Find $a, b$ such that $f(x) = ax - lfloor bx+crfloor$ is periodic and find its period, where $ab ne 0$

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Find $a, b$ such that $f(x) = ax - lfloor bx+crfloor$ is periodic and find its period, where $ab ne 0$




I've tried to do it the following way:



$$
f(x) = f(x+T) \
ax - lfloorbx + crfloor = a(x+T) - lfloorb(x+T) + crfloor iff \
iff ax - lfloorbx + crfloor - a(x+T) + lfloorb(x+T) + crfloor = 0 \
lfloorbx + c + bTrfloor - lfloorbx + crfloor = aT
$$



That means $aT in mathbb Z$, and $lfloorbx + c + bTrfloor - lfloorbx + crfloor = aT iff bT in mathbb Z$ and $bT = aT$. Therefore $a=b$ if $Tne 0$.



But i'm stuck at finding value of $T$. Is the above correct and how do i find the period of such function?




algebra-precalculus floor-function periodic-functions




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  • Can you just find specific ones? In that case, why not just choose $a = b = 1$, which has a period of 1?
    – Mees de Vries
    Jul 23 at 11:29














up vote
0
down vote

favorite













Find $a, b$ such that $f(x) = ax - lfloor bx+crfloor$ is periodic and find its period, where $ab ne 0$




I've tried to do it the following way:



$$
f(x) = f(x+T) \
ax - lfloorbx + crfloor = a(x+T) - lfloorb(x+T) + crfloor iff \
iff ax - lfloorbx + crfloor - a(x+T) + lfloorb(x+T) + crfloor = 0 \
lfloorbx + c + bTrfloor - lfloorbx + crfloor = aT
$$



That means $aT in mathbb Z$, and $lfloorbx + c + bTrfloor - lfloorbx + crfloor = aT iff bT in mathbb Z$ and $bT = aT$. Therefore $a=b$ if $Tne 0$.



But i'm stuck at finding value of $T$. Is the above correct and how do i find the period of such function?




algebra-precalculus floor-function periodic-functions




share|cite|improve this question



















  • Can you just find specific ones? In that case, why not just choose $a = b = 1$, which has a period of 1?
    – Mees de Vries
    Jul 23 at 11:29












up vote
0
down vote

favorite









up vote
0
down vote

favorite












Find $a, b$ such that $f(x) = ax - lfloor bx+crfloor$ is periodic and find its period, where $ab ne 0$




I've tried to do it the following way:



$$
f(x) = f(x+T) \
ax - lfloorbx + crfloor = a(x+T) - lfloorb(x+T) + crfloor iff \
iff ax - lfloorbx + crfloor - a(x+T) + lfloorb(x+T) + crfloor = 0 \
lfloorbx + c + bTrfloor - lfloorbx + crfloor = aT
$$



That means $aT in mathbb Z$, and $lfloorbx + c + bTrfloor - lfloorbx + crfloor = aT iff bT in mathbb Z$ and $bT = aT$. Therefore $a=b$ if $Tne 0$.



But i'm stuck at finding value of $T$. Is the above correct and how do i find the period of such function?




algebra-precalculus floor-function periodic-functions




share|cite|improve this question












Find $a, b$ such that $f(x) = ax - lfloor bx+crfloor$ is periodic and find its period, where $ab ne 0$




I've tried to do it the following way:



$$
f(x) = f(x+T) \
ax - lfloorbx + crfloor = a(x+T) - lfloorb(x+T) + crfloor iff \
iff ax - lfloorbx + crfloor - a(x+T) + lfloorb(x+T) + crfloor = 0 \
lfloorbx + c + bTrfloor - lfloorbx + crfloor = aT
$$



That means $aT in mathbb Z$, and $lfloorbx + c + bTrfloor - lfloorbx + crfloor = aT iff bT in mathbb Z$ and $bT = aT$. Therefore $a=b$ if $Tne 0$.



But i'm stuck at finding value of $T$. Is the above correct and how do i find the period of such function?





algebra-precalculus floor-function periodic-functions





share|cite|improve this question










share|cite|improve this question




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asked Jul 23 at 11:20









roman

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4391412











  • Can you just find specific ones? In that case, why not just choose $a = b = 1$, which has a period of 1?
    – Mees de Vries
    Jul 23 at 11:29
















  • Can you just find specific ones? In that case, why not just choose $a = b = 1$, which has a period of 1?
    – Mees de Vries
    Jul 23 at 11:29















Can you just find specific ones? In that case, why not just choose $a = b = 1$, which has a period of 1?
– Mees de Vries
Jul 23 at 11:29




Can you just find specific ones? In that case, why not just choose $a = b = 1$, which has a period of 1?
– Mees de Vries
Jul 23 at 11:29










1 Answer
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The fractional part function $x=x-lfloor xrfloor$ is known to be periodic with period $1$.



Then



$$ax-(bx+c)+bx+c$$ can only be periodic if $a=b$ (otherwise $(a-b)x$ is aperiodic), and the period is such that $bT=1$.






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    1






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    active

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    up vote
    3
    down vote



    accepted










    The fractional part function $x=x-lfloor xrfloor$ is known to be periodic with period $1$.



    Then



    $$ax-(bx+c)+bx+c$$ can only be periodic if $a=b$ (otherwise $(a-b)x$ is aperiodic), and the period is such that $bT=1$.






    share|cite|improve this answer



























      up vote
      3
      down vote



      accepted










      The fractional part function $x=x-lfloor xrfloor$ is known to be periodic with period $1$.



      Then



      $$ax-(bx+c)+bx+c$$ can only be periodic if $a=b$ (otherwise $(a-b)x$ is aperiodic), and the period is such that $bT=1$.






      share|cite|improve this answer

























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        The fractional part function $x=x-lfloor xrfloor$ is known to be periodic with period $1$.



        Then



        $$ax-(bx+c)+bx+c$$ can only be periodic if $a=b$ (otherwise $(a-b)x$ is aperiodic), and the period is such that $bT=1$.






        share|cite|improve this answer















        The fractional part function $x=x-lfloor xrfloor$ is known to be periodic with period $1$.



        Then



        $$ax-(bx+c)+bx+c$$ can only be periodic if $a=b$ (otherwise $(a-b)x$ is aperiodic), and the period is such that $bT=1$.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 23 at 11:49


























        answered Jul 23 at 11:29









        Yves Daoust

        111k665203




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