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GeometricMean definition
Using a PatternTest versus a Condition for pattern matchingMoving the location of the branch cut in MathematicaUse elements of Array inside function definitionExtract information from function definition returned by DefineDLLFunctionStrange behaviour of MMA in derivatives of some standard functionsRecursive definition of nested sumsWhy is the evaluation of f different using rules and using arguments to it as a function?Domain definition of a functionRecursion definition wont workImmediate Function Definition inside a Module
$begingroup$
According to the documentation for
GeometricMean
GeometricMean[Table[x[i],{i,1,n}]]==Product[x[i],{i,1,n}]^(1/n)
Presumably this is correct for x[i] positive real.
However, GeometricMean returns answers for some negative and complex inputs that are not consistent with this definition.
For example
GeometricMean[{-4, -4}]
(* -4 *)
(Perhaps it simply recognises a constant list as a special case, and shortcuts the definition to give an incorrect answer?)
functions
asked yesterday
mikadomikado
6,6071929
$endgroup$
add a comment |
$begingroup$
According to the documentation for
GeometricMean
GeometricMean[Table[x[i],{i,1,n}]]==Product[x[i],{i,1,n}]^(1/n)
Presumably this is correct for x[i] positive real.
However, GeometricMean returns answers for some negative and complex inputs that are not consistent with this definition.
For example
GeometricMean[{-4, -4}]
(* -4 *)
(Perhaps it simply recognises a constant list as a special case, and shortcuts the definition to give an incorrect answer?)
functions
asked yesterday
mikadomikado
6,6071929
$endgroup$
$begingroup$
Exp[Mean[Log[{-4, -4}]]]yields-4. The definition stated in the documentation doesn't seem to be the one used for numeric input.
$endgroup$
– Coolwater
yesterday
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
yesterday
1
$begingroup$
"The geometric mean applies only to positive numbers." (Wiki(Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100]evaluates toTrue
$endgroup$
– Bob Hanlon
yesterday
$begingroup$
GeometricMean[{-4, -4.}]yields positive4., andGeometricMean[{-4, Unevaluated[-2^2]}]yields positive4. Apparently a shortcut is taken when input is a list of identical elements, namelyGeometricMean[{a, a,...}]is assumed to bea. (AndGeometricMean[{-4, -4, -4, -4, -4.}]illustrates one of @Somos's points.)
$endgroup$
– Michael E2
yesterday
add a comment |
$begingroup$
According to the documentation for
GeometricMean
GeometricMean[Table[x[i],{i,1,n}]]==Product[x[i],{i,1,n}]^(1/n)
Presumably this is correct for x[i] positive real.
However, GeometricMean returns answers for some negative and complex inputs that are not consistent with this definition.
For example
GeometricMean[{-4, -4}]
(* -4 *)
(Perhaps it simply recognises a constant list as a special case, and shortcuts the definition to give an incorrect answer?)
functions
asked yesterday
mikadomikado
6,6071929
$endgroup$
According to the documentation for
GeometricMean
GeometricMean[Table[x[i],{i,1,n}]]==Product[x[i],{i,1,n}]^(1/n)
Presumably this is correct for x[i] positive real.
However, GeometricMean returns answers for some negative and complex inputs that are not consistent with this definition.
For example
GeometricMean[{-4, -4}]
(* -4 *)
(Perhaps it simply recognises a constant list as a special case, and shortcuts the definition to give an incorrect answer?)
functions
functions
asked yesterday
mikadomikado
6,6071929
asked yesterday
mikadomikado
6,6071929
edited yesterday
mikado
asked yesterday
mikadomikado
6,6071929
asked yesterday
mikadomikado
6,6071929
asked yesterday
mikadomikado
6,6071929
6,6071929
$begingroup$
Exp[Mean[Log[{-4, -4}]]]yields-4. The definition stated in the documentation doesn't seem to be the one used for numeric input.
$endgroup$
– Coolwater
yesterday
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
yesterday
1
$begingroup$
"The geometric mean applies only to positive numbers." (Wiki(Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100]evaluates toTrue
$endgroup$
– Bob Hanlon
yesterday
$begingroup$
GeometricMean[{-4, -4.}]yields positive4., andGeometricMean[{-4, Unevaluated[-2^2]}]yields positive4. Apparently a shortcut is taken when input is a list of identical elements, namelyGeometricMean[{a, a,...}]is assumed to bea. (AndGeometricMean[{-4, -4, -4, -4, -4.}]illustrates one of @Somos's points.)
$endgroup$
– Michael E2
yesterday
add a comment |
$begingroup$
Exp[Mean[Log[{-4, -4}]]]yields-4. The definition stated in the documentation doesn't seem to be the one used for numeric input.
$endgroup$
– Coolwater
yesterday
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
yesterday
1
$begingroup$
"The geometric mean applies only to positive numbers." (Wiki(Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100]evaluates toTrue
$endgroup$
– Bob Hanlon
yesterday
$begingroup$
GeometricMean[{-4, -4.}]yields positive4., andGeometricMean[{-4, Unevaluated[-2^2]}]yields positive4. Apparently a shortcut is taken when input is a list of identical elements, namelyGeometricMean[{a, a,...}]is assumed to bea. (AndGeometricMean[{-4, -4, -4, -4, -4.}]illustrates one of @Somos's points.)
