| Re: Math Question [message #51012] |
Wed, 01 November 2006 04:51 |
rkombiyil
Messages: 59
Registered: March 2006
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Member |
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And fwiw, here is what OCTAVE gives :-)
--
>> -0.1^2
ans = -0.010000
>> -0.1^2.1
ans = -0.0079433
>> .1^2.1
ans = 0.0079433
>> (-0.1)^2.1
ans = 0.0075545 + 0.0024546i
--
Cheers,
~rk
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| Re: Math Question [message #51093 is a reply to message #51012] |
Tue, 31 October 2006 04:33  |
Jo Klein
Messages: 54
Registered: January 2006
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Member |
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>> It depends indeed...
>>
>>>> -0.1^2.1
>> ans =
>> -0.0079
>>
>>>> (-0.1)^2.1
>> ans =
>> 0.0076 + 0.0025i
Ah, my mistake - operator precedence strikes again. The first statement
actually expands to something like 0-(0.1^2.1). D'uh.
Jo
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| Re: Math Question [message #51098 is a reply to message #51093] |
Mon, 30 October 2006 13:07 |
news.qwest.net
Messages: 137
Registered: September 2005
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Senior Member |
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"Jo Klein" <jo_kln@yahoo.co.uk> wrote in message
news:ei5mlj$sd3$1@frank-exchange-of-views.oucs.ox.ac.uk...
>> Just out of curiosity, has anyone tried this on Matlab? I'd expect the
>> same (or similar) results, but it'd be interesting if they weren't.
> Matlab returns the real part of the argument if it's a real number you put
> in:
>>> -0.1^2.0
>
> ans =
> -0.0100
I think you have an order of operations problem there.
Matlab has given you
>> -(0.1^2.0)
instead of the intended
>> (-0.1)^2.0
likewise in the other case
Cheers,
bob
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| Re: Math Question [message #51099 is a reply to message #51098] |
Mon, 30 October 2006 12:32 |
Jean H.
Messages: 472
Registered: July 2006
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Senior Member |
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Jo Klein wrote:
>> Just out of curiosity, has anyone tried this on Matlab? I'd expect the
>> same (or similar) results, but it'd be interesting if they weren't.
>
> Matlab returns the real part of the argument if it's a real number you
> put in:
>>> -0.1^2.0
>
> ans =
> -0.0100
>
>>> -0.1^2.01
> ans =
> -0.0098
>
>>> complex(-0.1,0)^2.01
> ans =
> 0.0098 + 0.0003i
> Hmm - I don't know if this is more desirable than IDL's behaviour, I
> think it's fair enough to warn people who try to do something like that
> with their data. In Matlab, try and invert the second operation ... this
> can't be good. Suppose there are arguments for both approaches.
> Jo
It depends indeed...
>> -0.1^2.1
ans =
-0.0079
>> (-0.1)^2.1
ans =
0.0076 + 0.0025i
You don't have to specify the complex() statement!
Jean
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| Re: Math Question [message #51100 is a reply to message #51099] |
Mon, 30 October 2006 12:35 |
Jean H.
Messages: 472
Registered: July 2006
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Senior Member |
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Jean H. wrote:
> Jo Klein wrote:
>
>>> Just out of curiosity, has anyone tried this on Matlab? I'd expect the
>>> same (or similar) results, but it'd be interesting if they weren't.
>>
>>
>> Matlab returns the real part of the argument if it's a real number you
>> put in:
>>>> -0.1^2.0
>>
>> ans =
>> -0.0100
>>
>>>> -0.1^2.01
>> ans =
>> -0.0098
>>
>>>> complex(-0.1,0)^2.01
>> ans =
>> 0.0098 + 0.0003i
>> Hmm - I don't know if this is more desirable than IDL's behaviour, I
>> think it's fair enough to warn people who try to do something like
>> that with their data. In Matlab, try and invert the second operation
>> ... this can't be good. Suppose there are arguments for both approaches.
>> Jo
>
>
> It depends indeed...
>
>>> -0.1^2.1
> ans =
> -0.0079
>
>>> (-0.1)^2.1
> ans =
> 0.0076 + 0.0025i
>
> You don't have to specify the complex() statement!
>
> Jean
for clarity:
>> a=-0.1
a =
-0.1000
>> a^2
ans =
0.0100
>> a^2.1
ans =
0.0076 + 0.0025i
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| Re: Math Question [message #51101 is a reply to message #51099] |
Mon, 30 October 2006 12:17 |
Jo Klein
Messages: 54
Registered: January 2006
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Member |
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> Just out of curiosity, has anyone tried this on Matlab? I'd expect the
> same (or similar) results, but it'd be interesting if they weren't.
