Lint changes from William Rucklidge.
Change-Id: I6592b61451ead8f0407bec134fcf4b56ba22ffb9
diff --git a/docs/source/gradient_solver.rst b/docs/source/gradient_solver.rst
index 1ef1515..ba7cbae 100644
--- a/docs/source/gradient_solver.rst
+++ b/docs/source/gradient_solver.rst
@@ -231,7 +231,7 @@
In each iteration of the line search,
- .. math:: \text{new_step_size} >= \text{max_line_search_step_contraction} * \text{step_size}
+ .. math:: \text{new_step_size} \geq \text{max_line_search_step_contraction} * \text{step_size}
Note that by definition, for contraction:
@@ -243,7 +243,7 @@
In each iteration of the line search,
- .. math:: \text{new_step_size} <= \text{min_line_search_step_contraction} * \text{step_size}
+ .. math:: \text{new_step_size} \leq \text{min_line_search_step_contraction} * \text{step_size}
Note that by definition, for contraction:
@@ -260,7 +260,7 @@
As this is an 'artificial' constraint (one imposed by the user, not
the underlying math), if ``WOLFE`` line search is being used, *and*
points satisfying the Armijo sufficient (function) decrease
- condition have been found during the current search (in :math:`<=`
+ condition have been found during the current search (in :math:`\leq`
``max_num_line_search_step_size_iterations``). Then, the step size
with the lowest function value which satisfies the Armijo condition
will be returned as the new valid step, even though it does *not*
@@ -289,7 +289,7 @@
decreases sufficiently. Precisely, this second condition
is that we seek a step size s.t.
- .. math:: \|f'(\text{step_size})\| <= \text{sufficient_curvature_decrease} * \|f'(0)\|
+ .. math:: \|f'(\text{step_size})\| \leq \text{sufficient_curvature_decrease} * \|f'(0)\|
Where :math:`f()` is the line search objective and :math:`f'()` is the derivative
of :math:`f` with respect to the step size: :math:`\frac{d f}{d~\text{step size}}`.
@@ -304,7 +304,7 @@
satisfying the conditions is found. Precisely, at each iteration
of the expansion:
- .. math:: \text{new_step_size} <= \text{max_step_expansion} * \text{step_size}
+ .. math:: \text{new_step_size} \leq \text{max_step_expansion} * \text{step_size}
By definition for expansion
@@ -327,7 +327,7 @@
Solver terminates if
- .. math:: \frac{|\Delta \text{cost}|}{\text{cost}} <= \text{function_tolerance}
+ .. math:: \frac{|\Delta \text{cost}|}{\text{cost}} \leq \text{function_tolerance}
where, :math:`\Delta \text{cost}` is the change in objective
function value (up or down) in the current iteration of the line search.
@@ -338,7 +338,7 @@
Solver terminates if
- .. math:: \|x - \Pi \boxplus(x, -g(x))\|_\infty <= \text{gradient_tolerance}
+ .. math:: \|x - \Pi \boxplus(x, -g(x))\|_\infty \leq \text{gradient_tolerance}
where :math:`\|\cdot\|_\infty` refers to the max norm, :math:`\Pi`
is projection onto the bounds constraints and :math:`\boxplus` is
@@ -351,7 +351,7 @@
Solver terminates if
- .. math:: \|\Delta x\| <= (\|x\| + \text{parameter_tolerance}) * \text{parameter_tolerance}
+ .. math:: \|\Delta x\| \leq (\|x\| + \text{parameter_tolerance}) * \text{parameter_tolerance}
where :math:`\Delta x` is the step computed by the linear solver in
the current iteration of the line search.
