Mixing
If Rolle's Theorem is assumed to be true, doesn't that prove the MVT?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
If we assume that Rolle's Theorem is true is it practical to say that it also proves the MVT?
My reasoning is that even though Rolle's Theorem is the special case for when $f(a)=f(b)$ and the secant line between $(a,f(a))$ and $(b,f(b))$ is horizontal, doesn't that mean you can 'rotate' and 'stretch' the function $f$ and thus 'stretch' and 'rotate' the straight line between $(a,f(a))$ and $(b,f(b))$ as well as the line tangent to $(c,f(c))$ to maintain its symmetry to the line secant between $(a,f(a))$ and $(b,f(b))$ to thus prove that for the straight line intersecting points $(a,f(a))$ and $(b,f(b))$, there must be a point $c$ on $[a,b]$ for which $f'(c)$ is parallel to the aforementioned line between $(a,f(a))$ and $(b, f(b))$?
I feel that this could be better explained with something like linear algebra. I do not know it well enough to use it.
EDIT: My question was just as unclear as I thought it was. I am going to try and explain it differently:
For a function $f$, if it is differentiable/continuous over $[a,b] $/$(a,b)$ etc. then there is a straight line between $(a, f(a))$ and $(b,f(b))$. You can rotate $f$ around the origin until this straight line is parallel to the x-axis. By Rolle's Theorem, there must exist a $c$ on the transformed $(a,b)$ such that $f'(c)=0$ and thus there is at least one line tangent to $f$ on $(a,b)$ that is parallel to the secant of $(a,f(a))$ and $(b,f(b))$. Then, you can 'undo' the original rotation of $f$ and know that the lines proven to exist by Rolle's Theorem exist with the same symmetries with respect to $f$, meaning the MVT must hold even when the secant line is not horizontal.
calculus linear-algebra linear-transformations rolles-theorem
asked 1 hour ago
Matt
546
 |Â
show 7 more comments
up vote
1
down vote
favorite
If we assume that Rolle's Theorem is true is it practical to say that it also proves the MVT?
My reasoning is that even though Rolle's Theorem is the special case for when $f(a)=f(b)$ and the secant line between $(a,f(a))$ and $(b,f(b))$ is horizontal, doesn't that mean you can 'rotate' and 'stretch' the function $f$ and thus 'stretch' and 'rotate' the straight line between $(a,f(a))$ and $(b,f(b))$ as well as the line tangent to $(c,f(c))$ to maintain its symmetry to the line secant between $(a,f(a))$ and $(b,f(b))$ to thus prove that for the straight line intersecting points $(a,f(a))$ and $(b,f(b))$, there must be a point $c$ on $[a,b]$ for which $f'(c)$ is parallel to the aforementioned line between $(a,f(a))$ and $(b, f(b))$?
I feel that this could be better explained with something like linear algebra. I do not know it well enough to use it.
EDIT: My question was just as unclear as I thought it was. I am going to try and explain it differently:
For a function $f$, if it is differentiable/continuous over $[a,b] $/$(a,b)$ etc. then there is a straight line between $(a, f(a))$ and $(b,f(b))$. You can rotate $f$ around the origin until this straight line is parallel to the x-axis. By Rolle's Theorem, there must exist a $c$ on the transformed $(a,b)$ such that $f'(c)=0$ and thus there is at least one line tangent to $f$ on $(a,b)$ that is parallel to the secant of $(a,f(a))$ and $(b,f(b))$. Then, you can 'undo' the original rotation of $f$ and know that the lines proven to exist by Rolle's Theorem exist with the same symmetries with respect to $f$, meaning the MVT must hold even when the secant line is not horizontal.
calculus linear-algebra linear-transformations rolles-theorem
asked 1 hour ago
Matt
546
If the question is "Can the mean value theorem be proved using Rolle's theorem?" the answer is "Yes." However, I don't see what this has to do with linear algebra, and I am very confused by your exposition.
– Xander Henderson
1 hour ago
Did you read any notes or books on a proof of MVT? It should be clear that Rolle's theorem is used in any proof. If it is not clear to you, I can write an outline for you.
– edm
1 hour ago
This is indeed how the MVT is often proved.
