Mixing
Why Is a Function Defined As Having Only One Y-Value Output?
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I know that it is defined such that there is no more than one y-value output for any given x-value input, but I"m wondering WHY it is defined that way? Why can't we apply everything we know about functions to equations that have more than one y-value output for any given x-value input?
In short, what is the benefit or reason to define a function this way? At the same time, a little historical context would also be appreciated (i.e., when the concept originated and why).
Thanks,
Moshe
functions
edited Jan 6 '13 at 17:21
Amzoti
51.3k125398
asked Jan 6 '13 at 16:08
MosheMoshe
61114
$endgroup$
|
show 4 more comments
$begingroup$
I know that it is defined such that there is no more than one y-value output for any given x-value input, but I"m wondering WHY it is defined that way? Why can't we apply everything we know about functions to equations that have more than one y-value output for any given x-value input?
In short, what is the benefit or reason to define a function this way? At the same time, a little historical context would also be appreciated (i.e., when the concept originated and why).
Thanks,
Moshe
functions
edited Jan 6 '13 at 17:21
Amzoti
51.3k125398
asked Jan 6 '13 at 16:08
MosheMoshe
61114
$endgroup$
$begingroup$
en.wikipedia.org/wiki/Multivalued_function
$endgroup$
– Eckhard
Jan 6 '13 at 16:28
2
$begingroup$
Note that the "one value" constraint is less restrictive than you might think. For example a function $f: mathbb{R} to wp(mathbb{R})$ assigns a set of real numbers to any input.
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– WimC
Jan 6 '13 at 16:30
$begingroup$
@WimC Ermm ...but a function which maps a real to a set assigns a single value (a set) to a given argument.
$endgroup$
– Peter Smith
Jan 6 '13 at 16:37
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@PeterSmith, I think WimC is pointing out if the codomain can be varied - and specifically to the powerset of the OP's "y-values"- then any number of y-values can be assigned while still having a function.
$endgroup$
– alancalvitti
Jan 6 '13 at 16:41
$begingroup$
Of course, a function from $X$ to $mathcal P^Y$ is the same thing as a relation between $X$ and $Y$. These are only the same thing as a multivalued function if you allow multivalued to include zero values, which is not usual.
$endgroup$
– Rahul
Jan 6 '13 at 16:56
|
show 4 more comments
$begingroup$
I know that it is defined such that there is no more than one y-value output for any given x-value input, but I"m wondering WHY it is defined that way? Why can't we apply everything we know about functions to equations that have more than one y-value output for any given x-value input?
In short, what is the benefit or reason to define a function this way? At the same time, a little historical context would also be appreciated (i.e., when the concept originated and why).
Thanks,
Moshe
functions
edited Jan 6 '13 at 17:21
Amzoti
51.3k125398
asked Jan 6 '13 at 16:08
MosheMoshe
61114
$endgroup$
I know that it is defined such that there is no more than one y-value output for any given x-value input, but I"m wondering WHY it is defined that way? Why can't we apply everything we know about functions to equations that have more than one y-value output for any given x-value input?
In short, what is the benefit or reason to define a function this way? At the same time, a little historical context would also be appreciated (i.e., when the concept originated and why).
Thanks,
Moshe
functions
functions
edited Jan 6 '13 at 17:21
Amzoti
51.3k125398
asked Jan 6 '13 at 16:08
MosheMoshe
61114
edited Jan 6 '13 at 17:21
Amzoti
51.3k125398
asked Jan 6 '13 at 16:08
MosheMoshe
61114
edited Jan 6 '13 at 17:21
Amzoti
51.3k125398
edited Jan 6 '13 at 17:21
Amzoti
51.3k125398
edited Jan 6 '13 at 17:21
Amzoti
51.3k125398
51.3k125398
asked Jan 6 '13 at 16:08
MosheMoshe
61114
asked Jan 6 '13 at 16:08
MosheMoshe
61114
asked Jan 6 '13 at 16:08
MosheMoshe
61114
61114
$begingroup$
en.wikipedia.org/wiki/Multivalued_function
$endgroup$
– Eckhard
Jan 6 '13 at 16:28
2
$begingroup$
Note that the "one value" constraint is less restrictive than you might think. For example a function $f: mathbb{R} to wp(mathbb{R})$ assigns a set of real numbers to any input.
$endgroup$
– WimC
Jan 6 '13 at 16:30
$begingroup$
@WimC Ermm ...but a function which maps a real to a set assigns a single value (a set) to a given argument.
$endgroup$
– Peter Smith
Jan 6 '13 at 16:37
$begingroup$
@PeterSmith, I think WimC is pointing out if the codomain can be varied - and specifically to the powerset of the OP's "y-values"- then any number of y-values can be assigned while still having a function.
$endgroup$
– alancalvitti
Jan 6 '13 at 16:41
$begingroup$
Of course, a function from $X$ to $mathcal P^Y$ is the same thing as a relation between $X$ and $Y$. These are only the same thing as a multivalued function if you allow multivalued to include zero values, which is not usual.
