Mixing
Show that the function $H(x, y) = x^2 + y^2 + |x − y|^{-1}$ achieves its global minimum somewhere on the...
$begingroup$
Show that the function $H(x, y) = x^2 + y^2 + |x − y|^{-1}$ achieves its global minimum somewhere on the set ${(x, y) in mathbb{R}^2 : x ne y}$.
I kind of understand the minimum cannot be $x=y$ since $1/(x-y)=1/0$, which is infinity. I am trying to prove it via contradiction, but I am confused about how to find the minimum of a function with two variables.
Any advice or hints will be appreciated. Thank you!
real-analysis optimization
edited Jan 25 at 12:35
daw
24.7k1645
asked Jan 25 at 6:09
Eric JEric J
163
$endgroup$
add a comment |
$begingroup$
Show that the function $H(x, y) = x^2 + y^2 + |x − y|^{-1}$ achieves its global minimum somewhere on the set ${(x, y) in mathbb{R}^2 : x ne y}$.
I kind of understand the minimum cannot be $x=y$ since $1/(x-y)=1/0$, which is infinity. I am trying to prove it via contradiction, but I am confused about how to find the minimum of a function with two variables.
Any advice or hints will be appreciated. Thank you!
real-analysis optimization
edited Jan 25 at 12:35
daw
24.7k1645
asked Jan 25 at 6:09
Eric JEric J
163
$endgroup$
$begingroup$
This function does not seem well defined. Note that $2! = 2$, $1!=1$.
$endgroup$
– Lucas Corrêa
Jan 25 at 6:15
1
$begingroup$
sorry I meant x is not equal to y
$endgroup$
– Eric J
Jan 25 at 6:25
add a comment |
$begingroup$
Show that the function $H(x, y) = x^2 + y^2 + |x − y|^{-1}$ achieves its global minimum somewhere on the set ${(x, y) in mathbb{R}^2 : x ne y}$.
I kind of understand the minimum cannot be $x=y$ since $1/(x-y)=1/0$, which is infinity. I am trying to prove it via contradiction, but I am confused about how to find the minimum of a function with two variables.
Any advice or hints will be appreciated. Thank you!
real-analysis optimization
edited Jan 25 at 12:35
daw
24.7k1645
asked Jan 25 at 6:09
Eric JEric J
163
$endgroup$
Show that the function $H(x, y) = x^2 + y^2 + |x − y|^{-1}$ achieves its global minimum somewhere on the set ${(x, y) in mathbb{R}^2 : x ne y}$.
I kind of understand the minimum cannot be $x=y$ since $1/(x-y)=1/0$, which is infinity. I am trying to prove it via contradiction, but I am confused about how to find the minimum of a function with two variables.
Any advice or hints will be appreciated. Thank you!
real-analysis optimization
real-analysis optimization
edited Jan 25 at 12:35
daw
24.7k1645
asked Jan 25 at 6:09
Eric JEric J
163
edited Jan 25 at 12:35
daw
24.7k1645
asked Jan 25 at 6:09
Eric JEric J
163
edited Jan 25 at 12:35
daw
24.7k1645
edited Jan 25 at 12:35
daw
24.7k1645
edited Jan 25 at 12:35
daw
24.7k1645
24.7k1645
asked Jan 25 at 6:09
Eric JEric J
163
asked Jan 25 at 6:09
Eric JEric J
163
asked Jan 25 at 6:09
Eric JEric J
163
163
$begingroup$
This function does not seem well defined. Note that $2! = 2$, $1!=1$.
$endgroup$
– Lucas Corrêa
Jan 25 at 6:15
1
$begingroup$
sorry I meant x is not equal to y
$endgroup$
– Eric J
Jan 25 at 6:25
add a comment |
$begingroup$
This function does not seem well defined. Note that $2! = 2$, $1!=1$.
$endgroup$
– Lucas Corrêa
Jan 25 at 6:15
1
$begingroup$
sorry I meant x is not equal to y
$endgroup$
– Eric J
Jan 25 at 6:25
$begingroup$
This function does not seem well defined. Note that $2! = 2$, $1!=1$.
$endgroup$
– Lucas Corrêa
Jan 25 at 6:15
$begingroup$
This function does not seem well defined. Note that $2! = 2$, $1!=1$.