Matlab returns the real part of the argument if it's a real number you
put in:
>> -0.1^2.0
ans =
-0.0100
>> -0.1^2.01
ans =
-0.0098
>> complex(-0.1,0)^2.01
ans =
0.0098 + 0.0003i
Hmm - I don't know if this is more desirable than IDL's behaviour, I
think it's fair enough to warn people who try to do something like that
with their data. In Matlab, try and invert the second operation ... this
can't be good. Suppose there are arguments for both approaches.
Jo
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| Re: Math Question [message #51104 is a reply to message #51101] |
Mon, 30 October 2006 10:22 |
Braedley
Messages: 57
Registered: September 2006
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Member |
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edward.s.meinel@aero.org wrote:
> Rob wrote:
>> In general, powers of negative numbers are complex so you should start
>> with that assumption in the expression:
>>
>> IDL> print,(dcomplex(-1.0,0.0))^2.01
>> ( 0.99950656, 0.031410729)
>> IDL> print,(dcomplex(-1.0,0.0))^2.0
>> ( 1.0000000, -2.4492127e-016)
>>
>> It would be nice if IDL told you this rather than throwing a NaN at
>> you.
>
> Actually, to be consistent, IDL should just return a complex number.
> One of the reasons that I've been using IDL is that it automatically
> changes the variable TYPE when necessary so that I don't have to keep
> track of whether I'm operating on a BYTE, LONG, DOUBLE, whatever... Now
> I find out that it's "mostly" true (but who can refuse a nice mutton
> sandwich?).
>
> I would consider this a bug.
>
> Ed "to blave" Meinel
Just out of curiosity, has anyone tried this on Matlab? I'd expect the
same (or similar) results, but it'd be interesting if they weren't.
Braedley
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| Re: Math Question [message #51105 is a reply to message #51104] |
Mon, 30 October 2006 10:15 |
James Kuyper
Messages: 425
Registered: March 2000
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Senior Member |
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I shouldn't attempt to remember advanced mathematics in public when I'm
less than fully awake. :-( It's been a long time since I actually used
any of this stuff.
kuyper@wizard.net wrote:
...
> Considered solely as a function whose domain and range are restricted
> to real numbers, x^(a/b), should have two solutions if b is even, and 1
> solution if b is odd, regardless of whether x is positive.
That should be "two solutions if b is even and x is positive, and 1
solution if b is odd, regardless of whether x is positive".
After realizing that mistake, I decided to double check my statement
about the case where the range includes complex values:
...
> If a and b are mutually prime integers, then mathematically, x^(a/b)
> has b different values, at most two of which are real, the others are
> complex. If b is odd, only one of the values is real. This is true,
That count of real values is based upon the assumption that the
imaginary part of x is 0, and that x is positive. There's no simple
statement that covers the general case.
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| Re: Math Question [message #51106 is a reply to message #51105] |
Mon, 30 October 2006 09:19 |
David Fanning
Messages: 11724
Registered: August 2001
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Senior Member |
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Folks (and especially Mr. Kuyper):
Much obliged for the discussion. Most enlightening. :-)
Cheers,
David
P.S. You will be pleased to know I put the pretty
day to good advantage and won all four sets of tennis!
Lots of sweat, but no dust to be found. :-)
--
David Fanning, Ph.D.
Fanning Software Consulting, Inc.
Coyote's Guide to IDL Programming: http://www.dfanning.com/
Sepore ma de ni thui. ("Perhaps thou speakest truth.")
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| Re: Math Question [message #51107 is a reply to message #51106] |
Mon, 30 October 2006 09:04 |
James Kuyper
Messages: 425
Registered: March 2000
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Senior Member |
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Rob wrote:
> In general, powers of negative numbers are complex so you should start
> with that assumption in the expression:
>
> IDL> print,(dcomplex(-1.0,0.0))^2.01
> ( 0.99950656, 0.031410729)
> IDL> print,(dcomplex(-1.0,0.0))^2.0
> ( 1.0000000, -2.4492127e-016)
>
> It would be nice if IDL told you this rather than throwing a NaN at
> you.
>
> Rob
>
> On Oct 29, 11:02 am, David Fanning <n...@dfanning.com> wrote:
>> Folks,
>>
>> OK, I get the feeling that I am going to be referred to
>> my own web page with this question:
>>
>> http://www.dfanning.com/math_tips/sky_is_falling.html
>>
>> And it is certainly true that I have been watching WAY
>> too much TV lately (World Series, you know), but here is
>> my question. How does one explain the following two
>> IDL commands?
