Differentiablity of $lfloor(xsin(pi x)rfloor$ in the interval $(-1,1)$
$begingroup$
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
edited Dec 8 '18 at 5:36
YiFan
2,8091422
asked Dec 8 '18 at 4:34
Onkar DahaleOnkar Dahale
155
$endgroup$
add a comment |
$begingroup$
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
edited Dec 8 '18 at 5:36
YiFan
2,8091422
asked Dec 8 '18 at 4:34
Onkar DahaleOnkar Dahale
155
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 8 '18 at 5:36
YiFan
2,8091422
asked Dec 8 '18 at 4:34
Onkar DahaleOnkar Dahale
155
$endgroup$
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
edited Dec 8 '18 at 5:36
YiFan
2,8091422
asked Dec 8 '18 at 4:34
Onkar DahaleOnkar Dahale
155
edited Dec 8 '18 at 5:36
YiFan
2,8091422
asked Dec 8 '18 at 4:34
Onkar DahaleOnkar Dahale
155
edited Dec 8 '18 at 5:36
YiFan
2,8091422
edited Dec 8 '18 at 5:36
YiFan
2,8091422
edited Dec 8 '18 at 5:36
YiFan
2,8091422
2,8091422
asked Dec 8 '18 at 4:34
Onkar DahaleOnkar Dahale
155
asked Dec 8 '18 at 4:34
Onkar DahaleOnkar Dahale
155
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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.
answered Dec 8 '18 at 8:59
Alex VongAlex Vong
1,284819
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add a comment |
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1 Answer
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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.
answered Dec 8 '18 at 8:59
Alex VongAlex Vong
1,284819
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 8 '18 at 8:59
Alex VongAlex Vong
1,284819
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 8 '18 at 8:59
Alex VongAlex Vong
1,284819
$endgroup$
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.
answered Dec 8 '18 at 8:59
Alex VongAlex Vong
1,284819
answered Dec 8 '18 at 8:59
Alex VongAlex Vong
1,284819
answered Dec 8 '18 at 8:59
Alex VongAlex Vong
1,284819
answered Dec 8 '18 at 8:59
Alex VongAlex Vong
1,284819
1,284819
add a comment |
add a comment |
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$begingroup$
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