By looking at this formula, we can see that $\fc{Gf}x$ is exactly the expectation value of $\fc f{x+Y}$, where $Y$ is a random vector distributed according to the multivariate normal distribution with mean $0$ and covariance matrix $A$. Let’s say, however, that instead of a multivariate normal distribution, we have a discrete distribution on a finite set of points $\B{a_i}$, each with probability $w_i$, satisfying $$\sum_i w_i=1,\quad\sum_i w_ia_i=0.$$ Then, for $Y$ distributed according to this discrete distribution, the expectation value of $\fc f{x+Y}$ is $$\bopc E{\fc f{x+Y}}=\sum_i w_i\fc f{x+a_i}.$$ This distribution has the covariance matrix $A_{jk}=\sum_i w_ia_{ij}a_{ik}$, but this expectation value is not the same as the result $\fc{Gf}x$ of the Gaussian blur filter. In order to approach the Gaussian distribution, we need to sum $N$ independent samples $\B{Y_\alpha}$, each distributed according the same discrete distribution with covariance matrix $A/N$. Without changing the relation between $A$ and $\B{a_i}$, we can achieve this by taking the distribution to be among the points $\B{a_i/\sqrt N}$, with the same probabilities $w_i$. Then, according to the central limit theorem, we have $$\fc{Gf}x=\lim_{N\to\infty}\bopc E{\fc f{x+\sum_\alpha Y_\alpha}} =\lim_{N\to\infty}\sum_{\B{i_\alpha}}\p{\prod_\alpha w_{i_\alpha}}\fc f{x+\fr{1}{\sqrt N}\sum_\alpha a_{i_\alpha}}.$$ By staring at this formula, one may notice that it is actually the result of the same linear transformation $P_N$ applied $N$ times to the function $f$, where $$\fc{P_Nf}x\ceq\sum_iw_i\fc f{x+\fr{a_i}{\sqrt N}}.$$ Thus, we can write $$G=\lim_{N\to\infty}P_N^N.$$

We can now see why we can implement a Gaussian blur filter as a multi-pass filter: each pass of the filter is equivalent to applying the linear transformation $P_N$ once, and the Gaussian blur filter is the limit of applying this transformation infinitely many times. After choosing the points $\B{a_i}$ and the weights $w_i$, we can easily implement the filter in a fragment shader. After implementing $P_N$, one can implement the multi-pass filter by flip-flopping between two textures $N$ times. Here is an example implementation of $P_N$ for a 2D ($d=2$) texture, with $$a_0=\begin{pmatrix}\sgm\\0\end{pmatrix},\quad a_1=\begin{pmatrix}-\sgm\\0\end{pmatrix},\quad w_0=w_1=\fr12,$$ where $\sgm$ would be the standard deviation of the resulting Gaussian blur filter:

          varying vec2 vTextureCoord;

          

          uniform sampler2D uTexture;

          uniform float uStrength; // sigma / sqrt(N)

          

          void main() {

          	gl_FragColor  = texture2D(uTexture, vTextureCoord + vec2( uStrength, 0.0)) * 0.5;

          	gl_FragColor += texture2D(uTexture, vTextureCoord + vec2(-uStrength, 0.0)) * 0.5;

          }

        

When using the shader, the texture is input as the uniform uTexture, and $\sgm/\sqrt N$ is input as the uniform uStrength. This example is then called the 2-tap horizontal Gaussian blur filter (because to find $\fc fx$ for some $x$ one needs to evaluate $f$ twice).

To get a good Gaussian blur effect, $N$ should be large enough, which can quickly become computationally expensive. We can reduce the number of passes by rewriting $P_N^N=\p{P_N^n}^{N/n}$ so that the number of passes is now reduced from $N$ to $N/n$ without changing the result, but the cost is to implement $P_N^n$ instead of $P_N$ in the shader. With the $P_N$ example above, the $P_N^n$ filter is then a $\p{n+1}$-tap horizontal Gaussian blur filter. The total number of times to fetch the texture is $\p{n+1}N/n$ to render the whole blurred texture, which decreases as $n$ increases.