$endgroup$
– Rahul
Jan 6 '13 at 16:56
|
show 4 more comments
$begingroup$
en.wikipedia.org/wiki/Multivalued_function
$endgroup$
– Eckhard
Jan 6 '13 at 16:28
2
$begingroup$
Note that the "one value" constraint is less restrictive than you might think. For example a function $f: mathbb{R} to wp(mathbb{R})$ assigns a set of real numbers to any input.
$endgroup$
– WimC
Jan 6 '13 at 16:30
$begingroup$
@WimC Ermm ...but a function which maps a real to a set assigns a single value (a set) to a given argument.
$endgroup$
– Peter Smith
Jan 6 '13 at 16:37
$begingroup$
@PeterSmith, I think WimC is pointing out if the codomain can be varied - and specifically to the powerset of the OP's "y-values"- then any number of y-values can be assigned while still having a function.
$endgroup$
– alancalvitti
Jan 6 '13 at 16:41
$begingroup$
Of course, a function from $X$ to $mathcal P^Y$ is the same thing as a relation between $X$ and $Y$. These are only the same thing as a multivalued function if you allow multivalued to include zero values, which is not usual.
$endgroup$
– Rahul
Jan 6 '13 at 16:56
$begingroup$
en.wikipedia.org/wiki/Multivalued_function
$endgroup$
– Eckhard
Jan 6 '13 at 16:28
$begingroup$
en.wikipedia.org/wiki/Multivalued_function
$endgroup$
– Eckhard
Jan 6 '13 at 16:28
2
2
$begingroup$
Note that the "one value" constraint is less restrictive than you might think. For example a function $f: mathbb{R} to wp(mathbb{R})$ assigns a set of real numbers to any input.
$endgroup$
– WimC
Jan 6 '13 at 16:30
$begingroup$
Note that the "one value" constraint is less restrictive than you might think. For example a function $f: mathbb{R} to wp(mathbb{R})$ assigns a set of real numbers to any input.
$endgroup$
– WimC
Jan 6 '13 at 16:30
$begingroup$
@WimC Ermm ...but a function which maps a real to a set assigns a single value (a set) to a given argument.
$endgroup$
– Peter Smith
Jan 6 '13 at 16:37
$begingroup$
@WimC Ermm ...but a function which maps a real to a set assigns a single value (a set) to a given argument.
$endgroup$
– Peter Smith
Jan 6 '13 at 16:37
$begingroup$
@PeterSmith, I think WimC is pointing out if the codomain can be varied - and specifically to the powerset of the OP's "y-values"- then any number of y-values can be assigned while still having a function.
$endgroup$
– alancalvitti
Jan 6 '13 at 16:41
$begingroup$
@PeterSmith, I think WimC is pointing out if the codomain can be varied - and specifically to the powerset of the OP's "y-values"- then any number of y-values can be assigned while still having a function.
$endgroup$
– alancalvitti
Jan 6 '13 at 16:41
$begingroup$
Of course, a function from $X$ to $mathcal P^Y$ is the same thing as a relation between $X$ and $Y$. These are only the same thing as a multivalued function if you allow multivalued to include zero values, which is not usual.
$endgroup$
– Rahul
Jan 6 '13 at 16:56
$begingroup$
Of course, a function from $X$ to $mathcal P^Y$ is the same thing as a relation between $X$ and $Y$. These are only the same thing as a multivalued function if you allow multivalued to include zero values, which is not usual.
$endgroup$
– Rahul
Jan 6 '13 at 16:56
|
show 4 more comments
4 Answers
4
active
oldest
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Strictly speaking, a "well-defined" function associates one, and only one, output to any particular input. A function would not be well-defined if say $y = f(x)$ such that $y$ can take on any number of values at any particular input. Suppose the input $;x = a;$ outputs more than one distinct value, so that $y = f(a) in{y_1, y_2, ..., y_k, ...} $. We'd never be able to say what, precisely, $y$ is when $x = a$.
If $y = f(a) = b$, and $y = f(a) = c$, and $,bne c$, then the value of the function $,,f(x) = y,$ at $,a,$ is ambiguous: $y$ would not specify any particular value at $a$. That is, its value is not well-defined at $a$, and perhaps not well-defined at other inputs, as well. Nor can we say much about the behavior of a function at a particular value, if it can take on many values at a given point.
E.g. How would we define continuity of, say, a real-valued multifunction?
The strict definition of a function, in terms of "outputing" exactly one value for any given input is really no more than an attempt to keeping functions well-defined, and thus properties of functions well-defined.
But you'd might like to explore the following:
See this entry on multi-valued "functions":
A multivalued function (shortly: multifunction, other names: many-valued function, set-valued function, set-valued map, multi-valued map, multimap, correspondence, carrier) is a left-total relation; that is, every input is associated with at least one output.
