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 4 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 4 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
4 hours ago
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
4 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
4 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
24 mins 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 4 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 4 hours ago
mikadomikado
6,6021929
asked 4 hours ago
mikadomikado
6,6021929
edited 4 hours ago
mikado
asked 4 hours ago
mikadomikado
6,6021929
asked 4 hours ago
mikadomikado
6,6021929
asked 4 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
4 hours ago
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
4 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
4 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
24 mins 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
4 hours ago
$begingroup$
@Coolwater, but note thatFullSimplify[Exp[Mean[Log[#]]] == GeometricMean[#]] &@{-4, -5}returnsFalse, so I don't think your explanation generalises.
$endgroup$
– mikado
4 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
4 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
24 mins 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
4 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
4 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
4 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
4 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
4 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
4 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
24 mins 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
24 mins ago
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 4 hours ago
SomosSomos
1,33519
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192613%2fgeometricmean-definition%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
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.
answered 4 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 4 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 4 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 4 hours ago
SomosSomos
1,33519
edited 3 hours ago
answered 4 hours ago
SomosSomos
1,33519
answered 4 hours ago
SomosSomos
1,33519
answered 4 hours ago
SomosSomos
1,33519
1,33519
add a comment |
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192613%2fgeometricmean-definition%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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
4 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
4 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
4 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
24 mins ago