From 32aa8296034126469ce8f019480e8d28d1f1c298 Mon Sep 17 00:00:00 2001
From: Aner Zakobar <aner@zakobar.com>
Date: Thu, 2 Jul 2026 22:17:37 +0300
Subject: [PATCH] Added a lot of proof.

---
 content/method/method.tex | 49 +++++++++++++++++++++++++++++++++------
 1 file changed, 42 insertions(+), 7 deletions(-)

diff --git a/content/method/method.tex b/content/method/method.tex
index 81991d9..70d6713 100644
--- a/content/method/method.tex
+++ b/content/method/method.tex
@@ -555,28 +555,63 @@ Unfortunately for us, this involves calculating the derivative of a calculation
 In order to derive this, we can utilize our earlier computation, along with the inverse derivative rule:
 
 \[
-  f^{-1}'(x) = \frac{1}{f'(f^{-1}(x))}
+  f'(x) = \frac{1}{f^{-1}'(f(x))}
 \]
 
-We can use this to calculate the derivative we need
+This equation shows the relationship between a function's derivative and it's inverse. It shows that we may use the derivative of the
+inverse function. However, in order to do this, we must first define what is the inverse we are looking for.
+
+As a reminder we are trying to calculate $\frac{\partial \hat{F}_0^{-1}(\alpha ; X_0)}{\partial f_\theta(\mathbf{x}_i)}$. That is, we have a fuction that takes $\alpha, \mathbf{X}_0$ as parameters and calculates
+our threshold $\tau$. We are trying to calculate the derivative of this function w.r.t $x_i$.
+
+Therefor, we must inverse the function in order to calculate $x_i$ given $\alpha, \tau$ and all $X_0 \backslash \{x_i\}$.
+
+This inverse equation is given by:
 
 \[
-    \frac{\partial \hat{F}_0^{-1}(\alpha)}{\partial f_\theta(\mathbf{x}_i)}
+  \sigma^{-1}\left( |\mathcal{B}_0|\cdot\alpha - \sum_{x_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right) + \tau
 \]
+
+This is the inverse function given, and so we must calculate:
+
 \[
-    \frac{\partial \hat{F}_0^{-1}(\alpha ; \mathbf{X}_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{\partial \tau}{\partial f_{\theta}(x_i)}
+  \frac{\partial \hat{F}_0^{-1}(\alpha ; X_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{\frac{\partial \sigma^{-1}\left( |\mathcal{B}_0|\cdot \alpha - \sum_{x_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(x_j))\right) + \tau}{\partial \tau}}
+\]
+
+Applying the chain rule:
+
+
+\[
+  = \frac{1}{\sigma^{-1}'\left( |\mathcal{B}_0|\cdot \alpha - \sum_{x_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(x_j))\right)\cdot\left( \sum_{x_j \in \mathcal{B}_0 ; j \neq i}\sigma'(\tau - f_{\theta}(\mathbf{x}_j))\right) + 1}
+\]
+
+Remembering the calculation of $\alpha$ we get:
+
+\[
+  \alpha = \sum_{x_i \in \mathcal{B}_0}\sigma(\tau - f_{\theta}(x_j)) \Rightarrow
 \]
 
 \[
-  = \frac{1}{\frac{\partial\hat{F}_0(\tau ; X_0)}{\partial x_i}}
+ |\mathcal{B}_0|\cdot \alpha - \sum_{x_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(x_j)) = \sigma(\tau - f_{\theta}(x_i))
 \]
 
-We must now calculate the derivative $\frac{\partial\hat{F}_0(\tau ; X_0)}{\partial x_i}$.
+Therefor, our derivative to calculate from before is equal to:
+
 
 \[
-  \frac{\partial\hat{F}_0(\tau ; X_0)}{\partial x_i} = \frac{\frac{1}{|\mathcal{B}_0|}\sum_{x_j \in \mathcal{B}_0}\sigma_0(\tau - x_j)}{\partial x_i}
+  \frac{\partial \hat{F}_0^{-1}(\alpha ; X_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{\sigma^{-1}'\left( \sigma(\tau - f_{\theta}(x_i) \right)\cdot\left( \sum_{x_j \in \mathcal{B}_0 ; j \neq i}\sigma'(\tau - f_{\theta}(\mathbf{x}_j))\right) \right) + 1 }
 \]
 
+And remembering our chain rule, we note that $\frac{1}{\sigma^{-1}'(\sigma(\tau - f_{\theta}(x_i)))} = \sigma'(\tau - f_{\theta}(x_i))$
+
+And so:
+
+\[
+  \frac{\partial \hat{F}_0^{-1}(\alpha ; X_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{\sigma'(\tau - f_{\theta}(x_i))}{\left( \sum_{x_j \in \mathcal{B}_0 ; j \neq i}\sigma'(\tau - f_{\theta}(\mathbf{x}_j))\right) + 1}
+\]
+
+Which we can plug back into the ROLL derivation framework.
+
 %------------------------------------------------
 
 \section{Implementation Considerations}