there exist two antipodal points on the equator that have the same temperature.
$begingroup$
Argue that there exist at any time two antipodal points on the equator that have the same temperature.
The temperature function can be assumed to be continuous.
I am supposed to use the mean value theorem, but I don't see how. Thanks in advance.
calculus
edited Mar 17 '14 at 18:21
vadim123
76.5k897191
asked Mar 17 '14 at 18:16
kristof2014kristof2014
216127
$endgroup$
add a comment |
$begingroup$
Argue that there exist at any time two antipodal points on the equator that have the same temperature.
The temperature function can be assumed to be continuous.
I am supposed to use the mean value theorem, but I don't see how. Thanks in advance.
calculus
edited Mar 17 '14 at 18:21
vadim123
76.5k897191
asked Mar 17 '14 at 18:16
kristof2014kristof2014
216127
$endgroup$
1
$begingroup$
Have a look at page 32 theorem 1.10 (Borsuk-Ulam) math.cornell.edu/~hatcher/AT/AT.pdf
$endgroup$
– drhab
Mar 17 '14 at 18:22
2
$begingroup$
If the temperature is constant, you are done. Otherwise, there is (at least one) point where the temperature has a global maximum, and a different point where the temperature has a global minimum. Any value in between these extremes must occur at least twice: once between the max and min "clockwise", and once between the max and min "counterclockwise."
$endgroup$
– user7530
Mar 17 '14 at 18:24
add a comment |
$begingroup$
Argue that there exist at any time two antipodal points on the equator that have the same temperature.
The temperature function can be assumed to be continuous.
I am supposed to use the mean value theorem, but I don't see how. Thanks in advance.
calculus
edited Mar 17 '14 at 18:21
vadim123
76.5k897191
asked Mar 17 '14 at 18:16
kristof2014kristof2014
216127
$endgroup$
Argue that there exist at any time two antipodal points on the equator that have the same temperature.
The temperature function can be assumed to be continuous.
I am supposed to use the mean value theorem, but I don't see how. Thanks in advance.
calculus
calculus
edited Mar 17 '14 at 18:21
vadim123
76.5k897191
asked Mar 17 '14 at 18:16
kristof2014kristof2014
216127
edited Mar 17 '14 at 18:21
vadim123
76.5k897191
asked Mar 17 '14 at 18:16
kristof2014kristof2014
216127
edited Mar 17 '14 at 18:21
vadim123
76.5k897191
edited Mar 17 '14 at 18:21
vadim123
76.5k897191
edited Mar 17 '14 at 18:21
vadim123
76.5k897191
76.5k897191
asked Mar 17 '14 at 18:16
kristof2014kristof2014
216127
asked Mar 17 '14 at 18:16
kristof2014kristof2014
216127
asked Mar 17 '14 at 18:16
kristof2014kristof2014
216127
216127
1
$begingroup$
Have a look at page 32 theorem 1.10 (Borsuk-Ulam) math.cornell.edu/~hatcher/AT/AT.pdf
$endgroup$
– drhab
Mar 17 '14 at 18:22
2
$begingroup$
If the temperature is constant, you are done. Otherwise, there is (at least one) point where the temperature has a global maximum, and a different point where the temperature has a global minimum. Any value in between these extremes must occur at least twice: once between the max and min "clockwise", and once between the max and min "counterclockwise."
$endgroup$
– user7530
Mar 17 '14 at 18:24
add a comment |
1
$begingroup$
Have a look at page 32 theorem 1.10 (Borsuk-Ulam) math.cornell.edu/~hatcher/AT/AT.pdf
$endgroup$
– drhab
Mar 17 '14 at 18:22
2
$begingroup$
If the temperature is constant, you are done. Otherwise, there is (at least one) point where the temperature has a global maximum, and a different point where the temperature has a global minimum. Any value in between these extremes must occur at least twice: once between the max and min "clockwise", and once between the max and min "counterclockwise."
