RE: RE:Working Precision

*To*: mathgroup at smc.vnet.net*Subject*: [mg24047] RE: [mg23928] RE:Working Precision*From*: "Ersek, Ted R" <ErsekTR at navair.navy.mil>*Date*: Wed, 21 Jun 2000 02:20:07 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

I want to clear up some things. The only problem I have with the examples based on x=1.11111111111111111111; Do[x=2*x-x, {100}]; is that a message is never posted warning that the results are not reliable due to lose of precision in the value of (x). I suspect many users would want to know when arbitrary precision arithmetic gives results with very low precision. ------------------------ I also thought members of the MathGroup would be interested to know that a definition for f[x] apply to a wide range of values when (x) is a numeric value with precision near zero and even more so when (x) has negative precision. I didn't mean to suggest that this is a bug. I agree the results are correct. ------------------------- Now lets take another look at a new variation of my second example using the default setting of $MinPrecision. In[1]:= lst={Pi, Sqrt[2], 11, -10.749999999999999`4.28, -0.25`20}; b1=Plus@@lst Out[2]= 4.55580621596289 In[3]:= Precision[b1] Out[3]= 15 How could we know the value of (b1) with 15 decimal places of precision when the value of -10.7499 .... is only good to 4.28 decimal places? (*------------------------------*) I suspect it is possible to write a Plus routine that would give the answer I expect in cases like the one above (without needing $MinPrecision <0 ). I suggest that when evaluating Plus[args__] and all of the following are True 1) None of (args) are machine precision 2) Each args is numeric 3) One or more of (args) is arbitrary precision 4) Length[{args}]>2 then we should be sure not to add terms with opposite sign until everything else is added together. The method below seems to work well. In[4]:= quad1 = Select[lst, (0 <= Arg[#1] < Pi/2)& ]; quad2 = Select[lst, (Pi/2 <= Arg[#1] < Pi)& ]; quad3 = Select[lst, (Pi <= Arg[#1] < 3*Pi/2)& ]; quad4 = Complement[lst, quad1, quad2, quad3]; a1=Complement[Accuracy/@lst,{Infinity}]; accur=Min[Max[a1],Min[a1]+100]; q1=quad1/.z_?(Accuracy[#]===Infinity&):>SetAccuracy[z,accur]; q2=quad2/.z_?(Accuracy[#]===Infinity&):>SetAccuracy[z,accur]; q3=quad3/.z_?(Accuracy[#]===Infinity&):>SetAccuracy[z,accur]; q4=quad4/.z_?(Accuracy[#]===Infinity&):>SetAccuracy[z,accur]; b1=Plus@@{Plus@@q1,Plus@@q2,Plus@@q3,Plus@@q4} Out[14]= 4.556 In[15]:= Precision[b1] Out[15]= 4 Of course I would like to see something like this built-in to the kernel's (Plus) routine where it would be transparent to the user and run much faster. Perhaps I overlooked something when I came up with the above method. In any case I think this is a step in the right direction. -------------------- Regards, Ted Ersek Download Mathematica tips, tricks at http://www.verbeia.com/mathematica/tips/Tricks.html