How is $lfloor sin x rfloor$ continous at $3pi/2$ when there is a jump continuity to it?
$begingroup$
My book says that $f(x) =lfloor sin xrfloor$ is continuous at $3pi/2$, but on drawing the graph of $lfloor sin xrfloor$, there was a jump continuity at all values of $x$ where $y=1$ or $-1$.
The discontinuity is not removable, so how is the function continuous at $3pi/2$?
Am I wrong somewhere?
(Editor's Note. I have replaced the original usage of "$[;;]$" to denote the "greatest integer function" with the unambiguous "$lfloor;;rfloor$". —@Blue)
calculus trigonometry continuity graphing-functions
edited Dec 6 '18 at 7:58
Blue
48.8k870156
asked Dec 6 '18 at 7:30
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
899
$endgroup$
|
show 3 more comments
$begingroup$
My book says that $f(x) =lfloor sin xrfloor$ is continuous at $3pi/2$, but on drawing the graph of $lfloor sin xrfloor$, there was a jump continuity at all values of $x$ where $y=1$ or $-1$.
The discontinuity is not removable, so how is the function continuous at $3pi/2$?
Am I wrong somewhere?
(Editor's Note. I have replaced the original usage of "$[;;]$" to denote the "greatest integer function" with the unambiguous "$lfloor;;rfloor$". —@Blue)
calculus trigonometry continuity graphing-functions
edited Dec 6 '18 at 7:58
Blue
48.8k870156
asked Dec 6 '18 at 7:30
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
899
$endgroup$
$begingroup$
To be clear: How is "$[;;]$" being used here?
$endgroup$
– Blue
Dec 6 '18 at 7:33
$begingroup$
you may drawn wrong graph...
$endgroup$
– neelkanth
Dec 6 '18 at 7:34
$begingroup$
$f(x)=-1$ in a neighborhood of $3pi /2$ so it is continuous at $3pi /2$.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:35
$begingroup$
To pile on with what Blue mentioned - the function $f(x) = [sin(x)]$ can be taken as "$sin(x)$ rounded to the nearest integer." If you just mean it as parentheses or whatever, I have no clue what's going on, honestly, because $sin(x)$ is continuous for all $x$, and certainly has no "jump" discontinuities.
$endgroup$
– Eevee Trainer
Dec 6 '18 at 7:35
$begingroup$
its $-1$ on $(pi , 2pi )$
$endgroup$
– neelkanth
Dec 6 '18 at 7:36
|
show 3 more comments
$begingroup$
My book says that $f(x) =lfloor sin xrfloor$ is continuous at $3pi/2$, but on drawing the graph of $lfloor sin xrfloor$, there was a jump continuity at all values of $x$ where $y=1$ or $-1$.
The discontinuity is not removable, so how is the function continuous at $3pi/2$?
Am I wrong somewhere?
(Editor's Note. I have replaced the original usage of "$[;;]$" to denote the "greatest integer function" with the unambiguous "$lfloor;;rfloor$". —@Blue)
calculus trigonometry continuity graphing-functions
edited Dec 6 '18 at 7:58
Blue
48.8k870156
asked Dec 6 '18 at 7:30
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
899
$endgroup$
My book says that $f(x) =lfloor sin xrfloor$ is continuous at $3pi/2$, but on drawing the graph of $lfloor sin xrfloor$, there was a jump continuity at all values of $x$ where $y=1$ or $-1$.
The discontinuity is not removable, so how is the function continuous at $3pi/2$?
Am I wrong somewhere?
(Editor's Note. I have replaced the original usage of "$[;;]$" to denote the "greatest integer function" with the unambiguous "$lfloor;;rfloor$". —@Blue)
calculus trigonometry continuity graphing-functions
calculus trigonometry continuity graphing-functions
edited Dec 6 '18 at 7:58
Blue
48.8k870156
asked Dec 6 '18 at 7:30
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
899
edited Dec 6 '18 at 7:58
Blue
48.8k870156
asked Dec 6 '18 at 7:30
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
899
edited Dec 6 '18 at 7:58
Blue
48.8k870156
edited Dec 6 '18 at 7:58
Blue
48.8k870156
edited Dec 6 '18 at 7:58
Blue
48.8k870156
48.8k870156
asked Dec 6 '18 at 7:30
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
899
asked Dec 6 '18 at 7:30
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
899
asked Dec 6 '18 at 7:30
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
899
899
$begingroup$
To be clear: How is "$[;;]$" being used here?
$endgroup$
– Blue
Dec 6 '18 at 7:33
$begingroup$
you may drawn wrong graph...
