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Definition: expressions that can be evaluated versus those that can't
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Suppose I have an expression $y + f(x)$, and I know $y in mathbb R, x in X$ and $f$ is an arbitrary function from $X$ to the reals.
If I know $y$ and $x$ but don't know $f$, I can't evaluate the expression $y + f(x)$. At least, not completely.
But once I know $f$, (say $f(z) = z^2$), then I can evaluate the expression $y + f(x)$.
Is there a name for the difference between these two circumstances?
computability symbolic-computation
asked Aug 6 at 23:26
alexpghayes
1013
add a comment |Â
up vote
0
down vote
favorite
Suppose I have an expression $y + f(x)$, and I know $y in mathbb R, x in X$ and $f$ is an arbitrary function from $X$ to the reals.
If I know $y$ and $x$ but don't know $f$, I can't evaluate the expression $y + f(x)$. At least, not completely.
But once I know $f$, (say $f(z) = z^2$), then I can evaluate the expression $y + f(x)$.
Is there a name for the difference between these two circumstances?
computability symbolic-computation
asked Aug 6 at 23:26
alexpghayes
1013
In maths, there is no such instance where you can not evaluate a function. At he very least, I can always say that the evaluation of $f$ at $x$ is $f(x)$. Actually, I do not really see in what context your interrogation arises. Is there a context where this matters?
– Suzet
Aug 6 at 23:53
@Suzet That is not true. You can easily define a relation, for which you can prove that it is a function, but that you cannot prove any of its values.
– user580373
Aug 6 at 23:55
@spiralstotheleft What does "to prove a value" mean? Is my sentence "I can always say that the evaluation of $f$ at $x$ is $f(x)$" wrong? :o
– Suzet
Aug 6 at 23:57
@Suzet As having a function $f:Xto0,1$ but for a given $xin X$ you cannot prove whether $f(x)=0$ is true or false and whether $f(x)=1$ is true or false.
– user580373
Aug 6 at 23:59
Right, but if for any reason I need to consider the evaluation of $f$ at $x$, that I can always do it, by calling this number $f(x)$, whether it equals $0$ or $1$ does not matter. Conceptually, there is no problem with this. Well however, I think that it is indeed not what the OP @alexpghayes is looking for. I'll let someone else give a proper answer.
– Suzet
Aug 7 at 0:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose I have an expression $y + f(x)$, and I know $y in mathbb R, x in X$ and $f$ is an arbitrary function from $X$ to the reals.
If I know $y$ and $x$ but don't know $f$, I can't evaluate the expression $y + f(x)$. At least, not completely.
But once I know $f$, (say $f(z) = z^2$), then I can evaluate the expression $y + f(x)$.
Is there a name for the difference between these two circumstances?
computability symbolic-computation
asked Aug 6 at 23:26
alexpghayes
1013
Suppose I have an expression $y + f(x)$, and I know $y in mathbb R, x in X$ and $f$ is an arbitrary function from $X$ to the reals.
If I know $y$ and $x$ but don't know $f$, I can't evaluate the expression $y + f(x)$. At least, not completely.
But once I know $f$, (say $f(z) = z^2$), then I can evaluate the expression $y + f(x)$.
Is there a name for the difference between these two circumstances?
computability symbolic-computation
asked Aug 6 at 23:26
alexpghayes
1013
asked Aug 6 at 23:26
alexpghayes
1013
asked Aug 6 at 23:26
alexpghayes
1013
asked Aug 6 at 23:26
alexpghayes
1013
1013
In maths, there is no such instance where you can not evaluate a function. At he very least, I can always say that the evaluation of $f$ at $x$ is $f(x)$. Actually, I do not really see in what context your interrogation arises. Is there a context where this matters?
– Suzet
Aug 6 at 23:53
@Suzet That is not true. You can easily define a relation, for which you can prove that it is a function, but that you cannot prove any of its values.