– Bungo
1 hour ago
Perhaps what you mean by "rotate and stretch" is shear mapping.
– edm
1 hour ago
@edm I disagree with the statement that "It should be clear that Rolle's theorem is used in any proof." For example, my recollection of baby Rudin is that he proves the mean value theorem, then (maybe?) states Rolle's as a corollary or porism.
– Xander Henderson
1 hour ago
 |Â
show 7 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If we assume that Rolle's Theorem is true is it practical to say that it also proves the MVT?
My reasoning is that even though Rolle's Theorem is the special case for when $f(a)=f(b)$ and the secant line between $(a,f(a))$ and $(b,f(b))$ is horizontal, doesn't that mean you can 'rotate' and 'stretch' the function $f$ and thus 'stretch' and 'rotate' the straight line between $(a,f(a))$ and $(b,f(b))$ as well as the line tangent to $(c,f(c))$ to maintain its symmetry to the line secant between $(a,f(a))$ and $(b,f(b))$ to thus prove that for the straight line intersecting points $(a,f(a))$ and $(b,f(b))$, there must be a point $c$ on $[a,b]$ for which $f'(c)$ is parallel to the aforementioned line between $(a,f(a))$ and $(b, f(b))$?
I feel that this could be better explained with something like linear algebra. I do not know it well enough to use it.
EDIT: My question was just as unclear as I thought it was. I am going to try and explain it differently:
For a function $f$, if it is differentiable/continuous over $[a,b] $/$(a,b)$ etc. then there is a straight line between $(a, f(a))$ and $(b,f(b))$. You can rotate $f$ around the origin until this straight line is parallel to the x-axis. By Rolle's Theorem, there must exist a $c$ on the transformed $(a,b)$ such that $f'(c)=0$ and thus there is at least one line tangent to $f$ on $(a,b)$ that is parallel to the secant of $(a,f(a))$ and $(b,f(b))$. Then, you can 'undo' the original rotation of $f$ and know that the lines proven to exist by Rolle's Theorem exist with the same symmetries with respect to $f$, meaning the MVT must hold even when the secant line is not horizontal.
calculus linear-algebra linear-transformations rolles-theorem
asked 1 hour ago
Matt
546
If we assume that Rolle's Theorem is true is it practical to say that it also proves the MVT?
My reasoning is that even though Rolle's Theorem is the special case for when $f(a)=f(b)$ and the secant line between $(a,f(a))$ and $(b,f(b))$ is horizontal, doesn't that mean you can 'rotate' and 'stretch' the function $f$ and thus 'stretch' and 'rotate' the straight line between $(a,f(a))$ and $(b,f(b))$ as well as the line tangent to $(c,f(c))$ to maintain its symmetry to the line secant between $(a,f(a))$ and $(b,f(b))$ to thus prove that for the straight line intersecting points $(a,f(a))$ and $(b,f(b))$, there must be a point $c$ on $[a,b]$ for which $f'(c)$ is parallel to the aforementioned line between $(a,f(a))$ and $(b, f(b))$?
I feel that this could be better explained with something like linear algebra. I do not know it well enough to use it.
EDIT: My question was just as unclear as I thought it was. I am going to try and explain it differently:
For a function $f$, if it is differentiable/continuous over $[a,b] $/$(a,b)$ etc. then there is a straight line between $(a, f(a))$ and $(b,f(b))$. You can rotate $f$ around the origin until this straight line is parallel to the x-axis. By Rolle's Theorem, there must exist a $c$ on the transformed $(a,b)$ such that $f'(c)=0$ and thus there is at least one line tangent to $f$ on $(a,b)$ that is parallel to the secant of $(a,f(a))$ and $(b,f(b))$. Then, you can 'undo' the original rotation of $f$ and know that the lines proven to exist by Rolle's Theorem exist with the same symmetries with respect to $f$, meaning the MVT must hold even when the secant line is not horizontal.
calculus linear-algebra linear-transformations rolles-theorem
calculus linear-algebra linear-transformations rolles-theorem
asked 1 hour ago
Matt
546
asked 1 hour ago
Matt
546
edited 1 hour ago
asked 1 hour ago
Matt
546
asked 1 hour ago
Matt
546
asked 1 hour ago
Matt
546
546
If the question is "Can the mean value theorem be proved using Rolle's theorem?" the answer is "Yes." However, I don't see what this has to do with linear algebra, and I am very confused by your exposition.
