Seems like the DerivativeCompiler returns NaN.
IMHO it should return 0.
Is this worthy of an issue?
Thanks,
-Ajo
Hi!
IMHO, 0^x should equal 0. But note please, that 0^0 is not define, so
it should be NaN
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TIA, Rodion
Ajo Fod писал 23.08.2013 05:17 PM:
Seems like the DerivativeCompiler returns NaN.
IMHO it should return 0.
Is this worthy of an issue?
Thanks,
-Ajo
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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
-
To
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++) {
Search the archives for discussion of 0^0.
Phil
On 8/23/13 7:17 AM, Ajo Fod wrote:
> Seems like the DerivativeCompiler returns NaN.
>
> IMHO it should return 0.
>
> Is this worthy of an issue?
>
> Thanks,
> -Ajo
>
-
To unsubscr
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
Hello,
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");
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);
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;
>> Deriv
Are you saying patched the code? Can you provide the link?
-Ajo
On Sun, Aug 25, 2013 at 1:20 PM, Luc Maisonobe 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 deriv
Ajo Fod 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 metho
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%B4pi
On a side note. Given a derivative structure ds. Wouldn't it be nice to
generate a constant derivative structure with something like:
ds.getConstant(dobule value);
Currently I"m doing something like:
new DerivativeStructure(length, order, value); ... seesm more verbose than
necessary when I have
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 ru
Le 26/08/2013 22:37, Ajo Fod a écrit :
> On a side note. Given a derivative structure ds. Wouldn't it be nice to
> generate a constant derivative structure with something like:
>
> ds.getConstant(dobule value);
> Currently I"m doing something like:
> new DerivativeStructure(length, order, value);
Thanks for the constant structure.
No. The limit value when x->0+ is 1, not O.
I agree with this. I was just going for the derivatives = 0.
> 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 =
Hi Ajo,
Le 27/08/2013 16:44, Ajo Fod a écrit :
> Thanks for the constant structure.
>
> No. The limit value when x->0+ is 1, not O.
>
> I agree with this. I was just going for the derivatives = 0.
>
>
>> The nth derivative of a^x can be computed analytically as ln(a)^n a^x,
>> so the initial s
To define things precisely:
y = f(a,x) = |a|^x
Can we agree that:
df(a,x)/dx -> 0 when a->0 and x > 0 :[ NOTE: x > 0]
If this is acceptable, we get this very useful property that df (a,x)/dx is
defined and continuous for all a provided x>0 because we use the modulus of
a in the function definitio
Hi Ajo,
Le 28/08/2013 16:56, Ajo Fod a écrit :
> To define things precisely:
> y = f(a,x) = |a|^x
>
> Can we agree that:
> df(a,x)/dx -> 0 when a->0 and x > 0 :[ NOTE: x > 0]
Yes, of course, it is perfectly true.
>
> If this is acceptable, we get this very useful property that df (a,x)/dx is
>
Its a=0 that bothers me. x > 0 in my case.
In the code I use, the DerivativeStructure evaluates to NaN for a=0 when x
> 0 . I think we agree that in this condition the derivative should
evaluate to 0.
Perhaps I wrote something to mislead you on this detail.
-Ajo
On Wed, Aug 28, 2013 at 10:36
Le 28/08/2013 19:59, Ajo Fod a écrit :
> Its a=0 that bothers me. x > 0 in my case.
Then everything should be OK with the current code, which reads:
final double[] function = new double[1 + order];
if (a == 0) {
if (operand[operandOffset] == 0) {
functi
That works very well.
I missed the default value of 0 earlier
Thanks,
-Ajo
On Wed, Aug 28, 2013 at 11:05 AM, Luc Maisonobe wrote:
> Le 28/08/2013 19:59, Ajo Fod a écrit :
> > Its a=0 that bothers me. x > 0 in my case.
>
> Then everything should be OK with the current code, which reads:
>
>
BTW Luc. This derivatives structure is simply a godsent for big
optimization problems.
Cheers,
-Ajo.
On Wed, Aug 28, 2013 at 11:13 AM, Ajo Fod wrote:
> That works very well.
>
> I missed the default value of 0 earlier
>
> Thanks,
> -Ajo
>
>
> On Wed, Aug 28, 2013 at 11:05 AM, Luc Maisonobe wro
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