Is there more than one meaning of the notation “$f(x)=[x]$”?
$begingroup$
In my real analysis text book there is a question that says:
Decide whether $f(x)=[x]$ is bounded above or below on the interval $[0,a]$ where $a$ is arbitrary, and whether the function takes on it's maximum or minimum value within that same interval.
This question is very straightforward, assuming $[x]=x$. But if that is the case, then the choice of notation is very strange.
Is there another way to interpret the notation's meaning?
notation
edited Jan 4 at 6:14
Xander Henderson
14.8k103555
asked May 1 '11 at 22:42
objectiveseaobjectivesea
450413
$endgroup$
add a comment |
$begingroup$
In my real analysis text book there is a question that says:
Decide whether $f(x)=[x]$ is bounded above or below on the interval $[0,a]$ where $a$ is arbitrary, and whether the function takes on it's maximum or minimum value within that same interval.
This question is very straightforward, assuming $[x]=x$. But if that is the case, then the choice of notation is very strange.
Is there another way to interpret the notation's meaning?
notation
edited Jan 4 at 6:14
Xander Henderson
14.8k103555
asked May 1 '11 at 22:42
objectiveseaobjectivesea
450413
$endgroup$
1
$begingroup$
I have seen $[x]$ denote the fractional part of $x$, but that doesn't seem to be the desired meaning in this context.
$endgroup$
– Brandon Carter
May 1 '11 at 22:44
2
$begingroup$
I've seen that notation to be the nearest integer function before.
$endgroup$
– yunone
May 1 '11 at 22:44
2
$begingroup$
Could you specify what textbook you're using? I'm guessing there should be a place where the notation used is explained...
$endgroup$
– J. M. is not a mathematician
May 1 '11 at 22:47
$begingroup$
[x] is often used as the Iverson Bracket, 1 if is x is true, though that's clearly not the usage here. I've also seen it used for either floor or ceiling, when the right brackets (missing either the top or bottom) were difficult to produce.
$endgroup$
– wnoise
May 1 '11 at 23:00
$begingroup$
The question is from Michael Spivak's "Calculus" - 4th Edition (also 3rd edition). It's part (xii) of chapter 7 question 1.
$endgroup$
– objectivesea
May 1 '11 at 23:20
add a comment |
$begingroup$
In my real analysis text book there is a question that says:
Decide whether $f(x)=[x]$ is bounded above or below on the interval $[0,a]$ where $a$ is arbitrary, and whether the function takes on it's maximum or minimum value within that same interval.
This question is very straightforward, assuming $[x]=x$. But if that is the case, then the choice of notation is very strange.
Is there another way to interpret the notation's meaning?
notation
edited Jan 4 at 6:14
Xander Henderson
14.8k103555
asked May 1 '11 at 22:42
objectiveseaobjectivesea
450413
$endgroup$
In my real analysis text book there is a question that says:
Decide whether $f(x)=[x]$ is bounded above or below on the interval $[0,a]$ where $a$ is arbitrary, and whether the function takes on it's maximum or minimum value within that same interval.
This question is very straightforward, assuming $[x]=x$. But if that is the case, then the choice of notation is very strange.
Is there another way to interpret the notation's meaning?
notation
notation
edited Jan 4 at 6:14
Xander Henderson
14.8k103555
asked May 1 '11 at 22:42
objectiveseaobjectivesea
450413
edited Jan 4 at 6:14
Xander Henderson
14.8k103555
asked May 1 '11 at 22:42
objectiveseaobjectivesea
450413
edited Jan 4 at 6:14
Xander Henderson
14.8k103555
edited Jan 4 at 6:14
Xander Henderson
14.8k103555
edited Jan 4 at 6:14
Xander Henderson
14.8k103555
14.8k103555
asked May 1 '11 at 22:42
objectiveseaobjectivesea
450413
asked May 1 '11 at 22:42
objectiveseaobjectivesea
450413
asked May 1 '11 at 22:42
objectiveseaobjectivesea
450413
450413
1
$begingroup$
I have seen $[x]$ denote the fractional part of $x$, but that doesn't seem to be the desired meaning in this context.
$endgroup$
– Brandon Carter
May 1 '11 at 22:44
2
$begingroup$
I've seen that notation to be the nearest integer function before.
$endgroup$
– yunone
May 1 '11 at 22:44
2
$begingroup$
Could you specify what textbook you're using? I'm guessing there should be a place where the notation used is explained...
