exp_eps_for1¶

exp_eps: First Order Forward Sweep¶

First Order Expansion¶

We define \(x(t)\) and \(\varepsilon(t) ]\) near \(t = 0\) by the first order expansions

\begin{eqnarray} x(t) & = & x^{(0)} + x^{(1)} * t \\ \varepsilon(t) & = & \varepsilon^{(0)} + \varepsilon^{(1)} * t \end{eqnarray}

It follows that \(x^{(0)}\) (\(\varepsilon^{(0)}\)) is the zero, and \(x^{(1)}\) (\(\varepsilon^{(1)}\)) the first, order derivative of \(x(t)\) at \(t = 0\) (\(\varepsilon (t)\)) at \(t = 0\).

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 derivative with respect to \(x\), and the value of the derivative, are

\[\partial_x f ( x , \varepsilon ) = 1 + x = 1.5\]

Operation Sequence¶

Index¶

The Index column contains the index in the operation sequence of the corresponding atomic operation. A Forward sweep starts with the first operation and ends with the last.

Operation¶

The Operation column contains the mathematical function corresponding to each atomic operation in the sequence.

Zero Order¶

The Zero Order column contains the zero order derivatives for the corresponding variable in the operation sequence (see zero order sweep ).

Derivative¶

The Derivative column contains the mathematical function corresponding to the derivative with respect to \(t\), at \(t = 0\), for each variable in the sequence.

First Order¶

The First Order column contains the first order derivatives for the corresponding variable in the operation sequence; i.e.,

\[v_j (t) = v_j^{(0)} + v_j^{(1)} t\]

We use \(x^{(1)} = 1\) and \(\varepsilon^{(1)} = 0\), so that differentiation with respect to \(t\), at \(t = 0\), is the same partial differentiation with respect to \(x\) at \(x = x^{(0)}\).

Sweep¶

Index	Operation	Zero Order	Derivative	First Order
1	\(v_1 = x\)	0.5	\(v_1^{(1)} = x^{(1)}\)	\(v_1^{(1)} = 1\)
2	\(v_2 = 1 * v_1\)	0.5	\(v_2^{(1)} = 1 * v_1^{(1)}\)	\(v_2^{(1)} = 1\)
3	\(v_3 = v_2 / 1\)	0.5	\(v_3^{(1)} = v_2^{(1)} / 1\)	\(v_3^{(1)} = 1\)
4	\(v_4 = 1 + v_3\)	1.5	\(v_4^{(1)} = v_3^{(1)}\)	\(v_4^{(1)} = 1\)
5	\(v_5 = v_3 * v_1\)	0.25	\(v_5^{(1)} = v_3^{(1)} * v_1^{(0)} + v_3^{(0)} * v_1^{(1)}\)	\(v_5^{(1)} = 1\)
6	\(v_6 = v_5 / 2\)	0.125	\(v_6^{(1)} = v_5^{(1)} / 2\)	\(v_6^{(1)} = 0.5\)
7	\(v_7 = v_4 + v_6\)	1.625	\(v_7^{(1)} = v_4^{(1)} + v_6^{(1)}\)	\(v_7^{(1)} = 1.5\)

Return Value¶

The derivative of the return value for this case is

\begin{eqnarray} 1.5 & = & v_7^{(1)} = \left[ \D{v_7}{t} \right]_{t=0} = \left[ \D{}{t} f( x^{(0)} + x^{(1)} * t , \varepsilon^{(0)} ) \right]_{t=0} \\ & = & \partial_x f ( x^{(0)} , \varepsilon^{(0)} ) * x^{(1)} = \partial_x f ( x^{(0)} , \varepsilon^{(0)} ) \end{eqnarray}

(We have used the fact that \(x^{(1)} = 1\) and \(\varepsilon^{(1)} = 0\).)

Verification¶

The file exp_eps_for1.cpp contains a routine that verifies the values computed above.

Exercises¶

Suppose that \(x = .1\), what are the results of a zero and first order forward mode sweep for the operation sequence above; i.e., what are the corresponding values for \(v_1^{(0)}, v_2^{(0)}, \cdots , v_7^{(0)}\) and \(v_1^{(1)}, v_2^{(1)}, \cdots , v_7^{(1)}\) ?
Create a modified version of exp_eps_for1.cpp that verifies the derivative values from the previous exercise. Also create and run a main program that reports the result of calling the modified version of exp_eps_for1.cpp .
Suppose that \(x = .1\) and \(\epsilon = .2\), what is the operation sequence corresponding to

exp_eps ( x , epsilon )