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From Luc Maisonobe <Luc.Maison...@free.fr>
Subject Re: [MATH] What is the derivative of 0^x
Date Tue, 27 Aug 2013 11:33:45 GMT
Le 26/08/2013 17:23, Ajo Fod a écrit :
> With regards to what is happening in DsCompiler.pow():
> IMHO, when a==0 and x>=0 the function is well behaved because log|a| -> Inf
> slower than a^x -> 0. I got to this by simulation.
> One could probably get to something more conclusive using L'Hopital rule :
> http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule.
> There is one about xlog(x) behavior as x->0+.
>
> So, I propose this:
>
>         if (a == 0) {
>             if (operand[operandOffset] >= 0) {
>                 for (int i = 0; i < function.length; ++i) {
>                     function[i] = 0;

No. The limit value when x->0+ is 1, not O.
This can be seen by drawing the graphs for a^x for x between 0 and 2 for
example, and several values of a, say a=3, a=3, a=1, a=0.5, a=0.1,
a=0.01, a=0.001.

When a > 1, the graph starts at y=1 with a positive slope and increases
exponentially to infinity.

When a = 1, the graph is the horizontal line y=1.

When 0 < a < 1, the graph starts at y=1 with a negative slope and
decreases to reach asymptitically the line y=0. The smallest a is, the
more negative the initial slope is. For very small values of a, the
curve dives very quickly from its initial value y=1.

The nth derivative of a^x can be computed analytically as ln(a)^n a^x,
so the initial slope at x=0 is simply ln(a), positive for a > 1, zero
for a = 1, negative for 0 < a < 1 with a limit at -inifnity when a -> 0+.

The limit curve corresponding to a = 0 is therefore a singular function
with f(0) = 1 and f(x) = 0 for all x > 0. The fact f(0) = 1 and not 0 is
consistent with the derivative being negative infinity, as by definition
the derivative is the limit of [f(0+h) - f(0)] / h when h->0+, as the
finite difference is -1/h.

>                 }
>             }else{
>                 for (int i = 0; i < function.length; ++i) {
>                     function[i] = Double.NaN;
>                 }

This alternative case is a good improvement, thanks for it. I forgot to
handle negative cases properly. I have therefore changed the code
(committed as r1517788) with this improvement, together with several
test cases.

>             }
>         } else {
>
>
> in place of :
>
>         if (a == 0) {
>             if (operand[operandOffset] == 0) {
>                 function[0] = 1;
>                 double infinity = Double.POSITIVE_INFINITY;
>                 for (int i = 1; i < function.length; ++i) {
>                     infinity = -infinity;
>                     function[i] = infinity;
>                 }
>             }
>         } else {
>
>
> PS: I think you made a change to DSCompiler.pow too. If so, what happens
> when a=0 & x!=0  in that function?

No, I didn't change the other signatures of the pow function. So the
value should be OK (i.e. 1) but all derivatives, including the first
one, should be NaN. What the new function brings is a correct negetive
infinity first derivative at singularity point, better accuracy for
non-singular points, and possibly faster computation.

best regards,
Luc

>
>
> On Mon, Aug 26, 2013 at 12:38 AM, Luc Maisonobe <luc@spaceroots.org> wrote:
>
>>
>>
>>
>> Ajo Fod <ajo.fod@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
>>
>> 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 <luc@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.Maisonobe@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
>>> <gilles@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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