Higher Dimensional Mean Value theorem

By Brad Rodgers on Sunday, December 02, 2001 - 01:11 am:

Can anyone come up with a proof that for

X(t)

t₂ > t₁ , and X(t) continuous, there exists a c such that

t₁ < = c < = t₂

and X'(c)=X(t₂ )-X(t₁ )

This is simply a generalization of the mean value theorem, and seems relatively intuitional when looked at geometrically, however, when we consider it as a system of equations, it certainly doesn't seem so axiomatic. Any ideas for a proof (or disproof, though I strongly suspect it's true...)?

Thanks,

Brad

By Brad Rodgers on Sunday, December 02, 2001 - 02:55 am:

I've come up with a geometric proof. It's a proof by contradiction, and involves considering the "skewness" of the line between X(t₁ ) and X(t₂ ) and tangent lines of the curve X(t), and considering the distance between the two lines. It's somewhat difficult to explain though, and I'm not sure I'll be able to explain it tonight. I'm still looking for an algebraic proof though.

Brad

By Dan Goodman on Sunday, December 02, 2001 - 03:16 am:

I don't think it's even true. Consider in 2D X(t) defined to start at (0,0) move to (2,0) in a straight line, then loop round underneath the x-axis back to (2,0) pointing towards (1,1) and then continue in a straight line onwards to (1,1). The derivative of this function always lies in the second to fourth quadrants of the plane, whereas (1,1)-(0,0) lies in the first quadrant, so they cannot be equal. You might be able to prove that for some c you have that X'(c) is parallel to X(t₂ )-X(t₁ ), I'll think about it tomorrow. I seem to recall something about a higher dimensional mean value theorem in an analysis course many years ago - any second year Nrich team members remember this from last year?

By Kerwin Hui on Sunday, December 02, 2001 - 06:24 am:

A counterexample: Let

X (t) = (\cos t, \sin t)

. Now we have

X (2 π) = X (0)

, but there are no

t

such that

X' (t) = 0

Kerwin

By Brad Rodgers on Sunday, December 02, 2001 - 05:37 pm:

Okay. Upon looking back at my geometric proof, I realize that it doesn't in fact come close to proving this, (in fact it doesn't really prove anything at all), but it does attempt to prove the statment you made, Dan, that X'(c) is parallel to X(t₁ )-X(t₂ ).

By Brad Rodgers on Sunday, December 02, 2001 - 08:29 pm:

Is there a way to put that into a precise mathematical form? i.e. is there a way to say that given a curve in any number of dimensions, and given any two points on the curve, there will be another point on the curve between the two other points such that the tangent to this point will be parallel to the line connecting the original two points (whew!). That's what I was trying to say above. I would think that it would be that if you take two corresponding x_a and x_b from the vector, then there exists a c such that x_a '(c)/x_b '(c)=(x_a (t₁ )-x_a (t₂ )/(x_b (t₁ )-x_b (t₂ )), but I can't be sure that's right. Is it?

Anyways, the two dimensional case follows very easily:

If (x,y) = (g(t),f(t)), then let

Mean Value

Note that h(t₁ )=h(t₂ )=0. Thus, by Rolle's Theorem there exists a t₁ < c < t₂ such that h'(c)=0. Differentiating h(t), setting equal to zero and rearranging, the result pops out nicely. So f'(c)/g'(c)=(f(t₁ )-f(t₂ ))/(g(t₁ )-g(t₂ )). It think for higher dimensions, a sort of higher dimensional Rolle's theorem would have to be worked out...

Brad

By Michael Doré on Sunday, December 02, 2001 - 09:33 pm:

Yes, that's right; alternatively let $n$ be any vector perpendicular to $X (t_{2}) - X (t_{1})$ and consider $f (t) = n . X (t)$ and apply Rolle's theorem to $f$ in $[t_{1}, t_{2}]$ .