The term "multivalued function" is, therefore, a misnomer because functions are single-valued. Multivalued functions often arise from functions which are not injective. Such functions do not have an inverse function, but they do have an inverse relation. The multivalued function corresponds to this inverse relation. [bold-face mine]
answered Jan 6 '13 at 16:17
NamasteNamaste
1
$endgroup$
$begingroup$
Yet some texts in Complex Analysis do very cheerfully talk of many-valued functions ... E.g. Schaum's Outline of Complex Variables, 2ed (Schaum's Outline Series) by Murray Spiegel, Seymour Lipschutz, John Schiller, Dennis Spellman. This (2009 2nd edn. but first goes back to 1964: surely aims to be "middle of the road") explicitly talks about multi-valued functions (see §2.2, §2.6).
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– Peter Smith
Jan 6 '13 at 16:30
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Yes, indeed. I was referring to the "strict" definition students are usually introduced to.
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– Namaste
Jan 6 '13 at 16:34
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I wonder how the floor function is called function since for any value of x where x<x+1 all the results will be x. Any possible reason?
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– arqam
Apr 18 '18 at 11:40
add a comment |
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There is an interesting relevant post by Tim Gowers on his blog http://gowers.wordpress.com/2009/06/08/why-arent-all-functions-well-defined/
It is worth noting though, as a footnote, that the insistence that functions (properly so-called) be single-valued is a relative late-comer. G.H. Hardy, for example, in his once-canonical A Course in Pure Mathematics says that the characteristic of taking just one value for a given argument "is by no means involved in the general idea of a function".
answered Jan 6 '13 at 16:29
Peter SmithPeter Smith
40.9k341120
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Could you please shed some light on what that "general idea of a function" might be? I'm very curious, since the definition that Hardy is dismissing is the definition I eat and breathe when doing mathematics.
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– Ovi
Dec 25 '18 at 5:51
add a comment |
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Multi-valued functions do exist, but then that greatly limits our ability to say much about them. For functions, we're not just concerned about the value that it takes, but also the behavior of the function around the point. If we have multi-valued functions, then what we can say about the behavior of the point will be highly restricted, as it needs to be the same regardless of what value we choose.
A common example would be to let $f(x,y) = x^2 + y^2$, and ask for the values such that $f(x, y) = 1$ (i.e. the unit circle). The implicit function theorem is used, for us to conclude that there are 2 functions, namely $F(x) = sqrt{1-y^2}$ and $F(x) = - sqrt{1-y^2}$. If we had said that $F(x) = pm sqrt{1-y^2}$, it would be hard to talk about the behavior of $F'(x)$ exactly.
answered Jan 6 '13 at 16:30
Calvin LinCalvin Lin
36.3k349114
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You may wish to read about Kakutani's Fixed Point Theorem. As I mentioned in another comment this result is used in Gerard Debreu's Theory of Value. This is a short book in mathematical economics. Anyone who knows undergraduate analysis should be able to read it. I will make no claim that the theory in this book applies to the real world.
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– Jay
Jan 6 '13 at 23:02
add a comment |
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Obviously this limitation of defining a function f(x) as giving only ONE value of y is specific to the English (or American?) Maths.
In French, we consider y^2 = x as a function, we call it horizontal parabole...
See http://www.alloprof.qc.ca/BV/pages/m1330.aspx
I would be tempted to use pejorative language toward English (or AMerican?) Maths!
answered Jun 19 '17 at 3:02
Michel MacéMichel Macé
1
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But that's not true: "En mathématiques, une fonction est une relation entre un ensemble d’entrées (variable) et un ensemble de sorties (image), avec la propriété que chaque entrée est liée à exactement une sortie."
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– Trevor Gunn
Jun 19 '17 at 3:54
1
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The word "fonction" does not appear at all on the website you link to. What you have there are called "relations".
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– Trevor Gunn
Jun 19 '17 at 3:58
add a comment |
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4 Answers
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4 Answers
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$begingroup$
Strictly speaking, a "well-defined" function associates one, and only one, output to any particular input. A function would not be well-defined if say $y = f(x)$ such that $y$ can take on any number of values at any particular input. Suppose the input $;x = a;$ outputs more than one distinct value, so that $y = f(a) in{y_1, y_2, ..., y_k, ...} $. We'd never be able to say what, precisely, $y$ is when $x = a$.
If $y = f(a) = b$, and $y = f(a) = c$, and $,bne c$, then the value of the function $,,f(x) = y,$ at $,a,$ is ambiguous: $y$ would not specify any particular value at $a$. That is, its value is not well-defined at $a$, and perhaps not well-defined at other inputs, as well. Nor can we say much about the behavior of a function at a particular value, if it can take on many values at a given point.
E.g. How would we define continuity of, say, a real-valued multifunction?
The strict definition of a function, in terms of "outputing" exactly one value for any given input is really no more than an attempt to keeping functions well-defined, and thus properties of functions well-defined.
But you'd might like to explore the following:
See this entry on multi-valued "functions":
A multivalued function (shortly: multifunction, other names: many-valued function, set-valued function, set-valued map, multi-valued map, multimap, correspondence, carrier) is a left-total relation; that is, every input is associated with at least one output.