$endgroup$
– user7530
Mar 17 '14 at 18:24
1
1
$begingroup$
Have a look at page 32 theorem 1.10 (Borsuk-Ulam) math.cornell.edu/~hatcher/AT/AT.pdf
$endgroup$
– drhab
Mar 17 '14 at 18:22
$begingroup$
Have a look at page 32 theorem 1.10 (Borsuk-Ulam) math.cornell.edu/~hatcher/AT/AT.pdf
$endgroup$
– drhab
Mar 17 '14 at 18:22
2
2
$begingroup$
If the temperature is constant, you are done. Otherwise, there is (at least one) point where the temperature has a global maximum, and a different point where the temperature has a global minimum. Any value in between these extremes must occur at least twice: once between the max and min "clockwise", and once between the max and min "counterclockwise."
$endgroup$
– user7530
Mar 17 '14 at 18:24
$begingroup$
If the temperature is constant, you are done. Otherwise, there is (at least one) point where the temperature has a global maximum, and a different point where the temperature has a global minimum. Any value in between these extremes must occur at least twice: once between the max and min "clockwise", and once between the max and min "counterclockwise."
$endgroup$
– user7530
Mar 17 '14 at 18:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Okay, so lets just consider points along the equator, and let $ t:left [ 0,2pi right ]rightarrow mathbb{R}$ be the temperature at a point with angle $theta$ from some predetermined point on the equator.
Now, we are given that t is continuous in $theta$ on $left [ 0,2pi right ]$, and we see that t is $2pi$ periodic.
Define $T:left [ 0,2pi right ]rightarrow mathbb{R}$ to be the antipodal difference in temperature, that is $T := t(theta + pi) - t(theta)$
Then T is also continuous on $left [ 0,2pi right ]$, and we have that:
$T(0) = t(pi) - t(0)$ and $T(pi) = t(2pi) - t(pi)$
So as t is $2pi$ periodic, we get that $T(0) = -T(pi)$
If $T(0) = 0$ then we are done and we have our antipodal points with equal temperature, otherwise if $T(0) neq 0$, then as $T$ is continuous on $left [ 0,pi right ] subset$ $left [ 0,2pi right ]$ and without loss of generality $T(0) < 0 < T(pi)$, then $exists alpha in left [ 0,pi right ]$ such that $T(alpha) = 0$.
And then $t(alpha) = t(alpha + pi)$ so we have found our antipodal points with the same temperature.
This is really just a specific case of a really interesting theorem called the Borsuk–Ulam Theorem, which makes similar sorts of statements for n-dimensional spheres mapping to n-dimensional planes. Here we have a 2 dimensional sphere mapping to a 1 dimensional plane, but we considered a 1 dimensional subsphere (our equator), and the Borsuk–Ulam Theorem says on any continuous mapping of an n-dimensional sphere to an n-dimensional plane, there will be two antipodal points who get mapped to the same point.
http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem
answered Mar 17 '14 at 18:44
CameronJWhiteheadCameronJWhitehead
1,436618
$endgroup$
add a comment |
$begingroup$
Let $theta$ denote longitude; it varies from $+180$ to $-180$. Define $f(theta)$ to be the temperature at longitude $theta$, minus the temperature at the antipodal point to longitude $theta$. If ever $f(theta)=0$, you are done. However, $f(theta)=-f(theta')$, where $theta'$ is antipodal to $theta$. So, continuously vary $theta$ to get $theta'$, and apply IVT(Intermediate Value Theorem).
edited Jan 9 at 4:11
51alpha
396
answered Mar 17 '14 at 18:21
vadim123vadim123
76.5k897191
$endgroup$
$begingroup$
User @51alpha has suggested you meant Intermediate Value Theorem rather than MVT, but I declined this suggested edit because it might better be proposed as a Comment. The user has enough reputation, I think, to post such a Comment but did not take that approach, but you can see the pending edit.
$endgroup$
– hardmath
Jan 9 at 3:29
add a comment |
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$begingroup$
Okay, so lets just consider points along the equator, and let $ t:left [ 0,2pi right ]rightarrow mathbb{R}$ be the temperature at a point with angle $theta$ from some predetermined point on the equator.
Now, we are given that t is continuous in $theta$ on $left [ 0,2pi right ]$, and we see that t is $2pi$ periodic.
