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Forum index » Science and Technology » Math » Undergraduate
Mean Value Theorem in Several Dimensions
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jezza
science forum beginner


Joined: 28 May 2005
Posts: 8

PostPosted: Sat May 28, 2005 3:51 pm    Post subject: Mean Value Theorem in Several Dimensions Reply with quote
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
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jezza
science forum beginner


Joined: 28 May 2005
Posts: 8

PostPosted: Sat May 28, 2005 4:01 pm    Post subject: Re: Mean Value Theorem in Several Dimensions Reply with quote
Sorry

that was supposed to be subtraction signs instead of question marks. i dont know why it did not work
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The World Wide Wade
science forum Guru


Joined: 24 Mar 2005
Posts: 790

PostPosted: Sat May 28, 2005 6:19 pm    Post subject: Re: Mean Value Theorem in Several Dimensions Reply with quote
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 Smile

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].
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William Elliot
science forum Guru


Joined: 24 Mar 2005
Posts: 1906

PostPosted: Sat May 28, 2005 7:05 pm    Post subject: Re: Mean Value Theorem in Several Dimensions Reply with quote
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.
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The World Wide Wade
science forum Guru


Joined: 24 Mar 2005
Posts: 790

PostPosted: Sat May 28, 2005 11:41 pm    Post subject: Re: Mean Value Theorem in Several Dimensions Reply with quote
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
equaling 0. Think about it.

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.
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jezza
science forum beginner


Joined: 28 May 2005
Posts: 8

PostPosted: Sun May 29, 2005 12:58 am    Post subject: Re: Mean Value Theorem in Several Dimensions Reply with 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?
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.
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jezza
science forum beginner


Joined: 28 May 2005
Posts: 8

PostPosted: Sun May 29, 2005 8:33 am    Post subject: Re: Mean Value Theorem in Several Dimensions Reply with quote
My proof is there and I agree that it is a counterexample. Is my proof correct?

THanks
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David C. Ullrich
science forum Guru


Joined: 28 Apr 2005
Posts: 2250

PostPosted: Sun May 29, 2005 9:21 am    Post subject: Re: Mean Value Theorem in Several Dimensions Reply with quote
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
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jezza
science forum beginner


Joined: 28 May 2005
Posts: 8

PostPosted: Sun May 29, 2005 11:50 am    Post subject: Re: Mean Value Theorem in Several Dimensions Reply with quote
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?
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The World Wide Wade
science forum Guru


Joined: 24 Mar 2005
Posts: 790

PostPosted: Mon May 30, 2005 4:01 am    Post subject: Re: Mean Value Theorem in Several Dimensions Reply with quote
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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Google
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