– user580373
Aug 6 at 23:55
@spiralstotheleft What does "to prove a value" mean? Is my sentence "I can always say that the evaluation of $f$ at $x$ is $f(x)$" wrong? :o
– Suzet
Aug 6 at 23:57
@Suzet As having a function $f:Xto0,1$ but for a given $xin X$ you cannot prove whether $f(x)=0$ is true or false and whether $f(x)=1$ is true or false.
– user580373
Aug 6 at 23:59
Right, but if for any reason I need to consider the evaluation of $f$ at $x$, that I can always do it, by calling this number $f(x)$, whether it equals $0$ or $1$ does not matter. Conceptually, there is no problem with this. Well however, I think that it is indeed not what the OP @alexpghayes is looking for. I'll let someone else give a proper answer.
– Suzet
Aug 7 at 0:04
add a comment |Â
In maths, there is no such instance where you can not evaluate a function. At he very least, I can always say that the evaluation of $f$ at $x$ is $f(x)$. Actually, I do not really see in what context your interrogation arises. Is there a context where this matters?
– Suzet
Aug 6 at 23:53
@Suzet That is not true. You can easily define a relation, for which you can prove that it is a function, but that you cannot prove any of its values.
– user580373
Aug 6 at 23:55
@spiralstotheleft What does "to prove a value" mean? Is my sentence "I can always say that the evaluation of $f$ at $x$ is $f(x)$" wrong? :o
– Suzet
Aug 6 at 23:57
@Suzet As having a function $f:Xto0,1$ but for a given $xin X$ you cannot prove whether $f(x)=0$ is true or false and whether $f(x)=1$ is true or false.
– user580373
Aug 6 at 23:59
Right, but if for any reason I need to consider the evaluation of $f$ at $x$, that I can always do it, by calling this number $f(x)$, whether it equals $0$ or $1$ does not matter. Conceptually, there is no problem with this. Well however, I think that it is indeed not what the OP @alexpghayes is looking for. I'll let someone else give a proper answer.
– Suzet
Aug 7 at 0:04
In maths, there is no such instance where you can not evaluate a function. At he very least, I can always say that the evaluation of $f$ at $x$ is $f(x)$. Actually, I do not really see in what context your interrogation arises. Is there a context where this matters?
– Suzet
Aug 6 at 23:53
In maths, there is no such instance where you can not evaluate a function. At he very least, I can always say that the evaluation of $f$ at $x$ is $f(x)$. Actually, I do not really see in what context your interrogation arises. Is there a context where this matters?
– Suzet
Aug 6 at 23:53
@Suzet That is not true. You can easily define a relation, for which you can prove that it is a function, but that you cannot prove any of its values.
– user580373
Aug 6 at 23:55
@Suzet That is not true. You can easily define a relation, for which you can prove that it is a function, but that you cannot prove any of its values.
– user580373
Aug 6 at 23:55
@spiralstotheleft What does "to prove a value" mean? Is my sentence "I can always say that the evaluation of $f$ at $x$ is $f(x)$" wrong? :o
– Suzet
Aug 6 at 23:57
@spiralstotheleft What does "to prove a value" mean? Is my sentence "I can always say that the evaluation of $f$ at $x$ is $f(x)$" wrong? :o
– Suzet
Aug 6 at 23:57
@Suzet As having a function $f:Xto0,1$ but for a given $xin X$ you cannot prove whether $f(x)=0$ is true or false and whether $f(x)=1$ is true or false.
– user580373
Aug 6 at 23:59
@Suzet As having a function $f:Xto0,1$ but for a given $xin X$ you cannot prove whether $f(x)=0$ is true or false and whether $f(x)=1$ is true or false.
– user580373
Aug 6 at 23:59
Right, but if for any reason I need to consider the evaluation of $f$ at $x$, that I can always do it, by calling this number $f(x)$, whether it equals $0$ or $1$ does not matter. Conceptually, there is no problem with this. Well however, I think that it is indeed not what the OP @alexpghayes is looking for. I'll let someone else give a proper answer.