By utilizing the linear sampling feature of the GPU, this number can be further reduced (by half) if $\sgm/\sqrt N=1$. One can read further about this in this article.

Let us now move on to an interesting relation to the heat equation. The heat equation is a partial differential equation that reads $$\partial_t\fc{f_t}x=\sum_j\partial_j\partial_j\fc{f_t}x.$$ The physical meaning of $\fc{f_t}x$ is the temperature at position $x$ at time $t$, and the solution to this equation gives the time evolution of the temperature distribution in a medium with unit thermal diffusivity. More generally, we can have some other thermal diffusivities, which may even be anisotropic, in which case the equation reads $$\partial_t\fc{f_t}x=\fr12\sum_{jk}A_{jk}\partial_j\partial_k\fc{f_t}x,$$ where $A$ is a symmetric positive definite matrix. Such an equation can always be reduced to the form with unit thermal diffusivity by a linear transformation of the coordinates $x\mapsto Lx$, where $L$ satisfies $A/2=L^\mrm TL$ (the Cholesky decomposition of $A/2$). By using operator exponentiation, we can write the solution to the heat equation as $\fc{f_t}x=G^t\fc{f_0}x$, where $$G^t\ceq\fc\exp{\fr t2\sum_{jk}A_{jk}\partial_j\partial_k}$$ is called the time evolution operator (which is not generally defined for $t<0$, which means that $\B{G^t}$ cannot form a group but only a monoid).

The reason that I denote it as $G^t$ is that $G^1$ gives exactly the Gaussian blur filter $G$ (when acting on real analytic functions). This is not immediately obvious, and I will justify its correctness by starting from the multi-pass expression of the Gaussian blur filter. For analytic functions, we can get the translation operator by exponentiating the derivative operator, so $$\fc f{x+\fr{a_i}{\sqrt N}}=\fc\exp{\sum_j\fr{a_{ij}}{\sqrt N}\partial_j}\fc fx.$$ Therefore, we can write $$\begin{align*} P_N&=\sum_iw_i\fc\exp{\sum_j\fr{a_{ij}}{\sqrt N}\partial_j}\\ &=\underbrace{\sum_iw_i}_1 +\fr1{\sqrt N}\sum_j\underbrace{\sum_iw_ia_{ij}}_0\partial_j +\fr1{2N}\sum_{jk}\underbrace{\sum_iw_ia_{ij}a_{ik}}_{A_{jk}}\partial_j\partial_k +\order{N^{-3/2}}. \end{align*}$$ Now, in the limit of infinite $N$, we have $$G=\lim_{N\to\infty}P_N^N =\lim_{N\to\infty}\p{1+\fr1{2N}\sum_{jk}A_{jk}\partial_j\partial_k}^N =\fc\exp{\fr12\sum_{jk}A_{jk}\partial_j\partial_k}=G^1.$$ This means that applying the Gaussian blur filter is equivalent to evolving one unit of time according to the heat equation. Therefore, the equation in the beginning of this article, the definition of the Gaussian blur filter, can then be used to express the unit time evolution under the heat equation in the form of a integral transformation. To get the general $G^t$, we can just replace $A$ with $tA$.

As a byproduct, we can then show that the heat kernel is a Gaussian kernel. The heat kernel $\fc{K_t}{x,x'}$ is defined as the solution to the heat equation with initial condition $\fc{K_0}{x,x'}=\fc\dlt{x-x'}$. Directly applying $G^t$ to $\fc\dlt{x-x'}$ gives $$\begin{align*} \fc{K_t}{x,x'}&=G^t\fc\dlt{x-x'}\\ &=\int\fr{\d^dy}{\sqrt{\p{2\pi t}^d\det A}}\fc\exp{-\fr1{2t}\p{A^{-1}}_{jk}y_jy_k}\fc\dlt{x+y-x'}\\ &=\fr{1}{\sqrt{\p{2\pi t}^d\det A}}\fc\exp{-\fr1{2t}\p{A^{-1}}_{jk}\p{x_j-x'_j}\p{x_k-x'_k}}, \end{align*}$$ which is the form that you would find in textbooks.

