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 $x\in[0,\pi]$. 

$\blacksquare$

18 comments:

  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.

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  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.

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  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.

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  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.

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

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  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!

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  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.

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  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.

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  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. :)

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  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.

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  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

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  12. @1over137
    Hi there!

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

    Paul

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  13. Periodic over 360 but what about 180?

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  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).

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  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.

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    1. Thanks! Those are some neat theorems!

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  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.

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    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

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