On Tue, Dec 15, 2009 at 09:49:28AM +0100, Martin Maechler wrote:
> lgamma(x) and lfactorial(x) are defined to return
>
> ln|Gamma(x)| {= log(abs(gamma(x)))} or ln|Gamma(x+1)| respectively.
>
> Unfortunately, we haven't chosen the analogous definition for
> lchoose().
>
> So, currently
>
> > lchoose(1/2, 1:10)
> [1] -0.6931472 -2.0794415 NaN -3.2425924 NaN -3.8869494
> [7] NaN -4.3357508 NaN -4.6805913
> Warning message:
> In lchoose(n, k) : NaNs produced
> >
>
> which (the NaN's) is not particularly useful.
> (I have use case where I really have to workaround those NaNs.)
>
> I herebey propose to *amend* the definition of lchoose() such
> that it behaves analogously to lgamma() and lfactorial(),
> i.e., to return
>
> log(abs(choose(.,.))
>
> Your comments are welcome.
> Martin Maechler, ETH Zurich
I do not have a strong opinion, whether lchoose() should be log(choose(.,.))
or log(abs(choose(.,.))), although i would slightly prefere the latter (with
abs()).
One of the reasons is that this would simplify the code of the function
lchoose() in src/nmath/choose.c. It produces the NA by explicit testing
the sign, so this test could be simply removed.
Let me also point out that the current implementation seems to contain
a bug, which causes that it is neither log(choose(.,.)) nor
log(abs(choose(.,.))).
x <- choose(1/2, 1:10)
y <- lchoose(1/2, 1:10)
cbind(choose=x, lchoose=y, log.choose=log(abs(x)))
choose lchoose log.choose
# [1,] 0.500000000 -0.6931472 -0.6931472
# [2,] -0.125000000 -2.0794415 -2.0794415
# [3,] 0.062500000 NaN -2.7725887
# [4,] -0.039062500 -3.2425924 -3.2425924
# [5,] 0.027343750 NaN -3.5992673
# [6,] -0.020507812 -3.8869494 -3.8869494
# [7,] 0.016113281 NaN -4.1281114
# [8,] -0.013092041 -4.3357508 -4.3357508
# [9,] 0.010910034 NaN -4.5180723
# [10,] -0.009273529 -4.6805913 -4.6805913
So, besides x[1], we get NA for those cases, when choose() is positive.
The reason seems to be the test at line 89 of src/nmath/choose.c, which is
if (fmod(floor(n-k+1), 2.) == 0) /* choose() < 0 */
return ML_NAN;
but it should be negated. I tested it using
if (fmod(floor(n-k+1), 2.) != 0) /* choose() < 0 */
return ML_NAN;
when we get
choose lchoose log.choose
[1,] 0.500000000 -0.6931472 -0.6931472
[2,] -0.125000000 NaN -2.0794415
[3,] 0.062500000 -2.7725887 -2.7725887
[4,] -0.039062500 NaN -3.2425924
[5,] 0.027343750 -3.5992673 -3.5992673
[6,] -0.020507812 NaN -3.8869494
[7,] 0.016113281 -4.1281114 -4.1281114
[8,] -0.013092041 NaN -4.3357508
[9,] 0.010910034 -4.5180723 -4.5180723
[10,] -0.009273529 NaN -4.6805913
Petr Savicky.
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