diff --git a/.gitignore b/.gitignore
index 5a396e8..25669e1 100644
--- a/.gitignore
+++ b/.gitignore
@@ -306,4 +306,5 @@ TSWLatexianTemp*
 F4D1-Analysis-and-Geometry-on-Manifolds/Analysis_on_Manifolds_Notes.pdf
 V4A1-Algebraic-Geometry-I/Algebraic_Geometry_I_Notes.pdf
 V5A2-Habiro-Rings/Habiro_Rings_Notes.pdf
-V5B1-Advanced-Topics-in-Complex-Analysis/Advanced_Complex_Analysis_Notes.pdf
\ No newline at end of file
+V5B1-Advanced-Topics-in-Complex-Analysis/Advanced_Complex_Analysis_Notes.pdf
+V5A2-Habiro-Rings/._wordcount_selection.tex
diff --git a/V5A2-Habiro-Rings/Lecture 4.tex b/V5A2-Habiro-Rings/Lecture 4.tex
index 9c98abe..0104500 100644
--- a/V5A2-Habiro-Rings/Lecture 4.tex	
+++ b/V5A2-Habiro-Rings/Lecture 4.tex	
@@ -89,12 +89,12 @@ \section{Lecture 4 -- 15th November 2024}\label{sec: lecture 4}
     \end{align*}
     as desired. 
 \end{proof}
-We can modify the Nahm sum to rid ourselves of the $q^{a/2}$ factor in (\ref{eqn: q-difference equation of 1x1 Nahm sum}).\marginpar{The exposition that follows is drawn from the erratum discussed at the begining of lecture 5. The discussion below is likely extremely error-prone and the reader is encouraged to consult \cite{NahmAsymptotics} for full details.}
+We can modify the Nahm sum to rid ourselves of the $q^{a/2}$ factor in (\ref{eqn: q-difference equation of 1x1 Nahm sum}).\marginpar{The exposition that follows is drawn from the erratum discussed at the begining of lecture 5. The reader is encouraged to consult \cite{NahmAsymptotics} for full details.}
 \begin{corollary}\label{corr: q-difference equation of modified 1x1 Nahm sum}
     Let 
     $$f_{a}^{\mathrm{mod}}(t,q)=f_{a}(q^{-a/2}t,q)=\sum_{n\geq0}(-1)^{an}\frac{q^{\frac{1}{2}an^{2}-\frac{1}{2}an}}{(q;q)_{n}}t^{n}$$
     Then the modified Nahm sum satisfies the $t$-deformed $q$-difference equation 
-    \begin{equation}\label{eqn: q-difference equation}
+    \begin{equation}\label{eqn: q-difference equation of modified 1x1 Nahm sum}
         f_{a}^{\mathrm{mod}}(t,q) - f_{a}^{\mathrm{mod}}(qt,q) = (-1)^{a}t\cdot f_{a}^{\mathrm{mod}}(q^{a}t,q).
     \end{equation}
 \end{corollary}
@@ -111,7 +111,7 @@ \section{Lecture 4 -- 15th November 2024}\label{sec: lecture 4}
     f_{a}(t,q)=\exp\left(\frac{V(t)}{h}\right)g_{a}(t,q)
 \end{equation}
 \begin{equation}\label{eqn: qt 1x1 Nahm sum ansatz}
-    f_{a}(t,q)=\exp\left(\frac{V(qt)}{h}\right)g_{a}(qt,q)
+    f_{a}(qt,q)=\exp\left(\frac{V(qt)}{h}\right)g_{a}(qt,q)
 \end{equation}
 in what follows, with $V(t)\in\QQ[[t]]$ satisfying $V(0)=0$. \Cref{eqn: 1x1 Nahm sum ansatz,eqn: qt 1x1 Nahm sum ansatz} takes are related by the formula we now describe. 
 \begin{proposition}\label{prop: V functional equation for 1x1 Nahm ansatz}
@@ -134,18 +134,22 @@ \section{Lecture 4 -- 15th November 2024}\label{sec: lecture 4}
 \end{proposition}
 \begin{corollary}\label{corr: exponent ratio for 1x1 Nahm sum antsatz}
     Let $V(t)$ be as in the Nahm equation ansatz. The ratios of the exponential factors satisfy 
-    $$\exp\left(\frac{V(qt)}{h}\right)/\exp\left(\frac{V(t)}{h}\right)\sim\exp((\partial^{\log}V)(t))(1+O(h)).$$ 
+    $$\exp\left(\frac{V(qt)}{h}\right)/\exp\left(\frac{V(t)}{h}\right)\sim\exp((\partial^{\log}V)(t))(1+O(h))$$
+    for $q=e^{h}$.  