$endgroup$
– J. M. is not a mathematician
May 1 '11 at 22:47
$begingroup$
[x] is often used as the Iverson Bracket, 1 if is x is true, though that's clearly not the usage here. I've also seen it used for either floor or ceiling, when the right brackets (missing either the top or bottom) were difficult to produce.
$endgroup$
– wnoise
May 1 '11 at 23:00
$begingroup$
The question is from Michael Spivak's "Calculus" - 4th Edition (also 3rd edition). It's part (xii) of chapter 7 question 1.
$endgroup$
– objectivesea
May 1 '11 at 23:20
add a comment |
1
$begingroup$
I have seen $[x]$ denote the fractional part of $x$, but that doesn't seem to be the desired meaning in this context.
$endgroup$
– Brandon Carter
May 1 '11 at 22:44
2
$begingroup$
I've seen that notation to be the nearest integer function before.
$endgroup$
– yunone
May 1 '11 at 22:44
2
$begingroup$
Could you specify what textbook you're using? I'm guessing there should be a place where the notation used is explained...
$endgroup$
– J. M. is not a mathematician
May 1 '11 at 22:47
$begingroup$
[x] is often used as the Iverson Bracket, 1 if is x is true, though that's clearly not the usage here. I've also seen it used for either floor or ceiling, when the right brackets (missing either the top or bottom) were difficult to produce.
$endgroup$
– wnoise
May 1 '11 at 23:00
$begingroup$
The question is from Michael Spivak's "Calculus" - 4th Edition (also 3rd edition). It's part (xii) of chapter 7 question 1.
$endgroup$
– objectivesea
May 1 '11 at 23:20
1
1
$begingroup$
I have seen $[x]$ denote the fractional part of $x$, but that doesn't seem to be the desired meaning in this context.
$endgroup$
– Brandon Carter
May 1 '11 at 22:44
$begingroup$
I have seen $[x]$ denote the fractional part of $x$, but that doesn't seem to be the desired meaning in this context.
$endgroup$
– Brandon Carter
May 1 '11 at 22:44
2
2
$begingroup$
I've seen that notation to be the nearest integer function before.
$endgroup$
– yunone
May 1 '11 at 22:44
$begingroup$
I've seen that notation to be the nearest integer function before.
$endgroup$
– yunone
May 1 '11 at 22:44
2
2
$begingroup$
Could you specify what textbook you're using? I'm guessing there should be a place where the notation used is explained...
$endgroup$
– J. M. is not a mathematician
May 1 '11 at 22:47
$begingroup$
Could you specify what textbook you're using? I'm guessing there should be a place where the notation used is explained...
$endgroup$
– J. M. is not a mathematician
May 1 '11 at 22:47
$begingroup$
[x] is often used as the Iverson Bracket, 1 if is x is true, though that's clearly not the usage here. I've also seen it used for either floor or ceiling, when the right brackets (missing either the top or bottom) were difficult to produce.
$endgroup$
– wnoise
May 1 '11 at 23:00
$begingroup$
[x] is often used as the Iverson Bracket, 1 if is x is true, though that's clearly not the usage here. I've also seen it used for either floor or ceiling, when the right brackets (missing either the top or bottom) were difficult to produce.
$endgroup$
– wnoise
May 1 '11 at 23:00
$begingroup$
The question is from Michael Spivak's "Calculus" - 4th Edition (also 3rd edition). It's part (xii) of chapter 7 question 1.
$endgroup$
– objectivesea
May 1 '11 at 23:20
$begingroup$
The question is from Michael Spivak's "Calculus" - 4th Edition (also 3rd edition). It's part (xii) of chapter 7 question 1.
$endgroup$
– objectivesea
May 1 '11 at 23:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It had been fairly standard for $[x]$ to represent "the greatest integer not greater than $x$" (aka, the "floor" function). With fancier type-setting options allowing for $lfloor x rfloor$ for a more suggestive "floor" notation ---as well as $lceil x rceil$ for the counterpart "ceiling" ("smallest integer not smaller than $x$")--- I've seen $[x]$ taking on the role of "nearest integer" (that is, the "rounding" function) although $lfloor x rceil$ is also available for this, freeing up $[x]$ for author's discretion.
With regard to @Brandon's "fractional part", I've seen that more often represented as ${ x }$, usually in conjunction with the floor interpretation of $[x]$, so that one would write $x = [x] + {x}$ (at least for non-negative $x$).
answered May 1 '11 at 23:15
BlueBlue
48.7k870156
$endgroup$
add a comment |
$begingroup$
It's certainly not $[x]=x$.
