GeometricMean definition
$begingroup$
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 6 hours ago
mikadomikado
6,6021929
$endgroup$
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 6 hours ago
mikadomikado
6,6021929
$endgroup$
$begingroup$
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
6 hours ago
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
6 hours ago
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
5 hours ago
$begingroup$
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
1 hour ago
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 6 hours ago
mikadomikado
6,6021929
$endgroup$
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 6 hours ago
mikadomikado
6,6021929
asked 6 hours ago
mikadomikado
6,6021929
edited 5 hours ago
mikado
asked 6 hours ago
mikadomikado
6,6021929
asked 6 hours ago
mikadomikado
6,6021929
asked 6 hours ago
mikadomikado
6,6021929
6,6021929
$begingroup$
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
6 hours ago
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
6 hours ago
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
5 hours ago
$begingroup$
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
1 hour ago
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
6 hours ago
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
6 hours ago
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
5 hours ago
$begingroup$
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
1 hour ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
@Coolwater, but note that
FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5} returns False, so I don't think your explanation generalises.$endgroup$
– mikado
6 hours ago
$begingroup$
@Coolwater, but note that
FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5} returns False, so I don't think your explanation generalises.$endgroup$
– mikado
6 hours ago
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
5 hours ago
$begingroup$
"The geometric mean applies only to positive numbers." (
Wiki (Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100] evaluates to True$endgroup$
– Bob Hanlon
5 hours ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
1 Answer
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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 5 hours ago
SomosSomos
1,33519
$endgroup$
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 5 hours ago
SomosSomos
1,33519
$endgroup$
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 5 hours ago
SomosSomos
1,33519
$endgroup$
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 5 hours ago
SomosSomos
1,33519
$endgroup$
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 5 hours ago
SomosSomos
1,33519
edited 4 hours ago
answered 5 hours ago
SomosSomos
1,33519
answered 5 hours ago
SomosSomos
1,33519
answered 5 hours ago
SomosSomos
1,33519
1,33519
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
6 hours ago
$begingroup$
@Coolwater, but note that
FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.$endgroup$
– mikado
6 hours ago
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
5 hours ago
$begingroup$
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
1 hour ago