 \end{corollary}
 \begin{proof}
-    We have 
+    We have by \Cref{prop: V functional equation for 1x1 Nahm ansatz}
     \begin{align*}
         \exp\left(\frac{V(qt)}{h}\right) &= \exp\left(\frac{V(t)}{h}+(\partial^{\log}V)(t)+\frac{h}{2}((\partial^{\log})^{2}V)(t)\right)\\
         &=\exp\left(\frac{V(t)}{h}\right)\exp\left((\partial^{\log}V)(t)\right)\exp\left(\frac{h}{2}((\partial^{\log})^{2}V)(t)\right)
     \end{align*}
-    where the first term of the product above cancels in the ratio and the final term of the product expands to a power series in $h$. 
+    where the first term of the product above cancels in the ratio and the final term of the product expands to a power series in $h$ with constant coefficient 1. 
 \end{proof}
-Of special interest to us will be the final terms of the product above, where we denote $Z(t)=\exp((\partial^{\log}V)(t))$ and $\widetilde{Z}(t,q)=\exp\left((\partial^{\log}V)(t)\right)\exp\left(\frac{h}{2}((\partial^{\log})^{2}V)(t)\right)$.
-This allows us to produce a functional equation of the $g_{a}$ appearing in \Cref{eqn: 1x1 Nahm sum ansatz,eqn: qt 1x1 Nahm sum ansatz} which we will soon revisit. Note specializing at $q-1$ produces 
+Denoting $Z(t)=\exp((\partial^{\log}V)(t))$ and $\widetilde{Z}(t,q)=\exp\left((\partial^{\log}V)(t)\right)\exp\left(\frac{h}{2}((\partial^{\log})^{2}V)(t)\right)$, we divide (\ref{eqn: q-difference equation of modified 1x1 Nahm sum}) by the exponential prefactor $\exp\left(\frac{V(t)}{h}\right)$ and the ansatz \Cref{eqn: 1x1 Nahm sum ansatz,eqn: qt 1x1 Nahm sum ansatz} to observe 
+$$g_{a}(t,q)-\widetilde{Z}(t,q)g_{a}(qt,q)=(-1)^{a}\cdot t\cdot\prod_{i=0}^{a-1}\widetilde{Z}(q^{i}t,q)\cdot g_{a}(q^{a},t)$$
+which we seek to show lies in $\QQ[[t,q-1]]$.
+
+Specializing at $q=1$ (ie. $h=0$) produces 
 \begin{equation}\label{eqn: specialized Z functional equation}
     1-Z(t)=(-1)^{a}t\cdot Z(t)^{a}
 \end{equation}
diff --git a/V5A2-Habiro-Rings/Lecture 5.tex b/V5A2-Habiro-Rings/Lecture 5.tex
index 431e330..e981e2f 100644
--- a/V5A2-Habiro-Rings/Lecture 5.tex	
+++ b/V5A2-Habiro-Rings/Lecture 5.tex	
@@ -1,26 +1,39 @@
 \section{Lecture 5 -- 22nd November 2024}\label{sec: lecture 5}
-We continue our discussion of $q$-series and in particular a property of the modified Nahm sum considered in \Cref{corr: q-difference equation of modified 1x1 Nahm sum}.\marginpar{As it stands, the proofs and structure for this lecture are more rough than usual, and will be updated in due course.}
+We continue our discussion of $q$-series and in particular a property of the modified Nahm sum considered in \Cref{corr: q-difference equation of modified 1x1 Nahm sum}.
 
 The following definition is due to Konsevich-Soibelman \cite{DTInvariants}.
 \begin{definition}[Admissable Series]\label{def: admissable series}
-    A series $f\in\ZZ((q))[[t]]$ is admissable if it can be written as 
+    A series $f\in\ZZ((q))[[t]]$ such that $f\equiv 1\pmod{(t)}$ is admissable if it can be written as 
     $$f=\prod_{n\geq1}\prod_{i\in\ZZ}(q^{i}t^{n};q)_{\infty}^{a_{n,i}}$$
     such that for each $n$ only finitely many $a_{n,i}$ are nonzero. 
 \end{definition}
 \begin{remark}
-    These $a_{n,i}$'s are precisely Donaldson-Thomas invariants. 
+    These $a_{n,i}$'s are precisely Donaldson-Thomas invariants that arise in Gromov-Witten theory and enumerative geometry. 
 \end{remark}
 Admissable series force an algebraicity condition on the $q$, allowing $f$ to be written as an element of $\ZZ[q][[t]]$. Up to a condition on the residue of the series $f$ mod $(t)$, series in $\ZZ((q))[[t]]$ admit such an expansion. 
 \begin{proposition}
-    Let $f\in\ZZ((q))[[t]]$. If $f\equiv1\pmod{(t)}$ then $f$ admits a unique expansion as an admissable series. 