Define $T:left [ 0,2pi right ]rightarrow mathbb{R}$ to be the antipodal difference in temperature, that is $T := t(theta + pi) - t(theta)$
Then T is also continuous on $left [ 0,2pi right ]$, and we have that:
$T(0) = t(pi) - t(0)$ and $T(pi) = t(2pi) - t(pi)$
So as t is $2pi$ periodic, we get that $T(0) = -T(pi)$
If $T(0) = 0$ then we are done and we have our antipodal points with equal temperature, otherwise if $T(0) neq 0$, then as $T$ is continuous on $left [ 0,pi right ] subset$ $left [ 0,2pi right ]$ and without loss of generality $T(0) < 0 < T(pi)$, then $exists alpha in left [ 0,pi right ]$ such that $T(alpha) = 0$.
And then $t(alpha) = t(alpha + pi)$ so we have found our antipodal points with the same temperature.
This is really just a specific case of a really interesting theorem called the Borsuk–Ulam Theorem, which makes similar sorts of statements for n-dimensional spheres mapping to n-dimensional planes. Here we have a 2 dimensional sphere mapping to a 1 dimensional plane, but we considered a 1 dimensional subsphere (our equator), and the Borsuk–Ulam Theorem says on any continuous mapping of an n-dimensional sphere to an n-dimensional plane, there will be two antipodal points who get mapped to the same point.
http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem
answered Mar 17 '14 at 18:44
CameronJWhiteheadCameronJWhitehead
1,436618
$endgroup$
add a comment |
$begingroup$
Okay, so lets just consider points along the equator, and let $ t:left [ 0,2pi right ]rightarrow mathbb{R}$ be the temperature at a point with angle $theta$ from some predetermined point on the equator.
Now, we are given that t is continuous in $theta$ on $left [ 0,2pi right ]$, and we see that t is $2pi$ periodic.
Define $T:left [ 0,2pi right ]rightarrow mathbb{R}$ to be the antipodal difference in temperature, that is $T := t(theta + pi) - t(theta)$
Then T is also continuous on $left [ 0,2pi right ]$, and we have that:
$T(0) = t(pi) - t(0)$ and $T(pi) = t(2pi) - t(pi)$
So as t is $2pi$ periodic, we get that $T(0) = -T(pi)$
If $T(0) = 0$ then we are done and we have our antipodal points with equal temperature, otherwise if $T(0) neq 0$, then as $T$ is continuous on $left [ 0,pi right ] subset$ $left [ 0,2pi right ]$ and without loss of generality $T(0) < 0 < T(pi)$, then $exists alpha in left [ 0,pi right ]$ such that $T(alpha) = 0$.
And then $t(alpha) = t(alpha + pi)$ so we have found our antipodal points with the same temperature.
This is really just a specific case of a really interesting theorem called the Borsuk–Ulam Theorem, which makes similar sorts of statements for n-dimensional spheres mapping to n-dimensional planes. Here we have a 2 dimensional sphere mapping to a 1 dimensional plane, but we considered a 1 dimensional subsphere (our equator), and the Borsuk–Ulam Theorem says on any continuous mapping of an n-dimensional sphere to an n-dimensional plane, there will be two antipodal points who get mapped to the same point.
http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem
answered Mar 17 '14 at 18:44
CameronJWhiteheadCameronJWhitehead
1,436618
$endgroup$
add a comment |
$begingroup$
Okay, so lets just consider points along the equator, and let $ t:left [ 0,2pi right ]rightarrow mathbb{R}$ be the temperature at a point with angle $theta$ from some predetermined point on the equator.
Now, we are given that t is continuous in $theta$ on $left [ 0,2pi right ]$, and we see that t is $2pi$ periodic.
Define $T:left [ 0,2pi right ]rightarrow mathbb{R}$ to be the antipodal difference in temperature, that is $T := t(theta + pi) - t(theta)$
Then T is also continuous on $left [ 0,2pi right ]$, and we have that:
$T(0) = t(pi) - t(0)$ and $T(pi) = t(2pi) - t(pi)$
So as t is $2pi$ periodic, we get that $T(0) = -T(pi)$
If $T(0) = 0$ then we are done and we have our antipodal points with equal temperature, otherwise if $T(0) neq 0$, then as $T$ is continuous on $left [ 0,pi right ] subset$ $left [ 0,2pi right ]$ and without loss of generality $T(0) < 0 < T(pi)$, then $exists alpha in left [ 0,pi right ]$ such that $T(alpha) = 0$.