The term "multivalued function" is, therefore, a misnomer because functions are single-valued. Multivalued functions often arise from functions which are not injective. Such functions do not have an inverse function, but they do have an inverse relation. The multivalued function corresponds to this inverse relation. [bold-face mine]
answered Jan 6 '13 at 16:17
NamasteNamaste
1
$endgroup$
$begingroup$
Yet some texts in Complex Analysis do very cheerfully talk of many-valued functions ... E.g. Schaum's Outline of Complex Variables, 2ed (Schaum's Outline Series) by Murray Spiegel, Seymour Lipschutz, John Schiller, Dennis Spellman. This (2009 2nd edn. but first goes back to 1964: surely aims to be "middle of the road") explicitly talks about multi-valued functions (see §2.2, §2.6).
$endgroup$
– Peter Smith
Jan 6 '13 at 16:30
$begingroup$
Yes, indeed. I was referring to the "strict" definition students are usually introduced to.
$endgroup$
– Namaste
Jan 6 '13 at 16:34
$begingroup$
I wonder how the floor function is called function since for any value of x where x<x+1 all the results will be x. Any possible reason?
$endgroup$
– arqam
Apr 18 '18 at 11:40
add a comment |
$begingroup$
Strictly speaking, a "well-defined" function associates one, and only one, output to any particular input. A function would not be well-defined if say $y = f(x)$ such that $y$ can take on any number of values at any particular input. Suppose the input $;x = a;$ outputs more than one distinct value, so that $y = f(a) in{y_1, y_2, ..., y_k, ...} $. We'd never be able to say what, precisely, $y$ is when $x = a$.
If $y = f(a) = b$, and $y = f(a) = c$, and $,bne c$, then the value of the function $,,f(x) = y,$ at $,a,$ is ambiguous: $y$ would not specify any particular value at $a$. That is, its value is not well-defined at $a$, and perhaps not well-defined at other inputs, as well. Nor can we say much about the behavior of a function at a particular value, if it can take on many values at a given point.
E.g. How would we define continuity of, say, a real-valued multifunction?
The strict definition of a function, in terms of "outputing" exactly one value for any given input is really no more than an attempt to keeping functions well-defined, and thus properties of functions well-defined.
But you'd might like to explore the following:
See this entry on multi-valued "functions":
A multivalued function (shortly: multifunction, other names: many-valued function, set-valued function, set-valued map, multi-valued map, multimap, correspondence, carrier) is a left-total relation; that is, every input is associated with at least one output.
The term "multivalued function" is, therefore, a misnomer because functions are single-valued. Multivalued functions often arise from functions which are not injective. Such functions do not have an inverse function, but they do have an inverse relation. The multivalued function corresponds to this inverse relation. [bold-face mine]
answered Jan 6 '13 at 16:17
NamasteNamaste
1
$endgroup$
$begingroup$
Yet some texts in Complex Analysis do very cheerfully talk of many-valued functions ... E.g. Schaum's Outline of Complex Variables, 2ed (Schaum's Outline Series) by Murray Spiegel, Seymour Lipschutz, John Schiller, Dennis Spellman. This (2009 2nd edn. but first goes back to 1964: surely aims to be "middle of the road") explicitly talks about multi-valued functions (see §2.2, §2.6).
$endgroup$
– Peter Smith
Jan 6 '13 at 16:30
$begingroup$
Yes, indeed. I was referring to the "strict" definition students are usually introduced to.
$endgroup$
– Namaste
Jan 6 '13 at 16:34
$begingroup$
I wonder how the floor function is called function since for any value of x where x<x+1 all the results will be x. Any possible reason?
$endgroup$
– arqam
Apr 18 '18 at 11:40
add a comment |
$begingroup$
Strictly speaking, a "well-defined" function associates one, and only one, output to any particular input. A function would not be well-defined if say $y = f(x)$ such that $y$ can take on any number of values at any particular input. Suppose the input $;x = a;$ outputs more than one distinct value, so that $y = f(a) in{y_1, y_2, ..., y_k, ...} $. We'd never be able to say what, precisely, $y$ is when $x = a$.
If $y = f(a) = b$, and $y = f(a) = c$, and $,bne c$, then the value of the function $,,f(x) = y,$ at $,a,$ is ambiguous: $y$ would not specify any particular value at $a$. That is, its value is not well-defined at $a$, and perhaps not well-defined at other inputs, as well. Nor can we say much about the behavior of a function at a particular value, if it can take on many values at a given point.
E.g. How would we define continuity of, say, a real-valued multifunction?
The strict definition of a function, in terms of "outputing" exactly one value for any given input is really no more than an attempt to keeping functions well-defined, and thus properties of functions well-defined.
But you'd might like to explore the following:
See this entry on multi-valued "functions":
A multivalued function (shortly: multifunction, other names: many-valued function, set-valued function, set-valued map, multi-valued map, multimap, correspondence, carrier) is a left-total relation; that is, every input is associated with at least one output.
The term "multivalued function" is, therefore, a misnomer because functions are single-valued. Multivalued functions often arise from functions which are not injective. Such functions do not have an inverse function, but they do have an inverse relation. The multivalued function corresponds to this inverse relation. [bold-face mine]
answered Jan 6 '13 at 16:17
NamasteNamaste
1
$endgroup$
Strictly speaking, a "well-defined" function associates one, and only one, output to any particular input. A function would not be well-defined if say $y = f(x)$ such that $y$ can take on any number of values at any particular input. Suppose the input $;x = a;$ outputs more than one distinct value, so that $y = f(a) in{y_1, y_2, ..., y_k, ...} $. We'd never be able to say what, precisely, $y$ is when $x = a$.
