### Fee Growth Outside at Tick i First, we will learn how *fee growth outside* at tick \(i\) is initialized and updated in an algorithm: \[ f_{o,i} := \text{Fee growth outside at tick } i \] ### **Initialize** ```plaintext f_o,i = f_g if i ≤ current tick i_c = 0 otherwise ``` ### **Update** ```plaintext when f_g crosses tick i: f_o,i = f_g - f_o,i ``` --- ## Fee Growth Below Now let's examine how this algorithm is used to calculate *fee growth below*, \( f_b \). We start by defining some variables: \[ f_g = \text{Fee growth of token Y at current time} \] \[ f_g^k = \text{Fee growth of token Y at time } t_k \] Here, our time variable \( t_k \) has the property: \[ t_k < t_{k+1} < \dots \] We define: \[ f_b = \text{Fee growth of token Y below tick } i \] We create a giant table with each row representing a time interval: - \( t_0 \le t < t_1 \) - \( t_1 \le t < t_2 \) - \(\dots\) - \( t_6 \le t < t_7 \) The first column will represent what \( f_b \) should be in each time interval, the second column will represent *fee growth inside* according to the algorithm, and lastly we rewrite \( f_b \) in terms of *fee growth outside*. ### **Row 1**: \( t_0 \le t < t_1 \) - At time \( t_0 \), fee growth is \( f_g^0 \). - Between \( t_0 \) and \( t_1 \), the fee growth *below* (the height of the bottom red rectangle) is \( f_g^0 \). Thus, \[ f_b = f_g^0. \] ### **Row 2**: \( t_1 \le t < t_2 \) - Fee growth crosses over tick \( i \). - The fee growth below will be the previous \( f_b \) plus the height of the second red rectangle, which is \((f_g^1 - f_g^0)\). Hence, \[ f_b = f_g^0 + (f_g^1 - f_g^0) = f_g^1. \] ### **Row 3**: \( t_2 \le t < t_3 \) - Fee growth is above tick \( i \). - We add the red rectangle \((f_g^2 - f_g^1)\) to the previous \( f_b \). So, \[ f_b = (f_g^0 + (f_g^1 - f_g^0)) + (f_g^2 - f_g^1) = f_g^2. \] ### **Row 4**: \( t_3 \le t < t_4 \) - Fee growth has gone below tick \( i \). - The red rectangle is \( f_g^3 - (f_g^2 - f_g^1 + f_g^0) \). So we add that to the previous fee growth: \[ f_b = f_g^2 + \Bigl[f_g^3 - (f_g^2 - f_g^1 + f_g^0)\Bigr] = f_g^3 - f_g^1 + f_g^0. \] ### **Row 5**: \( t_4 \le t < t_5 \) - Fee growth is now above tick \( i \) again. - We carry over the previous value and add the new red rectangle \(\bigl[f_g^4 - (f_g^3 - f_g^2 + f_g^1 - f_g^0)\bigr]\). Hence, \[ f_b = \bigl(f_g^3 - f_g^1 + f_g^0\bigr) + \Bigl[f_g^4 - (f_g^3 - f_g^2 + f_g^1 - f_g^0)\Bigr] = f_g^4 - f_g^2 + f_g^0. \] ### **Row 6**: \( t_5 \le t < t_6 \) - Fee growth is still above tick \( i \). - Again, we add \(\bigl[f_g^5 - (f_g^4 - f_g^3 + f_g^2 - f_g^1 + f_g^0)\bigr]\). Thus, \[ f_b = \bigl(f_g^4 - f_g^2 + f_g^0\bigr) + \Bigl[f_g^5 - (f_g^4 - f_g^3 + f_g^2 - f_g^1 + f_g^0)\Bigr] = f_g^5 - f_g^3 + f_g^1. \] ### **Row 7**: \( t_6 \le t < t_7 \) - Fee growth has crossed below tick \( i \). - We add \(\bigl[f_g^6 - (f_g^5 - f_g^4 + f_g^3 - f_g^2 + f_g^1 - f_g^0)\bigr]\). Hence, \[ f_b = \bigl(f_g^5 - f_g^3 + f_g^1\bigr) + \Bigl[f_g^6 - (f_g^5 - f_g^4 + f_g^3 - f_g^2 + f_g^1 - f_g^0)\Bigr] = f_g^6 - f_g^4 + f_g^2 - f_g^0. \] --- ## Fee Growth Inside Next, we fill out the *fee growth inside* column using the same example: - For \( t_0 \le t < t_1 \) (fee growth is above tick \( i \)): \[ f_{o,i} = f_g^0. \] - For \( t_1 \le t < t_2 \) (current tick > \( i \)): \[ f_{o,i} = f_g^1 - f_g^0. \] - For \( t_2 \le t < t_3 \): \[ f_{o,i} = f_g^2 - \bigl(f_g^1 - f_g^0\bigr) = f_g^2 - f_g^1 + f_g^0. \] - For \( t_3 \le t < t_4 \): \[ f_{o,i} = f_g^3 - \bigl(f_g^2 - f_g^1 + f_g^0\bigr) = f_g^3 - f_g^2 + f_g^1 - f_g^0. \] - For \( t_4 \le t < t_5 \): \[ f_{o,i} = f_g^4 - \bigl(f_g^3 - f_g^2 + f_g^1 - f_g^0\bigr) = f_g^4 - f_g^3 + f_g^2 - f_g^1 + f_g^0. \] - For \( t_5 \le t < t_6 \): \[ f_{o,i} = f_g^5 - \bigl(f_g^4 - f_g^3 + f_g^2 - f_g^1 + f_g^0\bigr) = f_g^5 - f_g^4 + f_g^3 - f_g^2 + f_g^1 - f_g^0. \] - For \( t_6 \le t < t_7 \): \[ f_{o,i} = f_g^6 - \bigl(f_g^5 - f_g^4 + f_g^3 - f_g^2 + f_g^1 - f_g^0\bigr) = f_g^6 - f_g^5 + f_g^4 - f_g^3 + f_g^2 - f_g^1 + f_g^0. \] --- ## Rewriting \(f_b\) in Terms of \(f_{o,i}\) Finally, we rewrite \( f_b \) in terms of *fee growth outside*, \( f_{o,i} \). - Compare the first \( f_b \) with \( f_{o,i} \): - Both are equal to \( f_g^0 \) in the first interval. - Thus, \( f_b = f_{o,i} \) (for that interval). - In the second interval, \( f_b = f_g^1 \). Rewriting in terms of \( f_{o,i} \), we also see it matches. - For the third interval, \( f_b = f_g^2 - f_g^1 + f_g^0 \), which again matches \( f_{o,i} \). Observing a pattern for *even rows* (where fee growth has just crossed over tick \( i \) to go above \( i \)), we have: \[ f_{o,i} = f_g - f_{o,i}. \] Hence, the generalized equation for *fee growth below*, \( f_b \), becomes: \[ f_b = \begin{cases} f_{o,i} & \text{if } i \le i_c,\\[6pt] f_g - f_{o,i} & \text{if } i_c < i. \end{cases} \] We will derive the equation for *fee growth above* in the next lesson.