$endgroup$
– Lucas Corrêa
Jan 25 at 6:15
1
1
$begingroup$
sorry I meant x is not equal to y
$endgroup$
– Eric J
Jan 25 at 6:25
$begingroup$
sorry I meant x is not equal to y
$endgroup$
– Eric J
Jan 25 at 6:25
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You don't actually need to find the minimum to prove that there is in fact a global minimum. You can set the stage to use the Weierstrass theorem (any continuous function reaches a maximum and a minimum value over a compact—i.e., closed and bounded—set).
Here, a sketch of the proof, based on the idea that the minimum won't be very close to the line $x=y$ (where the values go to $+infty$) nor very far from the origin (for the same reason).
So let's take any point in the domain, say $(1,0)$. Since $f(1,0)=2$, we know that if there is a minimum is not greater than $2$.
Now, let's define a compact set such that on it's complement the function always takes values greater than $2$.
Define
$$A=left{(x,y)colon |x-y|gefrac12right}$$
and
$$B=left{(x,y)colon x^2+y^2le2right}.$$
Is easy to see that the set $Acap B$ is closed and bounded, that is, is compact, and since $f$ is continuous, reaches a minimum in this set. Also, this minimum value is no more than $2$.
And if $(x,y)$ is not in $A$ (that is $|x-y|<frac12$) or if is not in $B$ (and so $x^2+y^2>2$), then is easy to prove that $f(x,y)>2$, and so the minimum reached in $Acap B$ is a global minimum.
answered Jan 25 at 6:50
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
What this question is really asking is if the function has a global minimum at all, as it isn't even defined on the set $x=y$ as you mentioned.
The set ${(x,y)in R^2:xne y}$ is open, and therefore ay local (and thus the global) minimum must have the partial derivatives equal $0$. The function is symmetric, so we assume WLOG $x>y$ and the function is now $H(x,y)=x^2+y^2+frac{1}{x-y}$. Thus $H_x(x,y)=2x-frac{1}{(x-y)^2}$, $H_y(x,y)=2y+frac{1}{(x-y)^2}$. These must both be $0$, so the sum of them must be $0$, so $2x+2y=0$ so $x=-y$.
Plugging this back into $H_x$ and setting equal to $0$ gives $2x=frac{1}{(2x)^2}$ so $8x^3=1$, so $x=frac12$, and thus $y=-frac12$. To see if this is actually a minimum, we find the Hessian at this point to be
$$
begin{bmatrix}
2.5 & -.5 \
-.5 & 2.5 \
end{bmatrix}
$$
which has eigenvalues $2$ and $3$ so is positive definite, thus this is a local minimum. As they are the only local minima, the global minimum is achieved at $H(frac12,-frac12)=frac32$, and by symmetry at $H(-frac12,frac12)=frac32$ as well.
answered Jan 25 at 6:31
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$begingroup$
You have found a critical point. You have not even checked that it is a local minimum. Do you have any reason to believe that it is a global minimum?
$endgroup$
– Misha Lavrov
Jan 25 at 6:36
$begingroup$
The function tends to infinity if any of $x$ or $y$ does and its domain can be written as an increasing union of compact sets.
$endgroup$
– Will M.
Jan 25 at 6:42
$begingroup$
@WillM. Exactly, and this (not the computation of critical points) is the actual argument needed.
$endgroup$
– Misha Lavrov
Jan 25 at 6:45
$begingroup$
@MishaLavrov Good catch, I totally forgot! I'll add that real quick.
$endgroup$
– Erik Parkinson
Jan 25 at 6:55
$begingroup$
@ErikParkinson It is still false that one of the local minima of a function must be a global minimum.
$endgroup$
– Misha Lavrov
Jan 25 at 15:02
add a comment |
$begingroup$
Consider the set
$$
S = left{(x,y) in mathbb R^2 : |x| le 10, |y| le 10, |x-y| ge frac{1}{100}right}.
$$
This set $S$ is closed and bounded, and therefore (by the extreme value theorem) $H$ has a global minimum on $S$.
Points outside $S$ are guaranteed to be worse than that minimum. If $(x,y) notin S$, then one of three things must be true:
$|x| > 10$, in which case $H(x,y) ge x^2 > 100$.
