Is $sqrt{x}$ an even function?
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From my Calculus class I don't think that I would say that the function $f:[0.infty) to mathbb{R}$ given by $f(x) = sqrt{x}$ is an even function. The graph isn't symmetric about the $y$-axis.
But according to my book, and to Wikipedia a function $f$ is even if $f(-x) = f(x)$ for all $x$ and $-x$ in the domain of $f$. So since the domain of the square root function is $[0,infty)$, the function would satisfy this.
Is that correct?
functions definition even-and-odd-functions
edited Dec 10 '18 at 22:16
José Carlos Santos
158k22126228
asked Aug 23 '17 at 16:10
John DoeJohn Doe
27021346
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|
show 1 more comment
$begingroup$
From my Calculus class I don't think that I would say that the function $f:[0.infty) to mathbb{R}$ given by $f(x) = sqrt{x}$ is an even function. The graph isn't symmetric about the $y$-axis.
But according to my book, and to Wikipedia a function $f$ is even if $f(-x) = f(x)$ for all $x$ and $-x$ in the domain of $f$. So since the domain of the square root function is $[0,infty)$, the function would satisfy this.
Is that correct?
functions definition even-and-odd-functions
edited Dec 10 '18 at 22:16
José Carlos Santos
158k22126228
asked Aug 23 '17 at 16:10
John DoeJohn Doe
27021346
$endgroup$
7
$begingroup$
No, since $-x$ isn't even in the domain of $f$ for all values $x>0$.
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– Guillermo Angeris
Aug 23 '17 at 16:12
3
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No. It isn't even defined for $x<0$, so it can't be even. Even if you extend to the complex plane, $sqrt{-x} = isqrt{x} neq sqrt{x}$. They are on different axes!
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– Brevan Ellefsen
Aug 23 '17 at 16:12
3
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I think that's what OP is getting at.. it sounds like the function is "vacuously even."
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– tilper
Aug 23 '17 at 16:13
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@tilper I suppose we could modify the definition of an even function to say that, at whatever $x$ where $f(-x)$ is defined, $f(x) = f(-x)$, but I can't think of any nice properties this has
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– Brevan Ellefsen
Aug 23 '17 at 16:15
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@BrevanEllefsen, I think, keeping in mind the geometric definition (graph is symmetric over the $y$-axis), it would be helpful to add the restriction that $f$ must be defined on an interval centered at $x=0$. A function not defined on such an interval will then not be even. But then I guess even that is "too strict" since $f$ could be defined on, say, a union of intervals "centered" at $x=0$, e.g. $f : [-4,-3] cup [-2,-1] cup [1,2] cup [3,4] to {1}$ is even. Meh.
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– tilper
Aug 23 '17 at 16:21
|
show 1 more comment
$begingroup$
From my Calculus class I don't think that I would say that the function $f:[0.infty) to mathbb{R}$ given by $f(x) = sqrt{x}$ is an even function. The graph isn't symmetric about the $y$-axis.
But according to my book, and to Wikipedia a function $f$ is even if $f(-x) = f(x)$ for all $x$ and $-x$ in the domain of $f$. So since the domain of the square root function is $[0,infty)$, the function would satisfy this.
Is that correct?
functions definition even-and-odd-functions
edited Dec 10 '18 at 22:16
José Carlos Santos
158k22126228
asked Aug 23 '17 at 16:10
John DoeJohn Doe
27021346
$endgroup$
From my Calculus class I don't think that I would say that the function $f:[0.infty) to mathbb{R}$ given by $f(x) = sqrt{x}$ is an even function. The graph isn't symmetric about the $y$-axis.
But according to my book, and to Wikipedia a function $f$ is even if $f(-x) = f(x)$ for all $x$ and $-x$ in the domain of $f$. So since the domain of the square root function is $[0,infty)$, the function would satisfy this.
Is that correct?
functions definition even-and-odd-functions
functions definition even-and-odd-functions
edited Dec 10 '18 at 22:16
José Carlos Santos
158k22126228
asked Aug 23 '17 at 16:10
John DoeJohn Doe
27021346
edited Dec 10 '18 at 22:16
José Carlos Santos
158k22126228
asked Aug 23 '17 at 16:10
John DoeJohn Doe
27021346
edited Dec 10 '18 at 22:16
José Carlos Santos
158k22126228
edited Dec 10 '18 at 22:16
José Carlos Santos
158k22126228
edited Dec 10 '18 at 22:16
José Carlos Santos
158k22126228
158k22126228
asked Aug 23 '17 at 16:10
John DoeJohn Doe
27021346
asked Aug 23 '17 at 16:10
John DoeJohn Doe
27021346
asked Aug 23 '17 at 16:10
John DoeJohn Doe
27021346
27021346
7
$begingroup$
No, since $-x$ isn't even in the domain of $f$ for all values $x>0$.
$endgroup$
– Guillermo Angeris
Aug 23 '17 at 16:12
3
$begingroup$
No. It isn't even defined for $x<0$, so it can't be even. Even if you extend to the complex plane, $sqrt{-x} = isqrt{x} neq sqrt{x}$. They are on different axes!