– Suzet
Aug 7 at 0:04
Right, but if for any reason I need to consider the evaluation of $f$ at $x$, that I can always do it, by calling this number $f(x)$, whether it equals $0$ or $1$ does not matter. Conceptually, there is no problem with this. Well however, I think that it is indeed not what the OP @alexpghayes is looking for. I'll let someone else give a proper answer.
– Suzet
Aug 7 at 0:04
add a comment |Â
2 Answers
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active
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up vote
1
down vote
Here is the terminology from predicate logic. An expression such as $x + f(y)$ is called a term. In this particular term there are several free variables (e.g. $x$ and $y$ at least, and also $f$ if we view that as a variable for a function). If we replace the free variables by constants (also known sometimes as parameters), so that no free variables remain, the term is then a closed term. A term that has one or more free variables is called an open term.
In the usual framework of predicate logic, each closed term with parameters from a model $M$ denotes a particular element of the model $M$. Not only do we need the model to have parameters to substitute for the variables, we also need the model to give meaning to symbols such as $+$ that appear in the term.
answered Aug 7 at 13:59
Carl Mummert
63.7k7128237
add a comment |Â
up vote
0
down vote
The following is informal, but sufficient for my purposes.
The key thing to understand is the difference between an expression and a value. Values are the atomic elements under consideration. Expressions are a combination of values, functions, and other expressions (or whatever a given grammar allows). All values are expressions.
For the case that I'm interested in, reals are values, and functions are not.
Examples of values: 1, 2, $pi$
Examples of expressions: 1, 2, $pi$, $x$, $x + 2$, $y + f(x)$
Expressions can be evaluated, resulting in either a value, or a new expression. If we bind the value 3 to $x$, for example, we can substitute 3 for $x$ wherever we see it in an expression:
$2 to 2$
$x to 3$
$x + 2 to 3 + 2 to 5$
$y + f(3)$
If the result of this evaluation is a value, we say that evaluation is complete. In some cases, it isn't possible to reduce an expression any further given the existing definitions:
$y + f(x) to y + f(3)$
Here the result is still an expression. In other cases, evaluation of the expression never terminates. Let $f(x) := g(x)$, $g(x) := h(x)$, $h(x) := f(x)$.
$f(x) to g(x) to h(x) to f(x) to ...$
I was looking for a definition that meant an expression that can be evaluated to completion, i.e. an expression that can be reduced to a value rather than an expression. I believe the appropriate term is a closed-form expression.
A closed-form expression contains no free (unbound) variables. This means that all of the elements of the expression have been defined, and provided there are no problems with infinite recursion, the expression can evaluated by a Turing machine in a finite number of steps.
A reference.
answered Aug 7 at 5:08
alexpghayes
1013
$int_-1^1 e^-x^2,dx$ has no free variables but I think that an explicit integral is not usually called a closed form expression.
– Carl Mummert
Aug 7 at 15:51
It seems analytical expression is slightly more appropriate in this context: en.wikipedia.org/wiki/…? I think the usage varies in math and CS.
– alexpghayes
Aug 7 at 21:38
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here is the terminology from predicate logic. An expression such as $x + f(y)$ is called a term. In this particular term there are several free variables (e.g. $x$ and $y$ at least, and also $f$ if we view that as a variable for a function). If we replace the free variables by constants (also known sometimes as parameters), so that no free variables remain, the term is then a closed term. A term that has one or more free variables is called an open term.
In the usual framework of predicate logic, each closed term with parameters from a model $M$ denotes a particular element of the model $M$. Not only do we need the model to have parameters to substitute for the variables, we also need the model to give meaning to symbols such as $+$ that appear in the term.
answered Aug 7 at 13:59
Carl Mummert
63.7k7128237
add a comment |Â
up vote
1
down vote
Here is the terminology from predicate logic. An expression such as $x + f(y)$ is called a term. In this particular term there are several free variables (e.g. $x$ and $y$ at least, and also $f$ if we view that as a variable for a function). If we replace the free variables by constants (also known sometimes as parameters), so that no free variables remain, the term is then a closed term. A term that has one or more free variables is called an open term.