diff --git a/include/ceres/jet.h b/include/ceres/jet.h
index 96b6dcf..d84dca3 100644
--- a/include/ceres/jet.h
+++ b/include/ceres/jet.h
@@ -576,11 +576,12 @@
// We have various special cases, see the comment for pow(Jet, Jet) for
// analysis:
//
-// 1. For a > 0 we have: (a)^(p + dp) ~= a^p + a^p log(a) dp
+// 1. For f > 0 we have: (f)^(g + dg) ~= f^g + f^g log(f) dg
//
-// 2. For a == 0 and p > 0 we have: (a)^(p + dp) ~= a^p
+// 2. For f == 0 and g > 0 we have: (f)^(g + dg) ~= f^g
//
-// 3. For a < 0 and integer p we have: (a)^(p + dp) ~= a^p
+// 3. For f < 0 and integer g we have: (f)^(g + dg) ~= f^g but if dg
+// != 0, the derivatives are not defined and we return NaN.
template <typename T, int N> inline
Jet<T, N> pow(double f, const Jet<T, N>& g) {
@@ -607,37 +608,37 @@
// pow -- both base and exponent are differentiable functions. This has a
// variety of special cases that require careful handling.
//
-// 1. For a > 0: (a + da)^(b + db) ~= a^b + a^(b - 1) * (b*da + a*log(a)*db)
-// The numerical evaluation of a*log(a) for a > 0 is well behaved, even for
+// 1. For f > 0: (f + df)^(g + dg) ~= f^g + f^(g - 1) * (g * df + f * log(f) * dg)
+// The numerical evaluation of f * log(f) for f > 0 is well behaved, even for
// extremely small values (e.g. 1e-99).
//
-// 2. For a == 0 and b > 1: (a + da)^(b + db) ~= 0
-// This cases is needed because log(0) can not be evaluated in the a > 0
-// expression. However the function a*log(a) is well behaved around a == 0
-// and its limit as a-->0 is zero.
+// 2. For f == 0 and g > 1: (f + df)^(g + dg) ~= 0
+// This cases is needed because log(0) can not be evaluated in the f > 0
+// expression. However the function f*log(f) is well behaved around f == 0
+// and its limit as f-->0 is zero.
//
-// 3. For a == 0 and b == 1: (a + da)^(b + db) ~= 0 + da
+// 3. For f == 0 and g == 1: (f + df)^(g + dg) ~= 0 + df
//
-// 4. For a == 0 and 0 < b < 1: The value is finite but the derivatives are not.
+// 4. For f == 0 and 0 < g < 1: The value is finite but the derivatives are not.
//
-// 5. For a == 0 and b < 0: The value and derivatives of a^b are not finite.
+// 5. For f == 0 and g < 0: The value and derivatives of f^g are not finite.
//
-// 6. For a == 0 and b == 0: The C standard incorrectly defines 0^0 to be 1
+// 6. For f == 0 and g == 0: The C standard incorrectly defines 0^0 to be 1
// "because there are applications that can exploit this definition". We
// (arbitrarily) decree that derivatives here will be nonfinite, since that
-// is consistent with the behavior for a==0, b < 0 and 0 < b < 1. Practically
+// is consistent with the behavior for f == 0, g < 0 and 0 < g < 1. Practically
// any definition could have been justified because mathematical consistency
// has been lost at this point.
//
-// 7. For a < 0, b integer, db == 0: (a + da)^(b + db) ~= a^b + b * a^(b - 1) da
-// This is equivalent to the case where a is a differentiable function and b
+// 7. For f < 0, g integer, dg == 0: (f + df)^(g + dg) ~= f^g + g * f^(g - 1) df
+// This is equivalent to the case where f is a differentiable function and g
// is a constant (to first order).
//
-// 8. For a < 0, b integer, db != 0: The value is finite but the derivatives are
-// not, because any change in the value of b moves us away from the point
+// 8. For f < 0, g integer, dg != 0: The value is finite but the derivatives are
+// not, because any change in the value of g moves us away from the point
// with a real-valued answer into the region with complex-valued answers.
//
-// 9. For a < 0, b noninteger: The value and derivatives of a^b are not finite.
+// 9. For f < 0, g noninteger: The value and derivatives of f^g are not finite.
template <typename T, int N> inline
Jet<T, N> pow(const Jet<T, N>& f, const Jet<T, N>& g) {