– Xander Henderson
1 hour ago
Did you read any notes or books on a proof of MVT? It should be clear that Rolle's theorem is used in any proof. If it is not clear to you, I can write an outline for you.
– edm
1 hour ago
This is indeed how the MVT is often proved.
– Bungo
1 hour ago
Perhaps what you mean by "rotate and stretch" is shear mapping.
– edm
1 hour ago
@edm I disagree with the statement that "It should be clear that Rolle's theorem is used in any proof." For example, my recollection of baby Rudin is that he proves the mean value theorem, then (maybe?) states Rolle's as a corollary or porism.
– Xander Henderson
1 hour ago
 |Â
show 7 more comments
If the question is "Can the mean value theorem be proved using Rolle's theorem?" the answer is "Yes." However, I don't see what this has to do with linear algebra, and I am very confused by your exposition.
– Xander Henderson
1 hour ago
Did you read any notes or books on a proof of MVT? It should be clear that Rolle's theorem is used in any proof. If it is not clear to you, I can write an outline for you.
– edm
1 hour ago
This is indeed how the MVT is often proved.
– Bungo
1 hour ago
Perhaps what you mean by "rotate and stretch" is shear mapping.
– edm
1 hour ago
@edm I disagree with the statement that "It should be clear that Rolle's theorem is used in any proof." For example, my recollection of baby Rudin is that he proves the mean value theorem, then (maybe?) states Rolle's as a corollary or porism.
– Xander Henderson
1 hour ago
If the question is "Can the mean value theorem be proved using Rolle's theorem?" the answer is "Yes." However, I don't see what this has to do with linear algebra, and I am very confused by your exposition.
– Xander Henderson
1 hour ago
If the question is "Can the mean value theorem be proved using Rolle's theorem?" the answer is "Yes." However, I don't see what this has to do with linear algebra, and I am very confused by your exposition.
– Xander Henderson
1 hour ago
Did you read any notes or books on a proof of MVT? It should be clear that Rolle's theorem is used in any proof. If it is not clear to you, I can write an outline for you.
– edm
1 hour ago
Did you read any notes or books on a proof of MVT? It should be clear that Rolle's theorem is used in any proof. If it is not clear to you, I can write an outline for you.
– edm
1 hour ago
This is indeed how the MVT is often proved.
– Bungo
1 hour ago
This is indeed how the MVT is often proved.
– Bungo
1 hour ago
Perhaps what you mean by "rotate and stretch" is shear mapping.
– edm
1 hour ago
Perhaps what you mean by "rotate and stretch" is shear mapping.
– edm
1 hour ago
@edm I disagree with the statement that "It should be clear that Rolle's theorem is used in any proof." For example, my recollection of baby Rudin is that he proves the mean value theorem, then (maybe?) states Rolle's as a corollary or porism.
– Xander Henderson
1 hour ago
@edm I disagree with the statement that "It should be clear that Rolle's theorem is used in any proof." For example, my recollection of baby Rudin is that he proves the mean value theorem, then (maybe?) states Rolle's as a corollary or porism.
– Xander Henderson
1 hour ago
 |Â
show 7 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
You don't need any linear transformations in $mathbbR^2$; you can do this by subtracting a linear polynomial from your function (which is, itself, a shear map). Simply take your continuous function $f : [a, b] to mathbbR$ that is differentiable on $(a, b)$, and let $g(x) = f(x) - fracf(b) - f(a)b - a (x - a) - f(a)$. In particular $f(a) = f(b) = 0$, and when you apply Rolle's theorem to find $c in (a, b)$ such that $g'(c) = 0$, you'll find that
$$f'(c) = fracf(b)- f(a)b - a.$$
answered 1 hour ago
Theo Bendit
14.4k12045
add a comment |Â
up vote
2
down vote
A proof of MVT used Rolle's theorem explicitly.
