Proof existence of minimum and maximum on discontinuous function
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I have the function $g(x) = begin{cases} e^{-x}cdotsin(frac{1}{x}), &x>0 ;\ 0, & x=0 end{cases}$ and I have to determine if the function has a min/max in the interval $[0,1]$. I can't use the EVT because the function isn't continuous. How would you do this?
functional-analysis
asked Dec 11 '18 at 21:06
fujafuja
11
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add a comment |
$begingroup$
I have the function $g(x) = begin{cases} e^{-x}cdotsin(frac{1}{x}), &x>0 ;\ 0, & x=0 end{cases}$ and I have to determine if the function has a min/max in the interval $[0,1]$. I can't use the EVT because the function isn't continuous. How would you do this?
functional-analysis
asked Dec 11 '18 at 21:06
fujafuja
11
$endgroup$
1
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Something is not right. Is your function only defined on rationals?
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– Anurag A
Dec 11 '18 at 21:08
1
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You don't have a minimum or a maximum, but you have an infimum and supremum. You can get infinitely close to $pm 1$, without reaching those values
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– Andrei
Dec 11 '18 at 21:27
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@AnuragA sry, typo. I corrected it!
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– fuja
Dec 12 '18 at 5:44
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@Andrei I don't think so... The function has a minimum and maximum in my oppinion. And otherwise: how would you proof this?
$endgroup$
– fuja
Dec 12 '18 at 8:11
add a comment |
$begingroup$
I have the function $g(x) = begin{cases} e^{-x}cdotsin(frac{1}{x}), &x>0 ;\ 0, & x=0 end{cases}$ and I have to determine if the function has a min/max in the interval $[0,1]$. I can't use the EVT because the function isn't continuous. How would you do this?
functional-analysis
asked Dec 11 '18 at 21:06
fujafuja
11
$endgroup$
I have the function $g(x) = begin{cases} e^{-x}cdotsin(frac{1}{x}), &x>0 ;\ 0, & x=0 end{cases}$ and I have to determine if the function has a min/max in the interval $[0,1]$. I can't use the EVT because the function isn't continuous. How would you do this?
functional-analysis
functional-analysis
asked Dec 11 '18 at 21:06
fujafuja
11
asked Dec 11 '18 at 21:06
fujafuja
11
edited Dec 12 '18 at 5:44
fuja
asked Dec 11 '18 at 21:06
fujafuja
11
asked Dec 11 '18 at 21:06
fujafuja
11
asked Dec 11 '18 at 21:06
fujafuja
11
11
1
$begingroup$
Something is not right. Is your function only defined on rationals?
$endgroup$
– Anurag A
Dec 11 '18 at 21:08
1
$begingroup$
You don't have a minimum or a maximum, but you have an infimum and supremum. You can get infinitely close to $pm 1$, without reaching those values
$endgroup$
– Andrei
Dec 11 '18 at 21:27
$begingroup$
@AnuragA sry, typo. I corrected it!
$endgroup$
– fuja
Dec 12 '18 at 5:44
$begingroup$
@Andrei I don't think so... The function has a minimum and maximum in my oppinion. And otherwise: how would you proof this?
$endgroup$
– fuja
Dec 12 '18 at 8:11
add a comment |
1
$begingroup$
Something is not right. Is your function only defined on rationals?
$endgroup$
– Anurag A
Dec 11 '18 at 21:08
1
$begingroup$
You don't have a minimum or a maximum, but you have an infimum and supremum. You can get infinitely close to $pm 1$, without reaching those values
$endgroup$
– Andrei
Dec 11 '18 at 21:27
$begingroup$
@AnuragA sry, typo. I corrected it!
$endgroup$
– fuja
Dec 12 '18 at 5:44
$begingroup$
@Andrei I don't think so... The function has a minimum and maximum in my oppinion. And otherwise: how would you proof this?
$endgroup$
– fuja
Dec 12 '18 at 8:11
1
1
$begingroup$
Something is not right. Is your function only defined on rationals?
$endgroup$
– Anurag A
Dec 11 '18 at 21:08
$begingroup$
Something is not right. Is your function only defined on rationals?
$endgroup$
– Anurag A
Dec 11 '18 at 21:08
1
1
$begingroup$
You don't have a minimum or a maximum, but you have an infimum and supremum. You can get infinitely close to $pm 1$, without reaching those values
$endgroup$
– Andrei
Dec 11 '18 at 21:27
$begingroup$
You don't have a minimum or a maximum, but you have an infimum and supremum. You can get infinitely close to $pm 1$, without reaching those values
$endgroup$
– Andrei
Dec 11 '18 at 21:27
$begingroup$
@AnuragA sry, typo. I corrected it!
$endgroup$
– fuja
Dec 12 '18 at 5:44
$begingroup$
@AnuragA sry, typo. I corrected it!
$endgroup$
– fuja
Dec 12 '18 at 5:44
$begingroup$
@Andrei I don't think so... The function has a minimum and maximum in my oppinion. And otherwise: how would you proof this?
$endgroup$
– fuja
Dec 12 '18 at 8:11
$begingroup$
@Andrei I don't think so... The function has a minimum and maximum in my oppinion. And otherwise: how would you proof this?