+    Let $f\in\ZZ((q))[[t]]$. If $f\equiv1\pmod{(t)}$ then $f$ admits a unique expansion as a series of the form 
+    $$f=\prod_{n\geq1}\prod_{i\in\ZZ}(q^{i}t^{n};q)_{\infty}^{a_{n,i}}.$$
 \end{proposition}
+\begin{proof}[Proof Outline]
+    This can be solved for truncated polynomials so for $f\equiv 1\pmod{(t)}$, it suffices to consider $f\equiv 1\pmod{(t^{m})}$ and solve inductively to give an expression algebraic in $t$.
+\end{proof}
 This result is in fact much more general and it can be shown that the modified Nahm sum 
 $$f_{a}(t,q)=\sum_{n\geq0}(-1)^{an}\frac{q^{\frac{1}{2}an^{2}-\frac{1}{2}an}}{(q;q)_{n}}t^{n}$$
 as previously defined is admissable. The original proof is highly involved, and we will instead offer a simpler exposition of the same result. Recall from \Cref{prop: logarithm at worst simple poles at roots of unity}, we have 
 \begin{equation}\label{eqn: modified Nahm sum as exponent of sum}
-    (q^{i}t;q)_{\infty}=\exp\left(-\sum_{\ell\geq 1}\frac{1}{\ell}\cdot\frac{q^{i\ell}t^{n\ell}}{1-q^{\ell}}\right)
+    (q^{i}t;q)_{\infty}=\exp\left(-\sum_{\ell\geq 1}\frac{1}{\ell}\cdot\frac{q^{i\ell}t^{n\ell}}{1-q^{\ell}}\right). 
 \end{equation}
-and further recall that $\ZZ((q))[[t]]$ is a $\lambda$-ring -- admits an action with $\NN$ as a multiplicative monoid -- and admits Adams operations $\psi_{n}:\ZZ((q))[[t]]\to\ZZ((q))[[t]]$ by $t\mapsto t^{n},q\mapsto q^{n}$. The Adams operations allow us to rewrite (\ref{eqn: modified Nahm sum as exponent of sum}) as 
+Furthermore, $\ZZ((q))[[t]]$ is a $\lambda$-ring as we now define. 
+\begin{definition}[$\lambda$-Ring]\label{def: lambda-ring}
+    A $\lambda$-ring $A$ is a ring equipped with set maps $\lambda^{k}:A\to A$ for $0\leq k\leq\infty$ such that 
+    \begin{enumerate}[label=(\roman*)]
+        \item $\lambda^{0}(a)=1$ for all $a\in A$.
+        \item $\lambda^{1}(a)=a$ for all $a\in A$. 
+        \item $\lambda^{n}(a+b)=\sum_{i+j=n}\lambda^{i}(a)\lambda^{i}(b)$ for all $a,b\in A$. 
+    \end{enumerate}
+\end{definition}
+Such a ring admits Adams operations $\psi_{n}:\ZZ((q))[[t]]\to\ZZ((q))[[t]]$ by $t\mapsto t^{n},q\mapsto q^{n}$. The Adams operations allow us to rewrite (\ref{eqn: modified Nahm sum as exponent of sum}) as 
 \begin{equation}\label{eqn: modified Nahm sum as exponent of Adams sum}
     \exp\left(-\sum_{\ell\geq1}\psi_{\ell}\left(\frac{q^{i}t^{n}}{1-q}\right)\right).
 \end{equation}
@@ -31,13 +44,22 @@ \section{Lecture 5 -- 22nd November 2024}\label{sec: lecture 5}
 \end{definition}
 Now taking 
 $$\phi(t,q)=-\sum_{n\geq1}\sum_{i\in\ZZ}a_{n,i}q^{i}t^{n}\in\ZZ((q))[[t]]$$
-writing $\frac{\phi(t,q)}{1-q}$ as the plethystic logarithm of $f_{a}(t,q)$, inverse to the plethystic exponential, the coefficients of the expansion as an admissable series will be those coefficients of the plethystic logarithm since the plethystic exponential gives an isomorphism $t\QQ[q^{\pm}][[t]]\to 1+t\QQ[q^{\pm}][[t]]$. It thus suffices to show that the plethystic logarithm of $f_{a}$ is a function that is a sum $\frac{\phi_{0}(t)}{1-q}$ with an element of $\QQ[q^{\pm}][[t]]$. 
