docs/source/numerical_derivatives.rst - ceres-solver - Git at Google

 .. default-domain:: cpp

 .. cpp:namespace:: ceres

 .. _chapter-numerical_derivatives:

 ===================
 Numeric derivatives
 ===================

 The other extreme from using analytic derivatives is to use numeric
 derivatives. The key observation here is that the process of
 differentiating a function :math:`f(x)` w.r.t :math:`x` can be written
 as the limiting process:

 .. math::
    Df(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}


 Forward Differences
 ===================

 Now of course one cannot perform the limiting operation numerically on
 a computer so we do the next best thing, which is to choose a small
 value of :math:`h` and approximate the derivative as

 .. math::
    Df(x) \approx \frac{f(x + h) - f(x)}{h}


 The above formula is the simplest most basic form of numeric
 differentiation. It is known as the *Forward Difference* formula.

 So how would one go about constructing a numerically differentiated
 version of ``Rat43Analytic`` (`Rat43
 <http://www.itl.nist.gov/div898/strd/nls/data/ratkowsky3.shtml>`_) in
 Ceres Solver. This is done in two steps:

   1. Define *Functor* that given the parameter values will evaluate the
      residual for a given :math:`(x,y)`.
   2. Construct a :class:`CostFunction` by using
      :class:`NumericDiffCostFunction` to wrap an instance of
      ``Rat43CostFunctor``.

 .. code-block:: c++

   struct Rat43CostFunctor {
     Rat43CostFunctor(const double x, const double y) : x_(x), y_(y) {}

     bool operator()(const double* parameters, double* residuals) const {
       const double b1 = parameters[0];
       const double b2 = parameters[1];
       const double b3 = parameters[2];
       const double b4 = parameters[3];
       residuals[0] = b1 * pow(1.0 + exp(b2 -  b3 * x_), -1.0 / b4) - y_;
       return true;
     }

     const double x_;
     const double y_;
   }

   CostFunction* cost_function =
     new NumericDiffCostFunction<Rat43CostFunctor, FORWARD, 1, 4>(
       new Rat43CostFunctor(x, y));

 This is about the minimum amount of work one can expect to do to
 define the cost function. The only thing that the user needs to do is
 to make sure that the evaluation of the residual is implemented
 correctly and efficiently.

 Before going further, it is instructive to get an estimate of the
 error in the forward difference formula. We do this by considering the
 `Taylor expansion <https://en.wikipedia.org/wiki/Taylor_series>`_ of
 :math:`f` near :math:`x`.

 .. math::
    \begin{align}
    f(x+h) &= f(x) + h Df(x) + \frac{h^2}{2!} D^2f(x) +
    \frac{h^3}{3!}D^3f(x) + \cdots \\
    Df(x) &= \frac{f(x + h) - f(x)}{h} - \left [\frac{h}{2!}D^2f(x) +
    \frac{h^2}{3!}D^3f(x) + \cdots  \right]\\
    Df(x) &= \frac{f(x + h) - f(x)}{h} + O(h)
    \end{align}

 i.e., the error in the forward difference formula is
 :math:`O(h)` [#f4]_.


 Implementation Details
 ----------------------

 :class:`NumericDiffCostFunction` implements a generic algorithm to
 numerically differentiate a given functor. While the actual
 implementation of :class:`NumericDiffCostFunction` is complicated, the
 net result is a :class:`CostFunction` that roughly looks something
 like the following:

 .. code-block:: c++

   class Rat43NumericDiffForward : public SizedCostFunction<1,4> {
      public:
        Rat43NumericDiffForward(const Rat43Functor* functor) : functor_(functor) {}
        virtual ~Rat43NumericDiffForward() {}
        virtual bool Evaluate(double const* const* parameters,
                              double* residuals,
 			     double** jacobians) const {
  	 functor_(parameters[0], residuals);
 	 if (!jacobians) return true;
 	 double* jacobian = jacobians[0];
 	 if (!jacobian) return true;

 	 const double f = residuals[0];
 	 double parameters_plus_h[4];
 	 for (int i = 0; i < 4; ++i) {
 	   std::copy(parameters, parameters + 4, parameters_plus_h);
 	   const double kRelativeStepSize = 1e-6;
 	   const double h = std::abs(parameters[i]) * kRelativeStepSize;
 	   parameters_plus_h[i] += h;
            double f_plus;
   	   functor_(parameters_plus_h, &f_plus);
 	   jacobian[i] = (f_plus - f) / h;
          }
 	 return true;
        }

      private:
        std::unique_ptr<Rat43Functor> functor_;
    };


