Re: Re: SetPrecision vs N

*To*: mathgroup at smc.vnet.net*Subject*: [mg71742] Re: [mg71696] Re: [mg71683] SetPrecision vs N*From*: Bruce Colletti <vze269bv at verizon.net>*Date*: Tue, 28 Nov 2006 06:03:55 -0500 (EST)

Why does the code below return True? Bruce ----------------- x=N[Pi-22./7,30] y=SetPrecision[Pi-22./7,40] Precision/@{x,y} x==y -0.00126449 -0.001264489267349677703577981446869671344757 {MachinePrecision,40.} True ===================== From: Andrzej Kozlowski <akoz at mimuw.edu.pl> Date: 2006/11/27 Mon AM 03:03:57 CST To: mathgroup at smc.vnet.net Subject: [mg71742] [mg71696] Re: [mg71683] SetPrecision vs N 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, > Andrew > They 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