For higher dimensions it's false, e.g. consider

$x = \cos t$

$y = \sin t$

$z = t^{2} (π - t)$

in the region $[0, π]$

The way you stated the result is fine I think - although technically you need to deal separately with cases such as $x_{b} (t_{1}) - x_{b} (t_{2}) = 0$ otherwise your fraction is undefined. I think it's probably simpler just to say that (in my notation) $X (t_{2}) - X (t_{1})$ is parallel to $X' (t)$ for some $t$ in $(t_{1}, t_{2})$ (where parallel is easily defined, e.g. by saying two vectors $x$ , $y$ are parallel iff there exists $α$ , $β$ not both zero with $α x = β y$ ).

By the way, the result I think Dan is referring to is the mean value inequality. For functions $X : R \to R^{n}$ this says that if $X$ is differentiable in some open interval $S$ of $R$ and has $| X' (t) | \leq K$ for all $t$ in $S$ then for any $t_{1}$ , $t_{2}$ in $S$ we have:

$| X (t_{2}) - Z (t_{1}) | \leq K | t_{2} - t_{1} |$

Of course this is intuitively obvious and is easily proved by reducing it to the mean value theorem in one dimension.

By Brad Rodgers on Monday, December 03, 2001 - 01:09 am:

Okay, I see your counterexample. Could we say that there is some sort of mean value theorem for hypersurfaces: Given a surface of m-1 dimensions contained in m dimensions, and given some (m-1) dimension plane intersecting the surface in someway, there exists another tangent (m-1)D plane on the surface in the area bounded by the first plane (I'm not quite sure how to be precise with "bounded", although it's meaning is hopefully clear). Here's an example with a paraboloid

3-D Mean Value

Brad

By Brad Rodgers on Monday, December 03, 2001 - 10:48 pm:

I have an idea that shows promise for a proof for the 3rd dimension, but I can't finish it. I'll go ahead and take it as far as I can though.

It proves that for a function z(x,y) and a plane going through 0 for that function, there is always some other function

First of all, a Lemma:

Let z=z(x,y) and be subject to the following conditions

a)z is continuous (I've never heard a formal definition for 3 dimensional continuity, so I'll use the definition z(x,y+d )=z(x+d ,y)=z(x,y) where the d denotes a small number as normal)

b) there exists a closed curve satisfying 0=z(x,y).

It then follows that some point in (x,y) yields

\partial z / \partial y = \partial z / \partial x = 0

.
First, we know there is a certain section of z set off from the rest, and this section is bounded by 0=f(x,y).

Suppose that we were to first examine this section at x being some constant. This subsection in the yz plane would look like this:

mean value 2-d

Which, by Rolle's theorem, has some point at which

\partial z / \partial y = 0

. Thus, at all

x

, there is some

y

point at which

\partial z / \partial y = 0

. By a similar argument, at all

y

, there is some

x

for which

\partial z / \partial x = 0

. Also note that the points at which

\partial z / \partial x

must fall into one continuous curve as there is only infinitesimal distortion in changing from one

x

x + d

(this reasoning is slightly unrigorous, but a rigorous explanation for this exists, and isn't too hard to derive). We then know that our cross section would look something like this, where the red line represents (x,y) where

\partial z / \partial y = 0

, and the blue where

\partial z / \partial x = 0

(and of course, the green the curve 0=z(x,y)).

cross-section

What we want to prove is that the red and blue lines cross, but I can't seem to prove this. Why, for example, should the diagram not lool like this

impossibility?

Anyways, if someone can establish this last step, then we'd have a sort of three dimensional Rolle's theorem, which could be applied to prove my conjecture in three dimensions in much the same way as Michael or I did above to prove the two dimensional version.

Brad

By Michael Doré on Tuesday, December 04, 2001 - 02:09 pm:

Brad, I don't really understand your conjecture. We have a m-1 D surface in R^m ; a m-1 D plane intersects this surface, and the surface and plane enclose a finite amount of volume, right?. Now your conjecture is that the original surface has a m-1 D tangential plane "in the area bounded by the first plane". What do you mean by this last bit? Do you mean the tangential plane intersects the bounded region? Or am I totally misunderstanding here...

