error in N // Floor? (corrected version)

Questions
,
1

Hello,

my question is easy to understand:
Shouldn’t Mathematica produce the same output for the following two inputs?

  1. Log[10^3]/Log[10] // FullSimplify // N // Floor
  2. Log[10^3]/Log[10] // N // Floor

The corresponding output of mathematica version 8 and 9 are
3 (correct) and 2 (false) respectively.
You can see the problem mathematica has, when you use RealDigits instead of floor:

  • Log[10^3]/Log[10] // FullSimplify // N // RealDigits gives
    {{3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, 1}

  • Log[10^3]/Log[10] // N // RealDigits gives
    {{2, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9}, 1}

From my understanding a CAS-system shouldn give here the same outputs.
What do you think about it? Thanks in advance for your answer.

Peter Quedens

2

Hi,

Shouldn’t Mathematica produce the same output for the following two inputs?

It should not, since you use N[] this kind of behavior (round off error) is normal.

I.M.

3

Hi,

yes, N is used, but it is used in both expressions and so it should give the same result, since the only difference ist the FullSimplify.
In any case: I think mathematica should know, that Log[10^3]/Log[10] is 3 without a FullSimplify.
P.Q.

4

Even it is used in both expressions it is applied to different ones:

Log[10^3]/Log[10] // FullSimplify // N // Floor // Trace
Log[10^3]/Log[10] // N // Floor // Trace

I.M.

5

From my understanding a CAS-system shouldn give here the same outputs.

And it does so. This problem arises solely from the mindless usage of N[]

In[27]:= Union[(Floor[N[FullSimplify[Log[10^3]/Log[10]], #]] - Floor[N[Log[10^3]/Log[10], #]]) & /@ RandomReal[{1., 100.}, 1000]]
Out[27]= {0}

if you consider that a sloppy argument, look at this

In[28]:= Floor[N[FullSimplify[Log[10^3]/Log[10]], 1.]] - Floor[N[Log[10^3]/Log[10], 1.]]
Out[28]= 0

even that works

In[31]:= Floor[N[FullSimplify[Log[10^3]/Log[10]], .0]] - Floor[N[Log[10^3]/Log[10], .0]]
Out[31]= 0

and this

In[32]:= Floor[N[FullSimplify[Log[10^3]/Log[10]], 0]] - Floor[N[Log[10^3]/Log[10], 0]]
Out[32]= 0

and this

 In[35]:= Floor[N[FullSimplify[Log[10^3]/Log[10]], $MachinePrecision]] - Floor[N[Log[10^3]/Log[10], $MachinePrecision]]
 Out[35]= 0

You hit seemingly the only case not working

In[34]:= Floor[N[FullSimplify[Log[10^3]/Log[10]]]] - Floor[N[Log[10^3]/Log[10]]]
Out[34]= 1

despite the description in the help, that In[35] is equivalent to In[34].

6

Carrying out the division with machine precision numbers is subject to floating point error.

This is not specific to Mathematica at all, for example numpy gives the exact same result:

>>> a=numpy.log(1000)
>>> b=numpy.log(10)
>>> a
6.9077552789821368
>>> b
2.3025850929940459
>>> a/b
2.9999999999999996

But in Mathematica we can also use arbitrary precision arithmetic, to any desired number of digits:

N[Log[1000]/Log[10], 200]