Re: Precision in Mathematica 6

*To*: mathgroup at smc.vnet.net*Subject*: [mg92281] Re: [mg92265] Precision in Mathematica 6*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Fri, 26 Sep 2008 06:24:40 -0400 (EDT)*Reply-to*: hanlonr at cox.net

int = Integrate[Sin[x]/x, {x, 0, t}] Si(t) soln = FindRoot[int == 1, {t, 1}] {t->1.06484} The output defaults to displaying six digits; however, there are actually many more soln // InputForm {t -> 1.0648397255365585} int - 1 /. soln -1.77636*10^-15 If you need more precision than machine precision soln2 = FindRoot[int == 1, {t, 1}, WorkingPrecision -> 20] {t->1.0648397255365608896} int - 1 /. soln2 0.*10^-19 N is used to force numerical evaluation of an exact number SinIntegral[1] Si(1) SinIntegral[1] // N 0.946083 % // InputForm 0.946083070367183 If you want to change the display of numbers, look up NumberForm Bob Hanlon ---- CaveSnow <cavesnow at gmail.com> wrote: ============= Hey guys! I have just started using Mathematica (I'm using v. 6) and I hate to admit that there are some strange things happening. I have already tried for quit a lot to figure things out but everything works in a strange way. this is the things I written in my notebook FindRoot[\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(t\)] FractionBox[\(Sin[x]\), \(x\)] \[DifferentialD]x\) == 1, {t, 1}] N[t, 10] /. % \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(t\)] FractionBox[\(Sin[x]\), \(x\)] \[DifferentialD]x\) - 1 /. %% \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1.0648397255365585`\)] FractionBox[\(Sin[x]\), \(x\)] \[DifferentialD]x\) - 1 \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1.06484\)] FractionBox[\(Sin[x]\), \(x\)] \[DifferentialD]x\) - 1 In other words I used FindRoot to find the t that makes the definite integral from 0 to t of sinx/x be 1. As a result I got a certain rule, that had a small amount of digits (only 6 of them). Then I issued the second command ti get more digits from the result. But the result, even if I asked 10 digits remained the same 1.06484. Then of course I supposed that the problem was that it was actually the findroot procedure that calculated only those digits. So then I did this: first I actually calculated the value of the integral using the /.%% substitution, thus applying the rule, and it gave me one answer. Then I triedo to copy and paste the value displayed in the rule but when I did the paste command it had me written many more digits than the original 6, and the answer was a little different (it actually displayed 17 digits). And after that I tried to evaluate the integral with the originally displayed 1.06484 value and the result of the integral was much more off than in the other two cases. The only thing that now I can say that actually the command N[t,10]/.% actually dowsn't write me those aditional digits and neither does the sequence of commands t/.% N[%,10]. What the Hell!!! I do not understand why these strange things happen. Mathematica is absolutely unpredictable! Anyway I could find no notice on why using N doesn't work in this case in the documentation. Is this a bug? Anyway if it is hard to read the commands in the way I written them I can send you the nb file. Please help! I already encountered this seemingly random behavior of mathematica in other occasions and not having to do only with the precision of numbers. It is rather annoying, and I am getting rather fed up with this. It may as well be my fault but there is no documentation to understand why. Thanks in advance -- Bob Hanlon

**Re: Precision in Mathematica 6**

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**Re: Precision in Mathematica 6**

**Re: Precision in Mathematica 6**