$endgroup$
– neelkanth
Dec 6 '18 at 7:34
$begingroup$
$f(x)=-1$ in a neighborhood of $3pi /2$ so it is continuous at $3pi /2$.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:35
$begingroup$
To pile on with what Blue mentioned - the function $f(x) = [sin(x)]$ can be taken as "$sin(x)$ rounded to the nearest integer." If you just mean it as parentheses or whatever, I have no clue what's going on, honestly, because $sin(x)$ is continuous for all $x$, and certainly has no "jump" discontinuities.
$endgroup$
– Eevee Trainer
Dec 6 '18 at 7:35
$begingroup$
its $-1$ on $(pi , 2pi )$
$endgroup$
– neelkanth
Dec 6 '18 at 7:36
|
show 3 more comments
$begingroup$
To be clear: How is "$[;;]$" being used here?
$endgroup$
– Blue
Dec 6 '18 at 7:33
$begingroup$
you may drawn wrong graph...
$endgroup$
– neelkanth
Dec 6 '18 at 7:34
$begingroup$
$f(x)=-1$ in a neighborhood of $3pi /2$ so it is continuous at $3pi /2$.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:35
$begingroup$
To pile on with what Blue mentioned - the function $f(x) = [sin(x)]$ can be taken as "$sin(x)$ rounded to the nearest integer." If you just mean it as parentheses or whatever, I have no clue what's going on, honestly, because $sin(x)$ is continuous for all $x$, and certainly has no "jump" discontinuities.
$endgroup$
– Eevee Trainer
Dec 6 '18 at 7:35
$begingroup$
its $-1$ on $(pi , 2pi )$
$endgroup$
– neelkanth
Dec 6 '18 at 7:36
$begingroup$
To be clear: How is "$[;;]$" being used here?
$endgroup$
– Blue
Dec 6 '18 at 7:33
$begingroup$
To be clear: How is "$[;;]$" being used here?
$endgroup$
– Blue
Dec 6 '18 at 7:33
$begingroup$
you may drawn wrong graph...
$endgroup$
– neelkanth
Dec 6 '18 at 7:34
$begingroup$
you may drawn wrong graph...
$endgroup$
– neelkanth
Dec 6 '18 at 7:34
$begingroup$
$f(x)=-1$ in a neighborhood of $3pi /2$ so it is continuous at $3pi /2$.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:35
$begingroup$
$f(x)=-1$ in a neighborhood of $3pi /2$ so it is continuous at $3pi /2$.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:35
$begingroup$
To pile on with what Blue mentioned - the function $f(x) = [sin(x)]$ can be taken as "$sin(x)$ rounded to the nearest integer." If you just mean it as parentheses or whatever, I have no clue what's going on, honestly, because $sin(x)$ is continuous for all $x$, and certainly has no "jump" discontinuities.
$endgroup$
– Eevee Trainer
Dec 6 '18 at 7:35
$begingroup$
To pile on with what Blue mentioned - the function $f(x) = [sin(x)]$ can be taken as "$sin(x)$ rounded to the nearest integer." If you just mean it as parentheses or whatever, I have no clue what's going on, honestly, because $sin(x)$ is continuous for all $x$, and certainly has no "jump" discontinuities.
$endgroup$
– Eevee Trainer
Dec 6 '18 at 7:35
$begingroup$
its $-1$ on $(pi , 2pi )$
$endgroup$
– neelkanth
Dec 6 '18 at 7:36
$begingroup$
its $-1$ on $(pi , 2pi )$
$endgroup$
– neelkanth
Dec 6 '18 at 7:36
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
As $[sin(x)]=-1$ on $(pi , 2pi)$ so it is continuous at $dfrac{3pi}{2}$.
edited Dec 6 '18 at 7:46
Philippe Malot
2,276824
answered Dec 6 '18 at 7:43
neelkanthneelkanth
2,28321129
$endgroup$
$begingroup$
thankyou sir i have drawn the graph wrong
$endgroup$
– HOME WORK AND EXERCISES
Dec 6 '18 at 7:44
$begingroup$
ok ....you are welcome ....
$endgroup$
– neelkanth
Dec 6 '18 at 7:45
add a comment |
$begingroup$
The function “greatest integer” is not continuous at integers. So if you consider $g(x)=lfloor f(x)rfloor$, where $f$ is continuous, then $g$ may be not continuous where $f$ takes on an integer value.
However, the fact that $xmapstolfloor xrfloor$ is not continuous at an integer $n$ stems from the fact that $xmapsto x$ can take on values greater and less than $n$ in every neighborhood of $n$.
