Hallo
I am new to mathematica. I have mathematica 5.1.
If I have the expresion a^b where a is an arbitrary double precision
decimal number and b is an exact number say a list from 1-10. How can
I see that the precision of this expression changes when raising a to
these powers?
Many thanks in advance
Sonya

sonyaBab...@yahoo.de wrote:
Hallo
I am new tomathematica. I havemathematica5.1.
If I have the expresion a^b where a is an arbitrary doubleprecision

decimal number and b is an exact number say a list from 1-10. How can
I see that theprecisionof this expression changes when raising a to
these powers?
Many thanks in advance
Sonya

Hi Sonya,

Theprecisionof the results does not change: since you have a mix of
hardwareprecisionnumbers and exact numbers the overallprecisionis
lowered to machineprecisionand the computations are done using the
functions implemented in the hardware. Detailed explanations are given in
"TheMathematicaBook Online / Advanced Mathematics inMathematica/
Numbers / 3.1.4 NumericalPrecision", available at

http://documents.wolfram.com/mathematica/book/section-3.1.4

To see theprecisionof aMathematicaexpression, you can use the
built-in functionPrecision[1]. For instance,

In[1]:> a = 2.;

In[2]:=Precision[a]

Out[2]> MachinePrecision

In[3]:> b = Range[10]

Out[3]> {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

In[4]:=Precision/@ b

Out[4]> {Infinity, Infinity, Infinity, Infinity, Infinity,

Infinity, Infinity, Infinity, Infinity, Infinity}

In[5]:> a^b

Out[5]> {2., 4., 8., 16., 32., 64., 128., 256., 512., 1024.}

In[6]:=Precision/@ %

Out[6]> {MachinePrecision, MachinePrecision, MachinePrecision,

MachinePrecision, MachinePrecision,

MachinePrecision, MachinePrecision,

MachinePrecision, MachinePrecision, MachinePrecision}

In[7]:> $MachinePrecision

Out[7]> 15.9546

In[8]:> $Version

Out[8]> 5.2 for Microsoft Windows (June 20, 2005)

Regards,
Jean-Marc

[1]http://documents.wolfram.com/mathematica/functions/Precision

On 20 Feb., 12:05, Jean-Marc Gulliet <jeanmarc.gull...@gmail.com
wrote:



sonyaBab...@yahoo.de wrote:
Hallo
I am new tomathematica. I havemathematica5.1.
If I have the expresion a^b where a is an arbitrary doubleprecision
decimal number and b is an exact number say a list from 1-10. How can
I see that theprecisionof this expression changes when raising a to
these powers?
Many thanks in advance
Sonya

Hi Sonya,

Theprecisionof the results does not change: since you have a mix of
hardwareprecisionnumbers and exact numbers the overallprecisionis
lowered to machineprecisionand the computations are done using the
functions implemented in the hardware. Detailed explanations are given in
"TheMathematicaBook Online / Advanced Mathematics inMathematica/
Numbers / 3.1.4 NumericalPrecision", available at

http://documents.wolfram.com/mathematica/book/section-3.1.4

To see theprecisionof aMathematicaexpression, you can use the
built-in functionPrecision[1]. For instance,

In[1]:> > a = 2.;

In[2]:=Precision[a]

Out[2]> > MachinePrecision

In[3]:> > b = Range[10]

Out[3]> > {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

In[4]:=Precision/@ b

Out[4]> > {Infinity, Infinity, Infinity, Infinity, Infinity,

Infinity, Infinity, Infinity, Infinity, Infinity}

In[5]:> > a^b

Out[5]> > {2., 4., 8., 16., 32., 64., 128., 256., 512., 1024.}

In[6]:=Precision/@ %

Out[6]> > {MachinePrecision, MachinePrecision, MachinePrecision,

MachinePrecision, MachinePrecision,

MachinePrecision, MachinePrecision,

MachinePrecision, MachinePrecision, MachinePrecision}

In[7]:> > $MachinePrecision

Out[7]> > 15.9546

In[8]:> > $Version

Out[8]> > 5.2 for Microsoft Windows (June 20, 2005)

Regards,
Jean-Marc

[1]http://documents.wolfram.com/mathematica/functions/Precision

Hi Jean
Thank you very much for quick response. I have done exactly the above
and as you could see

In[1]:> a=2.;

In[2]:=Precision[a]

Out[2]> MachinePrecision

In[3]:> b=Range[10]

Out[3]> {1,2,3,4,5,6,7,8,9,10}

In[4]:=Precision/@b

Out[4]> {∞,∞,∞,∞,∞,∞,∞,∞,∞,∞}

In[5]:=Precision/@%

Out[5]> {∞,∞,∞,∞,∞,∞,∞,∞,∞,∞}

In[6]:> $Version

Out[6]> 5.1 for Microsoft Windows (October 25, 2004)

What does this difference mean?
Many thanks

I am new to mathematica. I have mathematica 5.1.
If I have the expresion a^b where a is an arbitrary double precision
decimal number and b is an exact number say a list from 1-10. How can
I see that the precision of this expression changes when raising a to
these powers?

In article <1171965456.017093.124290@m58g2000cwm.googlegroups.com>,
sonyaBabken@yahoo.de wrote:


I am new to mathematica. I have mathematica 5.1.
If I have the expresion a^b where a is an arbitrary double precision
decimal number and b is an exact number say a list from 1-10. How can
I see that the precision of this expression changes when raising a to
these powers?


If a is a machine precision number, the precision does not change:

mp = 1.3 ^ Range[6,8]

{4.826809000000003, 6.274851700000004, 8.157307210000003}

Precision /@ mp

{MachinePrecision, MachinePrecision, MachinePrecision}

If a is an arbitrary precision number, the precision does change:

ap = 1.3`30 ^ Range[6,8];

Precision /@ ap

{29.221848749616356, 29.15490195998574, 29.096910013008056}

Note that 1.3`30 is a shorthand way of entering 1.3 followed by 28 zeros.

1.3`30

1.300000000000000000000000000000

Cheers,
Paul

_______________________________________________________________________
Paul Abbott Phone: 61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
AUSTRALIA http://physics.uwa.edu.au/~paul

The story is not so simple for arbitrary powers. Consider 10.0 ^ 1000000

which comes out as
1.00000000000155 * 10*^1000000
with precision 14.1484 and accuracy -999986.

Perhaps others can explain what this means. It might be a bug, since the
two inputs can be represented (easily) exactly. The output can be
represented exactly in Mathematica, too. Choosing an appropriate
precision and attributing some accuracy to it seems to me to be somewhat
arbitrary: what is the default precision used when a number overflows?
Or perhaps something else is going on. (in Mathematica 5.1)

RJF
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