If $y = f(a) = b$, and $y = f(a) = c$, and $,bne c$, then the value of the function $,,f(x) = y,$ at $,a,$ is ambiguous: $y$ would not specify any particular value at $a$. That is, its value is not well-defined at $a$, and perhaps not well-defined at other inputs, as well. Nor can we say much about the behavior of a function at a particular value, if it can take on many values at a given point.
E.g. How would we define continuity of, say, a real-valued multifunction?
The strict definition of a function, in terms of "outputing" exactly one value for any given input is really no more than an attempt to keeping functions well-defined, and thus properties of functions well-defined.
But you'd might like to explore the following:
See this entry on multi-valued "functions":
A multivalued function (shortly: multifunction, other names: many-valued function, set-valued function, set-valued map, multi-valued map, multimap, correspondence, carrier) is a left-total relation; that is, every input is associated with at least one output.
The term "multivalued function" is, therefore, a misnomer because functions are single-valued. Multivalued functions often arise from functions which are not injective. Such functions do not have an inverse function, but they do have an inverse relation. The multivalued function corresponds to this inverse relation. [bold-face mine]
answered Jan 6 '13 at 16:17
NamasteNamaste
1
edited Jan 6 '13 at 22:14
answered Jan 6 '13 at 16:17
NamasteNamaste
1
answered Jan 6 '13 at 16:17
NamasteNamaste
1
answered Jan 6 '13 at 16:17
NamasteNamaste
1
1
$begingroup$
Yet some texts in Complex Analysis do very cheerfully talk of many-valued functions ... E.g. Schaum's Outline of Complex Variables, 2ed (Schaum's Outline Series) by Murray Spiegel, Seymour Lipschutz, John Schiller, Dennis Spellman. This (2009 2nd edn. but first goes back to 1964: surely aims to be "middle of the road") explicitly talks about multi-valued functions (see §2.2, §2.6).
$endgroup$
– Peter Smith
Jan 6 '13 at 16:30
$begingroup$
Yes, indeed. I was referring to the "strict" definition students are usually introduced to.
$endgroup$
– Namaste
Jan 6 '13 at 16:34
$begingroup$
I wonder how the floor function is called function since for any value of x where x<x+1 all the results will be x. Any possible reason?
$endgroup$
– arqam
Apr 18 '18 at 11:40
add a comment |
$begingroup$
Yet some texts in Complex Analysis do very cheerfully talk of many-valued functions ... E.g. Schaum's Outline of Complex Variables, 2ed (Schaum's Outline Series) by Murray Spiegel, Seymour Lipschutz, John Schiller, Dennis Spellman. This (2009 2nd edn. but first goes back to 1964: surely aims to be "middle of the road") explicitly talks about multi-valued functions (see §2.2, §2.6).
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– Peter Smith
Jan 6 '13 at 16:30
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Yes, indeed. I was referring to the "strict" definition students are usually introduced to.
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– Namaste
Jan 6 '13 at 16:34
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I wonder how the floor function is called function since for any value of x where x<x+1 all the results will be x. Any possible reason?
$endgroup$
– arqam
Apr 18 '18 at 11:40
$begingroup$
Yet some texts in Complex Analysis do very cheerfully talk of many-valued functions ... E.g. Schaum's Outline of Complex Variables, 2ed (Schaum's Outline Series) by Murray Spiegel, Seymour Lipschutz, John Schiller, Dennis Spellman. This (2009 2nd edn. but first goes back to 1964: surely aims to be "middle of the road") explicitly talks about multi-valued functions (see §2.2, §2.6).
$endgroup$
– Peter Smith
Jan 6 '13 at 16:30
$begingroup$
Yet some texts in Complex Analysis do very cheerfully talk of many-valued functions ... E.g. Schaum's Outline of Complex Variables, 2ed (Schaum's Outline Series) by Murray Spiegel, Seymour Lipschutz, John Schiller, Dennis Spellman. This (2009 2nd edn. but first goes back to 1964: surely aims to be "middle of the road") explicitly talks about multi-valued functions (see §2.2, §2.6).
$endgroup$
– Peter Smith
Jan 6 '13 at 16:30
$begingroup$
Yes, indeed. I was referring to the "strict" definition students are usually introduced to.
$endgroup$
– Namaste
Jan 6 '13 at 16:34
$begingroup$
Yes, indeed. I was referring to the "strict" definition students are usually introduced to.
$endgroup$
– Namaste
Jan 6 '13 at 16:34
$begingroup$
I wonder how the floor function is called function since for any value of x where x<x+1 all the results will be x. Any possible reason?
$endgroup$
– arqam
Apr 18 '18 at 11:40
$begingroup$
I wonder how the floor function is called function since for any value of x where x<x+1 all the results will be x. Any possible reason?