$endgroup$
– Brevan Ellefsen
Aug 23 '17 at 16:12
3
$begingroup$
I think that's what OP is getting at.. it sounds like the function is "vacuously even."
$endgroup$
– tilper
Aug 23 '17 at 16:13
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@tilper I suppose we could modify the definition of an even function to say that, at whatever $x$ where $f(-x)$ is defined, $f(x) = f(-x)$, but I can't think of any nice properties this has
$endgroup$
– Brevan Ellefsen
Aug 23 '17 at 16:15
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@BrevanEllefsen, I think, keeping in mind the geometric definition (graph is symmetric over the $y$-axis), it would be helpful to add the restriction that $f$ must be defined on an interval centered at $x=0$. A function not defined on such an interval will then not be even. But then I guess even that is "too strict" since $f$ could be defined on, say, a union of intervals "centered" at $x=0$, e.g. $f : [-4,-3] cup [-2,-1] cup [1,2] cup [3,4] to {1}$ is even. Meh.
$endgroup$
– tilper
Aug 23 '17 at 16:21
|
show 1 more comment
7
$begingroup$
No, since $-x$ isn't even in the domain of $f$ for all values $x>0$.
$endgroup$
– Guillermo Angeris
Aug 23 '17 at 16:12
3
$begingroup$
No. It isn't even defined for $x<0$, so it can't be even. Even if you extend to the complex plane, $sqrt{-x} = isqrt{x} neq sqrt{x}$. They are on different axes!
$endgroup$
– Brevan Ellefsen
Aug 23 '17 at 16:12
3
$begingroup$
I think that's what OP is getting at.. it sounds like the function is "vacuously even."
$endgroup$
– tilper
Aug 23 '17 at 16:13
$begingroup$
@tilper I suppose we could modify the definition of an even function to say that, at whatever $x$ where $f(-x)$ is defined, $f(x) = f(-x)$, but I can't think of any nice properties this has
$endgroup$
– Brevan Ellefsen
Aug 23 '17 at 16:15
$begingroup$
@BrevanEllefsen, I think, keeping in mind the geometric definition (graph is symmetric over the $y$-axis), it would be helpful to add the restriction that $f$ must be defined on an interval centered at $x=0$. A function not defined on such an interval will then not be even. But then I guess even that is "too strict" since $f$ could be defined on, say, a union of intervals "centered" at $x=0$, e.g. $f : [-4,-3] cup [-2,-1] cup [1,2] cup [3,4] to {1}$ is even. Meh.
$endgroup$
– tilper
Aug 23 '17 at 16:21
7
7
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No, since $-x$ isn't even in the domain of $f$ for all values $x>0$.
$endgroup$
– Guillermo Angeris
Aug 23 '17 at 16:12
$begingroup$
No, since $-x$ isn't even in the domain of $f$ for all values $x>0$.
$endgroup$
– Guillermo Angeris
Aug 23 '17 at 16:12
3
3
$begingroup$
No. It isn't even defined for $x<0$, so it can't be even. Even if you extend to the complex plane, $sqrt{-x} = isqrt{x} neq sqrt{x}$. They are on different axes!
$endgroup$
– Brevan Ellefsen
Aug 23 '17 at 16:12
$begingroup$
No. It isn't even defined for $x<0$, so it can't be even. Even if you extend to the complex plane, $sqrt{-x} = isqrt{x} neq sqrt{x}$. They are on different axes!
$endgroup$
– Brevan Ellefsen
Aug 23 '17 at 16:12
3
3
$begingroup$
I think that's what OP is getting at.. it sounds like the function is "vacuously even."
$endgroup$
– tilper
Aug 23 '17 at 16:13
$begingroup$
I think that's what OP is getting at.. it sounds like the function is "vacuously even."
$endgroup$
– tilper
Aug 23 '17 at 16:13
$begingroup$
@tilper I suppose we could modify the definition of an even function to say that, at whatever $x$ where $f(-x)$ is defined, $f(x) = f(-x)$, but I can't think of any nice properties this has
$endgroup$
– Brevan Ellefsen
Aug 23 '17 at 16:15
$begingroup$
@tilper I suppose we could modify the definition of an even function to say that, at whatever $x$ where $f(-x)$ is defined, $f(x) = f(-x)$, but I can't think of any nice properties this has
$endgroup$
– Brevan Ellefsen
Aug 23 '17 at 16:15
$begingroup$
@BrevanEllefsen, I think, keeping in mind the geometric definition (graph is symmetric over the $y$-axis), it would be helpful to add the restriction that $f$ must be defined on an interval centered at $x=0$. A function not defined on such an interval will then not be even. But then I guess even that is "too strict" since $f$ could be defined on, say, a union of intervals "centered" at $x=0$, e.g. $f : [-4,-3] cup [-2,-1] cup [1,2] cup [3,4] to {1}$ is even. Meh.