 Note the choice of step size :math:`h` in the above code, instead of
 an absolute step size which is the same for all parameters, we use a
 relative step size of :math:`\text{kRelativeStepSize} = 10^{-6}`. This
 gives better derivative estimates than an absolute step size [#f2]_
 [#f3]_. This choice of step size only works for parameter values that
 are not close to zero. So the actual implementation of
 :class:`NumericDiffCostFunction`, uses a more complex step size
 selection logic, where close to zero, it switches to a fixed step
 size.


 Central Differences
 ===================

 :math:`O(h)` error in the Forward Difference formula is okay but not
 great. A better method is to use the *Central Difference* formula:

 .. math::
    Df(x) \approx \frac{f(x + h) - f(x - h)}{2h}

 Notice that if the value of :math:`f(x)` is known, the Forward
 Difference formula only requires one extra evaluation, but the Central
 Difference formula requires two evaluations, making it twice as
 expensive. So is the extra evaluation worth it?

 To answer this question, we again compute the error of approximation
 in the central difference formula:

 .. math::
    \begin{align}
   f(x + h) &= f(x) + h Df(x) + \frac{h^2}{2!}
   D^2f(x) + \frac{h^3}{3!} D^3f(x) + \frac{h^4}{4!} D^4f(x) + \cdots\\
     f(x - h) &= f(x) - h Df(x) + \frac{h^2}{2!}
   D^2f(x) - \frac{h^3}{3!} D^3f(c_2) + \frac{h^4}{4!} D^4f(x) +
   \cdots\\
   Df(x) & =  \frac{f(x + h) - f(x - h)}{2h} + \frac{h^2}{3!}
   D^3f(x) +  \frac{h^4}{5!}
   D^5f(x) + \cdots \\
   Df(x) & =  \frac{f(x + h) - f(x - h)}{2h} + O(h^2)
    \end{align}

 The error of the Central Difference formula is :math:`O(h^2)`, i.e.,
 the error goes down quadratically whereas the error in the Forward
 Difference formula only goes down linearly.

 Using central differences instead of forward differences in Ceres
 Solver is a simple matter of changing a template argument to
 :class:`NumericDiffCostFunction` as follows:

 .. code-block:: c++

   CostFunction* cost_function =
     new NumericDiffCostFunction<Rat43CostFunctor, CENTRAL, 1, 4>(
       new Rat43CostFunctor(x, y));

 But what do these differences in the error mean in practice? To see
 this, consider the problem of evaluating the derivative of the
 univariate function

 .. math::
    f(x) = \frac{e^x}{\sin x - x^2},

 at :math:`x = 1.0`.

 It is easy to determine that :math:`Df(1.0) =
 140.73773557129658`. Using this value as reference, we can now compute
 the relative error in the forward and central difference formulae as a
 function of the absolute step size and plot them.

 .. figure:: forward_central_error.png
    :figwidth: 100%
    :align: center

 Reading the graph from right to left, a number of things stand out in
 the above graph:

  1. The graph for both formulae have two distinct regions. At first,
     starting from a large value of :math:`h` the error goes down as
     the effect of truncating the Taylor series dominates, but as the
     value of :math:`h` continues to decrease, the error starts
     increasing again as roundoff error starts to dominate the
     computation. So we cannot just keep on reducing the value of
     :math:`h` to get better estimates of :math:`Df`. The fact that we
     are using finite precision arithmetic becomes a limiting factor.
  2. Forward Difference formula is not a great method for evaluating
     derivatives. Central Difference formula converges much more
     quickly to a more accurate estimate of the derivative with
     decreasing step size. So unless the evaluation of :math:`f(x)` is
     so expensive that you absolutely cannot afford the extra
     evaluation required by central differences, **do not use the
     Forward Difference formula**.
  3. Neither formula works well for a poorly chosen value of :math:`h`.


 Ridders' Method
 ===============

 So, can we get better estimates of :math:`Df` without requiring such
 small values of :math:`h` that we start hitting floating point
 roundoff errors?

 One possible approach is to find a method whose error goes down faster
 than :math:`O(h^2)`. This can be done by applying `Richardson
 Extrapolation
 <https://en.wikipedia.org/wiki/Richardson_extrapolation>`_ to the
 problem of differentiation. This is also known as *Ridders' Method*
 [Ridders]_.