$|y| > 10$, in which case $H(x,y) ge y^2 > 100$.
$|x-y| < frac{1}{100}$, in which case $H(x,y) ge frac{1}{|x-y|} > 100$.
Therefore $H(x,y) > 100$ everywhere outside $S$, which can't beat for example $H(0,1) = 2$.
This means that the global minimum of $H$ on $S$ is actually the global minimum of $H$ on its entire domain ${(x,y) in mathbb R^2 : x ne y}$.
answered Jan 25 at 6:44
Misha LavrovMisha Lavrov
47.7k657107
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You don't actually need to find the minimum to prove that there is in fact a global minimum. You can set the stage to use the Weierstrass theorem (any continuous function reaches a maximum and a minimum value over a compact—i.e., closed and bounded—set).
Here, a sketch of the proof, based on the idea that the minimum won't be very close to the line $x=y$ (where the values go to $+infty$) nor very far from the origin (for the same reason).
So let's take any point in the domain, say $(1,0)$. Since $f(1,0)=2$, we know that if there is a minimum is not greater than $2$.
Now, let's define a compact set such that on it's complement the function always takes values greater than $2$.
Define
$$A=left{(x,y)colon |x-y|gefrac12right}$$
and
$$B=left{(x,y)colon x^2+y^2le2right}.$$
Is easy to see that the set $Acap B$ is closed and bounded, that is, is compact, and since $f$ is continuous, reaches a minimum in this set. Also, this minimum value is no more than $2$.
And if $(x,y)$ is not in $A$ (that is $|x-y|<frac12$) or if is not in $B$ (and so $x^2+y^2>2$), then is easy to prove that $f(x,y)>2$, and so the minimum reached in $Acap B$ is a global minimum.
answered Jan 25 at 6:50
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
You don't actually need to find the minimum to prove that there is in fact a global minimum. You can set the stage to use the Weierstrass theorem (any continuous function reaches a maximum and a minimum value over a compact—i.e., closed and bounded—set).
Here, a sketch of the proof, based on the idea that the minimum won't be very close to the line $x=y$ (where the values go to $+infty$) nor very far from the origin (for the same reason).
So let's take any point in the domain, say $(1,0)$. Since $f(1,0)=2$, we know that if there is a minimum is not greater than $2$.
Now, let's define a compact set such that on it's complement the function always takes values greater than $2$.
Define
$$A=left{(x,y)colon |x-y|gefrac12right}$$
and
$$B=left{(x,y)colon x^2+y^2le2right}.$$
Is easy to see that the set $Acap B$ is closed and bounded, that is, is compact, and since $f$ is continuous, reaches a minimum in this set. Also, this minimum value is no more than $2$.
And if $(x,y)$ is not in $A$ (that is $|x-y|<frac12$) or if is not in $B$ (and so $x^2+y^2>2$), then is easy to prove that $f(x,y)>2$, and so the minimum reached in $Acap B$ is a global minimum.
answered Jan 25 at 6:50
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
You don't actually need to find the minimum to prove that there is in fact a global minimum. You can set the stage to use the Weierstrass theorem (any continuous function reaches a maximum and a minimum value over a compact—i.e., closed and bounded—set).
Here, a sketch of the proof, based on the idea that the minimum won't be very close to the line $x=y$ (where the values go to $+infty$) nor very far from the origin (for the same reason).
So let's take any point in the domain, say $(1,0)$. Since $f(1,0)=2$, we know that if there is a minimum is not greater than $2$.
Now, let's define a compact set such that on it's complement the function always takes values greater than $2$.
Define
$$A=left{(x,y)colon |x-y|gefrac12right}$$
and
$$B=left{(x,y)colon x^2+y^2le2right}.$$
Is easy to see that the set $Acap B$ is closed and bounded, that is, is compact, and since $f$ is continuous, reaches a minimum in this set. Also, this minimum value is no more than $2$.
And if $(x,y)$ is not in $A$ (that is $|x-y|<frac12$) or if is not in $B$ (and so $x^2+y^2>2$), then is easy to prove that $f(x,y)>2$, and so the minimum reached in $Acap B$ is a global minimum.
answered Jan 25 at 6:50
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
You don't actually need to find the minimum to prove that there is in fact a global minimum. You can set the stage to use the Weierstrass theorem (any continuous function reaches a maximum and a minimum value over a compact—i.e., closed and bounded—set).
