Mixing
![Give the linear systems: $(i) x_1+2x_2=2,3x_1+7x_2=8\(ii) x_1+2x_2=1,3x_1+7x_2=7$ [closed]](https://lh4.googleusercontent.com/-704sC3B_mEI/AAAAAAAAAAI/AAAAAAAAAx4/DuAQiT4vCt8/s72-c/photo.jpg?sz=32)
Show that $f'(x)$ is not integrable on $[0,1]$ if $f(x) = x^2sin(frac{1}{x^2}))$
$begingroup$
Show that $f'(x)$ is not integrable on $[0,1]$ if $f(x) = x^2sin(frac{1}{x^2}))chi_{(0,1]}$ and $f(x) = 0$ if $x = 0$.
I took the derivative
$$f'(x) = x^2cos(frac{1}{x^2})(-2)frac{1}{x^3} + 2xsin(frac{1}{x^2}) mbox{ if } x neq 0, f'(x) = 0 mbox{ if } x = 0$$
I must show that
$$int_{0}^{1}|f'(x)| = infty$$
I can't show that the integral is not finite.
real-analysis measure-theory derivatives lebesgue-integral
asked Jan 9 at 2:31
Richard ClareRichard Clare
1,066314
$endgroup$
add a comment |
$begingroup$
Show that $f'(x)$ is not integrable on $[0,1]$ if $f(x) = x^2sin(frac{1}{x^2}))chi_{(0,1]}$ and $f(x) = 0$ if $x = 0$.
I took the derivative
$$f'(x) = x^2cos(frac{1}{x^2})(-2)frac{1}{x^3} + 2xsin(frac{1}{x^2}) mbox{ if } x neq 0, f'(x) = 0 mbox{ if } x = 0$$
I must show that
$$int_{0}^{1}|f'(x)| = infty$$
I can't show that the integral is not finite.
real-analysis measure-theory derivatives lebesgue-integral
asked Jan 9 at 2:31
Richard ClareRichard Clare
1,066314
$endgroup$
1
$begingroup$
The issue is the first term. Start by isolating subintervals where $|cos(1/x^2)|$ is not too small (say greater than $1/2$), and then consider what happens when you integrate $1/x$ on these subintervals.
$endgroup$
– Ian
Jan 9 at 2:33
1
$begingroup$
Could you use the fundamental theorem of calculus, instead of taking the derivative itself? In that case, $int_0^1 f'(x) dx = f(1) - f(0)$, so you're clearly getting undefined terms. Or is this too naive of an approach?
$endgroup$
– kkc
Jan 9 at 3:27
$begingroup$
I tried that but I think we can't do that because we need |f'(x)| not $f'(x)$.
$endgroup$
– Richard Clare
Jan 9 at 3:47
1
$begingroup$
But f'(x) not integrable => |f'(x)| is not integrable
$endgroup$
– Richard Clare
Jan 9 at 3:48
add a comment |
$begingroup$
Show that $f'(x)$ is not integrable on $[0,1]$ if $f(x) = x^2sin(frac{1}{x^2}))chi_{(0,1]}$ and $f(x) = 0$ if $x = 0$.
I took the derivative
$$f'(x) = x^2cos(frac{1}{x^2})(-2)frac{1}{x^3} + 2xsin(frac{1}{x^2}) mbox{ if } x neq 0, f'(x) = 0 mbox{ if } x = 0$$
I must show that
$$int_{0}^{1}|f'(x)| = infty$$
I can't show that the integral is not finite.
real-analysis measure-theory derivatives lebesgue-integral
asked Jan 9 at 2:31
Richard ClareRichard Clare
1,066314
$endgroup$
Show that $f'(x)$ is not integrable on $[0,1]$ if $f(x) = x^2sin(frac{1}{x^2}))chi_{(0,1]}$ and $f(x) = 0$ if $x = 0$.