$endgroup$
– fuja
Dec 12 '18 at 8:11
add a comment |
1 Answer
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The function has values between $pm 1$. So the maximum, if exists, it has to be at most $+1$. But the sine function varies between $-1$ and $1$, and the exponential is strictly less than $1$, so the maximum must be smaller than $1$. If you choose any $epsilon>0$, and want to show that the maximum is $1-epsilon$, you get a contradiction (I can get a value greater than the maximum, no matter what positive $epsilon$ you choose). To show this, the exponential term is greater than $1-epsilon$ if $x<ln frac1{1-epsilon}$. There are an infinite number of $x$ values between $0$ and $ln frac1{1-epsilon}$ such that $1/x$ is of the form $2kpi+pi/2$ with $kinmathbb Z$. At those values the exponential part is greater that $1-epsilon$ and the sine part is equal to $1$. Therefore the function is greater than $1-epsilon$. So $1-epsilon$ is not a maximum.
answered Dec 12 '18 at 16:03
AndreiAndrei
13.2k21230
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$begingroup$
The function has values between $pm 1$. So the maximum, if exists, it has to be at most $+1$. But the sine function varies between $-1$ and $1$, and the exponential is strictly less than $1$, so the maximum must be smaller than $1$. If you choose any $epsilon>0$, and want to show that the maximum is $1-epsilon$, you get a contradiction (I can get a value greater than the maximum, no matter what positive $epsilon$ you choose). To show this, the exponential term is greater than $1-epsilon$ if $x<ln frac1{1-epsilon}$. There are an infinite number of $x$ values between $0$ and $ln frac1{1-epsilon}$ such that $1/x$ is of the form $2kpi+pi/2$ with $kinmathbb Z$. At those values the exponential part is greater that $1-epsilon$ and the sine part is equal to $1$. Therefore the function is greater than $1-epsilon$. So $1-epsilon$ is not a maximum.
answered Dec 12 '18 at 16:03
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
$begingroup$
The function has values between $pm 1$. So the maximum, if exists, it has to be at most $+1$. But the sine function varies between $-1$ and $1$, and the exponential is strictly less than $1$, so the maximum must be smaller than $1$. If you choose any $epsilon>0$, and want to show that the maximum is $1-epsilon$, you get a contradiction (I can get a value greater than the maximum, no matter what positive $epsilon$ you choose). To show this, the exponential term is greater than $1-epsilon$ if $x<ln frac1{1-epsilon}$. There are an infinite number of $x$ values between $0$ and $ln frac1{1-epsilon}$ such that $1/x$ is of the form $2kpi+pi/2$ with $kinmathbb Z$. At those values the exponential part is greater that $1-epsilon$ and the sine part is equal to $1$. Therefore the function is greater than $1-epsilon$. So $1-epsilon$ is not a maximum.
answered Dec 12 '18 at 16:03
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
$begingroup$
The function has values between $pm 1$. So the maximum, if exists, it has to be at most $+1$. But the sine function varies between $-1$ and $1$, and the exponential is strictly less than $1$, so the maximum must be smaller than $1$. If you choose any $epsilon>0$, and want to show that the maximum is $1-epsilon$, you get a contradiction (I can get a value greater than the maximum, no matter what positive $epsilon$ you choose). To show this, the exponential term is greater than $1-epsilon$ if $x<ln frac1{1-epsilon}$. There are an infinite number of $x$ values between $0$ and $ln frac1{1-epsilon}$ such that $1/x$ is of the form $2kpi+pi/2$ with $kinmathbb Z$. At those values the exponential part is greater that $1-epsilon$ and the sine part is equal to $1$. Therefore the function is greater than $1-epsilon$. So $1-epsilon$ is not a maximum.
answered Dec 12 '18 at 16:03
AndreiAndrei
13.2k21230
$endgroup$
The function has values between $pm 1$. So the maximum, if exists, it has to be at most $+1$. But the sine function varies between $-1$ and $1$, and the exponential is strictly less than $1$, so the maximum must be smaller than $1$. If you choose any $epsilon>0$, and want to show that the maximum is $1-epsilon$, you get a contradiction (I can get a value greater than the maximum, no matter what positive $epsilon$ you choose). To show this, the exponential term is greater than $1-epsilon$ if $x<ln frac1{1-epsilon}$. There are an infinite number of $x$ values between $0$ and $ln frac1{1-epsilon}$ such that $1/x$ is of the form $2kpi+pi/2$ with $kinmathbb Z$. At those values the exponential part is greater that $1-epsilon$ and the sine part is equal to $1$. Therefore the function is greater than $1-epsilon$. So $1-epsilon$ is not a maximum.
answered Dec 12 '18 at 16:03
AndreiAndrei
13.2k21230
answered Dec 12 '18 at 16:03
AndreiAndrei
13.2k21230
answered Dec 12 '18 at 16:03
AndreiAndrei
13.2k21230
answered Dec 12 '18 at 16:03
AndreiAndrei
13.2k21230
13.2k21230
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$begingroup$
Something is not right. Is your function only defined on rationals?
$endgroup$
– Anurag A
Dec 11 '18 at 21:08
1
$begingroup$
You don't have a minimum or a maximum, but you have an infimum and supremum. You can get infinitely close to $pm 1$, without reaching those values
$endgroup$
– Andrei
Dec 11 '18 at 21:27
$begingroup$
@AnuragA sry, typo. I corrected it!
$endgroup$
– fuja
Dec 12 '18 at 5:44
$begingroup$
@Andrei I don't think so... The function has a minimum and maximum in my oppinion. And otherwise: how would you proof this?
$endgroup$
– fuja
Dec 12 '18 at 8:11