Consider a continuous function $f:[a,b]toBbb R$ that is differentiable on $(a,b)$. We are to prove MVT, i.e. to prove there exists a point $cin(a,b)$ such that $f'(c)=fracf(b)-f(a)b-a$, assuming that Rolle's theorem is true.
Define a new function $g:[a,b]toBbb R$ by $g(x)=f(x)-fracf(b)-f(a)b-ax$. Verify yourself that $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Verify also that $g(a)=g(b)$. So we can apply Rolle's theorem to this function $g$, to obtain a point $cin(a,b)$ such that $g'(c)=0$. Note that the derivative of $g$ is $g'(x)=f'(x)-fracf(b)-f(a)b-a$. The point $c$ also satisfies $f'(c)-fracf(b)-f(a)b-a=0$.
answered 1 hour ago
edm
2,7461424
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You don't need any linear transformations in $mathbbR^2$; you can do this by subtracting a linear polynomial from your function (which is, itself, a shear map). Simply take your continuous function $f : [a, b] to mathbbR$ that is differentiable on $(a, b)$, and let $g(x) = f(x) - fracf(b) - f(a)b - a (x - a) - f(a)$. In particular $f(a) = f(b) = 0$, and when you apply Rolle's theorem to find $c in (a, b)$ such that $g'(c) = 0$, you'll find that
$$f'(c) = fracf(b)- f(a)b - a.$$
answered 1 hour ago
Theo Bendit
14.4k12045
add a comment |Â
up vote
2
down vote
You don't need any linear transformations in $mathbbR^2$; you can do this by subtracting a linear polynomial from your function (which is, itself, a shear map). Simply take your continuous function $f : [a, b] to mathbbR$ that is differentiable on $(a, b)$, and let $g(x) = f(x) - fracf(b) - f(a)b - a (x - a) - f(a)$. In particular $f(a) = f(b) = 0$, and when you apply Rolle's theorem to find $c in (a, b)$ such that $g'(c) = 0$, you'll find that
$$f'(c) = fracf(b)- f(a)b - a.$$
answered 1 hour ago
Theo Bendit
14.4k12045
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You don't need any linear transformations in $mathbbR^2$; you can do this by subtracting a linear polynomial from your function (which is, itself, a shear map). Simply take your continuous function $f : [a, b] to mathbbR$ that is differentiable on $(a, b)$, and let $g(x) = f(x) - fracf(b) - f(a)b - a (x - a) - f(a)$. In particular $f(a) = f(b) = 0$, and when you apply Rolle's theorem to find $c in (a, b)$ such that $g'(c) = 0$, you'll find that
$$f'(c) = fracf(b)- f(a)b - a.$$
answered 1 hour ago
Theo Bendit
14.4k12045
You don't need any linear transformations in $mathbbR^2$; you can do this by subtracting a linear polynomial from your function (which is, itself, a shear map). Simply take your continuous function $f : [a, b] to mathbbR$ that is differentiable on $(a, b)$, and let $g(x) = f(x) - fracf(b) - f(a)b - a (x - a) - f(a)$. In particular $f(a) = f(b) = 0$, and when you apply Rolle's theorem to find $c in (a, b)$ such that $g'(c) = 0$, you'll find that
$$f'(c) = fracf(b)- f(a)b - a.$$
answered 1 hour ago
Theo Bendit
14.4k12045
answered 1 hour ago
Theo Bendit
14.4k12045
answered 1 hour ago
Theo Bendit
14.4k12045
answered 1 hour ago
Theo Bendit
14.4k12045
14.4k12045
add a comment |Â
add a comment |Â
up vote
2
down vote
A proof of MVT used Rolle's theorem explicitly.
Consider a continuous function $f:[a,b]toBbb R$ that is differentiable on $(a,b)$. We are to prove MVT, i.e. to prove there exists a point $cin(a,b)$ such that $f'(c)=fracf(b)-f(a)b-a$, assuming that Rolle's theorem is true.
