\(\newcommand{\W}[1]{ \; #1 \; }\) \(\newcommand{\R}[1]{ {\rm #1} }\) \(\newcommand{\B}[1]{ {\bf #1} }\) \(\newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} }\) \(\newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} }\) \(\newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} }\) \(\newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }\)
pow_nan.cpp¶
View page sourcepow: Nan in Result of Pow Function: Example and Test¶
Purpose¶
The pow(x, y) function will work when \(x < 0\) and \(y\) is a parameter. It will often generate nan or infinity when \(x < 0\) and one tries to compute a derivatives (even if \(y\) is a positive integer). This is because the derivative of the log is \(1 / x\) and the power function uses the representation
Problem¶
There is a problem with this representation when \(y\) is a parameter and \(x = 0\). For example, when \(x = 0\) and \(y = 1\), it returns zero for the derivative, but the actual derivative w.r.t \(x\) is one.
# include <cppad/cppad.hpp>
# include <cmath>
bool pow_nan(void)
{ bool ok = true;
using std::cout;
using CppAD::AD;
using CppAD::vector;
//
vector<double> x(2), y(2), dx(2), dy(2), ddx(2), ddy(2);
vector< AD<double> > ax(2), ay(2);
//
// variable vector
ax[0] = x[0] = -1.0;
ax[1] = x[1] = 2.0;
//
CppAD::Independent(ax);
//
ay[0] = pow( ax[0], ax[1] );
ay[1] = pow( ax[0], 2.0 );
//
CppAD::ADFun<double> f(ax, ay);
//
// check_for_nan is set false so it does not generate an assert
// when compiling with debugging
f.check_for_nan(false);
//
// Zero order forward does not generate nan
y = f.Forward(0, x);
ok &= y[0] == 1.0;
ok &= y[1] == 1.0;
//
// First order forward generates a nan
dx[0] = 1.0;
dx[1] = 1.0;
dy = f.Forward(1, dx);
ok &= std::isnan( dy[0] );
ok &= dy[1] == -2.0;
//
// Second order Taylor coefficient is 1/2 times second derivative
ddx[0] = 0.0;
ddx[1] = 0.0;
ddy = f.Forward(2, ddx);
ok &= std::isnan( ddy[0] );
ok &= ddy[1] == 1.0;
//
return ok;
}