 Let us recall, the error in the central differences formula.

 .. math::
    \begin{align}
    Df(x) & =  \frac{f(x + h) - f(x - h)}{2h} + \frac{h^2}{3!}
    D^3f(x) +  \frac{h^4}{5!}
    D^5f(x) + \cdots\\
            & =  \frac{f(x + h) - f(x - h)}{2h} + K_2 h^2 + K_4 h^4 + \cdots
    \end{align}

 The key thing to note here is that the terms :math:`K_2, K_4, ...`
 are indepdendent of :math:`h` and only depend on :math:`x`.

 Let us now define:

 .. math::

    A(1, m) = \frac{f(x + h/2^{m-1}) - f(x - h/2^{m-1})}{2h/2^{m-1}}.

 Then observe that

 .. math::

    Df(x) = A(1,1) + K_2 h^2 + K_4 h^4 + \cdots

 and

 .. math::

    Df(x) = A(1, 2) + K_2 (h/2)^2 + K_4 (h/2)^4 + \cdots

 Here we have halved the step size to obtain a second central
 differences estimate of :math:`Df(x)`. Combining these two estimates,
 we get:

 .. math::

    Df(x) = \frac{4 A(1, 2) - A(1,1)}{4 - 1} + O(h^4)

 which is an approximation of :math:`Df(x)` with truncation error that
 goes down as :math:`O(h^4)`. But we do not have to stop here. We can
 iterate this process to obtain even more accurate estimates as
 follows:

 .. math::

    A(n, m) =  \begin{cases}
     \frac{\displaystyle f(x + h/2^{m-1}) - f(x -
     h/2^{m-1})}{\displaystyle 2h/2^{m-1}} & n = 1 \\
    \frac{\displaystyle 4^{n-1} A(n - 1, m + 1) - A(n - 1, m)}{\displaystyle 4^{n-1} - 1} & n > 1
    \end{cases}

 It is straightforward to show that the approximation error in
 :math:`A(n, 1)` is :math:`O(h^{2n})`. To see how the above formula can
 be implemented in practice to compute :math:`A(n,1)` it is helpful to
 structure the computation as the following tableau:

 .. math::
    \begin{array}{ccccc}
    A(1,1) & A(1, 2) & A(1, 3) & A(1, 4) & \cdots\\
           & A(2, 1) & A(2, 2) & A(2, 3) & \cdots\\
 	  &         & A(3, 1) & A(3, 2) & \cdots\\
 	  &         &         & A(4, 1) & \cdots \\
 	  &         &         &         & \ddots
    \end{array}

 So, to compute :math:`A(n, 1)` for increasing values of :math:`n` we
 move from the left to the right, computing one column at a
 time. Assuming that the primary cost here is the evaluation of the
 function :math:`f(x)`, the cost of computing a new column of the above
 tableau is two function evaluations. Since the cost of evaluating
 :math:`A(1, n)`, requires evaluating the central difference formula
 for step size of :math:`2^{1-n}h`

 Applying this method to :math:`f(x) = \frac{e^x}{\sin x - x^2}`
 starting with a fairly large step size :math:`h = 0.01`, we get:

 .. math::
    \begin{array}{rrrrr}
    141.678097131 &140.971663667 &140.796145400 &140.752333523 &140.741384778\\
    &140.736185846 &140.737639311 &140.737729564 &140.737735196\\
    & &140.737736209 &140.737735581 &140.737735571\\
    & & &140.737735571 &140.737735571\\
    & & & &140.737735571\\
    \end{array}

 Compared to the *correct* value :math:`Df(1.0) = 140.73773557129658`,
 :math:`A(5, 1)` has a relative error of :math:`10^{-13}`. For
 comparison, the relative error for the central difference formula with
 the same stepsize (:math:`0.01/2^4 = 0.000625`) is :math:`10^{-5}`.

 The above tableau is the basis of Ridders' method for numeric
 differentiation. The full implementation is an adaptive scheme that
 tracks its own estimation error and stops automatically when the
 desired precision is reached. Of course it is more expensive than the
 forward and central difference formulae, but is also significantly
 more robust and accurate.

