Show that a point is closest to another point with extreme value theoremExtreme value theorem - condition on...
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Show that a point is closest to another point with extreme value theorem
Extreme value theorem - condition on continuity for boundednessMultivariable version of the extreme value theoremContinuous functions/ Extreme value theoremExtreme value theorem examplesUnderstanding the Extreme Value TheoremApplying Extreme Value Theorem to prove existence of unique fixed pointWeierstrass' extreme value theorem: unbounded and not closedPremises of the extreme value theorem on “restricted” domainsTrouble understanding how the Transfer Principle is applied for the Extreme Value theorem.Show that a continuous function with infinite “tails” with value $0$ and $f(0)=1$ must have a maximum.
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Show there is a point of the plane ${x in mathbb{R^3} mid x_1 + 2x_2 + 3x_3 = 13}$ closest to the point $(1, 1, 1)$.
Let a function $f: A rightarrow mathbb{R}$ defined for all $x in A$ by $f(x_1, x_2, x_3) = x_1 + 2x_2 + 3x_3 - 13$.
We can take the domain $A = x_1 + 2x_2 + 3x_3 = 13 cap B[(1, 1, 1), r]$.
If we take $r = (1, 1, frac{10}{3})$, the plane intersects the ball.
So, this domain is bounded and closed.
We show that $f$ is continuous.
So, by the extreme value theorem, there is $a in A$ and $b in A$ such that for all $x in A$ : $f(a) le f(x) le f(b)$.
We showed that there is a point $a in A$ that is closest to the point $(1, 1, 1)$.
The basic idea behind this proof is that a plane is closed but not bounded. So we intersect the set with a closed ball then we can use the extreme value theorem.
Is it correct ?
real-analysis vectors extreme-value-theorem
$endgroup$
add a comment |
$begingroup$
Show there is a point of the plane ${x in mathbb{R^3} mid x_1 + 2x_2 + 3x_3 = 13}$ closest to the point $(1, 1, 1)$.
Let a function $f: A rightarrow mathbb{R}$ defined for all $x in A$ by $f(x_1, x_2, x_3) = x_1 + 2x_2 + 3x_3 - 13$.
We can take the domain $A = x_1 + 2x_2 + 3x_3 = 13 cap B[(1, 1, 1), r]$.
If we take $r = (1, 1, frac{10}{3})$, the plane intersects the ball.
So, this domain is bounded and closed.
We show that $f$ is continuous.
So, by the extreme value theorem, there is $a in A$ and $b in A$ such that for all $x in A$ : $f(a) le f(x) le f(b)$.
We showed that there is a point $a in A$ that is closest to the point $(1, 1, 1)$.
The basic idea behind this proof is that a plane is closed but not bounded. So we intersect the set with a closed ball then we can use the extreme value theorem.
Is it correct ?
real-analysis vectors extreme-value-theorem
$endgroup$
$begingroup$
Hm.The ball's radius $r$ should be a nonnegative number and not a point (you wrote $r=(1,1,10/3)$ )
$endgroup$
– Maksim
Mar 18 at 14:59
add a comment |
$begingroup$
Show there is a point of the plane ${x in mathbb{R^3} mid x_1 + 2x_2 + 3x_3 = 13}$ closest to the point $(1, 1, 1)$.
Let a function $f: A rightarrow mathbb{R}$ defined for all $x in A$ by $f(x_1, x_2, x_3) = x_1 + 2x_2 + 3x_3 - 13$.
We can take the domain $A = x_1 + 2x_2 + 3x_3 = 13 cap B[(1, 1, 1), r]$.
If we take $r = (1, 1, frac{10}{3})$, the plane intersects the ball.
So, this domain is bounded and closed.
We show that $f$ is continuous.
So, by the extreme value theorem, there is $a in A$ and $b in A$ such that for all $x in A$ : $f(a) le f(x) le f(b)$.
We showed that there is a point $a in A$ that is closest to the point $(1, 1, 1)$.
The basic idea behind this proof is that a plane is closed but not bounded. So we intersect the set with a closed ball then we can use the extreme value theorem.
Is it correct ?
real-analysis vectors extreme-value-theorem
$endgroup$
Show there is a point of the plane ${x in mathbb{R^3} mid x_1 + 2x_2 + 3x_3 = 13}$ closest to the point $(1, 1, 1)$.
Let a function $f: A rightarrow mathbb{R}$ defined for all $x in A$ by $f(x_1, x_2, x_3) = x_1 + 2x_2 + 3x_3 - 13$.
We can take the domain $A = x_1 + 2x_2 + 3x_3 = 13 cap B[(1, 1, 1), r]$.
If we take $r = (1, 1, frac{10}{3})$, the plane intersects the ball.
So, this domain is bounded and closed.
We show that $f$ is continuous.
So, by the extreme value theorem, there is $a in A$ and $b in A$ such that for all $x in A$ : $f(a) le f(x) le f(b)$.
We showed that there is a point $a in A$ that is closest to the point $(1, 1, 1)$.
The basic idea behind this proof is that a plane is closed but not bounded. So we intersect the set with a closed ball then we can use the extreme value theorem.
Is it correct ?
real-analysis vectors extreme-value-theorem
real-analysis vectors extreme-value-theorem
asked Mar 18 at 13:24
Mathieu RousseauMathieu Rousseau
62
62
$begingroup$
Hm.The ball's radius $r$ should be a nonnegative number and not a point (you wrote $r=(1,1,10/3)$ )
$endgroup$
– Maksim
Mar 18 at 14:59
add a comment |
$begingroup$
Hm.The ball's radius $r$ should be a nonnegative number and not a point (you wrote $r=(1,1,10/3)$ )
$endgroup$
– Maksim
Mar 18 at 14:59
$begingroup$
Hm.The ball's radius $r$ should be a nonnegative number and not a point (you wrote $r=(1,1,10/3)$ )
$endgroup$
– Maksim
Mar 18 at 14:59
$begingroup$
Hm.The ball's radius $r$ should be a nonnegative number and not a point (you wrote $r=(1,1,10/3)$ )
$endgroup$
– Maksim
Mar 18 at 14:59
add a comment |
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$begingroup$
Hm.The ball's radius $r$ should be a nonnegative number and not a point (you wrote $r=(1,1,10/3)$ )
$endgroup$
– Maksim
Mar 18 at 14:59