$endgroup$
– arqam
Apr 18 '18 at 11:40
add a comment |
$begingroup$
There is an interesting relevant post by Tim Gowers on his blog http://gowers.wordpress.com/2009/06/08/why-arent-all-functions-well-defined/
It is worth noting though, as a footnote, that the insistence that functions (properly so-called) be single-valued is a relative late-comer. G.H. Hardy, for example, in his once-canonical A Course in Pure Mathematics says that the characteristic of taking just one value for a given argument "is by no means involved in the general idea of a function".
answered Jan 6 '13 at 16:29
Peter SmithPeter Smith
40.9k341120
$endgroup$
$begingroup$
Could you please shed some light on what that "general idea of a function" might be? I'm very curious, since the definition that Hardy is dismissing is the definition I eat and breathe when doing mathematics.
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– Ovi
Dec 25 '18 at 5:51
add a comment |
$begingroup$
There is an interesting relevant post by Tim Gowers on his blog http://gowers.wordpress.com/2009/06/08/why-arent-all-functions-well-defined/
It is worth noting though, as a footnote, that the insistence that functions (properly so-called) be single-valued is a relative late-comer. G.H. Hardy, for example, in his once-canonical A Course in Pure Mathematics says that the characteristic of taking just one value for a given argument "is by no means involved in the general idea of a function".
answered Jan 6 '13 at 16:29
Peter SmithPeter Smith
40.9k341120
$endgroup$
$begingroup$
Could you please shed some light on what that "general idea of a function" might be? I'm very curious, since the definition that Hardy is dismissing is the definition I eat and breathe when doing mathematics.
$endgroup$
– Ovi
Dec 25 '18 at 5:51
add a comment |
$begingroup$
There is an interesting relevant post by Tim Gowers on his blog http://gowers.wordpress.com/2009/06/08/why-arent-all-functions-well-defined/
It is worth noting though, as a footnote, that the insistence that functions (properly so-called) be single-valued is a relative late-comer. G.H. Hardy, for example, in his once-canonical A Course in Pure Mathematics says that the characteristic of taking just one value for a given argument "is by no means involved in the general idea of a function".
answered Jan 6 '13 at 16:29
Peter SmithPeter Smith
40.9k341120
$endgroup$
There is an interesting relevant post by Tim Gowers on his blog http://gowers.wordpress.com/2009/06/08/why-arent-all-functions-well-defined/
It is worth noting though, as a footnote, that the insistence that functions (properly so-called) be single-valued is a relative late-comer. G.H. Hardy, for example, in his once-canonical A Course in Pure Mathematics says that the characteristic of taking just one value for a given argument "is by no means involved in the general idea of a function".
answered Jan 6 '13 at 16:29
Peter SmithPeter Smith
40.9k341120
edited Jan 6 '13 at 22:06
answered Jan 6 '13 at 16:29
Peter SmithPeter Smith
40.9k341120
answered Jan 6 '13 at 16:29
Peter SmithPeter Smith
40.9k341120
answered Jan 6 '13 at 16:29
Peter SmithPeter Smith
40.9k341120
40.9k341120
$begingroup$
Could you please shed some light on what that "general idea of a function" might be? I'm very curious, since the definition that Hardy is dismissing is the definition I eat and breathe when doing mathematics.
$endgroup$
– Ovi
Dec 25 '18 at 5:51
add a comment |
$begingroup$
Could you please shed some light on what that "general idea of a function" might be? I'm very curious, since the definition that Hardy is dismissing is the definition I eat and breathe when doing mathematics.
$endgroup$
– Ovi
Dec 25 '18 at 5:51
$begingroup$
Could you please shed some light on what that "general idea of a function" might be? I'm very curious, since the definition that Hardy is dismissing is the definition I eat and breathe when doing mathematics.
$endgroup$
– Ovi
Dec 25 '18 at 5:51
$begingroup$
Could you please shed some light on what that "general idea of a function" might be? I'm very curious, since the definition that Hardy is dismissing is the definition I eat and breathe when doing mathematics.
$endgroup$
– Ovi
Dec 25 '18 at 5:51
add a comment |
$begingroup$
Multi-valued functions do exist, but then that greatly limits our ability to say much about them. For functions, we're not just concerned about the value that it takes, but also the behavior of the function around the point. If we have multi-valued functions, then what we can say about the behavior of the point will be highly restricted, as it needs to be the same regardless of what value we choose.
A common example would be to let $f(x,y) = x^2 + y^2$, and ask for the values such that $f(x, y) = 1$ (i.e. the unit circle). The implicit function theorem is used, for us to conclude that there are 2 functions, namely $F(x) = sqrt{1-y^2}$ and $F(x) = - sqrt{1-y^2}$. If we had said that $F(x) = pm sqrt{1-y^2}$, it would be hard to talk about the behavior of $F'(x)$ exactly.
answered Jan 6 '13 at 16:30
Calvin LinCalvin Lin
36.3k349114
$endgroup$
$begingroup$
You may wish to read about Kakutani's Fixed Point Theorem. As I mentioned in another comment this result is used in Gerard Debreu's Theory of Value. This is a short book in mathematical economics. Anyone who knows undergraduate analysis should be able to read it. I will make no claim that the theory in this book applies to the real world.