$endgroup$
– Michael E2
yesterday
$begingroup$
Exp[Mean[Log[{-4, -4}]]] yields -4. The definition stated in the documentation doesn't seem to be the one used for numeric input.$endgroup$
– Coolwater
yesterday
$begingroup$
Exp[Mean[Log[{-4, -4}]]] yields -4. The definition stated in the documentation doesn't seem to be the one used for numeric input.$endgroup$
– Coolwater
yesterday
$begingroup$
@Coolwater, but note that
FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5} returns False, so I don't think your explanation generalises.$endgroup$
– mikado
yesterday
$begingroup$
@Coolwater, but note that
FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5} returns False, so I don't think your explanation generalises.$endgroup$
– mikado
yesterday
1
1
$begingroup$
"The geometric mean applies only to positive numbers." (
Wiki (Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100] evaluates to True$endgroup$
– Bob Hanlon
yesterday
$begingroup$
"The geometric mean applies only to positive numbers." (
Wiki (Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100] evaluates to True$endgroup$
– Bob Hanlon
yesterday
$begingroup$
GeometricMean[{-4, -4.}] yields positive 4., and GeometricMean[{-4, Unevaluated[-2^2]}] yields positive 4. Apparently a shortcut is taken when input is a list of identical elements, namely GeometricMean[{a, a,...}] is assumed to be a. (And GeometricMean[{-4, -4, -4, -4, -4.}] illustrates one of @Somos's points.)$endgroup$
– Michael E2
yesterday
$begingroup$
GeometricMean[{-4, -4.}] yields positive 4., and GeometricMean[{-4, Unevaluated[-2^2]}] yields positive 4. Apparently a shortcut is taken when input is a list of identical elements, namely GeometricMean[{a, a,...}] is assumed to be a. (And GeometricMean[{-4, -4, -4, -4, -4.}] illustrates one of @Somos's points.)$endgroup$
– Michael E2
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Similar to many other means, the geometric mean is homogenous. This means that
GeometricMean[ c data ] == c GeometricMean[ data ]
should be true for any number c. However, the problem is that $n$th roots are multi-valued in general and this causes no end of confusion. This is usually no problem for positive reals, but for negative reals it can be confusing. Do you want odd roots of negative reals be negative? Sometimes yes and sometimes no. For example, in version 10.2 the result of your example is $4$. Thus, the behavior has changed somewhat over time and there is no way to satisfy all expectations. The documentation for GeometricMean[] should be more clear in this regard.
answered yesterday
SomosSomos
1,36019
$endgroup$
add a comment |
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$begingroup$
Similar to many other means, the geometric mean is homogenous. This means that
GeometricMean[ c data ] == c GeometricMean[ data ]
should be true for any number c. However, the problem is that $n$th roots are multi-valued in general and this causes no end of confusion. This is usually no problem for positive reals, but for negative reals it can be confusing. Do you want odd roots of negative reals be negative? Sometimes yes and sometimes no. For example, in version 10.2 the result of your example is $4$. Thus, the behavior has changed somewhat over time and there is no way to satisfy all expectations. The documentation for GeometricMean[] should be more clear in this regard.
answered yesterday
SomosSomos
1,36019
$endgroup$
add a comment |
$begingroup$
Similar to many other means, the geometric mean is homogenous. This means that
GeometricMean[ c data ] == c GeometricMean[ data ]
should be true for any number c. However, the problem is that $n$th roots are multi-valued in general and this causes no end of confusion. This is usually no problem for positive reals, but for negative reals it can be confusing. Do you want odd roots of negative reals be negative? Sometimes yes and sometimes no. For example, in version 10.2 the result of your example is $4$. Thus, the behavior has changed somewhat over time and there is no way to satisfy all expectations. The documentation for GeometricMean[] should be more clear in this regard.
answered yesterday
SomosSomos
1,36019
$endgroup$
add a comment |
$begingroup$
Similar to many other means, the geometric mean is homogenous. This means that
GeometricMean[ c data ] == c GeometricMean[ data ]
should be true for any number c. However, the problem is that $n$th roots are multi-valued in general and this causes no end of confusion. This is usually no problem for positive reals, but for negative reals it can be confusing. Do you want odd roots of negative reals be negative? Sometimes yes and sometimes no. For example, in version 10.2 the result of your example is $4$. Thus, the behavior has changed somewhat over time and there is no way to satisfy all expectations. The documentation for GeometricMean[] should be more clear in this regard.
answered yesterday
SomosSomos
1,36019
$endgroup$
Similar to many other means, the geometric mean is homogenous. This means that
GeometricMean[ c data ] == c GeometricMean[ data ]
should be true for any number c. However, the problem is that $n$th roots are multi-valued in general and this causes no end of confusion. This is usually no problem for positive reals, but for negative reals it can be confusing. Do you want odd roots of negative reals be negative? Sometimes yes and sometimes no. For example, in version 10.2 the result of your example is $4$. Thus, the behavior has changed somewhat over time and there is no way to satisfy all expectations. The documentation for GeometricMean[] should be more clear in this regard.
answered yesterday
SomosSomos
1,36019
edited yesterday
answered yesterday
SomosSomos
1,36019
answered yesterday
SomosSomos
1,36019
answered yesterday
SomosSomos
1,36019
1,36019
add a comment |
add a comment |
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$begingroup$
Exp[Mean[Log[{-4, -4}]]]yields-4. The definition stated in the documentation doesn't seem to be the one used for numeric input.$endgroup$
– Coolwater
yesterday
$begingroup$
@Coolwater, but note that
FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.$endgroup$
– mikado
yesterday
1
$begingroup$
"The geometric mean applies only to positive numbers." (
Wiki(Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100]evaluates toTrue$endgroup$
– Bob Hanlon
yesterday
$begingroup$
GeometricMean[{-4, -4.}]yields positive4., andGeometricMean[{-4, Unevaluated[-2^2]}]yields positive4. Apparently a shortcut is taken when input is a list of identical elements, namelyGeometricMean[{a, a,...}]is assumed to bea. (AndGeometricMean[{-4, -4, -4, -4, -4.}]illustrates one of @Somos's points.)$endgroup$
– Michael E2
yesterday