       Re: Using Equal with Real Numbers

• To: mathgroup at smc.vnet.net
• Subject: [mg123184] Re: Using Equal with Real Numbers
• From: Bill Rowe <readnews at sbcglobal.net>
• Date: Fri, 25 Nov 2011 04:58:29 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com

```On 11/24/11 at 6:53 AM, gtlandi at gmail.com (Gabriel Landi) wrote:

>Consider:

>In:= list1 = Range[0, 1, 0.1] Out= {0., 0.1, 0.2, 0.3,
>0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.}

>Using InputForm we see that:

>In:= list1 // InputForm

>Out//InputForm={0., 0.1, 0.2, 0.30000000000000004, 0.4, 0.5,
>0.6000000000000001, 0.7000000000000001, 0.8, 0.9, 1.}

>That is, 0.3, 0.6 and 0.7 have some round-off error.

Actually, only three of these values can be expressed in a
finite number of digits. That is:

In:= FromDigits[First@#, 2] 2^(-Length@First@# + Last@#) & /@
RealDigits[Range[0, 1, .1], 2] - Range[0, 10]/10

Out= {0,1/180143985094819840,1/90071992547409920,1/22517998136852480,1/45035996273704960,0,1/11258999068426240,3/45035996273704960,1/22517998136852480,1/45035996273704960,0}

>Now:

>In:= {MemberQ[list1, 0.6], MemberQ[list1, 0.7]} Out >{True, False}

>(This actually depends on the OS and perhaps other things). The
>point is that he recognizes 0.6 as a member of list1 but not 0.7,
>even though both have the same InputForms. This issue, as you may
>imagine, prohibits one from using functions that implicitly make use
>of =, when dealing with real numbers.

>Here is my solution:

>range[xi_, xf_, df_] := N@Rationalize@Range[xi, xf, df]

It was somewhat surprising to me this actually worked.

It seems to me a simpler solution would be to use:

MemberQ[Rationalize[list1],6/10]

or

MemberQ[Rationalize[list1], Rationalize[.6]]

I believe the internal algorithm Mathematica uses for Range[xi,
xf, df] is add df to xi and repreat adding df to the sum until
xf is reached. If so, choosing a value for df that cannot be
represented in a finite number of binary digits means there will
be an accumulated error. I recall reading somewhere Mathematica
uses a few extra binary digits beyond the standard number used
for a machine precision value. If so, the accumulated error will
not be a simple monotonic sequence since at each step the sum is
converted to a machine precision value prior to the next sum
being computed. At least the following suggests my hypothesis is
close to correct.

In:= x =
FromDigits[First@#, 2] 2^(-Length@First@# + Last@#) & /@
RealDigits[Range[0, 1, .1], 2] - Range[0, 10]/10;

In:= x/x[]

Out= {0,1,2,8,4,0,16,12,8,4,0}

In any case, the underlying issue is one that is inherent to
using machine precision numbers rather than exact arithmetic.

```

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