Rather, it's undoubtedly the greatest integer function, aka floor function (also denoted $lfloor xrfloor$); that is, the greatest integer less than or equal to $x$. The ceiling function ($lceil xrceil$) is, correspondingly, the smallest integer greater than or equal to $x$..
According to this, Spanier and Oldham called it the "integer value" in $1987$. (Incidentally, I took a course from Spanier, and I consider him to have been an outstanding mathematician.
Of course, that puts me in a fairly large club. But I digress.)
The formula would be: $lfloor xrfloor =n$, where $n$ is the integer such that $nle xlt n+1$.
Another way of describing it, would be to take the number's decimal representation, and truncate it. That is, set the part after the decimal to zero.
answered Jan 4 at 7:11
Chris CusterChris Custer
14.1k3827
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It had been fairly standard for $[x]$ to represent "the greatest integer not greater than $x$" (aka, the "floor" function). With fancier type-setting options allowing for $lfloor x rfloor$ for a more suggestive "floor" notation ---as well as $lceil x rceil$ for the counterpart "ceiling" ("smallest integer not smaller than $x$")--- I've seen $[x]$ taking on the role of "nearest integer" (that is, the "rounding" function) although $lfloor x rceil$ is also available for this, freeing up $[x]$ for author's discretion.
With regard to @Brandon's "fractional part", I've seen that more often represented as ${ x }$, usually in conjunction with the floor interpretation of $[x]$, so that one would write $x = [x] + {x}$ (at least for non-negative $x$).
answered May 1 '11 at 23:15
BlueBlue
48.7k870156
$endgroup$
add a comment |
$begingroup$
It had been fairly standard for $[x]$ to represent "the greatest integer not greater than $x$" (aka, the "floor" function). With fancier type-setting options allowing for $lfloor x rfloor$ for a more suggestive "floor" notation ---as well as $lceil x rceil$ for the counterpart "ceiling" ("smallest integer not smaller than $x$")--- I've seen $[x]$ taking on the role of "nearest integer" (that is, the "rounding" function) although $lfloor x rceil$ is also available for this, freeing up $[x]$ for author's discretion.
With regard to @Brandon's "fractional part", I've seen that more often represented as ${ x }$, usually in conjunction with the floor interpretation of $[x]$, so that one would write $x = [x] + {x}$ (at least for non-negative $x$).
answered May 1 '11 at 23:15
BlueBlue
48.7k870156
$endgroup$
add a comment |
$begingroup$
It had been fairly standard for $[x]$ to represent "the greatest integer not greater than $x$" (aka, the "floor" function). With fancier type-setting options allowing for $lfloor x rfloor$ for a more suggestive "floor" notation ---as well as $lceil x rceil$ for the counterpart "ceiling" ("smallest integer not smaller than $x$")--- I've seen $[x]$ taking on the role of "nearest integer" (that is, the "rounding" function) although $lfloor x rceil$ is also available for this, freeing up $[x]$ for author's discretion.
With regard to @Brandon's "fractional part", I've seen that more often represented as ${ x }$, usually in conjunction with the floor interpretation of $[x]$, so that one would write $x = [x] + {x}$ (at least for non-negative $x$).
answered May 1 '11 at 23:15
BlueBlue
48.7k870156
$endgroup$
It had been fairly standard for $[x]$ to represent "the greatest integer not greater than $x$" (aka, the "floor" function). With fancier type-setting options allowing for $lfloor x rfloor$ for a more suggestive "floor" notation ---as well as $lceil x rceil$ for the counterpart "ceiling" ("smallest integer not smaller than $x$")--- I've seen $[x]$ taking on the role of "nearest integer" (that is, the "rounding" function) although $lfloor x rceil$ is also available for this, freeing up $[x]$ for author's discretion.
With regard to @Brandon's "fractional part", I've seen that more often represented as ${ x }$, usually in conjunction with the floor interpretation of $[x]$, so that one would write $x = [x] + {x}$ (at least for non-negative $x$).
answered May 1 '11 at 23:15
BlueBlue
48.7k870156
answered May 1 '11 at 23:15
BlueBlue
48.7k870156
answered May 1 '11 at 23:15
BlueBlue
48.7k870156
answered May 1 '11 at 23:15
BlueBlue
48.7k870156
48.7k870156
add a comment |
add a comment |
$begingroup$
It's certainly not $[x]=x$.
Rather, it's undoubtedly the greatest integer function, aka floor function (also denoted $lfloor xrfloor$); that is, the greatest integer less than or equal to $x$. The ceiling function ($lceil xrceil$) is, correspondingly, the smallest integer greater than or equal to $x$..
