Show $f(x) = x^2 sin (x^{-3/2}), xin (0,1] $, $f(0) = 0$ is of bounded variation without using improper...
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This question already has an answer here:
When is $F(x)=x^asin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?
1 answer
Show $f(x) = x^2 sin (x^{-3/2}), xin (0,1] $, $f(0) = 0$ is of bounded variation.
Try
$forall P = {0 = x_0 < cdots, < x_n = 1}$, $exists c_k in (x_{k-1}, x_k)$ s.t. $f(x_k) - f(x_{k-1}) = f'(c_k) (x_k - x_{k-1})$$(because MVT)$.
Thus,
$$
V(f, P) = sum |f(x_k) - f(x_{k-1})| = sum |f'(c_k)| (x_k - x_{k-1})
$$
I think the next step should allow me to prove the RHS is less than some value. However, observing the following,
$$
f'(x) = begin{cases} 2x sin(x^{-3/2}) - frac{3}{2} x^{-1/2} cos(x^{-3/2}) & (x neq 0 ) \ 0 & (x=0) end{cases}
$$
I have $|f'(x)| le 2x + frac{3}{2}x^{-1/2}$, but RHS is not bounded on $[0,1]$.
Any help, including one that excludes the use of improper integral, will be appreciated.
real-analysis bounded-variation
asked Dec 1 '18 at 16:36
MoreblueMoreblue
8741216
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marked as duplicate by Moreblue, Community♦ Dec 2 '18 at 0:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
When is $F(x)=x^asin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?
1 answer
Show $f(x) = x^2 sin (x^{-3/2}), xin (0,1] $, $f(0) = 0$ is of bounded variation.
Try
$forall P = {0 = x_0 < cdots, < x_n = 1}$, $exists c_k in (x_{k-1}, x_k)$ s.t. $f(x_k) - f(x_{k-1}) = f'(c_k) (x_k - x_{k-1})$$(because MVT)$.
Thus,
$$
V(f, P) = sum |f(x_k) - f(x_{k-1})| = sum |f'(c_k)| (x_k - x_{k-1})
$$
I think the next step should allow me to prove the RHS is less than some value. However, observing the following,
$$
f'(x) = begin{cases} 2x sin(x^{-3/2}) - frac{3}{2} x^{-1/2} cos(x^{-3/2}) & (x neq 0 ) \ 0 & (x=0) end{cases}
$$
I have $|f'(x)| le 2x + frac{3}{2}x^{-1/2}$, but RHS is not bounded on $[0,1]$.
Any help, including one that excludes the use of improper integral, will be appreciated.
real-analysis bounded-variation
asked Dec 1 '18 at 16:36
MoreblueMoreblue
8741216
$endgroup$
marked as duplicate by Moreblue, Community♦ Dec 2 '18 at 0:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
When is $F(x)=x^asin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?
1 answer
Show $f(x) = x^2 sin (x^{-3/2}), xin (0,1] $, $f(0) = 0$ is of bounded variation.
Try
$forall P = {0 = x_0 < cdots, < x_n = 1}$, $exists c_k in (x_{k-1}, x_k)$ s.t. $f(x_k) - f(x_{k-1}) = f'(c_k) (x_k - x_{k-1})$$(because MVT)$.
Thus,
$$
V(f, P) = sum |f(x_k) - f(x_{k-1})| = sum |f'(c_k)| (x_k - x_{k-1})
$$
I think the next step should allow me to prove the RHS is less than some value. However, observing the following,
$$
f'(x) = begin{cases} 2x sin(x^{-3/2}) - frac{3}{2} x^{-1/2} cos(x^{-3/2}) & (x neq 0 ) \ 0 & (x=0) end{cases}
$$
I have $|f'(x)| le 2x + frac{3}{2}x^{-1/2}$, but RHS is not bounded on $[0,1]$.
Any help, including one that excludes the use of improper integral, will be appreciated.
real-analysis bounded-variation
asked Dec 1 '18 at 16:36
MoreblueMoreblue
8741216
$endgroup$
This question already has an answer here:
When is $F(x)=x^asin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?
1 answer
Show $f(x) = x^2 sin (x^{-3/2}), xin (0,1] $, $f(0) = 0$ is of bounded variation.
Try
$forall P = {0 = x_0 < cdots, < x_n = 1}$, $exists c_k in (x_{k-1}, x_k)$ s.t. $f(x_k) - f(x_{k-1}) = f'(c_k) (x_k - x_{k-1})$$(because MVT)$.
Thus,
$$
V(f, P) = sum |f(x_k) - f(x_{k-1})| = sum |f'(c_k)| (x_k - x_{k-1})
$$
I think the next step should allow me to prove the RHS is less than some value. However, observing the following,
$$
f'(x) = begin{cases} 2x sin(x^{-3/2}) - frac{3}{2} x^{-1/2} cos(x^{-3/2}) & (x neq 0 ) \ 0 & (x=0) end{cases}
$$
I have $|f'(x)| le 2x + frac{3}{2}x^{-1/2}$, but RHS is not bounded on $[0,1]$.
Any help, including one that excludes the use of improper integral, will be appreciated.
This question already has an answer here:
When is $F(x)=x^asin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?
1 answer
real-analysis bounded-variation
real-analysis bounded-variation
asked Dec 1 '18 at 16:36
MoreblueMoreblue
8741216
asked Dec 1 '18 at 16:36
MoreblueMoreblue
8741216
asked Dec 1 '18 at 16:36
MoreblueMoreblue
8741216
asked Dec 1 '18 at 16:36
MoreblueMoreblue
8741216
asked Dec 1 '18 at 16:36
MoreblueMoreblue
8741216
8741216
marked as duplicate by Moreblue, Community♦ Dec 2 '18 at 0:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Moreblue, Community♦ Dec 2 '18 at 0:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
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$begingroup$
Here is the more general version of the question you are asking for:
When is $F(x)=x^asin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?
For your example, just take $a=2$ and $b=3/2$
answered Dec 1 '18 at 17:37
Kernel_DirichletKernel_Dirichlet
1,139416
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is the more general version of the question you are asking for:
When is $F(x)=x^asin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?
For your example, just take $a=2$ and $b=3/2$
answered Dec 1 '18 at 17:37
Kernel_DirichletKernel_Dirichlet
1,139416
$endgroup$
add a comment |
$begingroup$
Here is the more general version of the question you are asking for:
When is $F(x)=x^asin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?
For your example, just take $a=2$ and $b=3/2$
answered Dec 1 '18 at 17:37
Kernel_DirichletKernel_Dirichlet
1,139416
$endgroup$
add a comment |
$begingroup$
Here is the more general version of the question you are asking for:
When is $F(x)=x^asin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?
For your example, just take $a=2$ and $b=3/2$
answered Dec 1 '18 at 17:37
Kernel_DirichletKernel_Dirichlet
1,139416
$endgroup$
Here is the more general version of the question you are asking for:
When is $F(x)=x^asin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?
For your example, just take $a=2$ and $b=3/2$
answered Dec 1 '18 at 17:37
Kernel_DirichletKernel_Dirichlet
1,139416
answered Dec 1 '18 at 17:37
Kernel_DirichletKernel_Dirichlet
1,139416
answered Dec 1 '18 at 17:37
Kernel_DirichletKernel_Dirichlet
1,139416
answered Dec 1 '18 at 17:37
Kernel_DirichletKernel_Dirichlet
1,139416
1,139416
add a comment |
add a comment |