I took the derivative
$$f'(x) = x^2cos(frac{1}{x^2})(-2)frac{1}{x^3} + 2xsin(frac{1}{x^2}) mbox{ if } x neq 0, f'(x) = 0 mbox{ if } x = 0$$
I must show that
$$int_{0}^{1}|f'(x)| = infty$$
I can't show that the integral is not finite.
real-analysis measure-theory derivatives lebesgue-integral
real-analysis measure-theory derivatives lebesgue-integral
asked Jan 9 at 2:31
Richard ClareRichard Clare
1,066314
asked Jan 9 at 2:31
Richard ClareRichard Clare
1,066314
asked Jan 9 at 2:31
Richard ClareRichard Clare
1,066314
asked Jan 9 at 2:31
Richard ClareRichard Clare
1,066314
asked Jan 9 at 2:31
Richard ClareRichard Clare
1,066314
1,066314
1
$begingroup$
The issue is the first term. Start by isolating subintervals where $|cos(1/x^2)|$ is not too small (say greater than $1/2$), and then consider what happens when you integrate $1/x$ on these subintervals.
$endgroup$
– Ian
Jan 9 at 2:33
1
$begingroup$
Could you use the fundamental theorem of calculus, instead of taking the derivative itself? In that case, $int_0^1 f'(x) dx = f(1) - f(0)$, so you're clearly getting undefined terms. Or is this too naive of an approach?
$endgroup$
– kkc
Jan 9 at 3:27
$begingroup$
I tried that but I think we can't do that because we need |f'(x)| not $f'(x)$.
$endgroup$
– Richard Clare
Jan 9 at 3:47
1
$begingroup$
But f'(x) not integrable => |f'(x)| is not integrable
$endgroup$
– Richard Clare
Jan 9 at 3:48
add a comment |
1
$begingroup$
The issue is the first term. Start by isolating subintervals where $|cos(1/x^2)|$ is not too small (say greater than $1/2$), and then consider what happens when you integrate $1/x$ on these subintervals.
$endgroup$
– Ian
Jan 9 at 2:33
1
$begingroup$
Could you use the fundamental theorem of calculus, instead of taking the derivative itself? In that case, $int_0^1 f'(x) dx = f(1) - f(0)$, so you're clearly getting undefined terms. Or is this too naive of an approach?
$endgroup$
– kkc
Jan 9 at 3:27
$begingroup$
I tried that but I think we can't do that because we need |f'(x)| not $f'(x)$.
$endgroup$
– Richard Clare
Jan 9 at 3:47
1
$begingroup$
But f'(x) not integrable => |f'(x)| is not integrable
$endgroup$
– Richard Clare
Jan 9 at 3:48
1
1
$begingroup$
The issue is the first term. Start by isolating subintervals where $|cos(1/x^2)|$ is not too small (say greater than $1/2$), and then consider what happens when you integrate $1/x$ on these subintervals.
$endgroup$
– Ian
Jan 9 at 2:33
$begingroup$
The issue is the first term. Start by isolating subintervals where $|cos(1/x^2)|$ is not too small (say greater than $1/2$), and then consider what happens when you integrate $1/x$ on these subintervals.
$endgroup$
– Ian
Jan 9 at 2:33
1
1
$begingroup$
Could you use the fundamental theorem of calculus, instead of taking the derivative itself? In that case, $int_0^1 f'(x) dx = f(1) - f(0)$, so you're clearly getting undefined terms. Or is this too naive of an approach?
$endgroup$
– kkc
Jan 9 at 3:27
$begingroup$
Could you use the fundamental theorem of calculus, instead of taking the derivative itself? In that case, $int_0^1 f'(x) dx = f(1) - f(0)$, so you're clearly getting undefined terms. Or is this too naive of an approach?
$endgroup$
– kkc
Jan 9 at 3:27
$begingroup$
I tried that but I think we can't do that because we need |f'(x)| not $f'(x)$.
$endgroup$
– Richard Clare
Jan 9 at 3:47
$begingroup$
I tried that but I think we can't do that because we need |f'(x)| not $f'(x)$.