By Brad Rodgers on Tuesday, December 04, 2001 - 08:32 pm:

Suppose we had some surface bounded by a plane, call it plane X. Then my conjecture is that within the region the one plane intersects and bounds, there is some other plane, call it plane Y, such that this plane is both a tangent plane (it intersects the surface in only one point), and such that this plane (plane Y) is parallel to the original plane (plane X). I might try to make a 10 second or so video showing this in mathCAD, as it's very hard to convey with still images.

Brad

By Brad Rodgers on Wednesday, December 05, 2001 - 02:01 am:

I made a 5 second video demonstrating the idea pretty clearly....

[Editor: Kerwin put the video in his website...]

By Kerwin Hui on Wednesday, December 05, 2001 - 06:39 pm:

I just put a copy in my website under other interesting stuff. It should last for some years.

Kerwin

By Michael Doré on Thursday, December 06, 2001 - 01:23 am:

Ah! I think I see now. How about the following, in three dimensions.

Orient the axes so that (without loss of generality) the bounding plane is the plane z = 0. We can think intuitively of the bounding surface as the ground, and our surface "bulges" upwards out of the ground in some finite region of the ground call it R. (I say upwards - assuming the surface doesn't just co-incide with the bounding plane, in which case they'd be nothing to prove, without loss of generality we can say that for some point in (x,y) in R, z(x,y) > 0.)

We want to show that the surface in R has a tangential plane of the form z = c, in other words some point in the bulge has a level tangential plane.

Now it is a standard result in analysis that in a closed and bounded region, a continuous function has a supremum and it attains its supremum in this region. Consider the supremum of z(x,y) in R (intuitively z(x,y) is the height.) Since z is continuous and the region R is closed and bounded (can you see why?), the supremum of the height is attained at a point T in R. So T is at least as high as all other points in R.

What's more T is not on the boundary. Now if $\partial z / \partial x$ is not 0 at R then by perturbing the x co-ordinate we can either increase or decrease the height (depending on the direction in which we perturb the x co-ordinate). This is impossible since the point we are at has is at least as high as every other point in R. So $\partial z / \partial x = 0$ as is the partial derivative w.r.t. y. Therefore, since z is differentiable (as a function from R2 to R) we have that the tangential plane at T is parallel to z = 0.

I'll try and make this clearer tomorrow. Would you like a proof of the analysis result I mentioned?

By Brad Rodgers on Thursday, December 06, 2001 - 02:21 am:

Thanks Dan and Kerwin.

I have a reason why the two curves in my december 3, 10:48 post would have to cross. It's because all y coordinates have to have point where

\partial z / \partial y = 0

, and similar for x's. Thus, either there exists a point where the curves representing

\partial z / \partial y = 0

and

\partial z / \partial x = 0

touch at a point on

0 = z (x, y)

, or the curve for

\partial z / \partial y = 0

is above both above and below the curve

\partial z / \partial x = 0

. That is to say, if

P_{y} (x)

represents the

y

coordinate for a given

x

where

\partial z / \partial y = 0

, and

P_{x} (x)

represents likewise for where

\partial z / \partial x = 0

. There exist given points such that

P_{y} (x) - P_{x} (x) > 0

and where

P_{y} (x) - P_{x} (x) < 0

(I've implicitly assumed these are functions, but the proof can be easily modified if they aren't). Thus,

P_{y} - P_{x} = 0

by the intermediate value theorem. Thus, at some point in the contour there exists a

x

and

y

such that

\partial z / \partial x = \partial z / \partial y = 0

, which is what we wanted to prove. We can then apply this three dimensional variant of Rolle to surfaces in the same way that we applied Rolle's original theorem to curves, thereby giving proof of my conjecture for 3 dimensions. However, I'm not sure this technique will generalize, it's rather long, and I think holes could probably be poked in it (which I think could consequently be repaired...).
Comments? Ideas?

Brad

By Brad Rodgers on Thursday, December 06, 2001 - 10:37 pm:

I'd like the proof of the supremum result, though I think I've proved in my posts (in an unelegant way most likely). Anyways, can it be generalized?

