Saturday, July 7, 2007

What's Your Function? Topic: Algebra. Level: AMC/AIME.

Problem: Given two positive reals $ \alpha $ and $ \beta $, show that there is a continuous function $ f $ that satisfies $ f(x) = f(x+\alpha)+f(x+\beta) $.

Solution: There are several special cases that are interesting to look at before we make a guess as to what type of function $ f $ will be. First we consider the case $ \alpha = \beta = 1 $. This immediately gives

$ f(x) = 2f(x+1) $.

It should not be too difficult to guess that $ f(x) = 2^{-x} $ is a solution to this, as well as any constant multiple of it. Now try $ \alpha = -1 $ and $ \beta = -2 $, resulting in

$ f(x) = f(x-1)+f(x-2) $.

Looks a lot like Fibonacci, right? In fact, one possible function is just $ f(x) = \phi^x $. That's pretty convenient.

Notice how both of these illuminating examples are exponential functions, which leads us to guess that our function will be exponential as well. So, following this track, we set

$ f(x) = a^x $

so we simply need to solve

$ a^x = a^{x+\alpha}+a^{x+\beta} $

$ a^x = a^x\left(a^{\alpha}+a^{\beta}\right) $.

Unfortunately, the equation $ a^{\alpha}+a^{\beta} = 1 $ does not always have a solution (take $ \alpha = 1 $ and $ \beta = -1 $ for example). But that's ok and I'll worry about it some other time. In any case we have found a function for whenever the equation $ a^{\alpha}+a^{\beta} = 1 $ has a solution in the reals.

6 comments:

  1. AHH JEFFREY!!!
    I"M FAILING ADVANCED CALC!!!
    we have our final tomorrow and i know basically nothing!!!
    ok. i'm just surfing the web cuz i don' t want to study

    but i'll go to my emo corner and study now ...

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  2. Oh, Xuan.

    It wasn't that bad... for me >_>

    I was reminded of BATH because Andreea just now found out about the question written about her and dating Qiaochu. I wanted to look over the test again.

    I wonder if you ever look at this...

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  3. My god.

    I swear I already left a comment here. I must be going crazy.

    I came back a while ago to look at BATH tests, and now I'm looking up Mu Alpha Theta stuff.

    Advanced Calculus is, like... easy, coming from someone who actually reads the textbook >_>

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  4. Oh.

    Uhh... I didn't see that before. Now I just look stupid.

    Sorry about that.

    ReplyDelete
  5. Great blog, why don't you continue it?

    ReplyDelete