$endgroup$
– tilper
Aug 23 '17 at 16:21
$begingroup$
@BrevanEllefsen, I think, keeping in mind the geometric definition (graph is symmetric over the $y$-axis), it would be helpful to add the restriction that $f$ must be defined on an interval centered at $x=0$. A function not defined on such an interval will then not be even. But then I guess even that is "too strict" since $f$ could be defined on, say, a union of intervals "centered" at $x=0$, e.g. $f : [-4,-3] cup [-2,-1] cup [1,2] cup [3,4] to {1}$ is even. Meh.
$endgroup$
– tilper
Aug 23 '17 at 16:21
|
show 1 more comment
7 Answers
7
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oldest
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I think that by reading carefully and correctly you have found a subtle problem with the wikipedia definition. It is indeed true that "whenever both $x$ and $-x$ are in the domain of the square root function the function values agree" because (as you know quite well) only $x=0$ satisfies that hypothesis. So according to the strict reading of the definition, the square root function is even.
But (as you also realize) that's not the intent of the definition. Implicit in that definition is the assumption that the domain is symmetric about $0$, so that it contains $-x$ whenever it contains $x$.
I disagree with most (but by no means all) of the other answers and comments. They are just reminding you that the domain is $[0, infty)$.
Good for you for paying this kind of attention.
answered Aug 23 '17 at 16:17
Ethan BolkerEthan Bolker
42.5k549112
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Note. the wikepedia article did not say for all $(x,-x)$ pairs where $x$ and $-x$ are both in the domain. It said: for all $x$ in the domain for all $-x$ in the domain. Thus to be even $f(x) = f(-x)$ whenever $x$ is in the domain or if $-x$ is in the domain. And if one is in the domain and the other isn't this can not be true.
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– fleablood
Dec 10 '18 at 22:36
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Also does it matter what the Wikipedia definition is, if no other textbook uses it? Wikipedia is not a textbook and if it's definition disagrees with the accepted standard then we should simply say: Wikipedia is wrong. (if it is as you and the OP interpreted it)
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– fleablood
Dec 10 '18 at 22:38
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@fleablood Indeed wikipedia is sometimes wrong. It might be here - I don't want to spend the time to parse it really carefully. In general, I find it pretty reliable on mathematics. And textbooks can be wrong too.In this particular example I wanted to respond because the OP had thought carefully.
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– Ethan Bolker
Dec 10 '18 at 23:56
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He had. According to HIS/HER definition he was correct. But I don't think his/her definition is that universal. He did think carefully though.
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– fleablood
Dec 11 '18 at 0:26
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@fleablood I'm usually careful about pronouns. I think it safe that this OP was "he". If not, then we're supposed to use a singular "they" these days rather than "he/she". Mathematics is easier ...
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– Ethan Bolker
Dec 11 '18 at 0:34
add a comment |
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The language used at that Wikipedia article is innacurate. A good definition would be: if $Dsubsetmathbb R$, a function $fcolon Dlongrightarrowmathbb R$ is even if $(forall xin D):-xin Dtext{ and }f(-x)=f(x)$. Under this definition, I hope that you agree that the square root function (or, for that matter, any function whose domain is $[0,+infty)$) is not even.
answered Aug 23 '17 at 16:20
José Carlos SantosJosé Carlos Santos
158k22126228
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add a comment |
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I think the wikipedia definition is rescued by this sentence:
The concept of evenness or oddness is defined for functions whose domain and image both have an additive inverse.
The domain $[0,infty)$ does not have an additive inverse. So the remainder of the article doesn't even apply to it, and it's not surprising that attempting to apply it results in nonsense.
On the other hand, if $D$ is a subset of $Bbb{R}$ which is symmetric about $0$, we can define an additive inverse on $D$ by restriction from $Bbb{R}$, so is reasonable to ask whether functions from $D$ to $Bbb{R}$ are even.
answered Aug 23 '17 at 16:30
MicahMicah
30k1364106
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If $x = 4$, then it is not true that $f(x) = f(-x)$, because $f(-x)$ is not even defined. Therefore, $f$ is is not even. However the function $f(x) = sqrt{|x|}$ is even.
answered Aug 23 '17 at 16:13
Jake MirraJake Mirra
716
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add a comment |
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No because if $x>0$ then $f(-x)$ doesn't even make sense (as a real number). The definition given in Wikipedia is for functions of a real value, i.e, functions from $mathbb{R}$ to $mathbb{R}$. In your case that's not satisfied. You can, however, enlarge the definition as follows: "A function $f:Dtomathbb{R}$ is even if $forall xin D,-xin D,land f(-x)=f(x)$". In your case, the domain is not symmetric, so the function is not even.
answered Aug 23 '17 at 16:15
ScientificaScientifica
6,73141335
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HINT: $f(x)=sqrt{x}$ is defined for all $x$ with $$0le x<+infty$$ and $$sqrt{-x}=-sqrt{x}$$ makes no sense and $$sqrt{-x}=sqrt{x}$$ also
answered Aug 23 '17 at 16:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
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add a comment |
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It doesn't matter.
Function parity is useful for functions defined both in the negatives and positives.