Define a new function $g:[a,b]toBbb R$ by $g(x)=f(x)-fracf(b)-f(a)b-ax$. Verify yourself that $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Verify also that $g(a)=g(b)$. So we can apply Rolle's theorem to this function $g$, to obtain a point $cin(a,b)$ such that $g'(c)=0$. Note that the derivative of $g$ is $g'(x)=f'(x)-fracf(b)-f(a)b-a$. The point $c$ also satisfies $f'(c)-fracf(b)-f(a)b-a=0$.
answered 1 hour ago
edm
2,7461424
add a comment |Â
up vote
2
down vote
A proof of MVT used Rolle's theorem explicitly.
Consider a continuous function $f:[a,b]toBbb R$ that is differentiable on $(a,b)$. We are to prove MVT, i.e. to prove there exists a point $cin(a,b)$ such that $f'(c)=fracf(b)-f(a)b-a$, assuming that Rolle's theorem is true.
Define a new function $g:[a,b]toBbb R$ by $g(x)=f(x)-fracf(b)-f(a)b-ax$. Verify yourself that $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Verify also that $g(a)=g(b)$. So we can apply Rolle's theorem to this function $g$, to obtain a point $cin(a,b)$ such that $g'(c)=0$. Note that the derivative of $g$ is $g'(x)=f'(x)-fracf(b)-f(a)b-a$. The point $c$ also satisfies $f'(c)-fracf(b)-f(a)b-a=0$.
answered 1 hour ago
edm
2,7461424
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A proof of MVT used Rolle's theorem explicitly.
Consider a continuous function $f:[a,b]toBbb R$ that is differentiable on $(a,b)$. We are to prove MVT, i.e. to prove there exists a point $cin(a,b)$ such that $f'(c)=fracf(b)-f(a)b-a$, assuming that Rolle's theorem is true.
Define a new function $g:[a,b]toBbb R$ by $g(x)=f(x)-fracf(b)-f(a)b-ax$. Verify yourself that $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Verify also that $g(a)=g(b)$. So we can apply Rolle's theorem to this function $g$, to obtain a point $cin(a,b)$ such that $g'(c)=0$. Note that the derivative of $g$ is $g'(x)=f'(x)-fracf(b)-f(a)b-a$. The point $c$ also satisfies $f'(c)-fracf(b)-f(a)b-a=0$.
answered 1 hour ago
edm
2,7461424
A proof of MVT used Rolle's theorem explicitly.
Consider a continuous function $f:[a,b]toBbb R$ that is differentiable on $(a,b)$. We are to prove MVT, i.e. to prove there exists a point $cin(a,b)$ such that $f'(c)=fracf(b)-f(a)b-a$, assuming that Rolle's theorem is true.
Define a new function $g:[a,b]toBbb R$ by $g(x)=f(x)-fracf(b)-f(a)b-ax$. Verify yourself that $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Verify also that $g(a)=g(b)$. So we can apply Rolle's theorem to this function $g$, to obtain a point $cin(a,b)$ such that $g'(c)=0$. Note that the derivative of $g$ is $g'(x)=f'(x)-fracf(b)-f(a)b-a$. The point $c$ also satisfies $f'(c)-fracf(b)-f(a)b-a=0$.
answered 1 hour ago
edm
2,7461424
answered 1 hour ago
edm
2,7461424
answered 1 hour ago
edm
2,7461424
answered 1 hour ago
edm
2,7461424
2,7461424
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2937558%2fif-rolles-theorem-is-assumed-to-be-true-doesnt-that-prove-the-mvt%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
If the question is "Can the mean value theorem be proved using Rolle's theorem?" the answer is "Yes." However, I don't see what this has to do with linear algebra, and I am very confused by your exposition.
– Xander Henderson
1 hour ago
Did you read any notes or books on a proof of MVT? It should be clear that Rolle's theorem is used in any proof. If it is not clear to you, I can write an outline for you.
– edm
1 hour ago
This is indeed how the MVT is often proved.
– Bungo
1 hour ago
Perhaps what you mean by "rotate and stretch" is shear mapping.
– edm
1 hour ago
@edm I disagree with the statement that "It should be clear that Rolle's theorem is used in any proof." For example, my recollection of baby Rudin is that he proves the mean value theorem, then (maybe?) states Rolle's as a corollary or porism.
– Xander Henderson
1 hour ago