+we have 
+\begin{equation*}\label{eqn: adams admissable Nahm sum}
+    \prod_{n\geq1}\prod_{i\in\ZZ}(q^{i}t;q)_{\infty}=\exp\left(\sum_{\ell\geq1}\frac{1}{\ell}\psi_{\ell}\left(\frac{\phi(t,q)}{1-q}\right)\right).
+\end{equation*}
+We can define the plethystic logarithm as the inverse of the plethystic exponential and observe $\frac{\phi(t,q)}{1-q}$ is the plethystic logarithm of the modified Nahm sum $f_{a}(t,q)$. We then seek to show that this plethystic logarithm $\frac{\phi(t,q)}{1-q}\in\frac{1}{1-q}\ZZ[q^{\pm}][[t]]$ which has a single simple pole at $q=1$. Indeed, this suffices as the behavior at other roots of unity are determined by the Adams operations. 
+
+
+Observe the plethystic exponential gives an isomorphism $t\QQ[q^{\pm}][[t]]\to 1+t\QQ[q^{\pm}][[t]]$. It thus suffices to show that the plethystic logarithm of $f_{a}$ is a function that is a sum $\frac{\phi_{0}(t)}{1-q}$ with an element of $\QQ[q^{\pm}][[t]]$. 
 
 Now using the ansatz
 \begin{equation}\label{eqn: plethystic exponential ansatz}
     f_{a}(t,q)=\exp\left(\sum_{\ell\geq 1}\frac{1}{\ell}\frac{\phi_{0}(t^{\ell})}{1-q^{\ell}}\right)g_{a}(t,q)
 \end{equation}
-we show that there exists a chioce of function $\psi_{0}(t)$ so that the remainder $g_{a}(t,q)\in 1+t\QQ[q^{\pm}][[t]]$ and from which the result would follow by application of the Plethystic exponential. But a choice of $\phi_{0}\in t\cdot\QQ[[t]]$ can be made  such that $\sum_{\ell\geq 1}\frac{\phi_{0}(t^{\ell})}{\ell^{2}}=-V(t)$. 
+we show that there exists a chioce of function $\phi_{0}(t)$ satisfying the ansatz above would imply the factor $g_{a}(t,q)\in 1+t\QQ[q^{\pm}][[t]]$ and from which the result  of showing $g_{a}(t,q)$ lying in $1+t\QQ[t^{\pm}][[t]]$ would follow by application of the Plethystic exponential. 
+
+But a choice of $\phi_{0}\in t\cdot\QQ[[t]]$ can be made such that $\sum_{\ell\geq 1}\frac{\phi_{0}(t^{\ell})}{\ell^{2}}=-V(t)$ whose plethystic exponential has leading term asymptotics agreeing with $f_{a}(t,q)$ at all roots of unity.
 
 This is gives the desired result as stated below. 
 \begin{theorem}[Kontsevich-Soibelman, Efimov]\label{thm: modified 1x1 Nahm sum is admissable}
@@ -45,4 +67,14 @@ \section{Lecture 5 -- 22nd November 2024}\label{sec: lecture 5}
     $$f_{a}(t,q)=\sum_{n\geq0}(-1)^{an}\frac{q^{\frac{1}{2}an^{2}-\frac{1}{2}an}}{(q;q)_{n}}t^{n}$$
     is admissable. 
 \end{theorem}
+Let us unfurl some of the consequences of \Cref{thm: modified 1x1 Nahm sum is admissable}. 
 
+To understand the $\phi_{0}(t)$ function better, we use compute its logarithmic derivative so in conjunction with \Cref{prop: V functional equation for 1x1 Nahm ansatz} we have 
+$$\sum_{\ell\geq1}\frac{(\partial^{\log}\phi_{0})(t^{\ell})}{\ell}=-\log Z(t)$$
+with $Z(t)$ is the logarithmic derivative of $V(t)$. Note that $Z(t)\in 1+t\ZZ[[t]]$ so $(\partial^{\log}\phi_{0})(t)=-\sum_{n\geq1}c_{n}t^{n}$ for $Z(t)=\prod_{n\geq1}(1-t^{n})^{c_{n}}$ for $c_{n}\in\ZZ$. This shows $n|c_{n}$. 
+\begin{example}
+    For $a=0$, $f_{1}(t,q)=(t;q)^{-1}_{\infty}$ so $Z(t)=1-t$ and $\phi_{0}(t)=\pm t$ which agrees with $V(t)$ being the dilogarithm (up to a sign). 
+\end{example}
+\begin{example}
+    In the case $a=2$ which was discussed in \Cref{sec: lecture 1}, we recover $Z(t)$ as an alternating sum of the Catalan numbers. 
+\end{example}
\ No newline at end of file