>>
>> IDL> Help, (-0.1)^2.0
>> <Expression> FLOAT = 0.0100000
>> IDL> Help, (-0.1)^2.01
>> <Expression> FLOAT = -NaN
>>
>> In general, raising a negative number to any integer
>> power seems to produce a real number, whereas raising
>> a negative number to a non-integer power causes a NAN.
>>
>> I am sure this is explained in one of those textbooks
>> covered with dust in my garage, but I thought one of
>> you math guys could rescue me from a beautiful day
>> spent covered with dust. :-)
If a and b are mutually prime integers, then mathematically, x^(a/b)
has b different values, at most two of which are real, the others are
complex. If b is odd, only one of the values is real. This is true,
whether or not 'x' is negative. However, I don't think we want to have
(-0.10)^2.01 return a list of 200 different complex values. Even when
the operands are complex number IDL only attempts to return one of the
possible values.
More imporatantly, I don't think it should return any complex values at
all. In most cases, if the operands of the ^ operator aren't already
complex, there's a pretty good chance that the code wasn't written to
handle the possiblity of complex values from the result.
Considered solely as a function whose domain and range are restricted
to real numbers, x^(a/b), should have two solutions if b is even, and 1
solution if b is odd, regardless of whether x is positive. In practice,
for computer arithmetic that's not generally implementable. When you
calculate x^y, y can, in general, only be a floating point
approximation of a/b; it is an exact representation only if b is a
power of 2. I've used implementations of the power function that
attempt to recognize when y is an approximation of a/b for small values
of b, however, I gather that IDL is implemented using C code, expwhich
probably calls pow(x,y) to implement x^y. The pow() function does not
attempt to recognise approximations to rational numbers with small
denominators. It is required by the C standard to return a domain error
if x is negative and y is a finite non-integer. IDL handles this by
returning a NaN. This strikes me as a reasonable approach.
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| Re: Math Question [message #51108 is a reply to message #51107] |
Mon, 30 October 2006 08:54 |
news.qwest.net
Messages: 137
Registered: September 2005
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Senior Member |
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"David Fanning" <news@dfanning.com> wrote in message
news:MPG.1fae7e91b1ccf3e3989d78@news.frii.com...
> Folks,
...
> In general, raising a negative number to any integer
> power seems to produce a real number, whereas raising
> a negative number to a non-integer power causes a NAN.
Evidently IDL needs another value, NaRN (not a real number).
Cheers,
bob
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| Re: Math Question [message #51109 is a reply to message #51108] |
Mon, 30 October 2006 07:29 |
Foldy Lajos
Messages: 268
Registered: October 2001
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Senior Member |
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On Mon, 30 Oct 2006, edward.s.meinel@aero.org wrote:
>
> Rob wrote:
>> In general, powers of negative numbers are complex so you should start
>> with that assumption in the expression:
>>
>> IDL> print,(dcomplex(-1.0,0.0))^2.01
>> ( 0.99950656, 0.031410729)
>> IDL> print,(dcomplex(-1.0,0.0))^2.0
>> ( 1.0000000, -2.4492127e-016)
>>
>> It would be nice if IDL told you this rather than throwing a NaN at
>> you.
>
> Actually, to be consistent, IDL should just return a complex number.
> One of the reasons that I've been using IDL is that it automatically
> changes the variable TYPE when necessary so that I don't have to keep
> track of whether I'm operating on a BYTE, LONG, DOUBLE, whatever... Now
> I find out that it's "mostly" true (but who can refuse a nice mutton
> sandwich?).
>
> I would consider this a bug.
>
> Ed "to blave" Meinel
>
I think complex results should be returned only if complex input was
given, like in the example above. Otherwise, all your data would easily
become complex. Think of sqrt(-1.0), log(-1.0), asin(-2.0), etc.
regards,
lajos
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| Re: Math Question [message #51110 is a reply to message #51109] |
Mon, 30 October 2006 07:19 |
edward.s.meinel@aero.
Messages: 52
Registered: February 2005
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Member |
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Rob wrote:
> In general, powers of negative numbers are complex so you should start
> with that assumption in the expression:
>
> IDL> print,(dcomplex(-1.0,0.0))^2.01
> ( 0.99950656, 0.031410729)
> IDL> print,(dcomplex(-1.0,0.0))^2.0
> ( 1.0000000, -2.4492127e-016)
>
> It would be nice if IDL told you this rather than throwing a NaN at
> you.
Actually, to be consistent, IDL should just return a complex number.