The reason that I decided to study the Gaussian blur filter is that I spotted a flaw in the implementation of the blur filter in PixiJS, where the uniform uStrength in the example shader above is scaled by $1/N$ instead of $1/\sqrt N$. I was very happy to find the bug because this is the first time that I found a bug without actually producing an unexpected phenomenon first but by just staring at the source codes and deducing the mathematical formulas by hand. I opened an issue for my findings.

Suppose there is an annulus defined by the region $\set{\p{\rho,\vphi}}{R_\mrm{in}<\rho<R_\mrm{out}}$. What are the functions $\Phi$ defined on this region that satisfy $$\lmd\Phi=\nabla^2\Phi=\fr1\rho\partial_\rho\p{\rho\partial_\rho\Phi}+\fr1{\rho^2}\partial_\vphi^2\Phi$$ and the boundary conditions $$\partial_\rho\fc\Phi{\rho=R_\mrm{in},\vphi}=\partial_\rho\fc\Phi{\rho=R_\mrm{out},\vphi}=0,$$ i.e., eigenfunctions of the Laplacian on the annulus with homogeneous Neumann boundary conditions?

For any boundary condition with azimuthal symmetry, an easy separation of variable gives you solutions of the general form $$\fc\Phi{\rho,\vphi}=\p{\Xi^{\p J}\fc{J_m}{z\rho/R_\mrm{out}}+\Xi^{\p Y}\fc{Y_m}{z\rho/R_\mrm{out}}}\e^{\i m\vphi},$$ where $\Xi^{\p J}$, $\Xi^{\p Y}$, and $z$ are constants fixed by the boundary conditions, and $m$ is an integer and the azimuthal quantum number. The functions $J_m$ and $Y_m$ are Bessel functions of the first and second kind, respectively, which are two linearly independent solutions of the Bessel equation $$z^2\fc{y''}z+z\fc{y'}z+\p{z^2-m^2}\fc{y}z=0$$ $$(1)$$ with some particular normalization. Plugging the general form into the boundary condition, and we have $$\Xi^{\p J}\fc{J_m'}{z}+\Xi^{\p Y}\fc{Y_m'}{z}=0,\quad \Xi^{\p J}\fc{J_m'}{rz}+\Xi^{\p Y}\fc{Y_m'}{rz}=0,$$ where $r\ceq R_\mrm{in}/R_\mrm{out}$. This is regarded as a homogeneous linear system of equations for $\Xi^{\p J}$ and $\Xi^{\p Y}$. In order for it to have nontrivial solutions, the determinant of the coefficient matrix must vanish, which gives $$\fc gz\ceq\fc{J_m'}{rz}\fc{Y_m'}z-\fc{Y_m'}{rz}\fc{J_m'}z=0.$$ $$(2)$$ We can then assign a radial quantum number $n$ to the eigenfunctions by making the value of $z$ the $n$th root of $g$, which we may casually call $z_{mn}$.

We then have the final solution (up to normalization) $$\fc{\Phi_{mn}}{\rho,\vphi}=\p{\fc{Y_m'}{z_{mn}}\fc{J_m}{z_{mn}\rho/R_\mrm{out}} -\fc{J_m'}{z_{mn}}\fc{Y_m}{z_{mn}\rho/R_\mrm{out}}}\e^{\i m\vphi},$$ where $m$ is the azimuthal quantum number, and $n$ is the radial quantum number, and $z_{mn}$ is defined as the $n$th root of the function $g$. The eigenvalue is $\lmd_{mn}=-z_{mn}^2/R_\mrm{out}^2$. In order for the eigenfunctions to be complete, we need to allow $m$ to be any integer, but we will restrict ourselves to non-negative $m$ in the rest of the article because the negative ones are practically the same as the positive ones.

