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Analytical and/or Geometrical example of a certain function
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I need an example of a function that is continuous on some interval $I$, has a local minimum but doesn't have a global minimum. In other words, the problem of finding the global minimum of that function on an interval $I$ yields no solutions.
Does the function $Sinx^2, I=[-1;2)$ meet that condition?
optimization
asked Jan 26 at 19:02
Nick202Nick202
605
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add a comment |
$begingroup$
I need an example of a function that is continuous on some interval $I$, has a local minimum but doesn't have a global minimum. In other words, the problem of finding the global minimum of that function on an interval $I$ yields no solutions.
Does the function $Sinx^2, I=[-1;2)$ meet that condition?
optimization
asked Jan 26 at 19:02
Nick202Nick202
605
$endgroup$
add a comment |
$begingroup$
I need an example of a function that is continuous on some interval $I$, has a local minimum but doesn't have a global minimum. In other words, the problem of finding the global minimum of that function on an interval $I$ yields no solutions.
Does the function $Sinx^2, I=[-1;2)$ meet that condition?
optimization
asked Jan 26 at 19:02
Nick202Nick202
605
$endgroup$
I need an example of a function that is continuous on some interval $I$, has a local minimum but doesn't have a global minimum. In other words, the problem of finding the global minimum of that function on an interval $I$ yields no solutions.
Does the function $Sinx^2, I=[-1;2)$ meet that condition?
optimization
optimization
asked Jan 26 at 19:02
Nick202Nick202
605
asked Jan 26 at 19:02
Nick202Nick202
605
asked Jan 26 at 19:02
Nick202Nick202
605
asked Jan 26 at 19:02
Nick202Nick202
605
asked Jan 26 at 19:02
Nick202Nick202
605
605
add a comment |
add a comment |
1 Answer
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Yes, $sin(x^2)$ on $[-1,2)$ would work. $x=0$ is a local minimum, at which the value is $0$, but there is no global minimum as it never attains the infimum value of $sin(4)$.
answered Jan 26 at 19:28
Erik ParkinsonErik Parkinson
1,17519
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Yes, $sin(x^2)$ on $[-1,2)$ would work. $x=0$ is a local minimum, at which the value is $0$, but there is no global minimum as it never attains the infimum value of $sin(4)$.
answered Jan 26 at 19:28
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
$begingroup$
Yes, $sin(x^2)$ on $[-1,2)$ would work. $x=0$ is a local minimum, at which the value is $0$, but there is no global minimum as it never attains the infimum value of $sin(4)$.
answered Jan 26 at 19:28
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
$begingroup$
Yes, $sin(x^2)$ on $[-1,2)$ would work. $x=0$ is a local minimum, at which the value is $0$, but there is no global minimum as it never attains the infimum value of $sin(4)$.
answered Jan 26 at 19:28
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
Yes, $sin(x^2)$ on $[-1,2)$ would work. $x=0$ is a local minimum, at which the value is $0$, but there is no global minimum as it never attains the infimum value of $sin(4)$.
answered Jan 26 at 19:28
Erik ParkinsonErik Parkinson
1,17519
answered Jan 26 at 19:28
Erik ParkinsonErik Parkinson
1,17519
answered Jan 26 at 19:28
Erik ParkinsonErik Parkinson
1,17519
answered Jan 26 at 19:28
Erik ParkinsonErik Parkinson
1,17519
1,17519
add a comment |
add a comment |
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