 Using Ridder's method instead of forward or central differences in
 Ceres is again a simple matter of changing a template argument to
 :class:`NumericDiffCostFunction` as follows:

 .. code-block:: c++

   CostFunction* cost_function =
     new NumericDiffCostFunction<Rat43CostFunctor, RIDDERS, 1, 4>(
       new Rat43CostFunctor(x, y));

 The following graph shows the relative error of the three methods as a
 function of the absolute step size. For Ridders's method we assume
 that the step size for evaluating :math:`A(n,1)` is :math:`2^{1-n}h`.

 .. figure:: forward_central_ridders_error.png
    :figwidth: 100%
    :align: center

 Using the 10 function evaluations that are needed to compute
 :math:`A(5,1)` we are able to approximate :math:`Df(1.0)` about a 1000
 times better than the best central differences estimate. To put these
 numbers in perspective, machine epsilon for double precision
 arithmetic is :math:`\approx 2.22 \times 10^{-16}`.

 Going back to ``Rat43``, let us also look at the runtime cost of the
 various methods for computing numeric derivatives.

 ==========================   =========
 CostFunction                 Time (ns)
 ==========================   =========
 Rat43Analytic                      255
 Rat43AnalyticOptimized              92
 Rat43NumericDiffForward            262
 Rat43NumericDiffCentral            517
 Rat43NumericDiffRidders           3760
 ==========================   =========

 As expected, Central Differences is about twice as expensive as
 Forward Differences and the remarkable accuracy improvements of
 Ridders' method cost an order of magnitude more runtime.

 Recommendations
 ===============

 Numeric differentiation should be used when you cannot compute the
 derivatives either analytically or using automatic differention. This
 is usually the case when you are calling an external library or
 function whose analytic form you do not know or even if you do, you
 are not in a position to re-write it in a manner required to use
 :ref:`chapter-automatic_derivatives`.


 When using numeric differentiation, use at least Central Differences,
 and if execution time is not a concern or the objective function is
 such that determining a good static relative step size is hard,
 Ridders' method is recommended.

 .. rubric:: Footnotes

 .. [#f2] `Numerical Differentiation
 	 <https://en.wikipedia.org/wiki/Numerical_differentiation#Practical_considerations_using_floating_point_arithmetic>`_
 .. [#f3] [Press]_ Numerical Recipes, Section 5.7
 .. [#f4] In asymptotic error analysis, an error of :math:`O(h^k)`
 	 means that the absolute-value of the error is at most some
 	 constant times :math:`h^k` when :math:`h` is close enough to
 	 :math:`0`.
	.. default-domain:: cpp

	.. cpp:namespace:: ceres

	.. _chapter-numerical_derivatives:

	===================
	Numeric derivatives
	===================

	The other extreme from using analytic derivatives is to use numeric
	derivatives. The key observation here is that the process of
	differentiating a function :math:`f(x)` w.r.t :math:`x` can be written
	as the limiting process:

	.. math::
	Df(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}


	Forward Differences
	===================

	Now of course one cannot perform the limiting operation numerically on
	a computer so we do the next best thing, which is to choose a small
	value of :math:`h` and approximate the derivative as

	.. math::
	Df(x) \approx \frac{f(x + h) - f(x)}{h}


	The above formula is the simplest most basic form of numeric
	differentiation. It is known as the Forward Difference formula.

	So how would one go about constructing a numerically differentiated
	version of ``Rat43Analytic`` (`Rat43
	<http://www.itl.nist.gov/div898/strd/nls/data/ratkowsky3.shtml>`_) in
	Ceres Solver. This is done in two steps:

	1. Define Functor that given the parameter values will evaluate the
	residual for a given :math:`(x,y)`.
	2. Construct a :class:`CostFunction` by using
	:class:`NumericDiffCostFunction` to wrap an instance of
	``Rat43CostFunctor``.

	.. code-block:: c++

	struct Rat43CostFunctor {
	Rat43CostFunctor(const double x, const double y) : x_(x), y_(y) {}

	bool operator()(const double* parameters, double* residuals) const {
	const double b1 = parameters[0];
	const double b2 = parameters[1];
	const double b3 = parameters[2];
	const double b4 = parameters[3];
	residuals[0] = b1 * pow(1.0 + exp(b2 - b3 * x_), -1.0 / b4) - y_;
	return true;
	}

	const double x_;
	const double y_;
	}

	CostFunction* cost_function =
	new NumericDiffCostFunction<Rat43CostFunctor, FORWARD, 1, 4>(
	new Rat43CostFunctor(x, y));

	This is about the minimum amount of work one can expect to do to
	define the cost function. The only thing that the user needs to do is
	to make sure that the evaluation of the residual is implemented
	correctly and efficiently.

