Use the convention that a <weak modular form> is a <holomorphic function> $f$ on the <complex upper half-plane> satisfying
$$
f(\gamma z)=(cz+d)^k f(z)\quad(\gamma\in SL_2(\mathbb Z)),
$$
with no condition at infinity. In particular $f(z+1)=f(z)$, so it has a <Laurent series> $f(z)=\sum_{n\in\mathbb Z}a_nq^n$ on $0<|q|<1$, where $q=e^{2\pi iz}$. It is a <modular form> when it is <holomorphic at a cusp>, here the unique level-one cusp at infinity, equivalently $a_n=0$ for $n<0$. It is a <cusp form> when additionally $a_0=0$. If one uses the narrower weakly <holomorphic> convention, one also requires finitely many negative terms; the derivative argument below holds in that subspace as well. The matrix $-I$ forces odd-weight forms to be zero.

Put $(a)_r=a(a+1)\cdots(a+r-1)$, with $(a)_0=1$, and
$$
C_{\ell,j}=\binom\ell j(k+j)_{\ell-j}\quad(0\leq j\leq\ell).
$$
We prove the <derivative transformation of a weak modular form> in the uniform form
$$
f^{(\ell)}(-1/z)=\sum_{j=0}^{\ell}C_{\ell,j}z^{k+\ell+j}f^{(j)}(z).
$$
For $\ell=0$ this is the weight-$k$ transformation under $S$. Differentiating the formula and multiplying by $z^2$, since $d(-1/z)/dz=z^{-2}$, gives the recurrence
$$
C_{\ell+1,j}=(k+\ell+j)C_{\ell,j}+C_{\ell,j-1},
$$
where out-of-range coefficients are zero. The displayed explicit coefficients satisfy it. Indeed, for interior $j$, extract the common polynomial $(k+j)_{\ell-j}$; the remaining identity is
$$
\binom\ell j(k+\ell+j)+\binom\ell{j-1}(k+j-1)
=\binom{\ell+1}j(k+\ell).
$$
This follows from <Pascal's identity> and $j\binom\ell j=(\ell-j+1)\binom\ell{j-1}$. The endpoint coefficients satisfy the recurrence directly. This polynomial proof also applies when a factor is zero, avoiding division by that factor. As $C_{\ell,\ell}=1$, the result is
$$
\boxed{f^{(\ell)}(-1/z)=z^{k+2\ell}f^{(\ell)}(z)
+\sum_{j=0}^{\ell-1}\binom\ell j(k+\ell-1)\cdots(k+j)
 z^{k+\ell+j}f^{(j)}(z).}
$$
The letter $l$ in the printed exponent is the same derivative order $\ell$.

For integer $k<0$, take $\ell=1-k$. If $j<\ell$, the product runs from $k+j\leq0$ through $k+\ell-1=0$, and therefore vanishes. Consequently $f^{(1-k)}$ has weight $k+2(1-k)=2-k$ under $S$. It is also invariant under $T$, by differentiating $f(z+1)=f(z)$, and $S,T$ generate the <modular group>. Its holomorphy is preserved by differentiation. With $D=(2\pi i)^{-1}d/dz=q\,d/dq$, termwise differentiation of the locally convergent <Laurent series> gives the <Bol identity for modular forms>:
$$
\boxed{D^{1-k}f=\sum_{n\in\mathbb Z}n^{1-k}a_nq^n
\text{ is a weak modular form of weight }2-k.}
$$
For negative indices the exponent $1-k$ is a positive integer, and the constant term is killed, so every term is unambiguous.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 137 / 3 / ii / Solution

{c}
{title2=$D^{1-k}:M_k^!\longrightarrow M_{2-k}^!$}

For integer $k<0$, the operator $D=(2\pi i)^{-1}d/dz=q\,d/dq$ sends a weight-$k$ level-one <weak modular form> to a weight-$2-k$ one after $1-k$ iterations. In the <derivative transformation of a weak modular form>, every lower derivative coefficient then contains a zero factor. This also holds on the subspace $M_k^!$ of forms meromorphic at the cusp.

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 Incoming links: Bol identity for modular forms