One of the reasons that I've been using IDL is that it automatically
changes the variable TYPE when necessary so that I don't have to keep
track of whether I'm operating on a BYTE, LONG, DOUBLE, whatever... Now
I find out that it's "mostly" true (but who can refuse a nice mutton
sandwich?).
I would consider this a bug.
Ed "to blave" Meinel
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| Re: Math Question [message #51115 is a reply to message #51110] |
Sun, 29 October 2006 16:15 |
Rob Dimeo
Messages: 17
Registered: November 1999
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Junior Member |
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In general, powers of negative numbers are complex so you should start
with that assumption in the expression:
IDL> print,(dcomplex(-1.0,0.0))^2.01
( 0.99950656, 0.031410729)
IDL> print,(dcomplex(-1.0,0.0))^2.0
( 1.0000000, -2.4492127e-016)
It would be nice if IDL told you this rather than throwing a NaN at
you.
Rob
On Oct 29, 11:02 am, David Fanning <n...@dfanning.com> wrote:
> Folks,
>
> OK, I get the feeling that I am going to be referred to
> my own web page with this question:
>
> http://www.dfanning.com/math_tips/sky_is_falling.html
>
> And it is certainly true that I have been watching WAY
> too much TV lately (World Series, you know), but here is
> my question. How does one explain the following two
> IDL commands?
>
> IDL> Help, (-0.1)^2.0
> <Expression> FLOAT = 0.0100000
> IDL> Help, (-0.1)^2.01
> <Expression> FLOAT = -NaN
>
> In general, raising a negative number to any integer
> power seems to produce a real number, whereas raising
> a negative number to a non-integer power causes a NAN.
>
> I am sure this is explained in one of those textbooks
> covered with dust in my garage, but I thought one of
> you math guys could rescue me from a beautiful day
> spent covered with dust. :-)
>
> Cheers,
>
> David
> --
> David Fanning, Ph.D.
> Fanning Software Consulting, Inc.
> Coyote's Guide to IDL Programming:http://www.dfanning.com/
> Sepore ma de ni thui. ("Perhaps thou speakest truth.")
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| Re: Math Question [message #51116 is a reply to message #51115] |
Sun, 29 October 2006 15:40 |
Robbie
Messages: 165
Registered: February 2006
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Senior Member |
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Actually that reasoning is totally flawed, we should keep those
envelopes for experimental physicists.
Lookup "Mathematical Operators", subheading "Using Exponentiation" in
the IDL help.
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| Re: Math Question [message #51117 is a reply to message #51116] |
Sun, 29 October 2006 15:01 |
Robbie
Messages: 165
Registered: February 2006
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Senior Member |
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Consider
y = x^(a/b)
then
y^b = x^a
The result of a and y being real happens when a+b is even. I have some
scrawl on the back of an envelope, but the reasoning is probably buried
in that dusty textbook somewhere.
If we are using floating point arithmetic, then we can't really find
out what a or b is because the accuracy of the machine is not perfect.
i.e. 2.01 = 201/200 - but we really can't be sure.
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| Re: Math Question [message #51118 is a reply to message #51117] |
Sun, 29 October 2006 09:59 |
Foldy Lajos
Messages: 268
Registered: October 2001
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Senior Member |
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On Sun, 29 Oct 2006, David Fanning wrote:
> Folks,
>
> OK, I get the feeling that I am going to be referred to
> my own web page with this question:
>
> http://www.dfanning.com/math_tips/sky_is_falling.html
>
> And it is certainly true that I have been watching WAY
> too much TV lately (World Series, you know), but here is
> my question. How does one explain the following two
> IDL commands?
>
> IDL> Help, (-0.1)^2.0
> <Expression> FLOAT = 0.0100000
> IDL> Help, (-0.1)^2.01
> <Expression> FLOAT = -NaN
>
> In general, raising a negative number to any integer
> power seems to produce a real number, whereas raising
> a negative number to a non-integer power causes a NAN.
>
> I am sure this is explained in one of those textbooks
> covered with dust in my garage, but I thought one of
> you math guys could rescue me from a beautiful day
> spent covered with dust. :-)
>
> Cheers,
>
> David
> --
> David Fanning, Ph.D.
> Fanning Software Consulting, Inc.
> Coyote's Guide to IDL Programming: http://www.dfanning.com/
> Sepore ma de ni thui. ("Perhaps thou speakest truth.")
>
I had the opposite problem with FL: (-0.1)^2.0 was also a NaN. Power is
defined for floats as x^y = exp(log(x)*y), and log() is undefined for
non-positive numbers. Now FL examines y first, and if its is an integer,
then uses multiplications. I think IDL does the same.
regards,
lajos
ps: what will be the result for (-0.1)^2.0000001 ? :-)
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