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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

4

$begingroup$

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

share|improve this question

edited yesterday

mikado

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 that FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5} returns False, 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 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
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

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

share|improve this question

edited yesterday

mikado

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 that FullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5} returns False, 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 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
  
  
  

  
  
  
  

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

share|improve this question

edited yesterday

mikado

asked yesterday

mikadomikado

6,6071929

$endgroup$

GeometricMean[Table[x[i],{i,1,n}]]==Product[x[i],{i,1,n}]^(1/n)

GeometricMean[{-4, -4}]

(* -4 *)

share|improve this question

edited yesterday

mikado

asked yesterday

mikadomikado

6,6071929

share|improve this question

edited yesterday

mikado

asked yesterday

mikadomikado

6,6071929

asked yesterday

mikadomikado

6,6071929

asked yesterday

mikadomikado

6,6071929

1 Answer
1

active

oldest

votes

6

$begingroup$

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.

share|improve this answer

edited yesterday

answered yesterday

SomosSomos

1,36019

$endgroup$

add a comment | 

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6

$begingroup$

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.

share|improve this answer

edited yesterday

answered yesterday

SomosSomos

1,36019

$endgroup$

add a comment | 

6

$begingroup$

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.

share|improve this answer

edited yesterday

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.

share|improve this answer

edited yesterday

answered yesterday

SomosSomos

1,36019

$endgroup$

GeometricMean[ c data ] == c GeometricMean[ data ]

share|improve this answer

edited yesterday

answered yesterday

SomosSomos

1,36019

answered yesterday

SomosSomos

1,36019

answered yesterday

SomosSomos

1,36019