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Up-> CppAD AD ADValued Arithmetic compound_assign

AD-> ad_ctor ad_assign Convert ADValued BoolValued VecAD base_require

ADValued-> Arithmetic unary_standard_math binary_math CondExp Discrete numeric_limits atomic

Arithmetic-> UnaryPlus UnaryMinus ad_binary compound_assign

compound_assign-> AddEq.cpp sub_eq.cpp mul_eq.cpp div_eq.cpp

Headings-> Syntax Purpose Op Base x y Result Operation Sequence Example Derivative ---..Addition ---..Subtraction ---..Multiplication ---..Division

@(@\newcommand{\W}[1]{ \; #1 \; } \newcommand{\R}[1]{ {\rm #1} } \newcommand{\B}[1]{ {\bf #1} } \newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} } \newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} } \newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} } \newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }@)@AD Compound Assignment Operators
Syntax
x Op y

Purpose
Performs compound assignment operations where either x has type AD<Base> .

Op
The operator Op is one of the following

Op	Meaning
+=	x is assigned x plus y
-=	x is assigned x minus y
*=	x is assigned x times y
/=	x is assigned x divided by y

Base
The type Base is determined by the operand x .

x
The operand x has the following prototype
     AD<Base> &x

y
The operand y has the following prototype
     const Type &y
where Type is VecAD<Base>::reference , AD<Base> , Base , or double.

Result
The result of this assignment can be used as a reference to x . For example, if z has the following type
     AD<Base> z
then the syntax
     z = x += y
will compute x plus y and then assign this value to both x and z .

Operation Sequence
This is an atomic AD of Base operation and hence it is part of the current AD of Base operation sequence .

Example
The following files contain examples and tests of these functions. Each test returns true if it succeeds and false otherwise.

AddEq.cpp	AD Compound Assignment Addition: Example and Test
sub_eq.cpp	AD Compound Assignment Subtraction: Example and Test
mul_eq.cpp	AD Compound Assignment Multiplication: Example and Test
div_eq.cpp	AD Compound Assignment Division: Example and Test

Derivative
If @(@ f @)@ and @(@ g @)@ are Base functions

Addition
@[@ \D{[ f(x) + g(x) ]}{x} = \D{f(x)}{x} + \D{g(x)}{x} @]@
Subtraction
@[@ \D{[ f(x) - g(x) ]}{x} = \D{f(x)}{x} - \D{g(x)}{x} @]@
Multiplication
@[@ \D{[ f(x) * g(x) ]}{x} = g(x) * \D{f(x)}{x} + f(x) * \D{g(x)}{x} @]@
Division
@[@ \D{[ f(x) / g(x) ]}{x} = [1/g(x)] * \D{f(x)}{x} - [f(x)/g(x)^2] * \D{g(x)}{x} @]@

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Input File: cppad/core/compound_assign.hpp