Whatever the classification of $sqrt x$, it is of no use.
answered Aug 23 '17 at 17:28
Yves DaoustYves Daoust
126k672226
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7 Answers
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7 Answers
7
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$begingroup$
I think that by reading carefully and correctly you have found a subtle problem with the wikipedia definition. It is indeed true that "whenever both $x$ and $-x$ are in the domain of the square root function the function values agree" because (as you know quite well) only $x=0$ satisfies that hypothesis. So according to the strict reading of the definition, the square root function is even.
But (as you also realize) that's not the intent of the definition. Implicit in that definition is the assumption that the domain is symmetric about $0$, so that it contains $-x$ whenever it contains $x$.
I disagree with most (but by no means all) of the other answers and comments. They are just reminding you that the domain is $[0, infty)$.
Good for you for paying this kind of attention.
answered Aug 23 '17 at 16:17
Ethan BolkerEthan Bolker
42.5k549112
$endgroup$
$begingroup$
Note. the wikepedia article did not say for all $(x,-x)$ pairs where $x$ and $-x$ are both in the domain. It said: for all $x$ in the domain for all $-x$ in the domain. Thus to be even $f(x) = f(-x)$ whenever $x$ is in the domain or if $-x$ is in the domain. And if one is in the domain and the other isn't this can not be true.
$endgroup$
– fleablood
Dec 10 '18 at 22:36
$begingroup$
Also does it matter what the Wikipedia definition is, if no other textbook uses it? Wikipedia is not a textbook and if it's definition disagrees with the accepted standard then we should simply say: Wikipedia is wrong. (if it is as you and the OP interpreted it)
$endgroup$
– fleablood
Dec 10 '18 at 22:38
$begingroup$
@fleablood Indeed wikipedia is sometimes wrong. It might be here - I don't want to spend the time to parse it really carefully. In general, I find it pretty reliable on mathematics. And textbooks can be wrong too.In this particular example I wanted to respond because the OP had thought carefully.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 23:56
$begingroup$
He had. According to HIS/HER definition he was correct. But I don't think his/her definition is that universal. He did think carefully though.
$endgroup$
– fleablood
Dec 11 '18 at 0:26
$begingroup$
@fleablood I'm usually careful about pronouns. I think it safe that this OP was "he". If not, then we're supposed to use a singular "they" these days rather than "he/she". Mathematics is easier ...
$endgroup$
– Ethan Bolker
Dec 11 '18 at 0:34
add a comment |
$begingroup$
I think that by reading carefully and correctly you have found a subtle problem with the wikipedia definition. It is indeed true that "whenever both $x$ and $-x$ are in the domain of the square root function the function values agree" because (as you know quite well) only $x=0$ satisfies that hypothesis. So according to the strict reading of the definition, the square root function is even.
But (as you also realize) that's not the intent of the definition. Implicit in that definition is the assumption that the domain is symmetric about $0$, so that it contains $-x$ whenever it contains $x$.
I disagree with most (but by no means all) of the other answers and comments. They are just reminding you that the domain is $[0, infty)$.
Good for you for paying this kind of attention.
answered Aug 23 '17 at 16:17
Ethan BolkerEthan Bolker
42.5k549112
$endgroup$
$begingroup$
Note. the wikepedia article did not say for all $(x,-x)$ pairs where $x$ and $-x$ are both in the domain. It said: for all $x$ in the domain for all $-x$ in the domain. Thus to be even $f(x) = f(-x)$ whenever $x$ is in the domain or if $-x$ is in the domain. And if one is in the domain and the other isn't this can not be true.
$endgroup$
– fleablood
Dec 10 '18 at 22:36
$begingroup$
Also does it matter what the Wikipedia definition is, if no other textbook uses it? Wikipedia is not a textbook and if it's definition disagrees with the accepted standard then we should simply say: Wikipedia is wrong. (if it is as you and the OP interpreted it)
$endgroup$
– fleablood
Dec 10 '18 at 22:38
$begingroup$
@fleablood Indeed wikipedia is sometimes wrong. It might be here - I don't want to spend the time to parse it really carefully. In general, I find it pretty reliable on mathematics. And textbooks can be wrong too.In this particular example I wanted to respond because the OP had thought carefully.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 23:56
$begingroup$
He had. According to HIS/HER definition he was correct. But I don't think his/her definition is that universal. He did think carefully though.
$endgroup$
– fleablood
Dec 11 '18 at 0:26
$begingroup$
@fleablood I'm usually careful about pronouns. I think it safe that this OP was "he". If not, then we're supposed to use a singular "they" these days rather than "he/she". Mathematics is easier ...
$endgroup$
– Ethan Bolker
Dec 11 '18 at 0:34
add a comment |
$begingroup$
I think that by reading carefully and correctly you have found a subtle problem with the wikipedia definition. It is indeed true that "whenever both $x$ and $-x$ are in the domain of the square root function the function values agree" because (as you know quite well) only $x=0$ satisfies that hypothesis. So according to the strict reading of the definition, the square root function is even.
But (as you also realize) that's not the intent of the definition. Implicit in that definition is the assumption that the domain is symmetric about $0$, so that it contains $-x$ whenever it contains $x$.