In the usual framework of predicate logic, each closed term with parameters from a model $M$ denotes a particular element of the model $M$. Not only do we need the model to have parameters to substitute for the variables, we also need the model to give meaning to symbols such as $+$ that appear in the term.
answered Aug 7 at 13:59
Carl Mummert
63.7k7128237
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is the terminology from predicate logic. An expression such as $x + f(y)$ is called a term. In this particular term there are several free variables (e.g. $x$ and $y$ at least, and also $f$ if we view that as a variable for a function). If we replace the free variables by constants (also known sometimes as parameters), so that no free variables remain, the term is then a closed term. A term that has one or more free variables is called an open term.
In the usual framework of predicate logic, each closed term with parameters from a model $M$ denotes a particular element of the model $M$. Not only do we need the model to have parameters to substitute for the variables, we also need the model to give meaning to symbols such as $+$ that appear in the term.
answered Aug 7 at 13:59
Carl Mummert
63.7k7128237
Here is the terminology from predicate logic. An expression such as $x + f(y)$ is called a term. In this particular term there are several free variables (e.g. $x$ and $y$ at least, and also $f$ if we view that as a variable for a function). If we replace the free variables by constants (also known sometimes as parameters), so that no free variables remain, the term is then a closed term. A term that has one or more free variables is called an open term.
In the usual framework of predicate logic, each closed term with parameters from a model $M$ denotes a particular element of the model $M$. Not only do we need the model to have parameters to substitute for the variables, we also need the model to give meaning to symbols such as $+$ that appear in the term.
answered Aug 7 at 13:59
Carl Mummert
63.7k7128237
answered Aug 7 at 13:59
Carl Mummert
63.7k7128237
answered Aug 7 at 13:59
Carl Mummert
63.7k7128237
answered Aug 7 at 13:59
Carl Mummert
63.7k7128237
63.7k7128237
add a comment |Â
add a comment |Â
up vote
0
down vote
The following is informal, but sufficient for my purposes.
The key thing to understand is the difference between an expression and a value. Values are the atomic elements under consideration. Expressions are a combination of values, functions, and other expressions (or whatever a given grammar allows). All values are expressions.
For the case that I'm interested in, reals are values, and functions are not.
Examples of values: 1, 2, $pi$
Examples of expressions: 1, 2, $pi$, $x$, $x + 2$, $y + f(x)$
Expressions can be evaluated, resulting in either a value, or a new expression. If we bind the value 3 to $x$, for example, we can substitute 3 for $x$ wherever we see it in an expression:
$2 to 2$
$x to 3$
$x + 2 to 3 + 2 to 5$
$y + f(3)$
If the result of this evaluation is a value, we say that evaluation is complete. In some cases, it isn't possible to reduce an expression any further given the existing definitions:
$y + f(x) to y + f(3)$
Here the result is still an expression. In other cases, evaluation of the expression never terminates. Let $f(x) := g(x)$, $g(x) := h(x)$, $h(x) := f(x)$.
$f(x) to g(x) to h(x) to f(x) to ...$
I was looking for a definition that meant an expression that can be evaluated to completion, i.e. an expression that can be reduced to a value rather than an expression. I believe the appropriate term is a closed-form expression.
A closed-form expression contains no free (unbound) variables. This means that all of the elements of the expression have been defined, and provided there are no problems with infinite recursion, the expression can evaluated by a Turing machine in a finite number of steps.
A reference.
answered Aug 7 at 5:08
alexpghayes
1013
$int_-1^1 e^-x^2,dx$ has no free variables but I think that an explicit integral is not usually called a closed form expression.
– Carl Mummert
Aug 7 at 15:51
It seems analytical expression is slightly more appropriate in this context: en.wikipedia.org/wiki/…? I think the usage varies in math and CS.
