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jezza
science forum beginner

Joined: 28 May 2005
Posts: 8

Posted: Sat May 28, 2005 3:51 pm    Post subject: Mean Value Theorem in Several Dimensions

Hey guys,

I have been told that the mean values theorem
f'(c) = (f(b) − f(a))/(b − a) for f : [a, b] R -> Rn continuous does not hold true. Can anybody think of a simple counterexample that would prove that it is not true? My brain does not seem to be functioning properly :)

Thanks for the help

Jezza
jezza
science forum beginner

Joined: 28 May 2005
Posts: 8

 Posted: Sat May 28, 2005 4:01 pm    Post subject: Re: Mean Value Theorem in Several Dimensions Sorry that was supposed to be subtraction signs instead of question marks. i dont know why it did not work
science forum Guru

Joined: 24 Mar 2005
Posts: 790

Posted: Sat May 28, 2005 6:19 pm    Post subject: Re: Mean Value Theorem in Several Dimensions

In article
<6312973.1117295544931.JavaMail.jakarta@nitrogen.mathforum.org>,
jezza <big__jezz@hotmail.com> wrote:

 Quote: I have been told that the mean values theorem f'(c) = (f(b) - f(a))/(b - a) for f : [a, b] R -> Rn continuous

Why do you say continuous? f is differentiable, right?

 Quote: does not hold true. Can anybody think of a simple counterexample that would prove that it is not true? My brain does not seem to be functioning properly

The most famous example involves moving once around a circle. You
arrive at where you started, so f(b) - f(a) = 0. But there need
not be any time where the velocity (the derivative) is 0, right?

You could also look at examples like f(x) = (x^2, x^3) on [0,1].
William Elliot
science forum Guru

Joined: 24 Mar 2005
Posts: 1906

Posted: Sat May 28, 2005 7:05 pm    Post subject: Re: Mean Value Theorem in Several Dimensions

On Sat, 28 May 2005, jezza wrote:

 Quote: Sorry that was supposed to be subtraction signs instead of question marks. i dont know why it did not work Because there is a vast variety of news browsers that render special

character in various weird ways. For best results to be read by all
news browsers use ascii text only, ie just the characters on the
keyboard.
science forum Guru

Joined: 24 Mar 2005
Posts: 790

Posted: Sat May 28, 2005 11:41 pm    Post subject: Re: Mean Value Theorem in Several Dimensions

In article
<19135042.1117328335198.JavaMail.jakarta@nitrogen.mathforum.org>,
jezza <big__jezz@hotmail.com> wrote:

 Quote: hey, Thanks for the help. With the circle example, would the mean value theorem by undefined? Since f(b) - f(a) divided by b-a where b-a= 0 undefined?

How can a theorem be undefined? You're getting confused. You go
around a circle in, let's say, one minute, the velocity never

 Quote: For the other example, how could you proved that there is no point that it works for? f(x) = (x^2, x^3) on [0,1]. Can you say: let b=a, a=0 so f(1)-f(0) = (1,1)-(0,0)/(1-0) = (1,1) f'(x)=(2x,3x^2) so if f'(c)=(1,1) the first component 1 = 2c => c=1/2 only solution. But for the second component if c=1/2 f'(c) second component is 3/4 does not equal 1.

I gave you the example. You have to decide if it's right.
jezza
science forum beginner

Joined: 28 May 2005
Posts: 8

Posted: Sun May 29, 2005 12:58 am    Post subject: Re: Mean Value Theorem in Several Dimensions

hey,

Thanks for the help. With the circle example, would the mean value theorem by undefined? Since f(b) - f(a) divided by b-a where b-a= 0 undefined?
For the other example, how could you proved that there is no point that it works for? f(x) = (x^2, x^3) on [0,1]. Can you say:
let b=a, a=0
so f(1)-f(0) = (1,1)-(0,0)/(1-0)
= (1,1)
f'(x)=(2x,3x^2)

so if f'(c)=(1,1) the first component 1 = 2c => c=1/2 only solution. But for the second component if c=1/2 f'(c) second component is 3/4 does not equal 1.

Hope this is right. I will be very happy.
jezza
science forum beginner

Joined: 28 May 2005
Posts: 8

 Posted: Sun May 29, 2005 8:33 am    Post subject: Re: Mean Value Theorem in Several Dimensions My proof is there and I agree that it is a counterexample. Is my proof correct? THanks
David C. Ullrich
science forum Guru

Joined: 28 Apr 2005
Posts: 2250

Posted: Sun May 29, 2005 9:21 am    Post subject: Re: Mean Value Theorem in Several Dimensions

On Sat, 28 May 2005 20:58:24 EDT, jezza <big__jezz@hotmail.com> wrote:

 Quote: hey, Thanks for the help. With the circle example, would the mean value theorem by undefined? Since f(b) - f(a) divided by b-a where b-a= 0 undefined?

No, in that example b-a > 0. But f(b) - f(a) = 0, while there's no
point x with f'(x) = 0.

 Quote: For the other example, how could you proved that there is no point that it works for? f(x) = (x^2, x^3) on [0,1]. Can you say: let b=a, a=0 so f(1)-f(0) = (1,1)-(0,0)/(1-0) = (1,1) f'(x)=(2x,3x^2) so if f'(c)=(1,1) the first component 1 = 2c => c=1/2 only solution. But for the second component if c=1/2 f'(c) second component is 3/4 does not equal 1. Hope this is right. I will be very happy.

************************

David C. Ullrich
jezza
science forum beginner

Joined: 28 May 2005
Posts: 8

 Posted: Sun May 29, 2005 11:50 am    Post subject: Re: Mean Value Theorem in Several Dimensions If you walk around in a circle, arent 'b' and 'a' the same point? If not, what are the points such that f(a) - f(b) = 0? But more importantly, is my working out for the function f(x) = (x^2, x^3) on [0,1] correct?
science forum Guru

Joined: 24 Mar 2005
Posts: 790

Posted: Mon May 30, 2005 4:01 am    Post subject: Re: Mean Value Theorem in Several Dimensions

In article
<11321050.1117367463443.JavaMail.jakarta@nitrogen.mathforum.org>,
jezza <big__jezz@hotmail.com> wrote:

 Quote: If you walk around in a circle, arent 'b' and 'a' the same point?

No; f(a) and f(b) are the same, a and b are not.

 Quote: If not, what are the points such that f(a) - f(b) = 0?

This is very simple; review your trig.

 Quote: But more importantly, is my working out for the function f(x) = (x^2, x^3) on [0,1] correct?

I don't know, I didn't read it. I gave you the example; it's up
to you to do the rest.

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