$endgroup$
– Richard Clare
Jan 9 at 3:47
1
1
$begingroup$
But f'(x) not integrable => |f'(x)| is not integrable
$endgroup$
– Richard Clare
Jan 9 at 3:48
$begingroup$
But f'(x) not integrable => |f'(x)| is not integrable
$endgroup$
– Richard Clare
Jan 9 at 3:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$f[0,1]to mathbb{R}$ is defined by
f(x) =
begin{cases}
x^2sin(frac{1}{x^2}) & text{if $xneq 0$} \
0 & text{if $x=0$} \
end{cases}
WTS: $f'$ is not Riemann Integrable on $[0,1]$. Let us do this by showing that $f'(0)$ exists. We do this by $f'(0)= lim_{xto 0}frac{f(x)-f(0)}{x-0}=lim_{xto0}frac{x^2sin(frac{1}{x^2})}{x}=lim_{xto0} xsin(frac{1}{x^2})$. Now, $-xleq xsin(frac{1}{x^2}) leq x.$ By Squeeze Theorem, $lim_{xto0}(-x)leq lim_{xto0} xsin(frac{1}{x^2})leq lim_{xto0}x$. Therefore, $lim_{xto0}xsin(frac{1}{x^2})=0$. Thus, $f'(0)=0$ and does exist.
Now, $f'(x)=x^{2}cos(frac{1}{x^2}) cdot frac{d}{dx}(x^{-2})+ sin(frac{1}{x^{2}})cdot2x.$ = $x^2cos(frac{1}{x^2})(-2x^{-3})+2xsin(frac{1}{x^2})$ = $frac{-2}{x}cos(frac{1}{x^2})+2xsin(frac{1}{x^2})$. So $f'(x)$ is not bounded on [-1,1].
$|2xsin(frac{1}{x^2})| leq2$ for all $xin[-1,1]$. For ease of notation, Let $a_n = sqrt{frac{2}{(2n-1)pi}}$. Then $cos(frac{1}{a_n^2})=1$ for all n as $a_nto 0$ and$frac{2cos(frac{1}{a_n^2})}{a_n}$= $frac{2}{a_n}$= $sqrt{2(2n-1)pi} to infty$ as $ntoinfty$.
Hence $f'$ is unbounded on $[-1,1]$ and therefore $f'$ is unbounded on $[0,1]subset[-1,1,]$. As a result, we see that $f'$ is not Riemann integrable on $[0,1]$.
Note: I read your question in bold, I used rather elementary methods. Sorry if this isn't what you are looking for.
answered Jan 9 at 7:04
MathRocksMathRocks
362
$endgroup$
add a comment |
$begingroup$
Formally, the integral of $|f'(x)|$ means we separte the original function $f(x)$ into monotone picese and add them up. Note that the monotone picese are separeted into $$
dfrac{1}{x_k^2}:=dfrac{pi}{2}+kpi,quadforall kinmathbb{N}.
$$
This makes the integral $$
int_0^1|f'(x)|mathtt{d} xgeq sum_{k=1}^{infty}x_k^2
=sum_{k=1}^{infty}dfrac{1}{dfrac{pi}{2}+kpi}
$$
which is obviously infinite.
Some constanst in the calculates above may not be rigourous, but the idea is quite clear already. I hope this is right. :)
answered Jan 9 at 5:12
Zixiao_LiuZixiao_Liu
788
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
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votes
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oldest
votes
$begingroup$
$f[0,1]to mathbb{R}$ is defined by
f(x) =
begin{cases}
x^2sin(frac{1}{x^2}) & text{if $xneq 0$} \
0 & text{if $x=0$} \
end{cases}
WTS: $f'$ is not Riemann Integrable on $[0,1]$. Let us do this by showing that $f'(0)$ exists. We do this by $f'(0)= lim_{xto 0}frac{f(x)-f(0)}{x-0}=lim_{xto0}frac{x^2sin(frac{1}{x^2})}{x}=lim_{xto0} xsin(frac{1}{x^2})$. Now, $-xleq xsin(frac{1}{x^2}) leq x.$ By Squeeze Theorem, $lim_{xto0}(-x)leq lim_{xto0} xsin(frac{1}{x^2})leq lim_{xto0}x$. Therefore, $lim_{xto0}xsin(frac{1}{x^2})=0$. Thus, $f'(0)=0$ and does exist.