– alexpghayes
Aug 7 at 21:38
add a comment |Â
up vote
0
down vote
The following is informal, but sufficient for my purposes.
The key thing to understand is the difference between an expression and a value. Values are the atomic elements under consideration. Expressions are a combination of values, functions, and other expressions (or whatever a given grammar allows). All values are expressions.
For the case that I'm interested in, reals are values, and functions are not.
Examples of values: 1, 2, $pi$
Examples of expressions: 1, 2, $pi$, $x$, $x + 2$, $y + f(x)$
Expressions can be evaluated, resulting in either a value, or a new expression. If we bind the value 3 to $x$, for example, we can substitute 3 for $x$ wherever we see it in an expression:
$2 to 2$
$x to 3$
$x + 2 to 3 + 2 to 5$
$y + f(3)$
If the result of this evaluation is a value, we say that evaluation is complete. In some cases, it isn't possible to reduce an expression any further given the existing definitions:
$y + f(x) to y + f(3)$
Here the result is still an expression. In other cases, evaluation of the expression never terminates. Let $f(x) := g(x)$, $g(x) := h(x)$, $h(x) := f(x)$.
$f(x) to g(x) to h(x) to f(x) to ...$
I was looking for a definition that meant an expression that can be evaluated to completion, i.e. an expression that can be reduced to a value rather than an expression. I believe the appropriate term is a closed-form expression.
A closed-form expression contains no free (unbound) variables. This means that all of the elements of the expression have been defined, and provided there are no problems with infinite recursion, the expression can evaluated by a Turing machine in a finite number of steps.
A reference.
answered Aug 7 at 5:08
alexpghayes
1013
$int_-1^1 e^-x^2,dx$ has no free variables but I think that an explicit integral is not usually called a closed form expression.
– Carl Mummert
Aug 7 at 15:51
It seems analytical expression is slightly more appropriate in this context: en.wikipedia.org/wiki/…? I think the usage varies in math and CS.
– alexpghayes
Aug 7 at 21:38
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The following is informal, but sufficient for my purposes.
The key thing to understand is the difference between an expression and a value. Values are the atomic elements under consideration. Expressions are a combination of values, functions, and other expressions (or whatever a given grammar allows). All values are expressions.
For the case that I'm interested in, reals are values, and functions are not.
Examples of values: 1, 2, $pi$
Examples of expressions: 1, 2, $pi$, $x$, $x + 2$, $y + f(x)$
Expressions can be evaluated, resulting in either a value, or a new expression. If we bind the value 3 to $x$, for example, we can substitute 3 for $x$ wherever we see it in an expression:
$2 to 2$
$x to 3$
$x + 2 to 3 + 2 to 5$
$y + f(3)$
If the result of this evaluation is a value, we say that evaluation is complete. In some cases, it isn't possible to reduce an expression any further given the existing definitions:
$y + f(x) to y + f(3)$
Here the result is still an expression. In other cases, evaluation of the expression never terminates. Let $f(x) := g(x)$, $g(x) := h(x)$, $h(x) := f(x)$.
$f(x) to g(x) to h(x) to f(x) to ...$
I was looking for a definition that meant an expression that can be evaluated to completion, i.e. an expression that can be reduced to a value rather than an expression. I believe the appropriate term is a closed-form expression.
A closed-form expression contains no free (unbound) variables. This means that all of the elements of the expression have been defined, and provided there are no problems with infinite recursion, the expression can evaluated by a Turing machine in a finite number of steps.
A reference.
answered Aug 7 at 5:08
alexpghayes
1013
The following is informal, but sufficient for my purposes.
The key thing to understand is the difference between an expression and a value. Values are the atomic elements under consideration. Expressions are a combination of values, functions, and other expressions (or whatever a given grammar allows). All values are expressions.
For the case that I'm interested in, reals are values, and functions are not.
