GeometricMean definition
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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 Mar 4 at 20:35
mikadomikado
6,8421929
$endgroup$
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 Mar 4 at 20:35
mikadomikado
6,8421929
$endgroup$
$begingroup$
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
Mar 4 at 20:43
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@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@-4, -5returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
Mar 4 at 20:47
2
$begingroup$
"The geometric mean applies only to positive numbers." (Wiki(Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100]evaluates toTrue
$endgroup$
– Bob Hanlon
Mar 4 at 21:19
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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
Mar 5 at 1:08
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 Mar 4 at 20:35
mikadomikado
6,8421929
$endgroup$
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 Mar 4 at 20:35
mikadomikado
6,8421929
asked Mar 4 at 20:35
mikadomikado
6,8421929
edited Mar 4 at 21:03
mikado
asked Mar 4 at 20:35
mikadomikado
6,8421929
asked Mar 4 at 20:35
mikadomikado
6,8421929
asked Mar 4 at 20:35
mikadomikado
6,8421929
6,8421929
$begingroup$
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
Mar 4 at 20:43
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@-4, -5returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
Mar 4 at 20:47
2
$begingroup$
"The geometric mean applies only to positive numbers." (Wiki(Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100]evaluates toTrue
$endgroup$
– Bob Hanlon
Mar 4 at 21:19
$begingroup$
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
Mar 5 at 1:08
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
Mar 4 at 20:43
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@-4, -5returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
Mar 4 at 20:47
2
$begingroup$
"The geometric mean applies only to positive numbers." (Wiki(Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100]evaluates toTrue
$endgroup$
– Bob Hanlon
Mar 4 at 21:19
$begingroup$
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
Mar 5 at 1:08
$begingroup$
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
Mar 4 at 20:43
$begingroup$
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
Mar 4 at 20:43
$begingroup$
@Coolwater, but note that
FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@-4, -5 returns False, so I don't think your explanation generalises.$endgroup$
– mikado
Mar 4 at 20:47
$begingroup$
@Coolwater, but note that
FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@-4, -5 returns False, so I don't think your explanation generalises.$endgroup$
– mikado
Mar 4 at 20:47
2
2
$begingroup$
"The geometric mean applies only to positive numbers." (
Wiki (Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100] evaluates to True$endgroup$
– Bob Hanlon
Mar 4 at 21:19
$begingroup$
"The geometric mean applies only to positive numbers." (
Wiki (Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100] evaluates to True$endgroup$
– Bob Hanlon
Mar 4 at 21:19
$begingroup$
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
Mar 5 at 1:08
$begingroup$
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
Mar 5 at 1:08
add a comment |
2 Answers
2
active
oldest
votes
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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. In particular it probably makes little sense to take the Geomtric mean of numbers which do not all all have the same sign..
answered Mar 4 at 21:09
SomosSomos
1,670110
$endgroup$
add a comment |
$begingroup$
For lists of positive values, all of the usual ways of calculating the GeometricMean are equivalent.
test[list_] := Simplify[
Equal @@
(#[list] & /@
GeometricMean,
(Times @@ #)^(1/Length[#]) &,
E^Mean[Log[#]] &,
Limit[Mean[#^t]^(1/t), t -> 0] &)]
Simplify[test[#], Thread[# > 0]] &@a, b, c, d
(* True *)
SeedRandom[0]
test /@ RandomReal[10^-10, 10, 10, 10]
(* True, True, True, True, True, True, True, True, True, True *)
answered Mar 6 at 17:33
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
$begingroup$
My hope, based on the initial observation thatGeometricMean[-4,-4]gave -4 was that Mathematica implemented a sensible generalisation ofGeometricMeanto values other than positive reals (in particular, complex numbers).
$endgroup$
– mikado
Mar 6 at 20:26
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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. In particular it probably makes little sense to take the Geomtric mean of numbers which do not all all have the same sign..
answered Mar 4 at 21:09
SomosSomos
1,670110
$endgroup$
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. In particular it probably makes little sense to take the Geomtric mean of numbers which do not all all have the same sign..
answered Mar 4 at 21:09
SomosSomos
1,670110
$endgroup$
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. In particular it probably makes little sense to take the Geomtric mean of numbers which do not all all have the same sign..
answered Mar 4 at 21:09
SomosSomos
1,670110
$endgroup$
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. In particular it probably makes little sense to take the Geomtric mean of numbers which do not all all have the same sign..
answered Mar 4 at 21:09
SomosSomos
1,670110
edited Mar 6 at 15:07
answered Mar 4 at 21:09
SomosSomos
1,670110
answered Mar 4 at 21:09
SomosSomos
1,670110
answered Mar 4 at 21:09
SomosSomos
1,670110
1,670110
add a comment |
add a comment |
$begingroup$
For lists of positive values, all of the usual ways of calculating the GeometricMean are equivalent.
test[list_] := Simplify[
Equal @@
(#[list] & /@
GeometricMean,
(Times @@ #)^(1/Length[#]) &,
E^Mean[Log[#]] &,
Limit[Mean[#^t]^(1/t), t -> 0] &)]
Simplify[test[#], Thread[# > 0]] &@a, b, c, d
(* True *)
SeedRandom[0]
test /@ RandomReal[10^-10, 10, 10, 10]
(* True, True, True, True, True, True, True, True, True, True *)
answered Mar 6 at 17:33
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
$begingroup$
My hope, based on the initial observation thatGeometricMean[-4,-4]gave -4 was that Mathematica implemented a sensible generalisation ofGeometricMeanto values other than positive reals (in particular, complex numbers).