	Before going further, it is instructive to get an estimate of the
	error in the forward difference formula. We do this by considering the
	`Taylor expansion <https://en.wikipedia.org/wiki/Taylor_series>`_ of
	:math:`f` near :math:`x`.

	.. math::
	\begin{align}
	f(x+h) &= f(x) + h Df(x) + \frac{h^2}{2!} D^2f(x) +
	\frac{h^3}{3!}D^3f(x) + \cdots \\
	Df(x) &= \frac{f(x + h) - f(x)}{h} - \left [\frac{h}{2!}D^2f(x) +
	\frac{h^2}{3!}D^3f(x) + \cdots \right]\\
	Df(x) &= \frac{f(x + h) - f(x)}{h} + O(h)
	\end{align}

	i.e., the error in the forward difference formula is
	:math:`O(h)` [#f4]_.


	Implementation Details
	----------------------

	:class:`NumericDiffCostFunction` implements a generic algorithm to
	numerically differentiate a given functor. While the actual
	implementation of :class:`NumericDiffCostFunction` is complicated, the
	net result is a :class:`CostFunction` that roughly looks something
	like the following:

	.. code-block:: c++

	class Rat43NumericDiffForward : public SizedCostFunction<1,4> {
	public:
	Rat43NumericDiffForward(const Rat43Functor* functor) : functor_(functor) {}
	virtual ~Rat43NumericDiffForward() {}
	virtual bool Evaluate(double const* const* parameters,
	double* residuals,
	double** jacobians) const {
	functor_(parameters[0], residuals);
	if (!jacobians) return true;
	double* jacobian = jacobians[0];
	if (!jacobian) return true;

	const double f = residuals[0];
	double parameters_plus_h[4];
	for (int i = 0; i < 4; ++i) {
	std::copy(parameters, parameters + 4, parameters_plus_h);
	const double kRelativeStepSize = 1e-6;
	const double h = std::abs(parameters[i]) * kRelativeStepSize;
	parameters_plus_h[i] += h;
	double f_plus;
	functor_(parameters_plus_h, &f_plus);
	jacobian[i] = (f_plus - f) / h;
	}
	return true;
	}

	private:
	std::unique_ptr<Rat43Functor> functor_;
	};


	Note the choice of step size :math:`h` in the above code, instead of
	an absolute step size which is the same for all parameters, we use a
	relative step size of :math:`\text{kRelativeStepSize} = 10^{-6}`. This
	gives better derivative estimates than an absolute step size [#f2]_
	[#f3]_. This choice of step size only works for parameter values that
	are not close to zero. So the actual implementation of
	:class:`NumericDiffCostFunction`, uses a more complex step size
	selection logic, where close to zero, it switches to a fixed step
	size.


	Central Differences
	===================

	:math:`O(h)` error in the Forward Difference formula is okay but not
	great. A better method is to use the Central Difference formula:

	.. math::
	Df(x) \approx \frac{f(x + h) - f(x - h)}{2h}

	Notice that if the value of :math:`f(x)` is known, the Forward
	Difference formula only requires one extra evaluation, but the Central
	Difference formula requires two evaluations, making it twice as
	expensive. So is the extra evaluation worth it?

	To answer this question, we again compute the error of approximation
	in the central difference formula:

	.. math::
	\begin{align}
	f(x + h) &= f(x) + h Df(x) + \frac{h^2}{2!}
	D^2f(x) + \frac{h^3}{3!} D^3f(x) + \frac{h^4}{4!} D^4f(x) + \cdots\\
	f(x - h) &= f(x) - h Df(x) + \frac{h^2}{2!}
	D^2f(x) - \frac{h^3}{3!} D^3f(c_2) + \frac{h^4}{4!} D^4f(x) +
	\cdots\\
	Df(x) & = \frac{f(x + h) - f(x - h)}{2h} + \frac{h^2}{3!}
	D^3f(x) + \frac{h^4}{5!}
	D^5f(x) + \cdots \\
	Df(x) & = \frac{f(x + h) - f(x - h)}{2h} + O(h^2)
	\end{align}

	The error of the Central Difference formula is :math:`O(h^2)`, i.e.,
	the error goes down quadratically whereas the error in the Forward
	Difference formula only goes down linearly.

