m=1 and m=2 the delta will be zero, won't it? If the text is correct in saying it's a Kronecker symbol, how do I calculate that? Have checked a few online sources but they're not clear to me (it's been a long time since high school math).
Thanks!
The notation means $\delta_{.,.}=1$ if the two indices are the same, $\delta_{.,.}=0$ if they differ. $$ \sum_{m=0}^2 (2-\delta_{0,m})\rho_x^m(\mu_s,\mu_\nu,\varphi-\varphi_0) \cos[m(\varphi-\varphi_0)] = [\sum_{m=0}^0+\sum_{m=1}^2] (2-\delta_{0,m})\rho_x^m(\mu_s,\mu_\nu,\varphi-\varphi_0) \cos[m(\varphi-\varphi_0)] $$ $$ = (2-1)\rho_x^0(\mu_s,\mu_\nu,\varphi-\varphi_0) \cos[0(\varphi-\varphi_0)] + \sum_{m=1}^2 (2-0)\rho_x^m(\mu_s,\mu_\nu,\varphi-\varphi_0) \cos[m(\varphi-\varphi_0)] $$ $$ = \rho_x^0(\mu_s,\mu_\nu,\varphi-\varphi_0) + 2\sum_{m=1}^2 \rho_x^m(\mu_s,\mu_\nu,\varphi-\varphi_0) \cos[m(\varphi-\varphi_0)] $$