Asymptotics of $j$-invariant at elliptic fixed points of $\text{PSL}(2,\mathbb Z)$ Let $j=12^3 E_4^3/(E_4^3-E_6^2)$ be the modular $j$-invariant
$$j(\tau)=q^{-1} + 744 + 196884q +...,$$
with $q=e^{2\pi i \tau}$ and $E_k$ the weight $k$ Eisenstein series. At the elliptic points $\tau=i$ and $\tau=\zeta_3=e^{2\pi i/3}$ of $\text{PSL}(2,\mathbb Z)$, it has a special behaviour: $\tau=\zeta_3$ is a third order zero of $j$, while $\tau=i$ is a second order zero of $j-12^3$. Thus near those elliptic points the behaviour of $j$ is determined through the first nonzero coefficients in the Taylor series, which are $j'''(\zeta_3)$ and $j''(i)$.
Using some experimental numerics, I found that
$$j'''(\zeta_3)=-\frac{162i \Gamma(\frac13)^{18}}{\pi^9}, \\
j''(i)=-\frac{162\Gamma(\frac14)^8}{\pi^4}.$$
I have not spotted these values in the literature, but they must have been found before. I would be very glad about any hints on the literature of such computations, as it would be great to confirm these numbers and learn about how they can be computed in similar examples (of say other modular forms). Many thanks!
 A: I will use the notation by Ramanujan with $E_2, E_4,E_6$ being denoted by $P, Q, R$. For clarity let us write
\begin{align}
P(q) & =1-24\sum_{n=1}^{\infty}\frac {nq^n} {1-q^n}\tag{1a}\\
Q(q) &=1+240\sum_{n=1}^{\infty}\frac{n^3q^n}{1-q^n}\tag{1b}\\
R(q) &=1-504\sum_{n=1}^{\infty}\frac{n^5q^n}{1-q^n}\tag{1c}\end{align}
where $q=\exp(2\pi i\tau) $ and $\tau$ has positive imaginary part. Let us also list Ramanujan's identities dealing with derivatives of these functions
\begin{align} 
q\frac{dP} {dq} &=\frac{P^2-Q}{12}\tag{2a}\\
q\frac{dQ}{dq}&=\frac{PQ-R}{3}\tag{2b}\\
q\frac{dR}{dq}&=\frac{PR-Q^2}{2}\tag{2c}
\end{align}
Further let us observe that $$\frac{dq} {d\tau} =2\pi iq$$ and hence $$\frac{d} {d\tau} =2\pi i q\frac{d} {dq} \tag{3}$$ We have by definition $$j(\tau) =\frac{12^3Q^3}{Q^3-R^2}\tag{4}$$ and one can prove using the equations $(2a),(2b),(2c)$ that $$q\frac{d} {dq} (Q^3-R^2)=P(Q^3-R^2)\tag{4}$$ Hence we have $$j'(\tau) =12^3\cdot 2\pi i q\frac{d} {dq} \frac{Q^3}{Q^3-R^2}=12^3(2\pi i) \cdot \dfrac{(Q^3-R^2)3Q^2q\dfrac{dQ}{dq}-Q^3q\dfrac{d}{dq}(Q^3-R^2)} {(Q^3-R^2)^2}=-12^3(2\pi i) \cdot\frac{Q^2R}{Q^3-R^2}$$ Differentiating again we get $$j''(\tau) =12^2(8\pi^2)\cdot\frac{PQ^2R-4QR^2-3Q^4}{Q^3-R^2}\tag{5}$$ One more round of differentiation gives us $$j'''(\tau) =12(64\pi^3 i)\cdot\frac{4R^3+\text{ terms containing }Q }{Q^3-R^2}\tag{6}$$ Next we use the link between $Q, R$ and elliptic integrals to evaluate them in closed form. Let $0<k<1$ be the elliptic modulus and $k'=\sqrt{1-k^2}$ be complementary modulus and let $K, K'$ be elliptic integrals defined by $$K=K(k) =\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}},K'=K(k')\tag{7}$$ Let $q'=\exp(-\pi K'/K) $ be the corresponding nome and then we have the standard formulas (proved here using the identities $(2a),(2b),(2c)$ mentioned above)
\begin{align}
Q(q'^2) &=\left(\frac{2K}{\pi}\right) ^4(1-k^2k'^2)\tag{8a}\\
R(q'^2) &=\left(\frac{2K}{\pi}\right)^6(1-2k^2)\left(1+\frac{k^2k'^2}{2}\right)\tag{8b}\\
Q(-q')&=\left(\frac{2K}{\pi}\right)^4(1-16k^2k'^2)\tag{8c}\\
R(-q')&=\left(\frac{2K}{\pi}\right)^6(1-2k^2)(1+32k^2k'^2)\tag{8d}
\end{align}
Further let us note that if $N$ is a positive rational number then there exists a unique positive value of modulus $k$ such that $K'/K=\sqrt{N} $ and this value $k$ is an algebraic number. Such $k$ are famously known as singular moduli. For small values of positive integer $N$ the values of $k=k_N$ as well as that of $K=K_N$ are well known. For our purposes we need to deal with $N=1,3$ and we have $$k_1=\frac{1}{\sqrt{2}},K_1=\frac {\Gamma ^2(1/4)}{4\sqrt{\pi}},k_3=\frac{\sqrt{3}-1}{2\sqrt{2}},K_3=\frac{3^{1/4}\Gamma ^{3}(1/3)}{2^{7/3}\pi} \tag{9}$$ (see details in this thread).
We can now evaluate $j''(i), j'''(\zeta_3)$ using the above given formulas. Let $\tau=i$ so that $q=e^{-2\pi}$ and then we can use the formulas $(8a),(8b),(9)$ with $q' =e^{-\pi}, k=k_1,K=K_1$ and note that $R=0$ and using $(5)$ we get desired value of $j''(i) $.
Next we use $\tau=\zeta_3$ so that $q=-e^{-\pi\sqrt{3}}$ and we use $(8c),(8d),(9)$ with $q' =e^{-\pi\sqrt{3}},k=k_3,K=K_3$ and note that $Q=0$ and using $(6)$ we get the desired value of $j''' (\zeta_3)$.