	Using central differences instead of forward differences in Ceres
	Solver is a simple matter of changing a template argument to
	:class:`NumericDiffCostFunction` as follows:

	.. code-block:: c++

	CostFunction* cost_function =
	new NumericDiffCostFunction<Rat43CostFunctor, CENTRAL, 1, 4>(
	new Rat43CostFunctor(x, y));

	But what do these differences in the error mean in practice? To see
	this, consider the problem of evaluating the derivative of the
	univariate function

	.. math::
	f(x) = \frac{e^x}{\sin x - x^2},

	at :math:`x = 1.0`.

	It is easy to determine that :math:`Df(1.0) =
	140.73773557129658`. Using this value as reference, we can now compute
	the relative error in the forward and central difference formulae as a
	function of the absolute step size and plot them.

	.. figure:: forward_central_error.png
	:figwidth: 100%
	:align: center

	Reading the graph from right to left, a number of things stand out in
	the above graph:

	1. The graph for both formulae have two distinct regions. At first,
	starting from a large value of :math:`h` the error goes down as
	the effect of truncating the Taylor series dominates, but as the
	value of :math:`h` continues to decrease, the error starts
	increasing again as roundoff error starts to dominate the
	computation. So we cannot just keep on reducing the value of
	:math:`h` to get better estimates of :math:`Df`. The fact that we
	are using finite precision arithmetic becomes a limiting factor.
	2. Forward Difference formula is not a great method for evaluating
	derivatives. Central Difference formula converges much more
	quickly to a more accurate estimate of the derivative with
	decreasing step size. So unless the evaluation of :math:`f(x)` is
	so expensive that you absolutely cannot afford the extra
	evaluation required by central differences, **do not use the
	Forward Difference formula**.
	3. Neither formula works well for a poorly chosen value of :math:`h`.


	Ridders' Method
	===============

	So, can we get better estimates of :math:`Df` without requiring such
	small values of :math:`h` that we start hitting floating point
	roundoff errors?

	One possible approach is to find a method whose error goes down faster
	than :math:`O(h^2)`. This can be done by applying `Richardson
	Extrapolation
	<https://en.wikipedia.org/wiki/Richardson_extrapolation>`_ to the
	problem of differentiation. This is also known as Ridders' Method
	[Ridders]_.

	Let us recall, the error in the central differences formula.

	.. math::
	\begin{align}
	Df(x) & = \frac{f(x + h) - f(x - h)}{2h} + \frac{h^2}{3!}
	D^3f(x) + \frac{h^4}{5!}
	D^5f(x) + \cdots\\
	& = \frac{f(x + h) - f(x - h)}{2h} + K_2 h^2 + K_4 h^4 + \cdots
	\end{align}

	The key thing to note here is that the terms :math:`K_2, K_4, ...`
	are indepdendent of :math:`h` and only depend on :math:`x`.

	Let us now define:

	.. math::

	A(1, m) = \frac{f(x + h/2^{m-1}) - f(x - h/2^{m-1})}{2h/2^{m-1}}.

	Then observe that

	.. math::

	Df(x) = A(1,1) + K_2 h^2 + K_4 h^4 + \cdots

	and

	.. math::

	Df(x) = A(1, 2) + K_2 (h/2)^2 + K_4 (h/2)^4 + \cdots

	Here we have halved the step size to obtain a second central
	differences estimate of :math:`Df(x)`. Combining these two estimates,
	we get:

	.. math::

	Df(x) = \frac{4 A(1, 2) - A(1,1)}{4 - 1} + O(h^4)

	which is an approximation of :math:`Df(x)` with truncation error that
	goes down as :math:`O(h^4)`. But we do not have to stop here. We can
	iterate this process to obtain even more accurate estimates as
	follows:

	.. math::

	A(n, m) = \begin{cases}
	\frac{\displaystyle f(x + h/2^{m-1}) - f(x -
	h/2^{m-1})}{\displaystyle 2h/2^{m-1}} & n = 1 \\
	\frac{\displaystyle 4^{n-1} A(n - 1, m + 1) - A(n - 1, m)}{\displaystyle 4^{n-1} - 1} & n > 1
	\end{cases}

	It is straightforward to show that the approximation error in
	:math:`A(n, 1)` is :math:`O(h^{2n})`. To see how the above formula can
	be implemented in practice to compute :math:`A(n,1)` it is helpful to
	structure the computation as the following tableau:

	.. math::
	\begin{array}{ccccc}
	A(1,1) & A(1, 2) & A(1, 3) & A(1, 4) & \cdots\\
	& A(2, 1) & A(2, 2) & A(2, 3) & \cdots\\
	& & A(3, 1) & A(3, 2) & \cdots\\
	& & & A(4, 1) & \cdots \\
	& & & & \ddots
	\end{array}

	So, to compute :math:`A(n, 1)` for increasing values of :math:`n` we
	move from the left to the right, computing one column at a
	time. Assuming that the primary cost here is the evaluation of the
	function :math:`f(x)`, the cost of computing a new column of the above
	tableau is two function evaluations. Since the cost of evaluating
	:math:`A(1, n)`, requires evaluating the central difference formula
	for step size of :math:`2^{1-n}h`

	Applying this method to :math:`f(x) = \frac{e^x}{\sin x - x^2}`
	starting with a fairly large step size :math:`h = 0.01`, we get:

	.. math::
	\begin{array}{rrrrr}
	141.678097131 &140.971663667 &140.796145400 &140.752333523 &140.741384778\\
	&140.736185846 &140.737639311 &140.737729564 &140.737735196\\
	& &140.737736209 &140.737735581 &140.737735571\\
	& & &140.737735571 &140.737735571\\
	& & & &140.737735571\\
	\end{array}

	Compared to the correct value :math:`Df(1.0) = 140.73773557129658`,
	:math:`A(5, 1)` has a relative error of :math:`10^{-13}`. For
	comparison, the relative error for the central difference formula with
	the same stepsize (:math:`0.01/2^4 = 0.000625`) is :math:`10^{-5}`.

	The above tableau is the basis of Ridders' method for numeric
	differentiation. The full implementation is an adaptive scheme that
	tracks its own estimation error and stops automatically when the
	desired precision is reached. Of course it is more expensive than the
	forward and central difference formulae, but is also significantly
	more robust and accurate.

	Using Ridder's method instead of forward or central differences in
	Ceres is again a simple matter of changing a template argument to
	:class:`NumericDiffCostFunction` as follows:

	.. code-block:: c++

	CostFunction* cost_function =
	new NumericDiffCostFunction<Rat43CostFunctor, RIDDERS, 1, 4>(
	new Rat43CostFunctor(x, y));

	The following graph shows the relative error of the three methods as a
	function of the absolute step size. For Ridders's method we assume
	that the step size for evaluating :math:`A(n,1)` is :math:`2^{1-n}h`.

	.. figure:: forward_central_ridders_error.png
	:figwidth: 100%
	:align: center

	Using the 10 function evaluations that are needed to compute
	:math:`A(5,1)` we are able to approximate :math:`Df(1.0)` about a 1000
	times better than the best central differences estimate. To put these
	numbers in perspective, machine epsilon for double precision
	arithmetic is :math:`\approx 2.22 \times 10^{-16}`.

	Going back to ``Rat43``, let us also look at the runtime cost of the
	various methods for computing numeric derivatives.

	========================== =========
	CostFunction Time (ns)
	========================== =========
	Rat43Analytic 255
	Rat43AnalyticOptimized 92
	Rat43NumericDiffForward 262
	Rat43NumericDiffCentral 517
	Rat43NumericDiffRidders 3760
	========================== =========

	As expected, Central Differences is about twice as expensive as
	Forward Differences and the remarkable accuracy improvements of
	Ridders' method cost an order of magnitude more runtime.

	Recommendations
	===============

	Numeric differentiation should be used when you cannot compute the
	derivatives either analytically or using automatic differention. This
	is usually the case when you are calling an external library or
	function whose analytic form you do not know or even if you do, you
	are not in a position to re-write it in a manner required to use
	:ref:`chapter-automatic_derivatives`.


	When using numeric differentiation, use at least Central Differences,
	and if execution time is not a concern or the objective function is
	such that determining a good static relative step size is hard,
	Ridders' method is recommended.

	.. rubric:: Footnotes

	.. [#f2] `Numerical Differentiation
	<https://en.wikipedia.org/wiki/Numerical_differentiation#Practical_considerations_using_floating_point_arithmetic>`_
	.. [#f3] [Press]_ Numerical Recipes, Section 5.7
	.. [#f4] In asymptotic error analysis, an error of :math:`O(h^k)`
	means that the absolute-value of the error is at most some
	constant times :math:`h^k` when :math:`h` is close enough to
	:math:`0`.