Examples of values: 1, 2, $pi$
Examples of expressions: 1, 2, $pi$, $x$, $x + 2$, $y + f(x)$
Expressions can be evaluated, resulting in either a value, or a new expression. If we bind the value 3 to $x$, for example, we can substitute 3 for $x$ wherever we see it in an expression:
$2 to 2$
$x to 3$
$x + 2 to 3 + 2 to 5$
$y + f(3)$
If the result of this evaluation is a value, we say that evaluation is complete. In some cases, it isn't possible to reduce an expression any further given the existing definitions:
$y + f(x) to y + f(3)$
Here the result is still an expression. In other cases, evaluation of the expression never terminates. Let $f(x) := g(x)$, $g(x) := h(x)$, $h(x) := f(x)$.
$f(x) to g(x) to h(x) to f(x) to ...$
I was looking for a definition that meant an expression that can be evaluated to completion, i.e. an expression that can be reduced to a value rather than an expression. I believe the appropriate term is a closed-form expression.
A closed-form expression contains no free (unbound) variables. This means that all of the elements of the expression have been defined, and provided there are no problems with infinite recursion, the expression can evaluated by a Turing machine in a finite number of steps.
A reference.
answered Aug 7 at 5:08
alexpghayes
1013
answered Aug 7 at 5:08
alexpghayes
1013
answered Aug 7 at 5:08
alexpghayes
1013
answered Aug 7 at 5:08
alexpghayes
1013
1013
$int_-1^1 e^-x^2,dx$ has no free variables but I think that an explicit integral is not usually called a closed form expression.
– Carl Mummert
Aug 7 at 15:51
It seems analytical expression is slightly more appropriate in this context: en.wikipedia.org/wiki/…? I think the usage varies in math and CS.
– alexpghayes
Aug 7 at 21:38
add a comment |Â
$int_-1^1 e^-x^2,dx$ has no free variables but I think that an explicit integral is not usually called a closed form expression.
– Carl Mummert
Aug 7 at 15:51
It seems analytical expression is slightly more appropriate in this context: en.wikipedia.org/wiki/…? I think the usage varies in math and CS.
– alexpghayes
Aug 7 at 21:38
$int_-1^1 e^-x^2,dx$ has no free variables but I think that an explicit integral is not usually called a closed form expression.
– Carl Mummert
Aug 7 at 15:51
$int_-1^1 e^-x^2,dx$ has no free variables but I think that an explicit integral is not usually called a closed form expression.
– Carl Mummert
Aug 7 at 15:51
It seems analytical expression is slightly more appropriate in this context: en.wikipedia.org/wiki/…? I think the usage varies in math and CS.
– alexpghayes
Aug 7 at 21:38
It seems analytical expression is slightly more appropriate in this context: en.wikipedia.org/wiki/…? I think the usage varies in math and CS.
– alexpghayes
Aug 7 at 21:38
add a comment |Â
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In maths, there is no such instance where you can not evaluate a function. At he very least, I can always say that the evaluation of $f$ at $x$ is $f(x)$. Actually, I do not really see in what context your interrogation arises. Is there a context where this matters?
– Suzet
Aug 6 at 23:53
@Suzet That is not true. You can easily define a relation, for which you can prove that it is a function, but that you cannot prove any of its values.
– user580373
Aug 6 at 23:55
@spiralstotheleft What does "to prove a value" mean? Is my sentence "I can always say that the evaluation of $f$ at $x$ is $f(x)$" wrong? :o
– Suzet
Aug 6 at 23:57
@Suzet As having a function $f:Xto0,1$ but for a given $xin X$ you cannot prove whether $f(x)=0$ is true or false and whether $f(x)=1$ is true or false.
– user580373
Aug 6 at 23:59
Right, but if for any reason I need to consider the evaluation of $f$ at $x$, that I can always do it, by calling this number $f(x)$, whether it equals $0$ or $1$ does not matter. Conceptually, there is no problem with this. Well however, I think that it is indeed not what the OP @alexpghayes is looking for. I'll let someone else give a proper answer.
– Suzet
Aug 7 at 0:04