How can I prove that this function is periodic :
$begingroup$
I know that a periodic function satisfies $F(x+a)=F(x)$
$y=leftlfloorfrac{sin x}xrightrfloor$
And the period of this fuction is $2pi$ But putting $f(x+2pi)$ I can't evaluate denominator inside the greatest integer function
So how can I prove this function periodic?
functional-analysis periodic-functions
edited Jan 11 at 20:08
Henrik
6,04492030
asked Jan 5 at 8:17
Shubham GawriShubham Gawri
286
$endgroup$
add a comment |
$begingroup$
I know that a periodic function satisfies $F(x+a)=F(x)$
$y=leftlfloorfrac{sin x}xrightrfloor$
And the period of this fuction is $2pi$ But putting $f(x+2pi)$ I can't evaluate denominator inside the greatest integer function
So how can I prove this function periodic?
functional-analysis periodic-functions
edited Jan 11 at 20:08
Henrik
6,04492030
asked Jan 5 at 8:17
Shubham GawriShubham Gawri
286
$endgroup$
$begingroup$
Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
$endgroup$
– Henrik
Jan 11 at 20:14
add a comment |
$begingroup$
I know that a periodic function satisfies $F(x+a)=F(x)$
$y=leftlfloorfrac{sin x}xrightrfloor$
And the period of this fuction is $2pi$ But putting $f(x+2pi)$ I can't evaluate denominator inside the greatest integer function
So how can I prove this function periodic?
functional-analysis periodic-functions
edited Jan 11 at 20:08
Henrik
6,04492030
asked Jan 5 at 8:17
Shubham GawriShubham Gawri
286
$endgroup$
I know that a periodic function satisfies $F(x+a)=F(x)$
$y=leftlfloorfrac{sin x}xrightrfloor$
And the period of this fuction is $2pi$ But putting $f(x+2pi)$ I can't evaluate denominator inside the greatest integer function
So how can I prove this function periodic?
functional-analysis periodic-functions
functional-analysis periodic-functions
edited Jan 11 at 20:08
Henrik
6,04492030
asked Jan 5 at 8:17
Shubham GawriShubham Gawri
286
edited Jan 11 at 20:08
Henrik
6,04492030
asked Jan 5 at 8:17
Shubham GawriShubham Gawri
286
edited Jan 11 at 20:08
Henrik
6,04492030
edited Jan 11 at 20:08
Henrik
6,04492030
edited Jan 11 at 20:08
Henrik
6,04492030
6,04492030
asked Jan 5 at 8:17
Shubham GawriShubham Gawri
286
asked Jan 5 at 8:17
Shubham GawriShubham Gawri
286
asked Jan 5 at 8:17
Shubham GawriShubham Gawri
286
286
$begingroup$
Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
$endgroup$
– Henrik
Jan 11 at 20:14
add a comment |
$begingroup$
Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
$endgroup$
– Henrik
Jan 11 at 20:14
$begingroup$
Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
$endgroup$
– Henrik
Jan 11 at 20:14
$begingroup$
Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
$endgroup$
– Henrik
Jan 11 at 20:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For any single variable function to be periodic,
$$ f(x) = f(x+ lambda)$$
should be strictly satisfied. Just having (at least) two sequence of its roots in arithmetic progression is not enough.
answered Jan 5 at 8:44
NarasimhamNarasimham
21k62158
$endgroup$
$begingroup$
Ok that means that this function cannot be said periodic It just follows a pattern..
$endgroup$
– Shubham Gawri
Jan 5 at 9:06
1
$begingroup$
Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
$endgroup$
– Narasimham
Jan 5 at 9:24
add a comment |
$begingroup$
The result is wrong
First of all you have an issue to deal with at $0$.
Second $y$ is equal to zero on $I=(-pi, pi)setminus {0}$. And there is no other translated subset of $I$ in $mathbb R$ on which $y$ takes such value.
answered Jan 5 at 8:37
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For any single variable function to be periodic,
$$ f(x) = f(x+ lambda)$$
should be strictly satisfied. Just having (at least) two sequence of its roots in arithmetic progression is not enough.
answered Jan 5 at 8:44
NarasimhamNarasimham
21k62158
$endgroup$
$begingroup$
Ok that means that this function cannot be said periodic It just follows a pattern..
