How to evaluate numerically $e^{-4 \cdotp 10^{35}}$ in the form $0,a_1a_2...\times 10^{-n}$
N[E^(-4*10^35)]
General::unfl: Underflow occurred in computation. >>
Underflow[]
doesn't work
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$\begingroup$ How many digits would you be satisfied with? $\endgroup$– Michael E2Feb 25, 2016 at 0:11
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2 Answers
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n = 35;
{pow, logdec} = {#[[1]][[;; #[[2]]]], #[[1]][[#[[2]] + 1 ;;]]} &@
RealDigits[N[Log[10, E^(-4 10^n)], 200]];
So there are
FromDigits@pow
(* 173717792761300731060451567566642032 *)
zeroes,followed by
dec = N[1/10^(FromDigits[logdec] 10^-Length[logdec]), 50]
(* 0.12084848148616706326389685430970719629910021783715 *)
... or may be I'm off by one :)
Checking with smaller numbers:
n=4;
{pow, logdec} = {#[[1]][[;; #[[2]]]], #[[1]][[#[[2]] + 1 ;;]]} &@
RealDigits[N[Log[10, E^(-4 10^n)], 200]];
FromDigits@pow
(* 17371 *)
dec = N[1/10^(FromDigits[logdec] 10^-Length[logdec]), 50]
0.16623553671520518223181112083319039297273582477854
E^(-4 10^n) // N
(* 1.66235536715*10^-17372 *)
answered Feb 25, 2016 at 1:06
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$\begingroup$ Somewhat simpler:
exp = -4*10^35; tenExp = Floor[exp/Log[10]]; rem = exp - tenExp Log[10]; N[Exp[rem], 50]*Inactivate@Power[10, tenExp]Also,remcan be written asMod[exp, Log[10]]but I wanted to make sure (and make it obvious) that the integer and fractional part add up right. $\endgroup$– The VeeMar 19, 2016 at 23:59
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$MinMachineNumber
$\text{2.2250738585072014$\grave{ }$*${}^{\wedge}$-308}$
and as Karsten 7 points out:
Log[$MinNumber]
$-3.121657384082590881601471993929\times 10^{15}$
Use Exp[x] == 10^(x/Log[10])
and Exp[a b] = Exp[a] + Exp[b] to find:
$0.0183156\ 10^{-10^{35}}$
answered Feb 24, 2016 at 23:22
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1$\begingroup$ That's for machine-precision.
$MinNumberis more relevant here. $\endgroup$– Karsten7Feb 24, 2016 at 23:25