An electricity and magnetism problem by Spandan Senapati

Assigning points with coordinates (m,n) and the point where current is injected be (0,0) with B being the point (1,1) we wish to find the equivalent resistance between A and B.Assign each of the resistances $1 \Omega$ so that finally the equivalent resistance can be scaled by a factor of $R$ .We superpose on this condition another in which current is ejected out from the node $B$ to get $R_{eq}=2(V_{1,1}-V_{0,0})$ It follows from Kirchoff's current law that $4V_{m,n}=V_{m,n+1}+V_{m,n-1}+V_{m+1,n}+V{m-1,n}$ Assuming a solution $V_{m,n}=A(\alpha ,\beta )\phi ^m\psi^n$ we have $4-(\phi +\frac {1}{\phi }+\psi +\frac {1}{\psi })=0$ letting $\phi =e^{i\alpha },\psi =e^{i\beta }$ it becomes $4-(e^{i\alpha }+e^{-i\alpha }+e^{i\beta }+e^{-i\beta })=0\implies 2-\cos \alpha -\cos \beta =0$ hence $V_{m,n}=A(\alpha ,\beta )e^{i(\alpha m+\beta n)}$ also it follows from physical symmetry that $V_{m,n}=V{m,-n}=V{-m,-n}=V_{-m,n}=A(\alpha ,\beta )e^{i(|m|\alpha +|n|\beta )}$ hence the general solution is given by the superposition over a period $(-\pi,\pi )$ , $V_{m,n}=\int _{-\pi }^{\pi }A(\alpha ,\beta )e^{i(|m|\alpha +|n|\beta )}$ further notice that $I_{m,0}=0$ so $4V_{m,0}-V_{m,1}-V{m,-1}-V_{m+1,0}-V{m-1,0}=0\implies -\int _{-\pi }^{\pi }2iA(\alpha ,\beta )\sin \beta e^{im\alpha }d\alpha =0$ Recalling the Fourier series expansion of a function $f(t)=\sum _{n=-\infty }^{\infty }c_{n}e^{in\omega t },c_{n}=\frac {1}{T}\int _{-T/2}^{T/2}f(t)e^{-in\omega t}dt$ for real values functions since the $i\sin$ terms sum to zero.Hence it follows that \sum _I_{m,0}e^{im\alpha }=I_{0,0}=-1 where we assume $1 A$ current is injected into the node at $A$ hence $A(\alpha ,\beta )=\frac {1}{4\pi i\sin \beta }$ ,this gives us the function $V{m,n}$ the equivalent resistance between the nodal points $0,0$ and $m,n$ , $m,n>0$ is hence $R_{m,m}=\frac {1}{2\pi i}\int _{-\pi }^{\pi }\frac {e^{i(m\alpha +n\beta )}-1}{\sin \beta }d\alpha$ split $\int _{-\pi }^{\pi }=\int _{-\pi }^0+\int _{0}^{\pi }$ and use De-Moivre's formula with change of variable in $\int _{-\pi }^{0}$ integral to get $R_{m,n}=\frac {1}{\pi i}\int _{0}^{\pi }\frac {1-\cos m\alpha e^{in\beta }}{\sin \beta }d\alpha$ it follows by integration that $R_{1,1}=\frac {2R}{\pi }$ and also yields the general relation $R_{m,m}=\frac {2R}{\pi }\sum _{k=1}^{n}\frac {1}{2k-1}$