Here, a sketch of the proof, based on the idea that the minimum won't be very close to the line $x=y$ (where the values go to $+infty$) nor very far from the origin (for the same reason).
So let's take any point in the domain, say $(1,0)$. Since $f(1,0)=2$, we know that if there is a minimum is not greater than $2$.
Now, let's define a compact set such that on it's complement the function always takes values greater than $2$.
Define
$$A=left{(x,y)colon |x-y|gefrac12right}$$
and
$$B=left{(x,y)colon x^2+y^2le2right}.$$
Is easy to see that the set $Acap B$ is closed and bounded, that is, is compact, and since $f$ is continuous, reaches a minimum in this set. Also, this minimum value is no more than $2$.
And if $(x,y)$ is not in $A$ (that is $|x-y|<frac12$) or if is not in $B$ (and so $x^2+y^2>2$), then is easy to prove that $f(x,y)>2$, and so the minimum reached in $Acap B$ is a global minimum.
answered Jan 25 at 6:50
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 25 at 6:50
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 25 at 6:50
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 25 at 6:50
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
add a comment |
add a comment |
$begingroup$
What this question is really asking is if the function has a global minimum at all, as it isn't even defined on the set $x=y$ as you mentioned.
The set ${(x,y)in R^2:xne y}$ is open, and therefore ay local (and thus the global) minimum must have the partial derivatives equal $0$. The function is symmetric, so we assume WLOG $x>y$ and the function is now $H(x,y)=x^2+y^2+frac{1}{x-y}$. Thus $H_x(x,y)=2x-frac{1}{(x-y)^2}$, $H_y(x,y)=2y+frac{1}{(x-y)^2}$. These must both be $0$, so the sum of them must be $0$, so $2x+2y=0$ so $x=-y$.
Plugging this back into $H_x$ and setting equal to $0$ gives $2x=frac{1}{(2x)^2}$ so $8x^3=1$, so $x=frac12$, and thus $y=-frac12$. To see if this is actually a minimum, we find the Hessian at this point to be
$$
begin{bmatrix}
2.5 & -.5 \
-.5 & 2.5 \
end{bmatrix}
$$
which has eigenvalues $2$ and $3$ so is positive definite, thus this is a local minimum. As they are the only local minima, the global minimum is achieved at $H(frac12,-frac12)=frac32$, and by symmetry at $H(-frac12,frac12)=frac32$ as well.
answered Jan 25 at 6:31
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$begingroup$
You have found a critical point. You have not even checked that it is a local minimum. Do you have any reason to believe that it is a global minimum?
$endgroup$
– Misha Lavrov
Jan 25 at 6:36
$begingroup$
The function tends to infinity if any of $x$ or $y$ does and its domain can be written as an increasing union of compact sets.
$endgroup$
– Will M.
Jan 25 at 6:42
$begingroup$
@WillM. Exactly, and this (not the computation of critical points) is the actual argument needed.
$endgroup$
– Misha Lavrov
Jan 25 at 6:45
$begingroup$
@MishaLavrov Good catch, I totally forgot! I'll add that real quick.
$endgroup$
– Erik Parkinson
Jan 25 at 6:55
$begingroup$
@ErikParkinson It is still false that one of the local minima of a function must be a global minimum.
$endgroup$
– Misha Lavrov
Jan 25 at 15:02
add a comment |
$begingroup$
What this question is really asking is if the function has a global minimum at all, as it isn't even defined on the set $x=y$ as you mentioned.
The set ${(x,y)in R^2:xne y}$ is open, and therefore ay local (and thus the global) minimum must have the partial derivatives equal $0$. The function is symmetric, so we assume WLOG $x>y$ and the function is now $H(x,y)=x^2+y^2+frac{1}{x-y}$. Thus $H_x(x,y)=2x-frac{1}{(x-y)^2}$, $H_y(x,y)=2y+frac{1}{(x-y)^2}$. These must both be $0$, so the sum of them must be $0$, so $2x+2y=0$ so $x=-y$.