$endgroup$
– Jay
Jan 6 '13 at 23:02
add a comment |
$begingroup$
Multi-valued functions do exist, but then that greatly limits our ability to say much about them. For functions, we're not just concerned about the value that it takes, but also the behavior of the function around the point. If we have multi-valued functions, then what we can say about the behavior of the point will be highly restricted, as it needs to be the same regardless of what value we choose.
A common example would be to let $f(x,y) = x^2 + y^2$, and ask for the values such that $f(x, y) = 1$ (i.e. the unit circle). The implicit function theorem is used, for us to conclude that there are 2 functions, namely $F(x) = sqrt{1-y^2}$ and $F(x) = - sqrt{1-y^2}$. If we had said that $F(x) = pm sqrt{1-y^2}$, it would be hard to talk about the behavior of $F'(x)$ exactly.
answered Jan 6 '13 at 16:30
Calvin LinCalvin Lin
36.3k349114
$endgroup$
$begingroup$
You may wish to read about Kakutani's Fixed Point Theorem. As I mentioned in another comment this result is used in Gerard Debreu's Theory of Value. This is a short book in mathematical economics. Anyone who knows undergraduate analysis should be able to read it. I will make no claim that the theory in this book applies to the real world.
$endgroup$
– Jay
Jan 6 '13 at 23:02
add a comment |
$begingroup$
Multi-valued functions do exist, but then that greatly limits our ability to say much about them. For functions, we're not just concerned about the value that it takes, but also the behavior of the function around the point. If we have multi-valued functions, then what we can say about the behavior of the point will be highly restricted, as it needs to be the same regardless of what value we choose.
A common example would be to let $f(x,y) = x^2 + y^2$, and ask for the values such that $f(x, y) = 1$ (i.e. the unit circle). The implicit function theorem is used, for us to conclude that there are 2 functions, namely $F(x) = sqrt{1-y^2}$ and $F(x) = - sqrt{1-y^2}$. If we had said that $F(x) = pm sqrt{1-y^2}$, it would be hard to talk about the behavior of $F'(x)$ exactly.
answered Jan 6 '13 at 16:30
Calvin LinCalvin Lin
36.3k349114
$endgroup$
Multi-valued functions do exist, but then that greatly limits our ability to say much about them. For functions, we're not just concerned about the value that it takes, but also the behavior of the function around the point. If we have multi-valued functions, then what we can say about the behavior of the point will be highly restricted, as it needs to be the same regardless of what value we choose.
A common example would be to let $f(x,y) = x^2 + y^2$, and ask for the values such that $f(x, y) = 1$ (i.e. the unit circle). The implicit function theorem is used, for us to conclude that there are 2 functions, namely $F(x) = sqrt{1-y^2}$ and $F(x) = - sqrt{1-y^2}$. If we had said that $F(x) = pm sqrt{1-y^2}$, it would be hard to talk about the behavior of $F'(x)$ exactly.
answered Jan 6 '13 at 16:30
Calvin LinCalvin Lin
36.3k349114
answered Jan 6 '13 at 16:30
Calvin LinCalvin Lin
36.3k349114
answered Jan 6 '13 at 16:30
Calvin LinCalvin Lin
36.3k349114
answered Jan 6 '13 at 16:30
Calvin LinCalvin Lin
36.3k349114
36.3k349114
$begingroup$
You may wish to read about Kakutani's Fixed Point Theorem. As I mentioned in another comment this result is used in Gerard Debreu's Theory of Value. This is a short book in mathematical economics. Anyone who knows undergraduate analysis should be able to read it. I will make no claim that the theory in this book applies to the real world.
$endgroup$
– Jay
Jan 6 '13 at 23:02
add a comment |
$begingroup$
You may wish to read about Kakutani's Fixed Point Theorem. As I mentioned in another comment this result is used in Gerard Debreu's Theory of Value. This is a short book in mathematical economics. Anyone who knows undergraduate analysis should be able to read it. I will make no claim that the theory in this book applies to the real world.
$endgroup$
– Jay
Jan 6 '13 at 23:02
$begingroup$
You may wish to read about Kakutani's Fixed Point Theorem. As I mentioned in another comment this result is used in Gerard Debreu's Theory of Value. This is a short book in mathematical economics. Anyone who knows undergraduate analysis should be able to read it. I will make no claim that the theory in this book applies to the real world.
$endgroup$
– Jay
Jan 6 '13 at 23:02
$begingroup$
You may wish to read about Kakutani's Fixed Point Theorem. As I mentioned in another comment this result is used in Gerard Debreu's Theory of Value. This is a short book in mathematical economics. Anyone who knows undergraduate analysis should be able to read it. I will make no claim that the theory in this book applies to the real world.
$endgroup$
– Jay
Jan 6 '13 at 23:02
add a comment |
$begingroup$
Obviously this limitation of defining a function f(x) as giving only ONE value of y is specific to the English (or American?) Maths.