By Michael Doré on Friday, February 15, 2002 - 11:00 pm:

This is yet another of the threads which I missed the followups to. I know it was a long time ago, but the supremum result is a very important theorem in analysis, so here is a proof. Firstly a useful lemma, often called the Bolzano-Weierstrass theorem.

If u₁ ,u₂ ,... is a bounded sequence of real numbers, then u_i has a convergent subsequence.

In other words we can find strictly increasing r₁ ,r₂ ,... such that u_r 1 ,u_r 2,... -> u for some real u.

We're given the sequence is bounded, so let's say (without loss of generality) the sequence if confined to the interval [0,1]. If the sequence u₁ ,u₂ ,... only has finitely many different values then we're done (can you see why?). So we assume the sequence hits infinitely many points in [0,1]. Well it must also hit infinitely many points in either [0,0.5] or [0.5,1]. (If it only hits finitely many points in [0,0.5] and [0.5,1] then it can only hit finitely many points in [0,1], a contradiction.) So we now have an interval (call it I₁ ) of size 1/2 (either [0,0.5] or [0.5,1]) in which the sequence hits infinitely many points. Now keep going. Partition I₁ into two equally-sized intervals. It is clear that the sequence must hit infinitely many points one of these subintervals of I₁ . Call such a subinterval I₂ . Now continue the process forever...

So I₁ , I₂ , I₃ , ... is a nested sequence of intervals. The length of I_n is 1/2ⁿ . We claim that there exists a number which is in I₁ , I₂ , ... How do we show this? Well the left endpoints of I₁ , I₂ , ... are increasing and they're bounded above by 1 so they converge to a number which we'll call u. This number u is certainly at least as big as all the left endpoints of each interval, and certainly no bigger than any of the right endpoints of each interval. Therefore it lies in I_n for all n.

We now claim that u is limit of a subsequence of u₁ ,u₂ ,... But we know that for each n there exists some member of the sequence u_i which is within 1/2ⁿ of u. In fact there are infinitely many of them - because u is in I_n which is of length 1/2ⁿ and has infinitely many members of the sequence u_i . So it should be clear now how to construct a subsequence of u_i tending to u.

Now the next thing we need to do is extend this up to sequences in any closed, bounded subset of Rⁿ . A closed subset S of Rⁿ is one in which for all sequences in S which converges to some number L in Rⁿ , L is necessarily in S. For example on the real line, the closed interval [0,1] is (as the name suggests) closed - it's not hard to check this. On the other hand (0,1] is not closed - because the sequence 1,1/2,1/4,1/8,... lies entirely inside (0,1] and converges to a real number 0, and yet 0 is not in (0,1].

If u₁ ,u₂ ,... is a sequence in a closed, bounded subset S of Rⁿ then if we label the x co-ordinate of u_i as u_i _x then u₁ _x , u₂ _x ,... is a bounded sequence so a subsequence of it converges to a real number. By the same argument, you can find a subsequence of the subsequence whose y co-ordinate converges to a real number. Now find a subsequence of this whose z co-ordinate converges. Keep going, until eventually you find a subsequence such that all the co-ordinates converge. If x,y,z,... are the limits of each co-ordinate of this subsequence then it is easy to check that the subsequence converges to the point (x,y,z,...) Furthermore as the subsequence is entirely within S, so is the limit.

Therefore any sequence in a closed bounded subset of Rⁿ has a convergent subsequence. How does this help? Well we want to show that if f is a bounded continuous function in a closed, bounded subset of Rⁿ then f attains its sup. Well if f is bounded in S, a closed and bounded subset of Rⁿ , then f is bounded. Otherwise there exists a sequence of points r₁ ,r₂ ,... in S such that f(r_i )-> infinity . But r_i has a convergent subsequence r_s i which tends to r in S. But f(r_s i) must tend to f(r) since f is continuous and this is a contradiction. Hence f is bounded in S.

So and by an axiom of the reals, the function f (restricted to S) therefore has a sup, call it s. Therefore we can find points x₁ ,x₂ ,... in S such that f(x_i ) -> s. We know x_i has a subsequence which converges to a number t, and since f is continuous we get f(t) = s. Hence f attains its sup.