In your case, there exists a neighborhood of $3pi/2$ where $xmapstosin x$ only takes on values $gesin(3pi/2)=-1$ (for the simple reason that $sin xge-1$ for every $x$).
More generally, if $c$ is a local maximum or minimum point for the function $f$ and $f(c)$ is an integer, then $g(x)=lfloor f(x)rfloor$ is continuous at $c$.
answered Dec 6 '18 at 9:35
egregegreg
183k1486205
$endgroup$
$begingroup$
Thankyou. I realised i had drawn the graph wrong.
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 16:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
As $[sin(x)]=-1$ on $(pi , 2pi)$ so it is continuous at $dfrac{3pi}{2}$.
edited Dec 6 '18 at 7:46
Philippe Malot
2,276824
answered Dec 6 '18 at 7:43
neelkanthneelkanth
2,28321129
$endgroup$
$begingroup$
thankyou sir i have drawn the graph wrong
$endgroup$
– HOME WORK AND EXERCISES
Dec 6 '18 at 7:44
$begingroup$
ok ....you are welcome ....
$endgroup$
– neelkanth
Dec 6 '18 at 7:45
add a comment |
$begingroup$
As $[sin(x)]=-1$ on $(pi , 2pi)$ so it is continuous at $dfrac{3pi}{2}$.
edited Dec 6 '18 at 7:46
Philippe Malot
2,276824
answered Dec 6 '18 at 7:43
neelkanthneelkanth
2,28321129
$endgroup$
$begingroup$
thankyou sir i have drawn the graph wrong
$endgroup$
– HOME WORK AND EXERCISES
Dec 6 '18 at 7:44
$begingroup$
ok ....you are welcome ....
$endgroup$
– neelkanth
Dec 6 '18 at 7:45
add a comment |
$begingroup$
As $[sin(x)]=-1$ on $(pi , 2pi)$ so it is continuous at $dfrac{3pi}{2}$.
edited Dec 6 '18 at 7:46
Philippe Malot
2,276824
answered Dec 6 '18 at 7:43
neelkanthneelkanth
2,28321129
$endgroup$
As $[sin(x)]=-1$ on $(pi , 2pi)$ so it is continuous at $dfrac{3pi}{2}$.
edited Dec 6 '18 at 7:46
Philippe Malot
2,276824
answered Dec 6 '18 at 7:43
neelkanthneelkanth
2,28321129
edited Dec 6 '18 at 7:46
Philippe Malot
2,276824
edited Dec 6 '18 at 7:46
Philippe Malot
2,276824
edited Dec 6 '18 at 7:46
Philippe Malot
2,276824
2,276824
answered Dec 6 '18 at 7:43
neelkanthneelkanth
2,28321129
answered Dec 6 '18 at 7:43
neelkanthneelkanth
2,28321129
answered Dec 6 '18 at 7:43
neelkanthneelkanth
2,28321129
2,28321129
$begingroup$
thankyou sir i have drawn the graph wrong
$endgroup$
– HOME WORK AND EXERCISES
Dec 6 '18 at 7:44
$begingroup$
ok ....you are welcome ....
$endgroup$
– neelkanth
Dec 6 '18 at 7:45
add a comment |
$begingroup$
thankyou sir i have drawn the graph wrong
$endgroup$
– HOME WORK AND EXERCISES
Dec 6 '18 at 7:44
$begingroup$
ok ....you are welcome ....
$endgroup$
– neelkanth
Dec 6 '18 at 7:45
$begingroup$
thankyou sir i have drawn the graph wrong
$endgroup$
– HOME WORK AND EXERCISES
Dec 6 '18 at 7:44
$begingroup$
thankyou sir i have drawn the graph wrong
$endgroup$
– HOME WORK AND EXERCISES
Dec 6 '18 at 7:44
$begingroup$
ok ....you are welcome ....
$endgroup$
– neelkanth
Dec 6 '18 at 7:45
$begingroup$
ok ....you are welcome ....
$endgroup$
– neelkanth
Dec 6 '18 at 7:45
add a comment |
$begingroup$
The function “greatest integer” is not continuous at integers. So if you consider $g(x)=lfloor f(x)rfloor$, where $f$ is continuous, then $g$ may be not continuous where $f$ takes on an integer value.
However, the fact that $xmapstolfloor xrfloor$ is not continuous at an integer $n$ stems from the fact that $xmapsto x$ can take on values greater and less than $n$ in every neighborhood of $n$.
In your case, there exists a neighborhood of $3pi/2$ where $xmapstosin x$ only takes on values $gesin(3pi/2)=-1$ (for the simple reason that $sin xge-1$ for every $x$).