In French, we consider y^2 = x as a function, we call it horizontal parabole...
See http://www.alloprof.qc.ca/BV/pages/m1330.aspx
I would be tempted to use pejorative language toward English (or AMerican?) Maths!
answered Jun 19 '17 at 3:02
Michel MacéMichel Macé
1
$endgroup$
$begingroup$
But that's not true: "En mathématiques, une fonction est une relation entre un ensemble d’entrées (variable) et un ensemble de sorties (image), avec la propriété que chaque entrée est liée à exactement une sortie."
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:54
1
$begingroup$
The word "fonction" does not appear at all on the website you link to. What you have there are called "relations".
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:58
add a comment |
$begingroup$
Obviously this limitation of defining a function f(x) as giving only ONE value of y is specific to the English (or American?) Maths.
In French, we consider y^2 = x as a function, we call it horizontal parabole...
See http://www.alloprof.qc.ca/BV/pages/m1330.aspx
I would be tempted to use pejorative language toward English (or AMerican?) Maths!
answered Jun 19 '17 at 3:02
Michel MacéMichel Macé
1
$endgroup$
$begingroup$
But that's not true: "En mathématiques, une fonction est une relation entre un ensemble d’entrées (variable) et un ensemble de sorties (image), avec la propriété que chaque entrée est liée à exactement une sortie."
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:54
1
$begingroup$
The word "fonction" does not appear at all on the website you link to. What you have there are called "relations".
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:58
add a comment |
$begingroup$
Obviously this limitation of defining a function f(x) as giving only ONE value of y is specific to the English (or American?) Maths.
In French, we consider y^2 = x as a function, we call it horizontal parabole...
See http://www.alloprof.qc.ca/BV/pages/m1330.aspx
I would be tempted to use pejorative language toward English (or AMerican?) Maths!
answered Jun 19 '17 at 3:02
Michel MacéMichel Macé
1
$endgroup$
Obviously this limitation of defining a function f(x) as giving only ONE value of y is specific to the English (or American?) Maths.
In French, we consider y^2 = x as a function, we call it horizontal parabole...
See http://www.alloprof.qc.ca/BV/pages/m1330.aspx
I would be tempted to use pejorative language toward English (or AMerican?) Maths!
answered Jun 19 '17 at 3:02
Michel MacéMichel Macé
1
answered Jun 19 '17 at 3:02
Michel MacéMichel Macé
1
answered Jun 19 '17 at 3:02
Michel MacéMichel Macé
1
answered Jun 19 '17 at 3:02
Michel MacéMichel Macé
1
1
$begingroup$
But that's not true: "En mathématiques, une fonction est une relation entre un ensemble d’entrées (variable) et un ensemble de sorties (image), avec la propriété que chaque entrée est liée à exactement une sortie."
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:54
1
$begingroup$
The word "fonction" does not appear at all on the website you link to. What you have there are called "relations".
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:58
add a comment |
$begingroup$
But that's not true: "En mathématiques, une fonction est une relation entre un ensemble d’entrées (variable) et un ensemble de sorties (image), avec la propriété que chaque entrée est liée à exactement une sortie."
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:54
1
$begingroup$
The word "fonction" does not appear at all on the website you link to. What you have there are called "relations".
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:58
$begingroup$
But that's not true: "En mathématiques, une fonction est une relation entre un ensemble d’entrées (variable) et un ensemble de sorties (image), avec la propriété que chaque entrée est liée à exactement une sortie."
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:54
$begingroup$
But that's not true: "En mathématiques, une fonction est une relation entre un ensemble d’entrées (variable) et un ensemble de sorties (image), avec la propriété que chaque entrée est liée à exactement une sortie."
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:54
1
1
$begingroup$
The word "fonction" does not appear at all on the website you link to. What you have there are called "relations".
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:58
$begingroup$
The word "fonction" does not appear at all on the website you link to. What you have there are called "relations".
$endgroup$
– Trevor Gunn
Jun 19 '17 at 3:58
add a comment |
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$begingroup$
en.wikipedia.org/wiki/Multivalued_function
$endgroup$
– Eckhard
Jan 6 '13 at 16:28
2
$begingroup$
Note that the "one value" constraint is less restrictive than you might think. For example a function $f: mathbb{R} to wp(mathbb{R})$ assigns a set of real numbers to any input.
$endgroup$
– WimC
Jan 6 '13 at 16:30
$begingroup$
@WimC Ermm ...but a function which maps a real to a set assigns a single value (a set) to a given argument.
$endgroup$
– Peter Smith
Jan 6 '13 at 16:37
$begingroup$
@PeterSmith, I think WimC is pointing out if the codomain can be varied - and specifically to the powerset of the OP's "y-values"- then any number of y-values can be assigned while still having a function.
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– alancalvitti
Jan 6 '13 at 16:41
$begingroup$
Of course, a function from $X$ to $mathcal P^Y$ is the same thing as a relation between $X$ and $Y$. These are only the same thing as a multivalued function if you allow multivalued to include zero values, which is not usual.
$endgroup$
– Rahul
Jan 6 '13 at 16:56