Annular harmonic interpolation factor ($H_n(r)=\frac{r^n-a^{2n}{r}^{-n}}{1-a^{2n}},K_n(r)=\frac{a^n(r^{-n}-r^n)}{1-a^{2n}}$)

For an annulus $a<r<1$ with $0<a<1$, these factors interpolate the two boundary coefficients of a Fourier harmonic of positive integer order $n$ in a harmonic function. They obey $H_n(1)=1$, $H_n(a)=0$, $K_n(1)=0$, $K_n(a)=1$. Thus a mode with outer coefficient $v$ and inner coefficient $w$ has radial factor $v{H}_n+w{K}_n$.

To derive them, separation of variables in Laplace's equation in polar coordinates gives $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, whose solutions are $r^n,r^{-n}$. Imposing the two Dirichlet boundary conditions on $R=\alpha r^n+\beta r^{-n}$ gives $\alpha +\beta =v$ and $a^n\alpha +a^{-n}\beta =w$. Solving yields $\alpha =(v-a^{n}w)/(1-a^{2n})$ and $\beta =a^n(w-a^{n}v)/(1-a^{2n})$, proving the formula. For the zero angular mode, the analogous factor is $v+(w-v)\log r/\log a$, since its radial solutions are $1$ and the logarithm $\log r$.