Ajo Fod <ajo....@gmail.com> a écrit :
>Are you saying patched the code? Can you provide the link?
I committed it in the development version. You just have to update your checked
out copy from either the official
Apache subversion repository or the git mirror we talked about in a previous
thread.
The new method is a static one called pow and taking a and x as arguments and
returning a^x. Not to
Be confused with the non-static methods that take only the power as argument
(either int, double or
DerivativeStructure) and use the instance as the base to apply power on.
Best regards,
Luc
>
>-Ajo
>
>
>On Sun, Aug 25, 2013 at 1:20 PM, Luc Maisonobe <l...@spaceroots.org>
>wrote:
>
>> Le 24/08/2013 11:24, Luc Maisonobe a écrit :
>> > Le 23/08/2013 19:20, Ajo Fod a écrit :
>> >> Hello,
>> >
>> > Hi Ajo,
>> >
>> >>
>> >> This shows one way of interpreting the derivative for strictly +ve
>> numbers.
>> >>
>> >> public static void main(final String[] args) {
>> >> final double x = 1d;
>> >> DerivativeStructure dsA = new DerivativeStructure(1, 1, 0,
>x);
>> >> System.out.println("Derivative of |a|^x wrt x");
>> >> for (int p = 10; p < 21; p++) {
>> >> double a;
>> >> if (p < 20) {
>> >> a = 1d / Math.pow(2d, p);
>> >> } else {
>> >> a = 0d;
>> >> }
>> >> final DerivativeStructure a_ds = new
>DerivativeStructure(1,
>> 1,
>> >> a);
>> >> final DerivativeStructure out = a_ds.pow(dsA);
>> >> final double calc = (Math.pow(a, x + EPS) -
>Math.pow(a, x))
>> /
>> >> EPS;
>> >> System.out.format("Derivative@%f=%f %f\n", a, calc,
>> >> out.getPartialDerivative(new int[]{1}));
>> >> }
>> >> }
>> >>
>> >> At this point I"m explicitly substituting the rule that
>> derivative(|a|^x) =
>> >> 0 for |a|=0.
>> >
>> > Yes, but this fails for x = 0, as the limit of the finite
>difference is
>> > -infinity and not 0.
>> >
>> > You can build your own function which explicitly assumes a is
>constant
>> > and takes care of special values as follows:
>> >
>> > public static DerivativeStructure aToX(final double a,
>> > final DerivativeStructure
>x) {
>> > final double lnA = (a == 0 && x.getValue() == 0) ?
>> > Double.NEGATIVE_INFINITY :
>> > FastMath.log(a);
>> > final double[] function = new double[1 + x.getOrder()];
>> > function[0] = FastMath.pow(a, x.getValue());
>> > for (int i = 1; i < function.length; ++i) {
>> > function[i] = lnA * function[i - 1];
>> > }
>> > return x.compose(function);
>> > }
>> >
>> > This will work and provides derivatives to any order for almost any
>> > values of a and x, including a=0, x=1 as in your exemple, but also
>> > slightly better for a=0, x=0. However, it still has an important
>> > drawback: it won't compute the n-th order derivative correctly for
>a=0,
>> > x=0 and n > 1. It will provide NaN for these higher order
>derivatives
>> > instead of +/-infinity according to parity of n.
>>
>> I have added a similar function to the DerivativeStructure class
>(with
>> some errors above corrected). The main interesting property of this
>> function is that it is more accurate that converting a to a
>> DerivativeStructure and using the general x^y function. It does its
>best
>> to handle the special case, but as written above, this does NOT work
>for
>> general combination (i.e. more than one variable or more than one
>> order). As soon as there is a combination, the derivative will
>involve
>> something like df/dx * dg/dy and as infinities and zeros are
>everywheren
>> NaN appears immediately for these partial derivatives. This cannot be
>> avoided.
>>
>> If you stay away from the singularity, the function behaves
>correctly.
>>
>> best regards,
>> Luc
>>
>> >
>> > This is a known problem that we already encountered when dealing
>with
>> > rootN. Here is an extract of a comment in the test case
>> > testRootNSingularity, where similar NaN appears instead of +/-
>infinity.
>> > The dsZero instance in the comment is simple the x parameter of the
>> > function, as a derivativeStructure with value 0.0 and depending on
>> > itself (dsZero = new DerivativeStructure(1, maxOrder, 0, 0.0)):
>> >
>> >
>> > // the following checks shows a LIMITATION of the current
>implementation
>> > // we have no way to tell dsZero is a pure linear variable x = 0
>> > // we only say: "dsZero is a structure with value = 0.0,
>> > // first derivative = 1.0, second and higher derivatives = 0.0".
