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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?)










share|improve this question











$endgroup$












  • $begingroup$
    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
    22 hours ago


















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?)










share|improve this question











$endgroup$












  • $begingroup$
    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
    22 hours ago
















4












4








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?)










share|improve this question











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday







mikado

















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 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
    22 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
    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
    22 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
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
22 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
22 hours ago












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











$endgroup$













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    active

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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











    $endgroup$


















      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











      $endgroup$
















        6












        6








        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











        $endgroup$



        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.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited yesterday

























        answered yesterday









        SomosSomos

        1,36019




        1,36019






























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