Plugging this back into $H_x$ and setting equal to $0$ gives $2x=frac{1}{(2x)^2}$ so $8x^3=1$, so $x=frac12$, and thus $y=-frac12$. To see if this is actually a minimum, we find the Hessian at this point to be
$$
begin{bmatrix}
2.5 & -.5 \
-.5 & 2.5 \
end{bmatrix}
$$
which has eigenvalues $2$ and $3$ so is positive definite, thus this is a local minimum. As they are the only local minima, the global minimum is achieved at $H(frac12,-frac12)=frac32$, and by symmetry at $H(-frac12,frac12)=frac32$ as well.
answered Jan 25 at 6:31
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$begingroup$
You have found a critical point. You have not even checked that it is a local minimum. Do you have any reason to believe that it is a global minimum?
$endgroup$
– Misha Lavrov
Jan 25 at 6:36
$begingroup$
The function tends to infinity if any of $x$ or $y$ does and its domain can be written as an increasing union of compact sets.
$endgroup$
– Will M.
Jan 25 at 6:42
$begingroup$
@WillM. Exactly, and this (not the computation of critical points) is the actual argument needed.
$endgroup$
– Misha Lavrov
Jan 25 at 6:45
$begingroup$
@MishaLavrov Good catch, I totally forgot! I'll add that real quick.
$endgroup$
– Erik Parkinson
Jan 25 at 6:55
$begingroup$
@ErikParkinson It is still false that one of the local minima of a function must be a global minimum.
$endgroup$
– Misha Lavrov
Jan 25 at 15:02
add a comment |
$begingroup$
What this question is really asking is if the function has a global minimum at all, as it isn't even defined on the set $x=y$ as you mentioned.
The set ${(x,y)in R^2:xne y}$ is open, and therefore ay local (and thus the global) minimum must have the partial derivatives equal $0$. The function is symmetric, so we assume WLOG $x>y$ and the function is now $H(x,y)=x^2+y^2+frac{1}{x-y}$. Thus $H_x(x,y)=2x-frac{1}{(x-y)^2}$, $H_y(x,y)=2y+frac{1}{(x-y)^2}$. These must both be $0$, so the sum of them must be $0$, so $2x+2y=0$ so $x=-y$.
Plugging this back into $H_x$ and setting equal to $0$ gives $2x=frac{1}{(2x)^2}$ so $8x^3=1$, so $x=frac12$, and thus $y=-frac12$. To see if this is actually a minimum, we find the Hessian at this point to be
$$
begin{bmatrix}
2.5 & -.5 \
-.5 & 2.5 \
end{bmatrix}
$$
which has eigenvalues $2$ and $3$ so is positive definite, thus this is a local minimum. As they are the only local minima, the global minimum is achieved at $H(frac12,-frac12)=frac32$, and by symmetry at $H(-frac12,frac12)=frac32$ as well.
answered Jan 25 at 6:31
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
What this question is really asking is if the function has a global minimum at all, as it isn't even defined on the set $x=y$ as you mentioned.
The set ${(x,y)in R^2:xne y}$ is open, and therefore ay local (and thus the global) minimum must have the partial derivatives equal $0$. The function is symmetric, so we assume WLOG $x>y$ and the function is now $H(x,y)=x^2+y^2+frac{1}{x-y}$. Thus $H_x(x,y)=2x-frac{1}{(x-y)^2}$, $H_y(x,y)=2y+frac{1}{(x-y)^2}$. These must both be $0$, so the sum of them must be $0$, so $2x+2y=0$ so $x=-y$.
Plugging this back into $H_x$ and setting equal to $0$ gives $2x=frac{1}{(2x)^2}$ so $8x^3=1$, so $x=frac12$, and thus $y=-frac12$. To see if this is actually a minimum, we find the Hessian at this point to be
$$
begin{bmatrix}
2.5 & -.5 \
-.5 & 2.5 \
end{bmatrix}
$$
which has eigenvalues $2$ and $3$ so is positive definite, thus this is a local minimum. As they are the only local minima, the global minimum is achieved at $H(frac12,-frac12)=frac32$, and by symmetry at $H(-frac12,frac12)=frac32$ as well.
answered Jan 25 at 6:31
Erik ParkinsonErik Parkinson
1,17519
edited Jan 25 at 6:58
answered Jan 25 at 6:31
Erik ParkinsonErik Parkinson
1,17519
answered Jan 25 at 6:31
Erik ParkinsonErik Parkinson
1,17519
answered Jan 25 at 6:31
Erik ParkinsonErik Parkinson
1,17519
1,17519
$begingroup$
You have found a critical point. You have not even checked that it is a local minimum. Do you have any reason to believe that it is a global minimum?