I disagree with most (but by no means all) of the other answers and comments. They are just reminding you that the domain is $[0, infty)$.
Good for you for paying this kind of attention.
answered Aug 23 '17 at 16:17
Ethan BolkerEthan Bolker
42.5k549112
$endgroup$
I think that by reading carefully and correctly you have found a subtle problem with the wikipedia definition. It is indeed true that "whenever both $x$ and $-x$ are in the domain of the square root function the function values agree" because (as you know quite well) only $x=0$ satisfies that hypothesis. So according to the strict reading of the definition, the square root function is even.
But (as you also realize) that's not the intent of the definition. Implicit in that definition is the assumption that the domain is symmetric about $0$, so that it contains $-x$ whenever it contains $x$.
I disagree with most (but by no means all) of the other answers and comments. They are just reminding you that the domain is $[0, infty)$.
Good for you for paying this kind of attention.
answered Aug 23 '17 at 16:17
Ethan BolkerEthan Bolker
42.5k549112
edited Aug 23 '17 at 20:10
answered Aug 23 '17 at 16:17
Ethan BolkerEthan Bolker
42.5k549112
answered Aug 23 '17 at 16:17
Ethan BolkerEthan Bolker
42.5k549112
answered Aug 23 '17 at 16:17
Ethan BolkerEthan Bolker
42.5k549112
42.5k549112
$begingroup$
Note. the wikepedia article did not say for all $(x,-x)$ pairs where $x$ and $-x$ are both in the domain. It said: for all $x$ in the domain for all $-x$ in the domain. Thus to be even $f(x) = f(-x)$ whenever $x$ is in the domain or if $-x$ is in the domain. And if one is in the domain and the other isn't this can not be true.
$endgroup$
– fleablood
Dec 10 '18 at 22:36
$begingroup$
Also does it matter what the Wikipedia definition is, if no other textbook uses it? Wikipedia is not a textbook and if it's definition disagrees with the accepted standard then we should simply say: Wikipedia is wrong. (if it is as you and the OP interpreted it)
$endgroup$
– fleablood
Dec 10 '18 at 22:38
$begingroup$
@fleablood Indeed wikipedia is sometimes wrong. It might be here - I don't want to spend the time to parse it really carefully. In general, I find it pretty reliable on mathematics. And textbooks can be wrong too.In this particular example I wanted to respond because the OP had thought carefully.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 23:56
$begingroup$
He had. According to HIS/HER definition he was correct. But I don't think his/her definition is that universal. He did think carefully though.
$endgroup$
– fleablood
Dec 11 '18 at 0:26
$begingroup$
@fleablood I'm usually careful about pronouns. I think it safe that this OP was "he". If not, then we're supposed to use a singular "they" these days rather than "he/she". Mathematics is easier ...
$endgroup$
– Ethan Bolker
Dec 11 '18 at 0:34
add a comment |
$begingroup$
Note. the wikepedia article did not say for all $(x,-x)$ pairs where $x$ and $-x$ are both in the domain. It said: for all $x$ in the domain for all $-x$ in the domain. Thus to be even $f(x) = f(-x)$ whenever $x$ is in the domain or if $-x$ is in the domain. And if one is in the domain and the other isn't this can not be true.
$endgroup$
– fleablood
Dec 10 '18 at 22:36
$begingroup$
Also does it matter what the Wikipedia definition is, if no other textbook uses it? Wikipedia is not a textbook and if it's definition disagrees with the accepted standard then we should simply say: Wikipedia is wrong. (if it is as you and the OP interpreted it)
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– fleablood
Dec 10 '18 at 22:38
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@fleablood Indeed wikipedia is sometimes wrong. It might be here - I don't want to spend the time to parse it really carefully. In general, I find it pretty reliable on mathematics. And textbooks can be wrong too.In this particular example I wanted to respond because the OP had thought carefully.
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– Ethan Bolker
Dec 10 '18 at 23:56
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He had. According to HIS/HER definition he was correct. But I don't think his/her definition is that universal. He did think carefully though.
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– fleablood
Dec 11 '18 at 0:26
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@fleablood I'm usually careful about pronouns. I think it safe that this OP was "he". If not, then we're supposed to use a singular "they" these days rather than "he/she". Mathematics is easier ...
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– Ethan Bolker
Dec 11 '18 at 0:34
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Note. the wikepedia article did not say for all $(x,-x)$ pairs where $x$ and $-x$ are both in the domain. It said: for all $x$ in the domain for all $-x$ in the domain. Thus to be even $f(x) = f(-x)$ whenever $x$ is in the domain or if $-x$ is in the domain. And if one is in the domain and the other isn't this can not be true.
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– fleablood
Dec 10 '18 at 22:36
$begingroup$
Note. the wikepedia article did not say for all $(x,-x)$ pairs where $x$ and $-x$ are both in the domain. It said: for all $x$ in the domain for all $-x$ in the domain. Thus to be even $f(x) = f(-x)$ whenever $x$ is in the domain or if $-x$ is in the domain. And if one is in the domain and the other isn't this can not be true.