And then $t(alpha) = t(alpha + pi)$ so we have found our antipodal points with the same temperature.
This is really just a specific case of a really interesting theorem called the Borsuk–Ulam Theorem, which makes similar sorts of statements for n-dimensional spheres mapping to n-dimensional planes. Here we have a 2 dimensional sphere mapping to a 1 dimensional plane, but we considered a 1 dimensional subsphere (our equator), and the Borsuk–Ulam Theorem says on any continuous mapping of an n-dimensional sphere to an n-dimensional plane, there will be two antipodal points who get mapped to the same point.
http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem
answered Mar 17 '14 at 18:44
CameronJWhiteheadCameronJWhitehead
1,436618
$endgroup$
Okay, so lets just consider points along the equator, and let $ t:left [ 0,2pi right ]rightarrow mathbb{R}$ be the temperature at a point with angle $theta$ from some predetermined point on the equator.
Now, we are given that t is continuous in $theta$ on $left [ 0,2pi right ]$, and we see that t is $2pi$ periodic.
Define $T:left [ 0,2pi right ]rightarrow mathbb{R}$ to be the antipodal difference in temperature, that is $T := t(theta + pi) - t(theta)$
Then T is also continuous on $left [ 0,2pi right ]$, and we have that:
$T(0) = t(pi) - t(0)$ and $T(pi) = t(2pi) - t(pi)$
So as t is $2pi$ periodic, we get that $T(0) = -T(pi)$
If $T(0) = 0$ then we are done and we have our antipodal points with equal temperature, otherwise if $T(0) neq 0$, then as $T$ is continuous on $left [ 0,pi right ] subset$ $left [ 0,2pi right ]$ and without loss of generality $T(0) < 0 < T(pi)$, then $exists alpha in left [ 0,pi right ]$ such that $T(alpha) = 0$.
And then $t(alpha) = t(alpha + pi)$ so we have found our antipodal points with the same temperature.
This is really just a specific case of a really interesting theorem called the Borsuk–Ulam Theorem, which makes similar sorts of statements for n-dimensional spheres mapping to n-dimensional planes. Here we have a 2 dimensional sphere mapping to a 1 dimensional plane, but we considered a 1 dimensional subsphere (our equator), and the Borsuk–Ulam Theorem says on any continuous mapping of an n-dimensional sphere to an n-dimensional plane, there will be two antipodal points who get mapped to the same point.
http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem
answered Mar 17 '14 at 18:44
CameronJWhiteheadCameronJWhitehead
1,436618
answered Mar 17 '14 at 18:44
CameronJWhiteheadCameronJWhitehead
1,436618
answered Mar 17 '14 at 18:44
CameronJWhiteheadCameronJWhitehead
1,436618
answered Mar 17 '14 at 18:44
CameronJWhiteheadCameronJWhitehead
1,436618
1,436618
add a comment |
add a comment |
$begingroup$
Let $theta$ denote longitude; it varies from $+180$ to $-180$. Define $f(theta)$ to be the temperature at longitude $theta$, minus the temperature at the antipodal point to longitude $theta$. If ever $f(theta)=0$, you are done. However, $f(theta)=-f(theta')$, where $theta'$ is antipodal to $theta$. So, continuously vary $theta$ to get $theta'$, and apply IVT(Intermediate Value Theorem).
edited Jan 9 at 4:11
51alpha
396
answered Mar 17 '14 at 18:21
vadim123vadim123
76.5k897191
$endgroup$
$begingroup$
User @51alpha has suggested you meant Intermediate Value Theorem rather than MVT, but I declined this suggested edit because it might better be proposed as a Comment. The user has enough reputation, I think, to post such a Comment but did not take that approach, but you can see the pending edit.
