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MathGroup Archive 2006

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Re: SetPrecision vs N

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71741] Re: SetPrecision vs N
  • From: "Andrew Moylan" <andrew.j.moylan at gmail.com>
  • Date: Tue, 28 Nov 2006 06:03:54 -0500 (EST)
  • References: <200611260849.DAA14687@smc.vnet.net><ekeasn$st1$1@smc.vnet.net>

Thanks Andrzej for your clear explanation.

Andrew


On Nov 27, 8:24 pm, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote:
> On 26 Nov 2006, at 17:49, Andrew Moylan wrote:
>
>
>
> > Hi all,
>
> > Suppose I want to evaluate an expression at a given precision. What is
> > the difference between using N[expr, precision] and using
> > SetPrecision[expr, precision]?
>
> > I've noticed that SetPrecision seems to be equivalent even in such
> > situations as e.g. N[Integrate[...]] automatically calling
> > NIntegrate[...] when the integral can't be done exactly:
>
> > SetPrecision[Integrate[x^x, {x, 0, 1}], 20]
> >   and
> > N[Integrate[x^x, {x, 0, 1}], 20]
> >   both give
> > 0.78343051071213440706
>
> > Are there important differences between SetPrecision and N that I
> > should be aware of?
>
> > Cheers,
> > AndrewThey are actually completely different (though sometimes the effect
> may look the same). N[expr,p] computes expression to precision p. If
> expr is an exact expression Mathematica uses significance arithmetic
> to determine what precision it should use for the input to give you
> the output with p digits of precision. SetPrecison does nothing but
> simply changes the precision of expression to p, either by padding it
> with 0 or truncating. To see the difference just compare these two
> examples:
>
> SetPrecision[2.2, 20]
>
> 2.20000000000000017763568394002504646778`19.999999999999996
>
> 2.2 was (essentially) padded to precision of around 20.
>
> N[2.2, 20]
>
> 2.2
>
> Nothing changed because N cannot compute a machine precision number
> with extended precision.
>
> Here is a different example:
>
> x = N[Pi - 22/7, 2]
>
> -0.0012644892673496187`1.9999999999999998
>
> y = SetPrecision[Pi - 22/7, 2]
>
> 0``1.2017327387937378
>
> Precision/@{x,y}
>
> {2.,0.}
>
> Here x is the value of Pi - 22/7 computed to give you two significant
> digits, whereas y has actually precision 0, because what it does is
> this: SetPrecision[Pi, 2] - SetPrecision[22/7, 2], which for obvious
> reasons has precision 0.
> 
> Andrzej Kozlowski
> 
> Tokyo, Japan


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