Now, $f'(x)=x^{2}cos(frac{1}{x^2}) cdot frac{d}{dx}(x^{-2})+ sin(frac{1}{x^{2}})cdot2x.$ = $x^2cos(frac{1}{x^2})(-2x^{-3})+2xsin(frac{1}{x^2})$ = $frac{-2}{x}cos(frac{1}{x^2})+2xsin(frac{1}{x^2})$. So $f'(x)$ is not bounded on [-1,1].
$|2xsin(frac{1}{x^2})| leq2$ for all $xin[-1,1]$. For ease of notation, Let $a_n = sqrt{frac{2}{(2n-1)pi}}$. Then $cos(frac{1}{a_n^2})=1$ for all n as $a_nto 0$ and$frac{2cos(frac{1}{a_n^2})}{a_n}$= $frac{2}{a_n}$= $sqrt{2(2n-1)pi} to infty$ as $ntoinfty$.
Hence $f'$ is unbounded on $[-1,1]$ and therefore $f'$ is unbounded on $[0,1]subset[-1,1,]$. As a result, we see that $f'$ is not Riemann integrable on $[0,1]$.
Note: I read your question in bold, I used rather elementary methods. Sorry if this isn't what you are looking for.
answered Jan 9 at 7:04
MathRocksMathRocks
362
$endgroup$
add a comment |
$begingroup$
$f[0,1]to mathbb{R}$ is defined by
f(x) =
begin{cases}
x^2sin(frac{1}{x^2}) & text{if $xneq 0$} \
0 & text{if $x=0$} \
end{cases}
WTS: $f'$ is not Riemann Integrable on $[0,1]$. Let us do this by showing that $f'(0)$ exists. We do this by $f'(0)= lim_{xto 0}frac{f(x)-f(0)}{x-0}=lim_{xto0}frac{x^2sin(frac{1}{x^2})}{x}=lim_{xto0} xsin(frac{1}{x^2})$. Now, $-xleq xsin(frac{1}{x^2}) leq x.$ By Squeeze Theorem, $lim_{xto0}(-x)leq lim_{xto0} xsin(frac{1}{x^2})leq lim_{xto0}x$. Therefore, $lim_{xto0}xsin(frac{1}{x^2})=0$. Thus, $f'(0)=0$ and does exist.
Now, $f'(x)=x^{2}cos(frac{1}{x^2}) cdot frac{d}{dx}(x^{-2})+ sin(frac{1}{x^{2}})cdot2x.$ = $x^2cos(frac{1}{x^2})(-2x^{-3})+2xsin(frac{1}{x^2})$ = $frac{-2}{x}cos(frac{1}{x^2})+2xsin(frac{1}{x^2})$. So $f'(x)$ is not bounded on [-1,1].
$|2xsin(frac{1}{x^2})| leq2$ for all $xin[-1,1]$. For ease of notation, Let $a_n = sqrt{frac{2}{(2n-1)pi}}$. Then $cos(frac{1}{a_n^2})=1$ for all n as $a_nto 0$ and$frac{2cos(frac{1}{a_n^2})}{a_n}$= $frac{2}{a_n}$= $sqrt{2(2n-1)pi} to infty$ as $ntoinfty$.
Hence $f'$ is unbounded on $[-1,1]$ and therefore $f'$ is unbounded on $[0,1]subset[-1,1,]$. As a result, we see that $f'$ is not Riemann integrable on $[0,1]$.
Note: I read your question in bold, I used rather elementary methods. Sorry if this isn't what you are looking for.
answered Jan 9 at 7:04
MathRocksMathRocks
362
$endgroup$
add a comment |
$begingroup$
$f[0,1]to mathbb{R}$ is defined by
f(x) =
begin{cases}
x^2sin(frac{1}{x^2}) & text{if $xneq 0$} \
0 & text{if $x=0$} \
end{cases}
WTS: $f'$ is not Riemann Integrable on $[0,1]$. Let us do this by showing that $f'(0)$ exists. We do this by $f'(0)= lim_{xto 0}frac{f(x)-f(0)}{x-0}=lim_{xto0}frac{x^2sin(frac{1}{x^2})}{x}=lim_{xto0} xsin(frac{1}{x^2})$. Now, $-xleq xsin(frac{1}{x^2}) leq x.$ By Squeeze Theorem, $lim_{xto0}(-x)leq lim_{xto0} xsin(frac{1}{x^2})leq lim_{xto0}x$. Therefore, $lim_{xto0}xsin(frac{1}{x^2})=0$. Thus, $f'(0)=0$ and does exist.