Distribution of eigenvalues

From the well-known asymptotic form of the Bessel functions (source) $$\fc{J_m}{z\to\infty}=\sqrt{\fr2{\pi z}}\fc\cos{z-\fr{m\pi}2-\fr\pi4},\quad \fc{Y_m}{z\to\infty}=\sqrt{\fr2{\pi z}}\fc\sin{z-\fr{m\pi}2-\fr\pi4},$$ we can easily conclude that the asymptotic form of $g$ is $$\fc g{z\to\infty}=\fr{2\fc\sin{\p{1-r}z}}{\pi\sqrt r\,z}.$$ This inspires us to define a function in companion with $g$ as $$\fc fz\ceq\fc{J_m'}{rz}\fc{J_m'}z+\fc{Y_m'}{rz}\fc{Y_m'}z,$$ $$(3)$$ which will have the asymptotic form $$\fc f{z\to\infty}=\fr{2\fc\cos{\p{1-r}z}}{\pi\sqrt r\,z}.$$ Therefore, $f\sim\cos\fc\tht z$ and $g\sim\sin\fc\tht z$ are like the cosine and sine pair of oscillatory functions, with a phase angle $\fc\tht{z\to\infty}=\p{1-r}z$ asymptotically directly proportional to $z$.

Here is a plot that shows how $\tht$ behaves:

The blue line is $\fc\arctan{\fc gz/\fc fz}$, and the orange line is $\arctan\fc\tan{\p{1-r}z}$. we can see that the two functions are asymptotically equal.

A good thing about this description is that we can now see that $z_{mn}$ as a root of $g$ is just the solution to the equation $\fc\tht z=n\pi$. In other words, $z_{mn}$ are just the points at which the blue line crosses the horizontal axis in the plot above. Therefore, if we can find the inverse function $\fc z\tht$ (maybe in a series expansion), we can directly get $z_{mn}=\fc z{n\pi}$. We can see from the plot that $\fc z\tht=\tht/\p{1-r}$ is probably not a good enough approximation. Therefore, we would like to seek higher order terms.

Kummer’s equation

For this part, we employ a method similar to a paper by Bremer.

First, for a general homogeneous second-order linear ODE in the form $y''+2\chi y'+\psi y=0$, where $\chi$ and $\psi$ are known functions of $z$, we can define $$y_\mrm n\ceq py,\quad p\ceq C\exp\int\chi\,\d z, \quad q\ceq p\psi-p'',$$ and the ODE will become a normal form $y_\mrm n''+qy_\mrm n=0$. Therefore, for any second-order linear ODE, we only need to consider those without the first-derivative term.

Then, one can prove that, for the ODE $y''+qy=0$, its two linearly independent solutions can be expressed as $$u=\fr{\cos\alp}{\sqrt{\alp'}},\quad v=\fr{\sin\alp}{\sqrt{\alp'}},$$ $$(4)$$ where $\alp$ satisfy Kummer’s equation $$\alp^{\prime2}+\fr{\alp'''}{2\alp'}-\fr{3\alp^{\prime\prime2}}{4\alp^{\prime2}}=q.$$ $$(5)$$ This may seem like converting a problem to a more complicated one, but the advantage is that $\alp$ is not oscillatory while $u$ and $v$ are oscillatory, which means that it would be easier to expand $\alp$ as a series for large $z$. Notice that $\alp'=1/\p{u^2+v^2}$, which provides a means of finding $\alp$ if the solutions of the original ODE are known.

