Differentiablity of $lfloor(xsin(pi x)rfloor$ in the interval $(-1,1)$












0












$begingroup$


My attempt:
here the $f(x)=lfloor(xsin(pi x)rfloor$.
For the function to be differentiable in an interval $(-1,1)$ the right hand derivative of $-1$ must be finite and the left hand derivative of $-2$ must be finite.
But i cannot figure out how to find the limit of $(f(-1+h)-f(-1))/h$ as h tends to $0^+$.






calculus limits derivatives







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$endgroup$












  • $begingroup$
    What do "left hand derivative" and "right hand derivative" mean?
    $endgroup$
    – YiFan
    Dec 8 '18 at 5:31










  • $begingroup$
    @YiFan Please refer to this math.stackexchange.com/questions/1158510/…
    $endgroup$
    – Onkar Dahale
    Dec 8 '18 at 5:36












  • $begingroup$
    Isn't $f(x) equiv 0$ since $f(x) in [0, 1)$ $forall x in (-1, 1)$? Can you prove it?
    $endgroup$
    – Alex Vong
    Dec 8 '18 at 6:05












  • $begingroup$
    @OnkarDahale okay, though I have to say that terminology is quite rare (usually I would just say left hand limit of [...] for example.)
    $endgroup$
    – YiFan
    Dec 8 '18 at 7:14
















0












$begingroup$


My attempt:
here the $f(x)=lfloor(xsin(pi x)rfloor$.
For the function to be differentiable in an interval $(-1,1)$ the right hand derivative of $-1$ must be finite and the left hand derivative of $-2$ must be finite.
But i cannot figure out how to find the limit of $(f(-1+h)-f(-1))/h$ as h tends to $0^+$.






calculus limits derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do "left hand derivative" and "right hand derivative" mean?
    $endgroup$
    – YiFan
    Dec 8 '18 at 5:31










  • $begingroup$
    @YiFan Please refer to this math.stackexchange.com/questions/1158510/…
    $endgroup$
    – Onkar Dahale
    Dec 8 '18 at 5:36












  • $begingroup$
    Isn't $f(x) equiv 0$ since $f(x) in [0, 1)$ $forall x in (-1, 1)$? Can you prove it?
    $endgroup$
    – Alex Vong
    Dec 8 '18 at 6:05












  • $begingroup$
    @OnkarDahale okay, though I have to say that terminology is quite rare (usually I would just say left hand limit of [...] for example.)
    $endgroup$
    – YiFan
    Dec 8 '18 at 7:14














0












0








0





$begingroup$


My attempt:
here the $f(x)=lfloor(xsin(pi x)rfloor$.
For the function to be differentiable in an interval $(-1,1)$ the right hand derivative of $-1$ must be finite and the left hand derivative of $-2$ must be finite.
But i cannot figure out how to find the limit of $(f(-1+h)-f(-1))/h$ as h tends to $0^+$.






calculus limits derivatives







share|cite|improve this question











$endgroup$




My attempt:
here the $f(x)=lfloor(xsin(pi x)rfloor$.
For the function to be differentiable in an interval $(-1,1)$ the right hand derivative of $-1$ must be finite and the left hand derivative of $-2$ must be finite.
But i cannot figure out how to find the limit of $(f(-1+h)-f(-1))/h$ as h tends to $0^+$.






calculus limits derivatives




calculus limits derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 5:36









YiFan

4,7261727




4,7261727










asked Dec 8 '18 at 4:34









Onkar DahaleOnkar Dahale

155




155












  • $begingroup$
    What do "left hand derivative" and "right hand derivative" mean?
    $endgroup$
    – YiFan
    Dec 8 '18 at 5:31










  • $begingroup$
    @YiFan Please refer to this math.stackexchange.com/questions/1158510/…
    $endgroup$
    – Onkar Dahale
    Dec 8 '18 at 5:36












  • $begingroup$
    Isn't $f(x) equiv 0$ since $f(x) in [0, 1)$ $forall x in (-1, 1)$? Can you prove it?
    $endgroup$
    – Alex Vong
    Dec 8 '18 at 6:05












  • $begingroup$
    @OnkarDahale okay, though I have to say that terminology is quite rare (usually I would just say left hand limit of [...] for example.)
    $endgroup$
    – YiFan
    Dec 8 '18 at 7:14


















  • $begingroup$
    What do "left hand derivative" and "right hand derivative" mean?
    $endgroup$
    – YiFan
    Dec 8 '18 at 5:31










  • $begingroup$
    @YiFan Please refer to this math.stackexchange.com/questions/1158510/…
    $endgroup$
    – Onkar Dahale
    Dec 8 '18 at 5:36












  • $begingroup$
    Isn't $f(x) equiv 0$ since $f(x) in [0, 1)$ $forall x in (-1, 1)$? Can you prove it?
    $endgroup$
    – Alex Vong
    Dec 8 '18 at 6:05












  • $begingroup$
    @OnkarDahale okay, though I have to say that terminology is quite rare (usually I would just say left hand limit of [...] for example.)
    $endgroup$
    – YiFan
    Dec 8 '18 at 7:14
















$begingroup$
What do "left hand derivative" and "right hand derivative" mean?
$endgroup$
– YiFan
Dec 8 '18 at 5:31




$begingroup$
What do "left hand derivative" and "right hand derivative" mean?
$endgroup$
– YiFan
Dec 8 '18 at 5:31












$begingroup$
@YiFan Please refer to this math.stackexchange.com/questions/1158510/…
$endgroup$
– Onkar Dahale
Dec 8 '18 at 5:36