$endgroup$
– Shubham Gawri
Jan 5 at 9:06
1
$begingroup$
Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
$endgroup$
– Narasimham
Jan 5 at 9:24
add a comment |
$begingroup$
For any single variable function to be periodic,
$$ f(x) = f(x+ lambda)$$
should be strictly satisfied. Just having (at least) two sequence of its roots in arithmetic progression is not enough.
answered Jan 5 at 8:44
NarasimhamNarasimham
21k62158
$endgroup$
$begingroup$
Ok that means that this function cannot be said periodic It just follows a pattern..
$endgroup$
– Shubham Gawri
Jan 5 at 9:06
1
$begingroup$
Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
$endgroup$
– Narasimham
Jan 5 at 9:24
add a comment |
$begingroup$
For any single variable function to be periodic,
$$ f(x) = f(x+ lambda)$$
should be strictly satisfied. Just having (at least) two sequence of its roots in arithmetic progression is not enough.
answered Jan 5 at 8:44
NarasimhamNarasimham
21k62158
$endgroup$
For any single variable function to be periodic,
$$ f(x) = f(x+ lambda)$$
should be strictly satisfied. Just having (at least) two sequence of its roots in arithmetic progression is not enough.
answered Jan 5 at 8:44
NarasimhamNarasimham
21k62158
answered Jan 5 at 8:44
NarasimhamNarasimham
21k62158
answered Jan 5 at 8:44
NarasimhamNarasimham
21k62158
answered Jan 5 at 8:44
NarasimhamNarasimham
21k62158
21k62158
$begingroup$
Ok that means that this function cannot be said periodic It just follows a pattern..
$endgroup$
– Shubham Gawri
Jan 5 at 9:06
1
$begingroup$
Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
$endgroup$
– Narasimham
Jan 5 at 9:24
add a comment |
$begingroup$
Ok that means that this function cannot be said periodic It just follows a pattern..
$endgroup$
– Shubham Gawri
Jan 5 at 9:06
1
$begingroup$
Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
$endgroup$
– Narasimham
Jan 5 at 9:24
$begingroup$
Ok that means that this function cannot be said periodic It just follows a pattern..
$endgroup$
– Shubham Gawri
Jan 5 at 9:06
$begingroup$
Ok that means that this function cannot be said periodic It just follows a pattern..
$endgroup$
– Shubham Gawri
Jan 5 at 9:06
1
1
$begingroup$
Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
$endgroup$
– Narasimham
Jan 5 at 9:24
$begingroup$
Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
$endgroup$
– Narasimham
Jan 5 at 9:24
add a comment |
$begingroup$
The result is wrong
First of all you have an issue to deal with at $0$.
Second $y$ is equal to zero on $I=(-pi, pi)setminus {0}$. And there is no other translated subset of $I$ in $mathbb R$ on which $y$ takes such value.
answered Jan 5 at 8:37
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
The result is wrong
First of all you have an issue to deal with at $0$.
Second $y$ is equal to zero on $I=(-pi, pi)setminus {0}$. And there is no other translated subset of $I$ in $mathbb R$ on which $y$ takes such value.
answered Jan 5 at 8:37
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
The result is wrong
First of all you have an issue to deal with at $0$.
Second $y$ is equal to zero on $I=(-pi, pi)setminus {0}$. And there is no other translated subset of $I$ in $mathbb R$ on which $y$ takes such value.
answered Jan 5 at 8:37
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
The result is wrong
First of all you have an issue to deal with at $0$.
Second $y$ is equal to zero on $I=(-pi, pi)setminus {0}$. And there is no other translated subset of $I$ in $mathbb R$ on which $y$ takes such value.
answered Jan 5 at 8:37
mathcounterexamples.netmathcounterexamples.net
27k22158
edited Jan 5 at 8:44
answered Jan 5 at 8:37
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 5 at 8:37
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 5 at 8:37
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
add a comment |
add a comment |
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$begingroup$
Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
$endgroup$
– Henrik
Jan 11 at 20:14