## 17 Jun 2011

### Proof - Equal temperatures for opposing points on a circle

If you would like a bit of intuition before we delve into the proof, see my last post.

Ready to go? Great.

Let the function $T(x)$ represent the temperature at some position $x$. Around a circle, $T(x)$ is periodic about $2\pi$ radians when $x$ lies on the circle's circumference. Thus, $T(x)=T(x+2\pi)$. Let us define a new function $D(x)=T(x)-T(x+\pi)$. Note that $D(x)$ is continuous as it is a difference of continuous functions $T(x)$ and $T(x+\pi)$.

Let:

$D(0)=L$ and $D(\pi)=-D(0)=-L$.

• If $L=-L=0,$ $T(x)-T(x+\pi)=0$ and we are done.
• If $L\neq0,$ then one of $L$ or $-L$ must be above zero and the other below zero.

Since $D(x)$ is a continuous function, we can apply the Intermediate Value Theorem to it. By the Intermediate Value Theorem, the function $D(x)=0$ for some $x\in[0,\pi]$.
Therefore, $T(x)-T(x+\pi)=0$, and $T(x)=T(x+\pi)$ for some $x\in[0,\pi]$.

$\blacksquare$

1. I like it. You have a note saying D(x) is continuous as its a composition of a continuous function T(X) and T(x+pi). I don't see in your proof stating T(X) is continuous and over what field?
I also don't see how a function being a composition of another continuous function proves its continuous.
Maybe you should have an assumption like this
Let the function T(x) where f: R->R is a continuous function. and do the same for D(X)
You do mention the use of the intermediate value theorem, but this requires your functions are real-valued continuous over said interval.

2. erm I mean I don't see how a function being a composition of another function proves its continuous. Unless you have proven the first one is continuous.

3. Anon #1 -- the function T(x) is the function mapping temperature on the circle form the first post. The function being continuous is one of the conditions of the proof.

4. @Anon #1 & #2
This post accompanies my last post where I explain that this proof applies to physical, continuous phenomenon only (things like temperature and pressure). You guys are quite right in saying that I need to prove the first function (T(x)) is continuous if I were doing this for some non-physical function, but in this case I think its alright to assume continuity.

5. D isn't a composition of functions, it's a difference of them. Composition would be f(g(x)) with functions f, g.

6. @Anonymous

Yes sorry, I was being a bit loose with my words (which is never a good thing to do in math). Fixed now. Thank you!

7. There is still a problem, although this does apply to a continous vector space, i.e every bounded set has a supremum. The actual physical world is discrete, so you can't get a circle, just some particles on a circle. So your right, but not right at the same time for the actual world.

8. @Anonymous

Yes. You're right heh. That's why I stated in the previous post that temperature could be approximated as continuous at macroscopic levels.

9. You have an ill posed problem on your hand.
Temperature is a physical phenomenon, the fact that it can be described as a mathematical expression is one thing. However proving it is periodical is entirely another, that is akin to modeling. It might be somewhat periodical within a certain error margin (say +/-5degrees Celsius) within the period of a year on a few days given conditions on those days were similar. But that's a probability problem.
And there is also the problem of sampling, as temperature fluctuates within a day not just a year on the same spot.
Hope this helps. :)

10. @tudor sabin

Hi Tudor, the temperature function is periodic because I am looking at temperature in a circle. If we sample the temperature at angle 0 deg, it must be the same as the temperature sampled at 360 deg (because we went full circle). Thus, the temperature function we're looking at here must be periodic.

11. Pretty cool! I love these types of problems; at first blush the conclusion seems completely unfounded, but once you think about it...

Anyway, I look forward to following your future posts :D

12. @1over137
Hi there!

Thank you for your kind words :) I look forward to your future blog posts as well.

Paul

13. Periodic over 360 but what about 180?

14. @Anonymous
There is no guarantee that the function is periodic over 180. Note that being periodic over 180 implies that on a circle of your choice, any point you choose will be a point opposite to it with the same temperature (instead of just one, which we've proven in this post).

15. Great Post Paul! Something similar that you might be interested in checking out with a similar flavor is the Brouwer Fixed-Point theorem. Also check out the Borsuk-Ulam theorem which again like your post makes use of antipodal points.

1. Thanks! Those are some neat theorems!

16. I found this as a result of looking for a better explanation of the recent Singing Banana video. Putting this together with that makes a more reasonable combo.

That said, I don't think it matters that T(x) is periodic. All you need to do is demonstrate that D(x) = L at 0 and -L at pi. That's James Grime's approach, but he doesn't make it clear enough that he's talking about the difference function.

1. Hi Paul,

Yes, you are right. We just need that D(x) is <= 0 at some point and >= 0 at some other point. However, we do use the periodicity somewhat of T to demonstrate that D(x) = L at x = 0, and -L at pi (using T(2pi) = T(0)).

Thanks for the comment!

Cheers,
Paul