Now, $f'(x)=x^{2}cos(frac{1}{x^2}) cdot frac{d}{dx}(x^{-2})+ sin(frac{1}{x^{2}})cdot2x.$ = $x^2cos(frac{1}{x^2})(-2x^{-3})+2xsin(frac{1}{x^2})$ = $frac{-2}{x}cos(frac{1}{x^2})+2xsin(frac{1}{x^2})$. So $f'(x)$ is not bounded on [-1,1].
$|2xsin(frac{1}{x^2})| leq2$ for all $xin[-1,1]$. For ease of notation, Let $a_n = sqrt{frac{2}{(2n-1)pi}}$. Then $cos(frac{1}{a_n^2})=1$ for all n as $a_nto 0$ and$frac{2cos(frac{1}{a_n^2})}{a_n}$= $frac{2}{a_n}$= $sqrt{2(2n-1)pi} to infty$ as $ntoinfty$.
Hence $f'$ is unbounded on $[-1,1]$ and therefore $f'$ is unbounded on $[0,1]subset[-1,1,]$. As a result, we see that $f'$ is not Riemann integrable on $[0,1]$.
Note: I read your question in bold, I used rather elementary methods. Sorry if this isn't what you are looking for.
answered Jan 9 at 7:04
MathRocksMathRocks
362
$endgroup$
$f[0,1]to mathbb{R}$ is defined by
f(x) =
begin{cases}
x^2sin(frac{1}{x^2}) & text{if $xneq 0$} \
0 & text{if $x=0$} \
end{cases}
WTS: $f'$ is not Riemann Integrable on $[0,1]$. Let us do this by showing that $f'(0)$ exists. We do this by $f'(0)= lim_{xto 0}frac{f(x)-f(0)}{x-0}=lim_{xto0}frac{x^2sin(frac{1}{x^2})}{x}=lim_{xto0} xsin(frac{1}{x^2})$. Now, $-xleq xsin(frac{1}{x^2}) leq x.$ By Squeeze Theorem, $lim_{xto0}(-x)leq lim_{xto0} xsin(frac{1}{x^2})leq lim_{xto0}x$. Therefore, $lim_{xto0}xsin(frac{1}{x^2})=0$. Thus, $f'(0)=0$ and does exist.
Now, $f'(x)=x^{2}cos(frac{1}{x^2}) cdot frac{d}{dx}(x^{-2})+ sin(frac{1}{x^{2}})cdot2x.$ = $x^2cos(frac{1}{x^2})(-2x^{-3})+2xsin(frac{1}{x^2})$ = $frac{-2}{x}cos(frac{1}{x^2})+2xsin(frac{1}{x^2})$. So $f'(x)$ is not bounded on [-1,1].
$|2xsin(frac{1}{x^2})| leq2$ for all $xin[-1,1]$. For ease of notation, Let $a_n = sqrt{frac{2}{(2n-1)pi}}$. Then $cos(frac{1}{a_n^2})=1$ for all n as $a_nto 0$ and$frac{2cos(frac{1}{a_n^2})}{a_n}$= $frac{2}{a_n}$= $sqrt{2(2n-1)pi} to infty$ as $ntoinfty$.
Hence $f'$ is unbounded on $[-1,1]$ and therefore $f'$ is unbounded on $[0,1]subset[-1,1,]$. As a result, we see that $f'$ is not Riemann integrable on $[0,1]$.
