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Using this nice method by MrWizard Global precision setting you can try $PreRead = (# /. s_String /; StringMatchQ[s, NumberString] && Precision@ToExpression@s == MachinePrecision :> s <> "`4." &); a = Table[0, {10}]; a[[1]] = 0.1834562432; Do[ a[[n]] = 1/n - 5 a[[n - 1]], {n, 2, 10} ]; and now a {0.1834562432`4., ...


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Not sure if this helps: $MaxExtraPrecision=4999 f[x_]:=-x^2-Log[Pi]/2-Log[x]-1/(2x^2)+5/(8x^4)-37/(24x^6) N[f[20],24] equals -403.569343334317232747174 N[Log[1-Erf[20]],24] equals -403.569343334104234962969 and N[Log[1-Erf[200]],24] returns Indeterminate but f[200] yields -40005.8706948090821358531 For what it's worth: Series[Log[1 - Erf[x]], ...


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I'm not sure whether this gives the correct answer, but you could use RealDigits with a high setting of $MaxExtraPrecision to get your answer. Since it is almost 1, we subtract Erf[200] from 1 to get a number close to zero and add later: Block[{$MaxExtraPrecision = 100000}, RealDigits[1 - Erf[200], 10, 10] ] (* {{4, 6, 8, 9, 3, 5, 9, 2, 9, 5}, -17374} *) ...


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How about 1 - N[1 - Erf[200], 20] (* 0.99999999999 <<17371 more 9s>> 99999999999531064070401639850196 *) In version 9: We can exploit the two argument Erf: Erf[z1] - Erf[z2] == Erf[z2, z1]. 1 - N[Erf[200, \[Infinity]], 20] (* 0.99999999999 <<17371 more 9s>> 99999999999531064070401639850196 *)



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