Now, we can try to apply this to $J_m'$ and $Y_m'$. First, we need to find the second-order ODE that those two functions satisfy. We already know that $J_m$ and $Y_m$ are solutions to Equation 1, so it would be straightforward to derive that $J_m'$ and $Y_m'$ satisfy $$\fc{y''}z+\fr1z\fr{z^2-3m^2}{z^2-m^2}\fc{y'}z +\p{1-\fr{m^2+1}{z^2}-\fr{2m^2}{z^2-m^2}}\fc yz=0.$$ To reduce this into the normal form without the first-derivative term, define $$\fc pz\ceq\sqrt{\fr\pi2}\fr{z^{3/2}}{\sqrt{z^2-m^2}},\quad \fc qz\ceq1-\fr{4m^2-1}{4z^2}-\fr{2m^2+z^2}{\p{m^2-z^2}^2},$$ $$(6)$$ and we have $\fc pz\fc{J_m'}z$ and $\fc pz\fc {Y_m'}z$ satisfy $\fc{y''}z+\fc qz\fc yz=0$. Equation 4 then gives $$\fc{Y_m'}z=\fr{\cos\fc\alp z}{\fc pz\sqrt{\fc{\alp'}z}},\quad -\fc{J_m'}z=\fr{\sin\fc\alp z}{\fc pz\sqrt{\fc{\alp'}z}}$$ $$(7)$$ (the normalization of $p$ and the initial value of $\alp$ are chosen to make sure they are correct). You may worry that $\fc p{z<m}$ is imaginary, but $\fc{\alp'}z$ will be negative in that region so that everything works out and is real in the end. Substitute Equation 7 into Equation 3 and 2, and we have $$\fc fz=\fr{\cos\fc\tht z}{\fc p{rz}\fc pz\sqrt{\fc{\alp'}{rz}\fc{\alp'}z}},\quad \fc gz=\fr{\sin\fc\tht z}{\fc p{rz}\fc pz\sqrt{\fc{\alp'}{rz}\fc{\alp'}z}},$$ where $\fc\tht z\ceq\fc\alp z-\fc\alp{rz}$. By this, we have a workable expression for $\fc\tht z$ so that we now just need to turn our attention to $\fc\alp z$.

We can substitute $\fc qz$ in Equation 6 into Equation 5 and expand on both sides as a series of $z$ at infinity and solve for the series coefficients to get $$\fc{\alp'}z=1+\fr{-4m^2-3}{8z^2}+\fr{-16m^4-184m^2+63}{128z^4}+\cdots,$$ $$(8)$$ which has a certain convergence radius which is not important at this stage.

Asymptotic expansion

Here we will work out another way of expanding $\fc\alp z$ as a series based on properties of the Bessel functions. From Equation 7, we have $$\fr1{\fc{\alp'}z}=\fc pz^2\p{\fc{J_m'}z^2+\fc{Y_m'}z^2}.$$ If we were working with $\fc{J_m}z^2+\fc{Y_m}z^2$ instead, we would be able to employ the handy Nicholson’s integral, but the same method cannot be applied here.

We can, however, use the asymptotic expansion of the Bessel functions directly (source): $$\begin{align*} \fc{Y_m'}z&=\sqrt{\fr2{\pi z}}\p{\fc\cos{z-\fr{m\pi}2-\fr\pi4} \sum_{k=0}^\infty\fr{\p{-1}^kc_{2k}}{z^{2k}} -\fc\sin{z-\fr{m\pi}2-\fr\pi4}\sum_{k=0}^\infty\fr{\p{-1}^kc_{2k+1}}{z^{2k+1}}},\\ -\fc{J_m'}z&=\sqrt{\fr2{\pi z}}\p{\fc\cos{z-\fr{m\pi}2-\fr\pi4} \sum_{k=0}^\infty\fr{\p{-1}^kc_{2k+1}}{z^{2k+1}} +\fc\sin{z-\fr{m\pi}2-\fr\pi4}\sum_{k=0}^\infty\fr{\p{-1}^kc_{2k}}{z^{2k}}}, \end{align*}$$ where $$c_k\ceq a_k+\p{k-\fr12}a_{k-1},\quad a_k\ceq\p{-1}^k\fr{\p{1/2-m}^{\overline k}\p{1/2+m}^{\overline k}}{2^kk!},$$ where the raising factorial notation is used. By combining them, we get $$\fc{J_m'}z^2+\fc{Y_m'}z^2=\fr2\pi\sum_{k=0}^\infty\fr{r_k}{z^{2k+1}},\quad r_k\ceq\p{-1}^k\sum_{l=0}^{2k}\p{-1}^lc_lc_{2k-l}.$$

