Re^3: Decimal Floating Point (DFP) and does Perl needs DFP?

would you mind stating the mathematically correct result (say, to six digits after decimal point)?

According to Math::Decimal128:

C:\>perl -MMath::Decimal128=":all" -le "$n1=Math::Decimal128->new('10'
+, 0);$n2=Math::Decimal128->new('1000001', -6);for(1..60000000){$n1 *=
+ $n2}print $n1;"
114197313013072744502959647597176e-5
[download]

Or, to 6 decimal places: 1141973130130727445029596475.971760 which is the same as flexvault got.

With 80-bit long doubles, the calculation came out as:

C:\>perl -MMath::LongDouble=":all" -le "$n1=Math::LongDouble->new('10'
+);$n2=Math::LongDouble->new('1.000001');for(1..60000000){$n1 *= $n2}p
+rint $n1;"
1.14197313013239314e+027
[download]

Or, printing to 28 decimal digits (which implies a level of precision that we don't have):

C:\>perl -MMath::LongDouble=":all" -le "$n1=Math::LongDouble->new('10'
+);$n2=Math::LongDouble->new('1.000001');for(1..60000000){$n1 *= $n2}p
+rint LDtoSTRP($n1, 28);"
1.141973130132393144424071168e+027
[download]

This figure is equivalent to the first one you quoted.

With __float128:

C:\>perl -MMath::Float128=":all" -le "$n1=Math::Float128->new('10');$n
+2=Math::Float128->new('1.000001');for(1..60000000){$n1 *= $n2}print $
+n1;"
1.14197313013072744502959647306684e+27
[download]

This differs from your second figure - and I don't think there's any valid reason that they should differ.
I'm inclined to think that mine is correct (surprise :-) because it matches up pretty well with the _Decimal128 calculation (where the respective precisions are reasonably close to each other).
Might you have assigned a long double or double value that got cast to a __float128, instead of assigning an actual __float128 value ?

As LanX points out, rounding is involved. The above one-liner programs all employ rounding to nearest with ties to even - the different results of course arising from the different precisions involved.

Cheers,
Rob

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Re^4: Decimal Floating Point (DFP) and does Perl needs DFP? by LanX (Saint) on Jan 19, 2015 at 01:44 UTC
> rounding is involved. Calculating 10 * 1.000001 ** 6e7 should minimize the propagation of rounding errors, if exponentiation is sufficiently well implemented. Cheers Rolf PS: Je suis Charlie!	[reply]
Re^5: Decimal Floating Point (DFP) and does Perl needs DFP? by syphilis (Archbishop) on Jan 19, 2015 at 03:48 UTC
Calculating 10 1.000001 ** 6e7 should minimize the propagation of rounding errors, if exponentiation is sufficiently well implemented.* I have found a case where that's not so for the calculation in question - namely, Windows 7, perl-5.20.0, nvtype is double. But the same perl version and configuration on Ubuntu supports your assertion. So it might just be that the Windows exponentiation (or something else there) is buggy. I haven't experimented much at all, and don't really intend to. What's the reasoning behind your assertion ? Cheers, Rob	[reply]
Re^6: Decimal Floating Point (DFP) and does Perl needs DFP? by LanX (Saint) on Jan 19, 2015 at 09:41 UTC
> What's the reasoning behind your assertion ? Exponentiation laws. E.g. `x^10 = x^8 * x^2` and `x^8 = ((x^2)^2)^2` Just because 10 is 1010 binary, hence < 42 multiplications. 6e7 should have a 26 bit representation so <52 multiplications needed. Of course there are even faster implementations which are less exact. I think that's what you are observing. edit Of course a binary calculation should be done with integer x=1000001 without fraction, dividing later thru 10^(660) should be easy enough. Cheers Rolf PS: Je suis Charlie! update proof of concept -> here	[reply] [d/l] [select]
Re^7: Decimal Floating Point (DFP) and does Perl needs DFP? by syphilis (Archbishop) on Jan 20, 2015 at 14:40 UTC
Re^8: Decimal Floating Point (DFP) and does Perl needs DFP? by LanX (Saint) on Jan 20, 2015 at 15:10 UTC
Some notes below your chosen depth have not been shown here
Re^4: Decimal Floating Point (DFP) and does Perl needs DFP? by Anonymous Monk on Jan 19, 2015 at 18:55 UTC
Right you are about the __float128 case. After rewriting the constant as `(__float128)1000001 / 1e6`, my result now matches your calculation. I'm inclined to think the bc-calculated value is accurate. But verifying that is rather laborious, for 1000001**6e7 is a number with ca 360M decimal digits.	[reply] [d/l]
Re^5: Decimal Floating Point (DFP) and does Perl needs DFP? by syphilis (Archbishop) on Jan 19, 2015 at 23:34 UTC
I'm inclined to think the bc-calculated value is accurate Yes, and the mpfr library agrees: `C:\>perl -MMath::MPFR=":mpfr" -le "Rmpfr_set_default_prec(1000000); $x +=Math::MPFR->new('1.000001'); $x *= 60000000; $x = 10; Rmpfr_out_st +r($x, 10,43,MPFR_RNDN); 1.141973130130727445029596475971653735685612e27` [download] That's calculated with a million bits of precision - and gives the same result as when calculated with a thousand bits of precision. So I think we can rest assured that that is the correct figure for 43 decimal digits. Cheers, Rob	[reply] [d/l]

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