Re: locating overlow/underflow

*To*: mathgroup at smc.vnet.net*Subject*: [mg112247] Re: locating overlow/underflow*From*: Leslaw Bieniasz <nbbienia at cyf-kr.edu.pl>*Date*: Mon, 6 Sep 2010 06:36:24 -0400 (EDT)*References*: <i5sucr$pej$1@smc.vnet.net>

Hi Bill, Thanks for your comments. I have reduced the range of y, and I am now obtaining results without any error messages. However, this exercise raises my doubts whether the desired "arbitrary precision" is indeed reached, especially when y approaches 100. The program output yields infinity for the number of valid significant digits, but how many digits can I really trust if I declare 70 digits? Shouldn't the output contain warnings if the number of valid digits is smaller than 70? I don't perhaps need as many as 70, something like 30 would be enough, but I need to be sure that I really have that accuracy. I don't know how all that works, but I can imagine MATHEMATICA should automatically allocate appropriately more bytes to the floating point representations used, if the current number of bytes is insufficient. There are algorithms that can do that, I believe, like interval arithmetics? How can I check whether the accuracy is indeed as I want it to be? Leslaw On Sat, 4 Sep 2010, Bill Rowe wrote: > On 9/3/10 at 6:10 AM, nbbienia at cyf-kr.edu.pl (Leslaw Bieniasz) wrote: > >> I am not very much familiar with MATHEMATICA, but I have to tabulate >> a certain mathematical expression in a possibly large range of >> independent variable, and with the highest possible number of >> accurate digits. I have a notebook file that does the job (included >> below in text form, together with the results) but during the cell >> evaluation I obtain error messages stating that underflow or >> overflow occurred. My question is: is there any way to rewrite the >> code in such a way so that I can obtain an information where exactly >> (that is for which values of independent variable) the errors >> actually occur? Or, is there any way to rewrite the code or change >> the program settings in such a way so that the errors do not occur >> at all? I am using MATHEMATICA 6.0. > >> In[1]:= Table[{N[y], >> SetPrecision[ 2 ((y - 1) Exp[y] Erfc[Sqrt[y]] + (y/3 - 1) 2 >> Sqrt[y/Pi] + 1)/(y y), 70], Precision[ 2 ((y - 1) Exp[y] >> Erfc[Sqrt[y]] + (y/3 - 1) 2 Sqrt[y/Pi] + 1)/(y y)] >> }, >> {y, { 1/10^19, ..., 5 10^15 >> }}] > > Where to start? > > SetPrecision simply doesn't do what you seem to think it does. > Using the first value for y (1/10^19) I can get 50 digit > precision by doing: > > In[18]:= f = > 2 ((y - 1) Exp[y] Erfc[Sqrt[y]] + (y/3 - 1) 2 Sqrt[y/Pi] + > 1)/(y y); > > In[19]:= N[f /. y -> 10^-19 // Simplify, 50] > > Out[19]= 0.99999999971454014148222328832683872045949424371145 > > There is no need to compute this number twice to get the > precision since doing N[expr, 50] will always return a number > with a precision of 50 if all of the values in expr are exact values. > > Strictly speaking, I did not need to simplify your expression > before converting it to a number with 50 digit precision. But > experience indicates this is a good idea for a complex > expression with extreme exact values. > > While this approach will work for several values of y, it will > not work for all values. > > Notice for y = 100 > > In[44]:= Log[10, Exp[{100, 200, 500}]] // N > > Out[44]= {43.4294,86.8589,217.147} > > and > > In[45]:= Log[10, Erfc[Sqrt[{100, 200, 500}]]] // N > > Out[45]= {-44.6802,-88.2591,-218.746} > > That is for y values of about 100, the first term > > (y - 1) Exp[y] Erfc[Sqrt[y]] > > will be about 1/10 the size of the second term > > (y/3 - 1) 2 Sqrt[y/Pi] + 1) > > But the next decade gives > > In[46]:= Log[10, Exp[{1000, 2000, 5000}]] // N > > Out[46]= {434.294,868.589,2171.47} > > and > > In[47]:= Log[10, Erfc[Sqrt[{1000, 2000, 5000}]]] // N > > Out[47]= {-436.043,-870.488,-2173.57} > > That is the first term decreases about one order of magnitude > while the second one increase by one order of magnitude. > Clearly, the first term can be dropped well before you get to > the maximum value you have for y without any significant error. > > The bottom line is if you truly need 70 digits of precision for > this expression with a value of y like 5x10^15 you will need to > have a far more sophisticated approach than what you are using. > > >