$endgroup$
– hardmath
Jan 9 at 3:29
add a comment |
$begingroup$
Let $theta$ denote longitude; it varies from $+180$ to $-180$. Define $f(theta)$ to be the temperature at longitude $theta$, minus the temperature at the antipodal point to longitude $theta$. If ever $f(theta)=0$, you are done. However, $f(theta)=-f(theta')$, where $theta'$ is antipodal to $theta$. So, continuously vary $theta$ to get $theta'$, and apply IVT(Intermediate Value Theorem).
edited Jan 9 at 4:11
51alpha
396
answered Mar 17 '14 at 18:21
vadim123vadim123
76.5k897191
$endgroup$
$begingroup$
User @51alpha has suggested you meant Intermediate Value Theorem rather than MVT, but I declined this suggested edit because it might better be proposed as a Comment. The user has enough reputation, I think, to post such a Comment but did not take that approach, but you can see the pending edit.
$endgroup$
– hardmath
Jan 9 at 3:29
add a comment |
$begingroup$
Let $theta$ denote longitude; it varies from $+180$ to $-180$. Define $f(theta)$ to be the temperature at longitude $theta$, minus the temperature at the antipodal point to longitude $theta$. If ever $f(theta)=0$, you are done. However, $f(theta)=-f(theta')$, where $theta'$ is antipodal to $theta$. So, continuously vary $theta$ to get $theta'$, and apply IVT(Intermediate Value Theorem).
edited Jan 9 at 4:11
51alpha
396
answered Mar 17 '14 at 18:21
vadim123vadim123
76.5k897191
$endgroup$
Let $theta$ denote longitude; it varies from $+180$ to $-180$. Define $f(theta)$ to be the temperature at longitude $theta$, minus the temperature at the antipodal point to longitude $theta$. If ever $f(theta)=0$, you are done. However, $f(theta)=-f(theta')$, where $theta'$ is antipodal to $theta$. So, continuously vary $theta$ to get $theta'$, and apply IVT(Intermediate Value Theorem).
edited Jan 9 at 4:11
51alpha
396
answered Mar 17 '14 at 18:21
vadim123vadim123
76.5k897191
edited Jan 9 at 4:11
51alpha
396
edited Jan 9 at 4:11
51alpha
396
edited Jan 9 at 4:11
51alpha
396
396
answered Mar 17 '14 at 18:21
vadim123vadim123
76.5k897191
answered Mar 17 '14 at 18:21
vadim123vadim123
76.5k897191
answered Mar 17 '14 at 18:21
vadim123vadim123
76.5k897191
76.5k897191
$begingroup$
User @51alpha has suggested you meant Intermediate Value Theorem rather than MVT, but I declined this suggested edit because it might better be proposed as a Comment. The user has enough reputation, I think, to post such a Comment but did not take that approach, but you can see the pending edit.
$endgroup$
– hardmath
Jan 9 at 3:29
add a comment |
$begingroup$
User @51alpha has suggested you meant Intermediate Value Theorem rather than MVT, but I declined this suggested edit because it might better be proposed as a Comment. The user has enough reputation, I think, to post such a Comment but did not take that approach, but you can see the pending edit.
$endgroup$
– hardmath
Jan 9 at 3:29
$begingroup$
User @51alpha has suggested you meant Intermediate Value Theorem rather than MVT, but I declined this suggested edit because it might better be proposed as a Comment. The user has enough reputation, I think, to post such a Comment but did not take that approach, but you can see the pending edit.
$endgroup$
– hardmath
Jan 9 at 3:29
$begingroup$
User @51alpha has suggested you meant Intermediate Value Theorem rather than MVT, but I declined this suggested edit because it might better be proposed as a Comment. The user has enough reputation, I think, to post such a Comment but did not take that approach, but you can see the pending edit.
$endgroup$
– hardmath
Jan 9 at 3:29
add a comment |
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Have a look at page 32 theorem 1.10 (Borsuk-Ulam) math.cornell.edu/~hatcher/AT/AT.pdf
$endgroup$
– drhab
Mar 17 '14 at 18:22
2
$begingroup$
If the temperature is constant, you are done. Otherwise, there is (at least one) point where the temperature has a global maximum, and a different point where the temperature has a global minimum. Any value in between these extremes must occur at least twice: once between the max and min "clockwise", and once between the max and min "counterclockwise."
$endgroup$
– user7530
Mar 17 '14 at 18:24