$endgroup$
– fleablood
Dec 10 '18 at 22:36
$begingroup$
Also does it matter what the Wikipedia definition is, if no other textbook uses it? Wikipedia is not a textbook and if it's definition disagrees with the accepted standard then we should simply say: Wikipedia is wrong. (if it is as you and the OP interpreted it)
$endgroup$
– fleablood
Dec 10 '18 at 22:38
$begingroup$
Also does it matter what the Wikipedia definition is, if no other textbook uses it? Wikipedia is not a textbook and if it's definition disagrees with the accepted standard then we should simply say: Wikipedia is wrong. (if it is as you and the OP interpreted it)
$endgroup$
– fleablood
Dec 10 '18 at 22:38
$begingroup$
@fleablood Indeed wikipedia is sometimes wrong. It might be here - I don't want to spend the time to parse it really carefully. In general, I find it pretty reliable on mathematics. And textbooks can be wrong too.In this particular example I wanted to respond because the OP had thought carefully.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 23:56
$begingroup$
@fleablood Indeed wikipedia is sometimes wrong. It might be here - I don't want to spend the time to parse it really carefully. In general, I find it pretty reliable on mathematics. And textbooks can be wrong too.In this particular example I wanted to respond because the OP had thought carefully.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 23:56
$begingroup$
He had. According to HIS/HER definition he was correct. But I don't think his/her definition is that universal. He did think carefully though.
$endgroup$
– fleablood
Dec 11 '18 at 0:26
$begingroup$
He had. According to HIS/HER definition he was correct. But I don't think his/her definition is that universal. He did think carefully though.
$endgroup$
– fleablood
Dec 11 '18 at 0:26
$begingroup$
@fleablood I'm usually careful about pronouns. I think it safe that this OP was "he". If not, then we're supposed to use a singular "they" these days rather than "he/she". Mathematics is easier ...
$endgroup$
– Ethan Bolker
Dec 11 '18 at 0:34
$begingroup$
@fleablood I'm usually careful about pronouns. I think it safe that this OP was "he". If not, then we're supposed to use a singular "they" these days rather than "he/she". Mathematics is easier ...
$endgroup$
– Ethan Bolker
Dec 11 '18 at 0:34
add a comment |
$begingroup$
The language used at that Wikipedia article is innacurate. A good definition would be: if $Dsubsetmathbb R$, a function $fcolon Dlongrightarrowmathbb R$ is even if $(forall xin D):-xin Dtext{ and }f(-x)=f(x)$. Under this definition, I hope that you agree that the square root function (or, for that matter, any function whose domain is $[0,+infty)$) is not even.
answered Aug 23 '17 at 16:20
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
add a comment |
$begingroup$
The language used at that Wikipedia article is innacurate. A good definition would be: if $Dsubsetmathbb R$, a function $fcolon Dlongrightarrowmathbb R$ is even if $(forall xin D):-xin Dtext{ and }f(-x)=f(x)$. Under this definition, I hope that you agree that the square root function (or, for that matter, any function whose domain is $[0,+infty)$) is not even.
answered Aug 23 '17 at 16:20
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
add a comment |
$begingroup$
The language used at that Wikipedia article is innacurate. A good definition would be: if $Dsubsetmathbb R$, a function $fcolon Dlongrightarrowmathbb R$ is even if $(forall xin D):-xin Dtext{ and }f(-x)=f(x)$. Under this definition, I hope that you agree that the square root function (or, for that matter, any function whose domain is $[0,+infty)$) is not even.
answered Aug 23 '17 at 16:20
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
The language used at that Wikipedia article is innacurate. A good definition would be: if $Dsubsetmathbb R$, a function $fcolon Dlongrightarrowmathbb R$ is even if $(forall xin D):-xin Dtext{ and }f(-x)=f(x)$. Under this definition, I hope that you agree that the square root function (or, for that matter, any function whose domain is $[0,+infty)$) is not even.
answered Aug 23 '17 at 16:20
José Carlos SantosJosé Carlos Santos
158k22126228
answered Aug 23 '17 at 16:20
José Carlos SantosJosé Carlos Santos
158k22126228
answered Aug 23 '17 at 16:20
José Carlos SantosJosé Carlos Santos
158k22126228
answered Aug 23 '17 at 16:20
José Carlos SantosJosé Carlos Santos
158k22126228
158k22126228
add a comment |
add a comment |
$begingroup$
I think the wikipedia definition is rescued by this sentence:
The concept of evenness or oddness is defined for functions whose domain and image both have an additive inverse.
The domain $[0,infty)$ does not have an additive inverse. So the remainder of the article doesn't even apply to it, and it's not surprising that attempting to apply it results in nonsense.
On the other hand, if $D$ is a subset of $Bbb{R}$ which is symmetric about $0$, we can define an additive inverse on $D$ by restriction from $Bbb{R}$, so is reasonable to ask whether functions from $D$ to $Bbb{R}$ are even.
answered Aug 23 '17 at 16:30
MicahMicah
30k1364106
$endgroup$
add a comment |
$begingroup$
I think the wikipedia definition is rescued by this sentence:
The concept of evenness or oddness is defined for functions whose domain and image both have an additive inverse.