More generally, if $c$ is a local maximum or minimum point for the function $f$ and $f(c)$ is an integer, then $g(x)=lfloor f(x)rfloor$ is continuous at $c$.
answered Dec 6 '18 at 9:35
egregegreg
183k1486205
$endgroup$
$begingroup$
Thankyou. I realised i had drawn the graph wrong.
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 16:33
add a comment |
$begingroup$
The function “greatest integer” is not continuous at integers. So if you consider $g(x)=lfloor f(x)rfloor$, where $f$ is continuous, then $g$ may be not continuous where $f$ takes on an integer value.
However, the fact that $xmapstolfloor xrfloor$ is not continuous at an integer $n$ stems from the fact that $xmapsto x$ can take on values greater and less than $n$ in every neighborhood of $n$.
In your case, there exists a neighborhood of $3pi/2$ where $xmapstosin x$ only takes on values $gesin(3pi/2)=-1$ (for the simple reason that $sin xge-1$ for every $x$).
More generally, if $c$ is a local maximum or minimum point for the function $f$ and $f(c)$ is an integer, then $g(x)=lfloor f(x)rfloor$ is continuous at $c$.
answered Dec 6 '18 at 9:35
egregegreg
183k1486205
$endgroup$
$begingroup$
Thankyou. I realised i had drawn the graph wrong.
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 16:33
add a comment |
$begingroup$
The function “greatest integer” is not continuous at integers. So if you consider $g(x)=lfloor f(x)rfloor$, where $f$ is continuous, then $g$ may be not continuous where $f$ takes on an integer value.
However, the fact that $xmapstolfloor xrfloor$ is not continuous at an integer $n$ stems from the fact that $xmapsto x$ can take on values greater and less than $n$ in every neighborhood of $n$.
In your case, there exists a neighborhood of $3pi/2$ where $xmapstosin x$ only takes on values $gesin(3pi/2)=-1$ (for the simple reason that $sin xge-1$ for every $x$).
More generally, if $c$ is a local maximum or minimum point for the function $f$ and $f(c)$ is an integer, then $g(x)=lfloor f(x)rfloor$ is continuous at $c$.
answered Dec 6 '18 at 9:35
egregegreg
183k1486205
$endgroup$
The function “greatest integer” is not continuous at integers. So if you consider $g(x)=lfloor f(x)rfloor$, where $f$ is continuous, then $g$ may be not continuous where $f$ takes on an integer value.
However, the fact that $xmapstolfloor xrfloor$ is not continuous at an integer $n$ stems from the fact that $xmapsto x$ can take on values greater and less than $n$ in every neighborhood of $n$.
In your case, there exists a neighborhood of $3pi/2$ where $xmapstosin x$ only takes on values $gesin(3pi/2)=-1$ (for the simple reason that $sin xge-1$ for every $x$).
More generally, if $c$ is a local maximum or minimum point for the function $f$ and $f(c)$ is an integer, then $g(x)=lfloor f(x)rfloor$ is continuous at $c$.
answered Dec 6 '18 at 9:35
egregegreg
183k1486205
answered Dec 6 '18 at 9:35
egregegreg
183k1486205
answered Dec 6 '18 at 9:35
egregegreg
183k1486205
answered Dec 6 '18 at 9:35
egregegreg
183k1486205
183k1486205
$begingroup$
Thankyou. I realised i had drawn the graph wrong.
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 16:33
add a comment |
$begingroup$
Thankyou. I realised i had drawn the graph wrong.
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 16:33
$begingroup$
Thankyou. I realised i had drawn the graph wrong.
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 16:33
$begingroup$
Thankyou. I realised i had drawn the graph wrong.
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 16:33
add a comment |
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$begingroup$
To be clear: How is "$[;;]$" being used here?
$endgroup$
– Blue
Dec 6 '18 at 7:33
$begingroup$
you may drawn wrong graph...
$endgroup$
– neelkanth
Dec 6 '18 at 7:34
$begingroup$
$f(x)=-1$ in a neighborhood of $3pi /2$ so it is continuous at $3pi /2$.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:35
$begingroup$
To pile on with what Blue mentioned - the function $f(x) = [sin(x)]$ can be taken as "$sin(x)$ rounded to the nearest integer." If you just mean it as parentheses or whatever, I have no clue what's going on, honestly, because $sin(x)$ is continuous for all $x$, and certainly has no "jump" discontinuities.
$endgroup$
– Eevee Trainer
Dec 6 '18 at 7:35
$begingroup$
its $-1$ on $(pi , 2pi )$
$endgroup$
– neelkanth
Dec 6 '18 at 7:36