>> > // Function composition rule for second derivatives is:
>> > // d2[f(g(x))]/dx2 = f''(g(x)) * [g'(x)]^2 + f'(g(x)) * g''(x)
>> > // when function f is the nth root and x = 0 we have:
>> > // f(0) = 0, f'(0) = +infinity, f''(0) = -infinity (and higher
>> > // derivatives keep switching between +infinity and -infinity)
>> > // so given that in our case dsZero represents g, we have g(x) = 0,
>> > // g'(x) = 1 and g''(x) = 0
>> > // applying the composition rules gives:
>> > // d2[f(g(x))]/dx2 = f''(g(x)) * [g'(x)]^2 + f'(g(x)) * g''(x)
>> > // = -infinity * 1^2 + +infinity * 0
>> > // = -infinity + NaN
>> > // = NaN
>> > // if we knew dsZero is really the x variable and not the identity
>> > // function applied to x, we would not have computed f'(g(x)) *
>g''(x)
>> > // and we would have found that the result was -infinity and not
>NaN
>> >
>> > Hope this helps
>> > Luc
>> >
>> >>
>> >> Thanks,
>> >> Ajo.
>> >>
>> >>
>> >>
>> >> On Fri, Aug 23, 2013 at 9:39 AM, Luc Maisonobe
><luc.maison...@free.fr
>> >wrote:
>> >>
>> >>> Hi Ajo,
>> >>>
>> >>> Le 23/08/2013 17:48, Ajo Fod a écrit :
>> >>>> Try this and I'm happy to explain if necessary:
>> >>>>
>> >>>> public class Derivative {
>> >>>>
>> >>>> public static void main(final String[] args) {
>> >>>> DerivativeStructure dsA = new DerivativeStructure(1, 1,
>0,
>> 1d);
>> >>>> System.out.println("Derivative of constant^x wrt x");
>> >>>> for (int a = -3; a < 3; a++) {
>> >>>
>> >>> We have chosen the classical definition which implies c^x is not
>> defined
>> >>> for real r and negative c.
>> >>>
>> >>> Our implementation is based on the decomposition c^r = exp(r *
>ln(c)),
>> >>> so the NaN comes from the logarithm when c <= 0.
>> >>>
>> >>> Noe also that as explained in the documentation here:
>> >>> <
>> >>>
>>
>http://commons.apache.org/proper/commons-math/userguide/analysis.html#a4.7_Differentiation
>> >>>> ,
>> >>> there are no concepts of "constants" and "variables" in this
>framework,
>> >>> so we cannot draw a line between c^r as seen as a univariate
>function
>> of
>> >>> r, or as a univariate function of c, or as a bivariate function
>of c
>> and
>> >>> r, or even as a pentavariate function of p1, p2, p3, p4, p5 with
>both c
>> >>> and r being computed elsewhere from p1...p5. So we don't make
>special
>> >>> cases for the case c = 0 for example.
>> >>>
>> >>> Does this explanation make sense to you?
>> >>>
>> >>> best regards,
>> >>> Luc
>> >>>
>> >>>
>> >>>> final DerivativeStructure a_ds = new
>> DerivativeStructure(1,
>> >>> 1,
>> >>>> a);
>> >>>> final DerivativeStructure out = a_ds.pow(dsA);
>> >>>> System.out.format("Derivative@%d=%f\n", a,
>> >>>> out.getPartialDerivative(new int[]{1}));
>> >>>> }
>> >>>> }
>> >>>> }
>> >>>>
>> >>>>
>> >>>>
>> >>>> On Fri, Aug 23, 2013 at 7:59 AM, Gilles
><gil...@harfang.homelinux.org
>> >>>> wrote:
>> >>>>
>> >>>>> On Fri, 23 Aug 2013 07:17:35 -0700, Ajo Fod wrote:
>> >>>>>
>> >>>>>> Seems like the DerivativeCompiler returns NaN.
>> >>>>>>
>> >>>>>> IMHO it should return 0.
>> >>>>>>
>> >>>>>
>> >>>>> What should be 0? And Why?
>> >>>>>
>> >>>>>
>> >>>>>
>> >>>>>> Is this worthy of an issue?
>> >>>>>>
>> >>>>>
>> >>>>> As is, no.
>> >>>>>
>> >>>>> Gilles
>> >>>>>
>> >>>>>
>> >>>>>> Thanks,
>> >>>>>> -Ajo
>> >>>>>>
>> >>>>>
>> >>>>>
>> >>>>>
>> >>>
>>
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>> >>>>>
>> >>>>
>> >>>
>> >>>
>> >>>
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>> >>>
>> >>
>> >
>> >
>> >
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