How do you measure the length from a to b?

]]>Kinks? What Kinks?

Correction, No kinks. Only, I haven't seen the IVT for only a 2D graph and not a 3D surface. Following up on it I was led to the Borsuk - Ulam theorem which may or may not apply.

]]>I have attempted to illustrate that ... please have a look at the revised page (at bottom).

Looks good, I suppose an attempt at proof would be just a little too much? It is rather surprising how elementary the proof is.

]]>True, that example does have some kinks in it, but does that mean there was a greater than 50% chance that it was an error from the start.

Kinks? What Kinks?

Ricky wrote:

The IVT says a lot more than "you will be at the same height". If you walk in a circle (this works for any parametrized continuous path, but circle is easiest to see), let your height at time t be described by h(t). Here we are walking around the circle in 1 unit time of time. Then there will exist at least one point which will be exactly the same height as the opposite side of the circle.

I have attempted to illustrate that ... please have a look at the revised page (at bottom).

]]>MathsisFun wrote:

But the "round trip" example was my own invention and is thus likely to be wrong.

We could argue for awhile about why it is "likely to be wrong" just because it is your own invention. True, that example does have some kinks in it, but does that mean there was a greater than 50% chance that it was an error from the start.

I thought the table problem was much more interesting, just because I have never seen even the handiest people level a table like that.

]]>Maybe "if you travel in a circle, then *at some point* the height opposite you on the circle will be exactly the same as where you are" ... or some such wording.

Or "when you have a circle with a wavy circumference, somewhere there will be two points opposite one another that are exactly the same height".

A rather-difficult 3d sketch could complement that.

]]>We can define a function:

g(t) = h(t) - h(t+1/2)

If g(0) = 0, then we're done: our starting point is at exactly the same height as the opposite side. Otherwise, g(0) is either positive or negative. But g(1/2) has exactly the opposite sign of g(0):

g(0) = h(0) - h(1/2)

g(1/2) = h(1/2) - h(0)

(Remember that h(0) = h(1). They are exactly the same point!)

Now the IVT says that g(t) must be zero at some point in between 0 and 1/2. And that's it.

Of course, probably too difficult to put in the page...

]]>[align=center]http://www.mathsisfun.com/graph/functio … =-2&ymax=2[/align]

In which case the IVT applies but Rolles theorem does not!

ie. You start and finish at the same height, so one part of your route must be flat.]]>

But the "round trip" example was my own invention and is thus likely to be wrong.

]]>MathsisFun wrote:

Anyone care to tear it to pieces?

Rather different tone than normal. Almost a challenge. You are proud of this one. You should be.

Very interesting page. You cleared up a point or two for me. I didn't know that about the table problem. Here is a link to the proof.

http://arxiv.org/abs/math/0511490

I guess I can only say the biggest compliment Is that I saved the page for my notebook.

]]>I have attempted to simplify it, and also included some examples which I may have got wrong, so I would appreciate a critical review before I let a wider audience know about the page.

Thanks!

]]>