The domain $[0,infty)$ does not have an additive inverse. So the remainder of the article doesn't even apply to it, and it's not surprising that attempting to apply it results in nonsense.
On the other hand, if $D$ is a subset of $Bbb{R}$ which is symmetric about $0$, we can define an additive inverse on $D$ by restriction from $Bbb{R}$, so is reasonable to ask whether functions from $D$ to $Bbb{R}$ are even.
answered Aug 23 '17 at 16:30
MicahMicah
30k1364106
$endgroup$
add a comment |
$begingroup$
I think the wikipedia definition is rescued by this sentence:
The concept of evenness or oddness is defined for functions whose domain and image both have an additive inverse.
The domain $[0,infty)$ does not have an additive inverse. So the remainder of the article doesn't even apply to it, and it's not surprising that attempting to apply it results in nonsense.
On the other hand, if $D$ is a subset of $Bbb{R}$ which is symmetric about $0$, we can define an additive inverse on $D$ by restriction from $Bbb{R}$, so is reasonable to ask whether functions from $D$ to $Bbb{R}$ are even.
answered Aug 23 '17 at 16:30
MicahMicah
30k1364106
$endgroup$
I think the wikipedia definition is rescued by this sentence:
The concept of evenness or oddness is defined for functions whose domain and image both have an additive inverse.
The domain $[0,infty)$ does not have an additive inverse. So the remainder of the article doesn't even apply to it, and it's not surprising that attempting to apply it results in nonsense.
On the other hand, if $D$ is a subset of $Bbb{R}$ which is symmetric about $0$, we can define an additive inverse on $D$ by restriction from $Bbb{R}$, so is reasonable to ask whether functions from $D$ to $Bbb{R}$ are even.
answered Aug 23 '17 at 16:30
MicahMicah
30k1364106
edited Aug 23 '17 at 17:18
answered Aug 23 '17 at 16:30
MicahMicah
30k1364106
answered Aug 23 '17 at 16:30
MicahMicah
30k1364106
answered Aug 23 '17 at 16:30
MicahMicah
30k1364106
30k1364106
add a comment |
add a comment |
$begingroup$
If $x = 4$, then it is not true that $f(x) = f(-x)$, because $f(-x)$ is not even defined. Therefore, $f$ is is not even. However the function $f(x) = sqrt{|x|}$ is even.
answered Aug 23 '17 at 16:13
Jake MirraJake Mirra
716
$endgroup$
add a comment |
$begingroup$
If $x = 4$, then it is not true that $f(x) = f(-x)$, because $f(-x)$ is not even defined. Therefore, $f$ is is not even. However the function $f(x) = sqrt{|x|}$ is even.
answered Aug 23 '17 at 16:13
Jake MirraJake Mirra
716
$endgroup$
add a comment |
$begingroup$
If $x = 4$, then it is not true that $f(x) = f(-x)$, because $f(-x)$ is not even defined. Therefore, $f$ is is not even. However the function $f(x) = sqrt{|x|}$ is even.
answered Aug 23 '17 at 16:13
Jake MirraJake Mirra
716
$endgroup$
If $x = 4$, then it is not true that $f(x) = f(-x)$, because $f(-x)$ is not even defined. Therefore, $f$ is is not even. However the function $f(x) = sqrt{|x|}$ is even.
answered Aug 23 '17 at 16:13
Jake MirraJake Mirra
716
answered Aug 23 '17 at 16:13
Jake MirraJake Mirra
716
answered Aug 23 '17 at 16:13
Jake MirraJake Mirra
716
answered Aug 23 '17 at 16:13
Jake MirraJake Mirra
716
716
add a comment |
add a comment |
$begingroup$
No because if $x>0$ then $f(-x)$ doesn't even make sense (as a real number). The definition given in Wikipedia is for functions of a real value, i.e, functions from $mathbb{R}$ to $mathbb{R}$. In your case that's not satisfied. You can, however, enlarge the definition as follows: "A function $f:Dtomathbb{R}$ is even if $forall xin D,-xin D,land f(-x)=f(x)$". In your case, the domain is not symmetric, so the function is not even.
answered Aug 23 '17 at 16:15
ScientificaScientifica
6,73141335
$endgroup$
add a comment |
$begingroup$
No because if $x>0$ then $f(-x)$ doesn't even make sense (as a real number). The definition given in Wikipedia is for functions of a real value, i.e, functions from $mathbb{R}$ to $mathbb{R}$. In your case that's not satisfied. You can, however, enlarge the definition as follows: "A function $f:Dtomathbb{R}$ is even if $forall xin D,-xin D,land f(-x)=f(x)$". In your case, the domain is not symmetric, so the function is not even.
answered Aug 23 '17 at 16:15
ScientificaScientifica
6,73141335
$endgroup$
add a comment |
$begingroup$
No because if $x>0$ then $f(-x)$ doesn't even make sense (as a real number). The definition given in Wikipedia is for functions of a real value, i.e, functions from $mathbb{R}$ to $mathbb{R}$. In your case that's not satisfied. You can, however, enlarge the definition as follows: "A function $f:Dtomathbb{R}$ is even if $forall xin D,-xin D,land f(-x)=f(x)$". In your case, the domain is not symmetric, so the function is not even.