According to this, Spanier and Oldham called it the "integer value" in $1987$. (Incidentally, I took a course from Spanier, and I consider him to have been an outstanding mathematician.
Of course, that puts me in a fairly large club. But I digress.)
The formula would be: $lfloor xrfloor =n$, where $n$ is the integer such that $nle xlt n+1$.
Another way of describing it, would be to take the number's decimal representation, and truncate it. That is, set the part after the decimal to zero.
answered Jan 4 at 7:11
Chris CusterChris Custer
14.1k3827
$endgroup$
add a comment |
$begingroup$
It's certainly not $[x]=x$.
Rather, it's undoubtedly the greatest integer function, aka floor function (also denoted $lfloor xrfloor$); that is, the greatest integer less than or equal to $x$. The ceiling function ($lceil xrceil$) is, correspondingly, the smallest integer greater than or equal to $x$..
According to this, Spanier and Oldham called it the "integer value" in $1987$. (Incidentally, I took a course from Spanier, and I consider him to have been an outstanding mathematician.
Of course, that puts me in a fairly large club. But I digress.)
The formula would be: $lfloor xrfloor =n$, where $n$ is the integer such that $nle xlt n+1$.
Another way of describing it, would be to take the number's decimal representation, and truncate it. That is, set the part after the decimal to zero.
answered Jan 4 at 7:11
Chris CusterChris Custer
14.1k3827
$endgroup$
add a comment |
$begingroup$
It's certainly not $[x]=x$.
Rather, it's undoubtedly the greatest integer function, aka floor function (also denoted $lfloor xrfloor$); that is, the greatest integer less than or equal to $x$. The ceiling function ($lceil xrceil$) is, correspondingly, the smallest integer greater than or equal to $x$..
According to this, Spanier and Oldham called it the "integer value" in $1987$. (Incidentally, I took a course from Spanier, and I consider him to have been an outstanding mathematician.
Of course, that puts me in a fairly large club. But I digress.)
The formula would be: $lfloor xrfloor =n$, where $n$ is the integer such that $nle xlt n+1$.
Another way of describing it, would be to take the number's decimal representation, and truncate it. That is, set the part after the decimal to zero.
answered Jan 4 at 7:11
Chris CusterChris Custer
14.1k3827
$endgroup$
It's certainly not $[x]=x$.
Rather, it's undoubtedly the greatest integer function, aka floor function (also denoted $lfloor xrfloor$); that is, the greatest integer less than or equal to $x$. The ceiling function ($lceil xrceil$) is, correspondingly, the smallest integer greater than or equal to $x$..
According to this, Spanier and Oldham called it the "integer value" in $1987$. (Incidentally, I took a course from Spanier, and I consider him to have been an outstanding mathematician.
Of course, that puts me in a fairly large club. But I digress.)
The formula would be: $lfloor xrfloor =n$, where $n$ is the integer such that $nle xlt n+1$.
Another way of describing it, would be to take the number's decimal representation, and truncate it. That is, set the part after the decimal to zero.
answered Jan 4 at 7:11
Chris CusterChris Custer
14.1k3827
answered Jan 4 at 7:11
Chris CusterChris Custer
14.1k3827
answered Jan 4 at 7:11
Chris CusterChris Custer
14.1k3827
answered Jan 4 at 7:11
Chris CusterChris Custer
14.1k3827
14.1k3827
add a comment |
add a comment |
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1
$begingroup$
I have seen $[x]$ denote the fractional part of $x$, but that doesn't seem to be the desired meaning in this context.
$endgroup$
– Brandon Carter
May 1 '11 at 22:44
2
$begingroup$
I've seen that notation to be the nearest integer function before.
$endgroup$
– yunone
May 1 '11 at 22:44
2
$begingroup$
Could you specify what textbook you're using? I'm guessing there should be a place where the notation used is explained...
$endgroup$
– J. M. is not a mathematician
May 1 '11 at 22:47
$begingroup$
[x] is often used as the Iverson Bracket, 1 if is x is true, though that's clearly not the usage here. I've also seen it used for either floor or ceiling, when the right brackets (missing either the top or bottom) were difficult to produce.
$endgroup$
– wnoise
May 1 '11 at 23:00
$begingroup$
The question is from Michael Spivak's "Calculus" - 4th Edition (also 3rd edition). It's part (xii) of chapter 7 question 1.
$endgroup$
– objectivesea
May 1 '11 at 23:20