$endgroup$
– mikado
Mar 6 at 20:26
add a comment |
$begingroup$
For lists of positive values, all of the usual ways of calculating the GeometricMean are equivalent.
test[list_] := Simplify[
Equal @@
(#[list] & /@
GeometricMean,
(Times @@ #)^(1/Length[#]) &,
E^Mean[Log[#]] &,
Limit[Mean[#^t]^(1/t), t -> 0] &)]
Simplify[test[#], Thread[# > 0]] &@a, b, c, d
(* True *)
SeedRandom[0]
test /@ RandomReal[10^-10, 10, 10, 10]
(* True, True, True, True, True, True, True, True, True, True *)
answered Mar 6 at 17:33
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
$begingroup$
My hope, based on the initial observation thatGeometricMean[-4,-4]gave -4 was that Mathematica implemented a sensible generalisation ofGeometricMeanto values other than positive reals (in particular, complex numbers).
$endgroup$
– mikado
Mar 6 at 20:26
add a comment |
$begingroup$
For lists of positive values, all of the usual ways of calculating the GeometricMean are equivalent.
test[list_] := Simplify[
Equal @@
(#[list] & /@
GeometricMean,
(Times @@ #)^(1/Length[#]) &,
E^Mean[Log[#]] &,
Limit[Mean[#^t]^(1/t), t -> 0] &)]
Simplify[test[#], Thread[# > 0]] &@a, b, c, d
(* True *)
SeedRandom[0]
test /@ RandomReal[10^-10, 10, 10, 10]
(* True, True, True, True, True, True, True, True, True, True *)
answered Mar 6 at 17:33
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
For lists of positive values, all of the usual ways of calculating the GeometricMean are equivalent.
test[list_] := Simplify[
Equal @@
(#[list] & /@
GeometricMean,
(Times @@ #)^(1/Length[#]) &,
E^Mean[Log[#]] &,
Limit[Mean[#^t]^(1/t), t -> 0] &)]
Simplify[test[#], Thread[# > 0]] &@a, b, c, d
(* True *)
SeedRandom[0]
test /@ RandomReal[10^-10, 10, 10, 10]
(* True, True, True, True, True, True, True, True, True, True *)
answered Mar 6 at 17:33
Bob HanlonBob Hanlon
61.1k33598
answered Mar 6 at 17:33
Bob HanlonBob Hanlon
61.1k33598
answered Mar 6 at 17:33
Bob HanlonBob Hanlon
61.1k33598
answered Mar 6 at 17:33
Bob HanlonBob Hanlon
61.1k33598
61.1k33598
$begingroup$
My hope, based on the initial observation thatGeometricMean[-4,-4]gave -4 was that Mathematica implemented a sensible generalisation ofGeometricMeanto values other than positive reals (in particular, complex numbers).
$endgroup$
– mikado
Mar 6 at 20:26
add a comment |
$begingroup$
My hope, based on the initial observation thatGeometricMean[-4,-4]gave -4 was that Mathematica implemented a sensible generalisation ofGeometricMeanto values other than positive reals (in particular, complex numbers).
$endgroup$
– mikado
Mar 6 at 20:26
$begingroup$
My hope, based on the initial observation that
GeometricMean[-4,-4] gave -4 was that Mathematica implemented a sensible generalisation of GeometricMean to values other than positive reals (in particular, complex numbers).$endgroup$
– mikado
Mar 6 at 20:26
$begingroup$
My hope, based on the initial observation that
GeometricMean[-4,-4] gave -4 was that Mathematica implemented a sensible generalisation of GeometricMean to values other than positive reals (in particular, complex numbers).$endgroup$
– mikado
Mar 6 at 20:26
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
Mar 4 at 20:43
$begingroup$
@Coolwater, but note that
FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@-4, -5returnsFalse, so I don't think your explanation generalises.$endgroup$
– mikado
Mar 4 at 20:47
2
$begingroup$
"The geometric mean applies only to positive numbers." (
Wiki(Times @@ #)^(1/Length@#) == Exp@Mean@Log@# &@ RandomReal[10, 100]evaluates toTrue$endgroup$
– Bob Hanlon
Mar 4 at 21:19
$begingroup$
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
Mar 5 at 1:08