answered Aug 23 '17 at 16:15
ScientificaScientifica
6,73141335
$endgroup$
No because if $x>0$ then $f(-x)$ doesn't even make sense (as a real number). The definition given in Wikipedia is for functions of a real value, i.e, functions from $mathbb{R}$ to $mathbb{R}$. In your case that's not satisfied. You can, however, enlarge the definition as follows: "A function $f:Dtomathbb{R}$ is even if $forall xin D,-xin D,land f(-x)=f(x)$". In your case, the domain is not symmetric, so the function is not even.
answered Aug 23 '17 at 16:15
ScientificaScientifica
6,73141335
answered Aug 23 '17 at 16:15
ScientificaScientifica
6,73141335
answered Aug 23 '17 at 16:15
ScientificaScientifica
6,73141335
answered Aug 23 '17 at 16:15
ScientificaScientifica
6,73141335
6,73141335
add a comment |
add a comment |
$begingroup$
HINT: $f(x)=sqrt{x}$ is defined for all $x$ with $$0le x<+infty$$ and $$sqrt{-x}=-sqrt{x}$$ makes no sense and $$sqrt{-x}=sqrt{x}$$ also
answered Aug 23 '17 at 16:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
$endgroup$
add a comment |
$begingroup$
HINT: $f(x)=sqrt{x}$ is defined for all $x$ with $$0le x<+infty$$ and $$sqrt{-x}=-sqrt{x}$$ makes no sense and $$sqrt{-x}=sqrt{x}$$ also
answered Aug 23 '17 at 16:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
$endgroup$
add a comment |
$begingroup$
HINT: $f(x)=sqrt{x}$ is defined for all $x$ with $$0le x<+infty$$ and $$sqrt{-x}=-sqrt{x}$$ makes no sense and $$sqrt{-x}=sqrt{x}$$ also
answered Aug 23 '17 at 16:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
$endgroup$
HINT: $f(x)=sqrt{x}$ is defined for all $x$ with $$0le x<+infty$$ and $$sqrt{-x}=-sqrt{x}$$ makes no sense and $$sqrt{-x}=sqrt{x}$$ also
answered Aug 23 '17 at 16:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
answered Aug 23 '17 at 16:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
answered Aug 23 '17 at 16:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
answered Aug 23 '17 at 16:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.8k42865
74.8k42865
add a comment |
add a comment |
$begingroup$
It doesn't matter.
Function parity is useful for functions defined both in the negatives and positives.
Whatever the classification of $sqrt x$, it is of no use.
answered Aug 23 '17 at 17:28
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
$begingroup$
It doesn't matter.
Function parity is useful for functions defined both in the negatives and positives.
Whatever the classification of $sqrt x$, it is of no use.
answered Aug 23 '17 at 17:28
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
$begingroup$
It doesn't matter.
Function parity is useful for functions defined both in the negatives and positives.
Whatever the classification of $sqrt x$, it is of no use.
answered Aug 23 '17 at 17:28
Yves DaoustYves Daoust
126k672226
$endgroup$
It doesn't matter.
Function parity is useful for functions defined both in the negatives and positives.
Whatever the classification of $sqrt x$, it is of no use.
answered Aug 23 '17 at 17:28
Yves DaoustYves Daoust
126k672226
answered Aug 23 '17 at 17:28
Yves DaoustYves Daoust
126k672226
answered Aug 23 '17 at 17:28
Yves DaoustYves Daoust
126k672226
answered Aug 23 '17 at 17:28
Yves DaoustYves Daoust
126k672226
126k672226
add a comment |
add a comment |
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$begingroup$
No, since $-x$ isn't even in the domain of $f$ for all values $x>0$.
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– Guillermo Angeris
Aug 23 '17 at 16:12
3
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No. It isn't even defined for $x<0$, so it can't be even. Even if you extend to the complex plane, $sqrt{-x} = isqrt{x} neq sqrt{x}$. They are on different axes!
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– Brevan Ellefsen
Aug 23 '17 at 16:12
3
$begingroup$
I think that's what OP is getting at.. it sounds like the function is "vacuously even."
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– tilper
Aug 23 '17 at 16:13
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@tilper I suppose we could modify the definition of an even function to say that, at whatever $x$ where $f(-x)$ is defined, $f(x) = f(-x)$, but I can't think of any nice properties this has
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– Brevan Ellefsen
Aug 23 '17 at 16:15
$begingroup$
@BrevanEllefsen, I think, keeping in mind the geometric definition (graph is symmetric over the $y$-axis), it would be helpful to add the restriction that $f$ must be defined on an interval centered at $x=0$. A function not defined on such an interval will then not be even. But then I guess even that is "too strict" since $f$ could be defined on, say, a union of intervals "centered" at $x=0$, e.g. $f : [-4,-3] cup [-2,-1] cup [1,2] cup [3,4] to {1}$ is even. Meh.
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– tilper
Aug 23 '17 at 16:21