= Annular harmonic interpolation factor
{title2=$H_n(r)=\frac{r^n-a^{2n}r^{-n}}{1-a^{2n}},\quad K_n(r)=\frac{a^n(r^{-n}-r^n)}{1-a^{2n}}$}
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 $vH_n+wK_n$.
To derive them, <separation of variables> in <Laplace's equation> in <polar coordinates> gives $r^2R''+rR'-n^2R=0$, 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 $\alpha=(v-a^nw)/(1-a^{2n})$ and $\beta=a^n(w-a^nv)/(1-a^{2n})$, 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> $\log r$.
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