Then, take the reciprocal of the series, and we have $$\fr1{\fc{J_m'}z^2+\fc{Y_m'}z^2}=\fr\pi2\sum_{k=0}^\infty\fr{r^\mrm r_k}{z^{2k-1}},$$ where $r^\mrm r_k$ is defined recursively as (noticing that $r_0=1$) $$r^\mrm r_0\ceq 1,\quad r^\mrm r_{k>0}\ceq-\sum_{l=1}^kr_lr^\mrm r_{k-l}.$$

Therefore, we get the asymptotic expansion of $\fc\alp z$ as $$\fc\alp z=\int\fr{\d z}{\fc pz^2\p{\fc{J_m'}z^2+\fc{Y_m'}z^2}} =-\fr{m\pi}2-\fr\pi4+\sum_{k=0}^\infty\fr{s_k}{z^{2k-1}},\quad s_k\ceq\fr{m^2r^\mrm r_{k-1}-r^\mrm r_k}{2k-1}$$ (with $r^\mrm r_{-1}$ understood as $0$; I will be assuming the reference to any expansion coefficients outside their designated range to be interpreted as zero in all occurrences in the rest of this article). The additive constant in $\fc\alp z$ is properly chosen so that Equation 7 is satisfied without being off by a phase (but it is actually unimportant anyway because it gets canceled in the expression of $\tht$). One can check that this agrees with Equation 8.

We can now find an expansion for $\tht$: $$\fc\tht z=\fc\alp z-\fc\alp{rz}=\p{1-r}\sum_{k=0}^\infty\fr{d_k}{z^{2k-1}},\quad d_k\ceq\p{1-\fr1{r^{2k-1}}}\fr{s_k}{1-r}.$$ Then it is the final step to find an expansion for the inverse function $\fc z\tht$. Take the reciprocal of $\fc\tht z$: $$\fr1{\fc\tht z}=\fr1{1-r}\sum_{k=0}^\infty\fr{d^\mrm r_k}{z^{2k+1}},\quad d^\mrm r_0\ceq1,\quad d^\mrm r_{k>0}\ceq-\sum_{l=1}^kd_l d^\mrm r_{k-l}.$$ Then, invert the series: $$\fr1{\fc z\tht}=\sum_{k=0}^\infty\fr{b^\mrm r_k}{\tht^{2k+1}},\quad b^\mrm r_k\ceq\fr{\p{1-r}^{2k+1}}{\p{2k+1}!}\begin{dcases} \sum_{l=1}^{2k}\p{-1}^l\p{2k+1}^{\overline l} \fc{B_{2k,l}}{0,2!d^\mrm r_1,0,4!d^\mrm r_2,0,\ldots},&k>0,\\ 1,&k=0, \end{dcases}$$ where $B_{2k,l}$ is the Bell polynomial. Finally, take the reciprocal: $$\fc z\tht=\sum_{k=0}^\infty\fr{b_k}{\tht^{2k-1}},\quad b_0\ceq\fr1{1-r},\quad b_{k>0}\ceq-\fr1{1-r}\sum_{l=1}^kb^\mrm r_lb_{k-l}.$$ The first few terms are $$\fc z\tht=\fr\tht{1-r}+\fr{\p{4m^2+3}\p{1-r}}{8r\tht}+\cdots$$ (I only write two terms here because the next term starts to be very long; it will turn out that only the first two terms are useful anyway).

