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exp_eps: First Order Reverse Sweep

Purpose
First order reverse mode uses the operation sequence , and zero order forward sweep values, to compute the first order derivative of one dependent variable with respect to all the independent variables. The computations are done in reverse of the order of the computations in the original algorithm.

Mathematical Form
Suppose that we use the algorithm exp_eps.hpp to compute exp_eps(x, epsilon) with x is equal to .5 and epsilon is equal to .2. For this case, the mathematical function for the operation sequence corresponding to exp_eps is $$f ( x , \varepsilon ) = 1 + x + x^2 / 2$$ The corresponding partial derivatives, and the value of the derivatives, are $$\begin{array}{rcl} \partial_x f ( x , \varepsilon ) & = & 1 + x = 1.5 \\ \partial_\varepsilon f ( x , \varepsilon ) & = & 0 \end{array}$$

epsilon
Since $\varepsilon$ is an independent variable, it could included as an argument to all of the $f_j$ functions below. The result would be that all the partials with respect to $\varepsilon$ would be zero and hence we drop it to simplify the presentation.

f_7
In reverse mode we choose one dependent variable and compute its derivative with respect to all the independent variables. For our example, we chose the value returned by exp_eps.hpp which is $v_7$. We begin with the function $f_7$ where $v_7$ is both an argument and the value of the function; i.e., $$\begin{array}{rcl} f_7 ( v_1 , v_2 , v_3 , v_4 , v_5 , v_6 , v_7 ) & = & v_7 \\ \D{f_7}{v_7} & = & 1 \end{array}$$ All the other partial derivatives of $f_7$ are zero.

Index 7: f_6
The last operation has index 7, $$v_7 = v_4 + v_6$$ We define the function $f_6 ( v_1 , v_2 , v_3 , v_4 , v_5 , v_6 )$ as equal to $f_7$ except that $v_7$ is eliminated using this operation; i.e. $$f_6 = f_7 [ v_1 , v_2 , v_3 , v_4 , v_5 , v_6 , v_7 ( v_4 , v_6 ) ]$$ It follows that $$\begin{array}{rcll} \D{f_6}{v_4} & = & \D{f_7}{v_4} + \D{f_7}{v_7} * \D{v_7}{v_4} & = 1 \\ \D{f_6}{v_6} & = & \D{f_7}{v_6} + \D{f_7}{v_7} * \D{v_7}{v_6} & = 1 \end{array}$$ All the other partial derivatives of $f_6$ are zero.

Index 6: f_5
The previous operation has index 6, $$v_6 = v_5 / 2$$ We define the function $f_5 ( v_1 , v_2 , v_3 , v_4 , v_5 )$ as equal to $f_6$ except that $v_6$ is eliminated using this operation; i.e., $$f_5 = f_6 [ v_1 , v_2 , v_3 , v_4 , v_5 , v_6 ( v_5 ) ]$$ It follows that $$\begin{array}{rcll} \D{f_5}{v_4} & = & \D{f_6}{v_4} & = 1 \\ \D{f_5}{v_5} & = & \D{f_6}{v_5} + \D{f_6}{v_6} * \D{v_6}{v_5} & = 0.5 \end{array}$$ All the other partial derivatives of $f_5$ are zero.

Index 5: f_4
The previous operation has index 5, $$v_5 = v_3 * v_1$$ We define the function $f_4 ( v_1 , v_2 , v_3 , v_4 )$ as equal to $f_5$ except that $v_5$ is eliminated using this operation; i.e., $$f_4 = f_5 [ v_1 , v_2 , v_3 , v_4 , v_5 ( v_3 , v_1 ) ]$$ Given the information from the forward sweep, we have $v_3 = 0.5$ and $v_1 = 0.5$. It follows that $$\begin{array}{rcll} \D{f_4}{v_1} & = & \D{f_5}{v_1} + \D{f_5}{v_5} * \D{v_5}{v_1} & = 0.25 \\ \D{f_4}{v_2} & = & \D{f_5}{v_2} & = 0 \\ \D{f_4}{v_3} & = & \D{f_5}{v_3} + \D{f_5}{v_5} * \D{v_5}{v_3} & = 0.25 \\ \D{f_4}{v_4} & = & \D{f_5}{v_4} & = 1 \end{array}$$

Index 4: f_3
The previous operation has index 4, $$v_4 = 1 + v_3$$ We define the function $f_3 ( v_1 , v_2 , v_3 )$ as equal to $f_4$ except that $v_4$ is eliminated using this operation; i.e., $$f_3 = f_4 [ v_1 , v_2 , v_3 , v_4 ( v_3 ) ]$$ It follows that $$\begin{array}{rcll} \D{f_3}{v_1} & = & \D{f_4}{v_1} & = 0.25 \\ \D{f_3}{v_2} & = & \D{f_4}{v_2} & = 0 \\ \D{f_3}{v_3} & = & \D{f_4}{v_3} + \D{f_4}{v_4} * \D{v_4}{v_3} & = 1.25 \end{array}$$

Index 3: f_2
The previous operation has index 3, $$v_3 = v_2 / 1$$ We define the function $f_2 ( v_1 , v_2 )$ as equal to $f_3$ except that $v_3$ is eliminated using this operation; i.e., $$f_2 = f_4 [ v_1 , v_2 , v_3 ( v_2 ) ]$$ It follows that $$\begin{array}{rcll} \D{f_2}{v_1} & = & \D{f_3}{v_1} & = 0.25 \\ \D{f_2}{v_2} & = & \D{f_3}{v_2} + \D{f_3}{v_3} * \D{v_3}{v_2} & = 1.25 \end{array}$$

Index 2: f_1
The previous operation has index 1, $$v_2 = 1 * v_1$$ We define the function $f_1 ( v_1 )$ as equal to $f_2$ except that $v_2$ is eliminated using this operation; i.e., $$f_1 = f_2 [ v_1 , v_2 ( v_1 ) ]$$ It follows that $$\begin{array}{rcll} \D{f_1}{v_1} & = & \D{f_2}{v_1} + \D{f_2}{v_2} * \D{v_2}{v_1} & = 1.5 \end{array}$$ Note that $v_1$ is equal to $x$, so the derivative of exp_eps(x, epsilon) at x equal to .5 and epsilon equal .2 is 1.5 in the x direction and zero in the epsilon direction. We also note that forward forward mode gave the same result for the partial in the x direction.

Verification
The file exp_eps_rev1.cpp contains a routine that verifies the values computed above. It returns true for success and false for failure. It only tests the partial derivatives of $f_j$ that might not be equal to the corresponding partials of $f_{j+1}$; i.e., the other partials of $f_j$ must be equal to the corresponding partials of $f_{j+1}$.

Exercises
1. Consider the case where $x = .1$ and we first preform a zero order forward mode sweep for the operation sequence used above (in reverse order). What are the results of a first order reverse mode sweep; i.e., what are the corresponding values for $\D{f_j}{v_k}$ for all $j, k$ such that $\D{f_j}{v_k} \neq 0$.
2. Create a modified version of exp_eps_rev1.cpp that verifies the values you obtained for the previous exercise. Also create and run a main program that reports the result of calling the modified version of exp_eps_rev1.cpp .

Input File: introduction/exp_eps.omh