$endgroup$
– Misha Lavrov
Jan 25 at 6:36
$begingroup$
The function tends to infinity if any of $x$ or $y$ does and its domain can be written as an increasing union of compact sets.
$endgroup$
– Will M.
Jan 25 at 6:42
$begingroup$
@WillM. Exactly, and this (not the computation of critical points) is the actual argument needed.
$endgroup$
– Misha Lavrov
Jan 25 at 6:45
$begingroup$
@MishaLavrov Good catch, I totally forgot! I'll add that real quick.
$endgroup$
– Erik Parkinson
Jan 25 at 6:55
$begingroup$
@ErikParkinson It is still false that one of the local minima of a function must be a global minimum.
$endgroup$
– Misha Lavrov
Jan 25 at 15:02
add a comment |
$begingroup$
You have found a critical point. You have not even checked that it is a local minimum. Do you have any reason to believe that it is a global minimum?
$endgroup$
– Misha Lavrov
Jan 25 at 6:36
$begingroup$
The function tends to infinity if any of $x$ or $y$ does and its domain can be written as an increasing union of compact sets.
$endgroup$
– Will M.
Jan 25 at 6:42
$begingroup$
@WillM. Exactly, and this (not the computation of critical points) is the actual argument needed.
$endgroup$
– Misha Lavrov
Jan 25 at 6:45
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@MishaLavrov Good catch, I totally forgot! I'll add that real quick.
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– Erik Parkinson
Jan 25 at 6:55
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@ErikParkinson It is still false that one of the local minima of a function must be a global minimum.
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– Misha Lavrov
Jan 25 at 15:02
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You have found a critical point. You have not even checked that it is a local minimum. Do you have any reason to believe that it is a global minimum?
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– Misha Lavrov
Jan 25 at 6:36
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You have found a critical point. You have not even checked that it is a local minimum. Do you have any reason to believe that it is a global minimum?
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– Misha Lavrov
Jan 25 at 6:36
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The function tends to infinity if any of $x$ or $y$ does and its domain can be written as an increasing union of compact sets.
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– Will M.
Jan 25 at 6:42
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The function tends to infinity if any of $x$ or $y$ does and its domain can be written as an increasing union of compact sets.
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– Will M.
Jan 25 at 6:42
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@WillM. Exactly, and this (not the computation of critical points) is the actual argument needed.
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– Misha Lavrov
Jan 25 at 6:45
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@WillM. Exactly, and this (not the computation of critical points) is the actual argument needed.
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– Misha Lavrov
Jan 25 at 6:45
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@MishaLavrov Good catch, I totally forgot! I'll add that real quick.
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– Erik Parkinson
Jan 25 at 6:55
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@MishaLavrov Good catch, I totally forgot! I'll add that real quick.
$endgroup$
– Erik Parkinson
Jan 25 at 6:55
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@ErikParkinson It is still false that one of the local minima of a function must be a global minimum.
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– Misha Lavrov
Jan 25 at 15:02
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@ErikParkinson It is still false that one of the local minima of a function must be a global minimum.
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– Misha Lavrov
Jan 25 at 15:02
add a comment |
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Consider the set
$$
S = left{(x,y) in mathbb R^2 : |x| le 10, |y| le 10, |x-y| ge frac{1}{100}right}.
$$
This set $S$ is closed and bounded, and therefore (by the extreme value theorem) $H$ has a global minimum on $S$.
Points outside $S$ are guaranteed to be worse than that minimum. If $(x,y) notin S$, then one of three things must be true:
$|x| > 10$, in which case $H(x,y) ge x^2 > 100$.
$|y| > 10$, in which case $H(x,y) ge y^2 > 100$.
$|x-y| < frac{1}{100}$, in which case $H(x,y) ge frac{1}{|x-y|} > 100$.