$begingroup$
@YiFan Please refer to this math.stackexchange.com/questions/1158510/…
$endgroup$
– Onkar Dahale
Dec 8 '18 at 5:36














$begingroup$
Isn't $f(x) equiv 0$ since $f(x) in [0, 1)$ $forall x in (-1, 1)$? Can you prove it?
$endgroup$
– Alex Vong
Dec 8 '18 at 6:05






$begingroup$
Isn't $f(x) equiv 0$ since $f(x) in [0, 1)$ $forall x in (-1, 1)$? Can you prove it?
$endgroup$
– Alex Vong
Dec 8 '18 at 6:05














$begingroup$
@OnkarDahale okay, though I have to say that terminology is quite rare (usually I would just say left hand limit of [...] for example.)
$endgroup$
– YiFan
Dec 8 '18 at 7:14




$begingroup$
@OnkarDahale okay, though I have to say that terminology is quite rare (usually I would just say left hand limit of [...] for example.)
$endgroup$
– YiFan
Dec 8 '18 at 7:14










1 Answer
1






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2












$begingroup$

Firstly, I want to clarify that since we are talking about the differentiability of $f$ on $(-1, 1)$, not $[-1, 1]$, we don't need to worry about the differentiability of $f$ at the end points.



With this in mind, let's see what we can say about $x sin(pi x)$. It is not hard to see that $(-x) sin(-pi x) = x sin(pi x)$. Therefore, $x sin(pi x)$ is an even function on $(-1, 1)$. Furthermore, $0 le x < 1$ and $0 le sin(pi x) le 1$ on $[0, 1)$. Hence $0 le x sin(pi x) < 1$ on $[0, 1)$. Finally, using the fact that $x sin(pi x)$ is an even function, we conclude $0 le x sin(pi x) < 1$ on $(-1, 1)$.



This means $lfloor x sin(pi x) rfloor = 0$ on $(-1, 1)$, which is clearly differentiable.






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    1 Answer
    1






    active

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    active

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    $begingroup$

    Firstly, I want to clarify that since we are talking about the differentiability of $f$ on $(-1, 1)$, not $[-1, 1]$, we don't need to worry about the differentiability of $f$ at the end points.



    With this in mind, let's see what we can say about $x sin(pi x)$. It is not hard to see that $(-x) sin(-pi x) = x sin(pi x)$. Therefore, $x sin(pi x)$ is an even function on $(-1, 1)$. Furthermore, $0 le x < 1$ and $0 le sin(pi x) le 1$ on $[0, 1)$. Hence $0 le x sin(pi x) < 1$ on $[0, 1)$. Finally, using the fact that $x sin(pi x)$ is an even function, we conclude $0 le x sin(pi x) < 1$ on $(-1, 1)$.



    This means $lfloor x sin(pi x) rfloor = 0$ on $(-1, 1)$, which is clearly differentiable.






    share|cite|improve this answer









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      2












      $begingroup$

      Firstly, I want to clarify that since we are talking about the differentiability of $f$ on $(-1, 1)$, not $[-1, 1]$, we don't need to worry about the differentiability of $f$ at the end points.



      With this in mind, let's see what we can say about $x sin(pi x)$. It is not hard to see that $(-x) sin(-pi x) = x sin(pi x)$. Therefore, $x sin(pi x)$ is an even function on $(-1, 1)$. Furthermore, $0 le x < 1$ and $0 le sin(pi x) le 1$ on $[0, 1)$. Hence $0 le x sin(pi x) < 1$ on $[0, 1)$. Finally, using the fact that $x sin(pi x)$ is an even function, we conclude $0 le x sin(pi x) < 1$ on $(-1, 1)$.



      This means $lfloor x sin(pi x) rfloor = 0$ on $(-1, 1)$, which is clearly differentiable.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Firstly, I want to clarify that since we are talking about the differentiability of $f$ on $(-1, 1)$, not $[-1, 1]$, we don't need to worry about the differentiability of $f$ at the end points.



        With this in mind, let's see what we can say about $x sin(pi x)$. It is not hard to see that $(-x) sin(-pi x) = x sin(pi x)$. Therefore, $x sin(pi x)$ is an even function on $(-1, 1)$. Furthermore, $0 le x < 1$ and $0 le sin(pi x) le 1$ on $[0, 1)$. Hence $0 le x sin(pi x) < 1$ on $[0, 1)$. Finally, using the fact that $x sin(pi x)$ is an even function, we conclude $0 le x sin(pi x) < 1$ on $(-1, 1)$.



        This means $lfloor x sin(pi x) rfloor = 0$ on $(-1, 1)$, which is clearly differentiable.






        share|cite|improve this answer









        $endgroup$



        Firstly, I want to clarify that since we are talking about the differentiability of $f$ on $(-1, 1)$, not $[-1, 1]$, we don't need to worry about the differentiability of $f$ at the end points.



        With this in mind, let's see what we can say about $x sin(pi x)$. It is not hard to see that $(-x) sin(-pi x) = x sin(pi x)$. Therefore, $x sin(pi x)$ is an even function on $(-1, 1)$. Furthermore, $0 le x < 1$ and $0 le sin(pi x) le 1$ on $[0, 1)$. Hence $0 le x sin(pi x) < 1$ on $[0, 1)$. Finally, using the fact that $x sin(pi x)$ is an even function, we conclude $0 le x sin(pi x) < 1$ on $(-1, 1)$.



        This means $lfloor x sin(pi x) rfloor = 0$ on $(-1, 1)$, which is clearly differentiable.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 8:59









        Alex VongAlex Vong

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