a[m_,k_]:= (-1)^k Pochhammer[1/2-m,k]Pochhammer[1/2+m,k]/(2^k k!)
c[m_,k_]:=a[m,k]+(k-1/2)a[m,k-1]
r[m_,k_]:=(-1)^k Sum[(-1)^l c[m,l]c[m,2k-l],{l,0,2k}]
rr[m_,k_]:=rr[m,k]=If[k<0,0,If[k==0,1,-Sum[r[m,l]rr[m,k-l],{l,1,k}]]]
s[m_,k_]:=(m^2 rr[m,k-1]-rr[m,k])/(2k-1)
d[m_,k_,R_]:=(1-1/R^(2k-1))s[m,k]/(1-R)
dr[m_,k_,R_]:=dr[m,k,R]=If[k==0,1,-Sum[d[m,l,R]dr[m,k-l,R],{l,1,k}]]
br[m_,k_,R_]:=(1-R)^(2k+1)/(2k+1)!If[k==0,1,Sum[(-1)^l Pochhammer[2k+1,l]BellY[2k,l,Table[If[EvenQ[j],dr[m,j/2,R]j!,0],{j,1,2k-l+1}]],{l,1,2k}]]
b[m_,k_,R_]:=b[m,k,R]=If[k==0,1/(1-R),-1/(1-R)Sum[br[m,l,R]b[m,k-l,R],{l,1,k}]]

The result is shown in the plot below.

The black line is the exact function $\fc\tht z$ for $m=4$, $r=1/3$, and the blue, green, orange, and red lines are the truncated asymptotic series of the inverse function $\fc z\tht$ with 1, 2, 3, and 4 terms, respectively. The horizontal grid lines are the integer multiples of $\pi$, whose intersections with the black line gives the wanted roots $z_{mn}$. To be honest, the result is a bit anticlimactic because it turns out that truncating the series to only the first two terms gives the best approximation. This is within expectation, though, because the asymptotic series is never guaranteed to give better approximations with more terms.

The $n=0$ mode

For all the previous plots, I have only shown $m=4$. For $m$ being other positive integers, they all look similar, but there is a special feature for $m=0$. To see this, first study the limiting form of $\fc fz$ and $\fc gz$ for small $z$. The limiting forms of the Bessel functions are (source) $$\begin{align*} \fc{J_0'}{z\to0}&=-\fr z2,& \fc{J_{m>0}'}{z\to0}&=\fr{1}{2^m\p{m-1}!}z^{m-1},\\ \fc{Y_0'}{z\to0}&=\fr2{\pi z},& \fc{Y_{m>0}'}{z\to0}&=\fr{2^mm!}{\pi}\fr1{z^{m+1}}. \end{align*}$$ $$(9)$$ We can then get $$\begin{align*} \fc f{z\to0}&=\begin{dcases} \fr4{\pi^2rz^2},&m=0,\\ \fr1{r^{m+1}}\p{\fr{2^mm!}\pi}^2\fr1{z^{2m+2}},&m>0, \end{dcases}\\ \fc g{z\to0}&=\begin{dcases} \fr{1-r^2}{\pi r},&m=0,\\ \fr m\pi\p{r^{m-1}-\fr1{r^{m+1}}}\fr1{z^2},&m>0. \end{dcases} \end{align*}$$ From this, we can get the limiting form of $\fc\tht z$: $$\fc\tht{z\to0}=\begin{dcases} \fr{\pi\p{1-r^2}}4z^2,&m=0,\\ -\fr{\pi m\p{1-r^{2m}}}{\p{m!2^m}^2}z^{2m},&m>0. \end{dcases}$$ The particularly interesting thing to note is the negative sign for the case of $m>0$. Remember that $\fc\tht{z\to\infty}=\p{1-r}z$, which is a positive thing, we would conclude that $\fc\tht z$ would become zero for some positive $z$ if $m>0$, while that may not be the case for $m=0$. Indeed, there is no positive $z$ for $\fc\tht z=0$ if $m=0$, as we can see from the plot below.

The red line is $\fc\tht z$ for $m=1$, which we can see that crosses the horizontal axis, while the blue line is $\fc\tht z$ for $m=0$, which does not cross the horizontal axis. The more lightly colored lines are the limiting forms of $\fc\tht z$ for small $z$.