Therefore $H(x,y) > 100$ everywhere outside $S$, which can't beat for example $H(0,1) = 2$.
This means that the global minimum of $H$ on $S$ is actually the global minimum of $H$ on its entire domain ${(x,y) in mathbb R^2 : x ne y}$.
answered Jan 25 at 6:44
Misha LavrovMisha Lavrov
47.7k657107
$endgroup$
add a comment |
$begingroup$
Consider the set
$$
S = left{(x,y) in mathbb R^2 : |x| le 10, |y| le 10, |x-y| ge frac{1}{100}right}.
$$
This set $S$ is closed and bounded, and therefore (by the extreme value theorem) $H$ has a global minimum on $S$.
Points outside $S$ are guaranteed to be worse than that minimum. If $(x,y) notin S$, then one of three things must be true:
$|x| > 10$, in which case $H(x,y) ge x^2 > 100$.
$|y| > 10$, in which case $H(x,y) ge y^2 > 100$.
$|x-y| < frac{1}{100}$, in which case $H(x,y) ge frac{1}{|x-y|} > 100$.
Therefore $H(x,y) > 100$ everywhere outside $S$, which can't beat for example $H(0,1) = 2$.
This means that the global minimum of $H$ on $S$ is actually the global minimum of $H$ on its entire domain ${(x,y) in mathbb R^2 : x ne y}$.
answered Jan 25 at 6:44
Misha LavrovMisha Lavrov
47.7k657107
$endgroup$
add a comment |
$begingroup$
Consider the set
$$
S = left{(x,y) in mathbb R^2 : |x| le 10, |y| le 10, |x-y| ge frac{1}{100}right}.
$$
This set $S$ is closed and bounded, and therefore (by the extreme value theorem) $H$ has a global minimum on $S$.
Points outside $S$ are guaranteed to be worse than that minimum. If $(x,y) notin S$, then one of three things must be true:
$|x| > 10$, in which case $H(x,y) ge x^2 > 100$.
$|y| > 10$, in which case $H(x,y) ge y^2 > 100$.
$|x-y| < frac{1}{100}$, in which case $H(x,y) ge frac{1}{|x-y|} > 100$.
Therefore $H(x,y) > 100$ everywhere outside $S$, which can't beat for example $H(0,1) = 2$.
This means that the global minimum of $H$ on $S$ is actually the global minimum of $H$ on its entire domain ${(x,y) in mathbb R^2 : x ne y}$.
answered Jan 25 at 6:44
Misha LavrovMisha Lavrov
47.7k657107
$endgroup$
Consider the set
$$
S = left{(x,y) in mathbb R^2 : |x| le 10, |y| le 10, |x-y| ge frac{1}{100}right}.
$$
This set $S$ is closed and bounded, and therefore (by the extreme value theorem) $H$ has a global minimum on $S$.
Points outside $S$ are guaranteed to be worse than that minimum. If $(x,y) notin S$, then one of three things must be true:
$|x| > 10$, in which case $H(x,y) ge x^2 > 100$.
$|y| > 10$, in which case $H(x,y) ge y^2 > 100$.
$|x-y| < frac{1}{100}$, in which case $H(x,y) ge frac{1}{|x-y|} > 100$.
Therefore $H(x,y) > 100$ everywhere outside $S$, which can't beat for example $H(0,1) = 2$.
This means that the global minimum of $H$ on $S$ is actually the global minimum of $H$ on its entire domain ${(x,y) in mathbb R^2 : x ne y}$.
answered Jan 25 at 6:44
Misha LavrovMisha Lavrov
47.7k657107
answered Jan 25 at 6:44
Misha LavrovMisha Lavrov
47.7k657107
answered Jan 25 at 6:44
Misha LavrovMisha Lavrov
47.7k657107
answered Jan 25 at 6:44
Misha LavrovMisha Lavrov
47.7k657107
47.7k657107
add a comment |
add a comment |
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$begingroup$
This function does not seem well defined. Note that $2! = 2$, $1!=1$.
$endgroup$
– Lucas Corrêa
Jan 25 at 6:15
1
$begingroup$
sorry I meant x is not equal to y
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– Eric J
Jan 25 at 6:25