Note: I read your question in bold, I used rather elementary methods. Sorry if this isn't what you are looking for.
answered Jan 9 at 7:04
MathRocksMathRocks
362
answered Jan 9 at 7:04
MathRocksMathRocks
362
answered Jan 9 at 7:04
MathRocksMathRocks
362
answered Jan 9 at 7:04
MathRocksMathRocks
362
362
add a comment |
add a comment |
$begingroup$
Formally, the integral of $|f'(x)|$ means we separte the original function $f(x)$ into monotone picese and add them up. Note that the monotone picese are separeted into $$
dfrac{1}{x_k^2}:=dfrac{pi}{2}+kpi,quadforall kinmathbb{N}.
$$
This makes the integral $$
int_0^1|f'(x)|mathtt{d} xgeq sum_{k=1}^{infty}x_k^2
=sum_{k=1}^{infty}dfrac{1}{dfrac{pi}{2}+kpi}
$$
which is obviously infinite.
Some constanst in the calculates above may not be rigourous, but the idea is quite clear already. I hope this is right. :)
answered Jan 9 at 5:12
Zixiao_LiuZixiao_Liu
788
$endgroup$
add a comment |
$begingroup$
Formally, the integral of $|f'(x)|$ means we separte the original function $f(x)$ into monotone picese and add them up. Note that the monotone picese are separeted into $$
dfrac{1}{x_k^2}:=dfrac{pi}{2}+kpi,quadforall kinmathbb{N}.
$$
This makes the integral $$
int_0^1|f'(x)|mathtt{d} xgeq sum_{k=1}^{infty}x_k^2
=sum_{k=1}^{infty}dfrac{1}{dfrac{pi}{2}+kpi}
$$
which is obviously infinite.
Some constanst in the calculates above may not be rigourous, but the idea is quite clear already. I hope this is right. :)
answered Jan 9 at 5:12
Zixiao_LiuZixiao_Liu
788
$endgroup$
add a comment |
$begingroup$
Formally, the integral of $|f'(x)|$ means we separte the original function $f(x)$ into monotone picese and add them up. Note that the monotone picese are separeted into $$
dfrac{1}{x_k^2}:=dfrac{pi}{2}+kpi,quadforall kinmathbb{N}.
$$
This makes the integral $$
int_0^1|f'(x)|mathtt{d} xgeq sum_{k=1}^{infty}x_k^2
=sum_{k=1}^{infty}dfrac{1}{dfrac{pi}{2}+kpi}
$$
which is obviously infinite.
Some constanst in the calculates above may not be rigourous, but the idea is quite clear already. I hope this is right. :)
answered Jan 9 at 5:12
Zixiao_LiuZixiao_Liu
788
$endgroup$
Formally, the integral of $|f'(x)|$ means we separte the original function $f(x)$ into monotone picese and add them up. Note that the monotone picese are separeted into $$
dfrac{1}{x_k^2}:=dfrac{pi}{2}+kpi,quadforall kinmathbb{N}.
$$
This makes the integral $$
int_0^1|f'(x)|mathtt{d} xgeq sum_{k=1}^{infty}x_k^2
=sum_{k=1}^{infty}dfrac{1}{dfrac{pi}{2}+kpi}
$$
which is obviously infinite.
Some constanst in the calculates above may not be rigourous, but the idea is quite clear already. I hope this is right. :)
answered Jan 9 at 5:12
Zixiao_LiuZixiao_Liu
788
answered Jan 9 at 5:12
Zixiao_LiuZixiao_Liu
788
answered Jan 9 at 5:12
Zixiao_LiuZixiao_Liu
788
answered Jan 9 at 5:12
Zixiao_LiuZixiao_Liu
788
788
add a comment |
add a comment |
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1
$begingroup$
The issue is the first term. Start by isolating subintervals where $|cos(1/x^2)|$ is not too small (say greater than $1/2$), and then consider what happens when you integrate $1/x$ on these subintervals.
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– Ian
Jan 9 at 2:33
1
$begingroup$
Could you use the fundamental theorem of calculus, instead of taking the derivative itself? In that case, $int_0^1 f'(x) dx = f(1) - f(0)$, so you're clearly getting undefined terms. Or is this too naive of an approach?
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– kkc
Jan 9 at 3:27
$begingroup$
I tried that but I think we can't do that because we need |f'(x)| not $f'(x)$.
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– Richard Clare
Jan 9 at 3:47
1
$begingroup$
But f'(x) not integrable => |f'(x)| is not integrable
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– Richard Clare
Jan 9 at 3:48