The basic input form for floating-point numbers?

Question

I just have discovered the fundamental difference between two input forms for floating-point numbers:

In[8]:= 1.5*^-334355//Hold//FullForm
1.5*10^-334355//Hold//FullForm
Out[8]//FullForm= Hold[1.5000000000000000000000000000000001`15.954589770191005*^-334355]
Out[9]//FullForm= Hold[Times[1.5`,Power[10,-334355]]]

These two forms differ very much in memory and time consumption:

In[7]:= start = MaxMemoryUsed[];
1.5*^-33432242 // Timing
start = MaxMemoryUsed[] - start
1.5*10^-33432242 // Timing
MaxMemoryUsed[] - start

Out[8]= {1.67401*10^-16, 1.500000000000000*10^-33432242}

Out[9]= 0

Out[10]= {7.741, 1.500000000000000*10^-33432242}

Out[11]= 34274192

But I cannot find out where the form *^ is documented. Is it a real basic input form for floating-point numbers? How is about numbers in other bases?

And why the second form is so much expensive?

From the help `The *^ form for scientific notation is always used in InputForm, and is independent of NumberMarks. ` Under **NumberMarks** _More Information_ — Dr. belisarius, Feb 08 '11 at 23:39
@belisarius Thanks but that note sounds ambiguously as being related only to the ScientificForm. — Alexey Popkov, Feb 09 '11 at 00:21
I think it's not ambiguous. It's telling you that using it doesn't affect evaluation, as it's only an InputForm artifact. — Dr. belisarius, Feb 09 '11 at 00:27
@belisarius Are there other ways to input such numbers directly? — Alexey Popkov, Feb 09 '11 at 00:32
The problem has already stated by @Leonid (10.) != (10) _or I don't understand your question_ — Dr. belisarius, Feb 09 '11 at 00:38
@belisarius I mean is the form `*^` the only computation-free way to input floating-point numbers in Mathematica? Is there another form for doing it? — Alexey Popkov, Feb 09 '11 at 00:50
I think you'll find your answer here http://reference.wolfram.com/mathematica/tutorial/NumericalPrecision.html where the *^ is explained — Dr. belisarius, Feb 09 '11 at 01:05

score 4 · Accepted Answer · answered Feb 08 '11 at 23:52

4

Regarding the time and memory consumption - these are the consequences of evaluation, have nothing to do with different forms. You use integer arithmetic for the power of 10 when 10 is present explicitly, thus the time/memory inefficiency. When we use machine precision from the start, the effect disappears:

In[1]:= MaxMemoryUsed[]
1.5*^-33432242 // Timing
MaxMemoryUsed[]
1.5*10.^-33432242 // Timing
MaxMemoryUsed[]

Out[1]= 17417696

Out[2]= {0., 1.500000000000000*10^-33432242}

Out[3]= 17417696

Out[4]= {0., 1.500000000043239*10^-33432242}

Out[5]= 17417696

HTH

answered Feb 08 '11 at 23:52

Leonid Shifrin

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I understand but `1.5*10.^-33432242` is not equivalent to `1.5*^-33432242`. I'm wondering why such an important thing as the `*^` form is not emphasized in the documentation? – Alexey Popkov Feb 09 '11 at 00:28
1

Well, I agree, but docs can not be perfect. Why aren't the two numbers equivalent? Both are machine precision, and since both are way smaller than `$MinMachineNumber`, I don't think that the actual values displayed make much sense - from the computational viewpoint, I think they are equivalent, and we could as well replace both by zero. – Leonid Shifrin Feb 09 '11 at 00:36

The basic input form for floating-point numbers?

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