Functional Equations: A Beauty in Themselves

by math_pi_rate, Sep 17, 2018, 2:56 PM

Even though the name of the blog suggests only geo (and I support it by heart as usual :P), but I'll post some of the FE's which I found out to be really interesting (Another reason for posting something apart from geo is to keep the interests of guys like kayak going #seemyshoutbox)

I'll start with IMO 2017 Problem 2, which is the most beautiful FE I have ever seen (Before doing this problem, remember that looks can be deceiving). First of all notice that the equation is symmetric. Also on seeing an FE on reals, one obvious thing is to plug in $x=y=0$, although I personally prefer plugging just $y=0$, as that helps build a condition on $f(x)$. This same trick can be seen in IMO 2010 Problem 1 also, but we'll return to that later. Anyway, so on plugging $y=0$, we get $f(f(x) \cdot f(0))+f(x) = f(0)$, which is not as helpful as we thought. So to make things easier we make cases on $f(0)$, i.e. when it is zero and when it is not (making these cases is one of the most important part of this question, and I remember using this trick in some other problem, although I can't remember which). Anyway, this trick of making cases is very important. (For eg. See here). Now $f(0)=0$ trivially gives $f(x)=0$. So now we focus on $f(0) \neq 0$. After some playing around with $f$, I couldn't get much, and so I decided to make another function $g$, defining $g(x)=f(x)-x$. This is another major trick (you can try comparing this trick with phantom points in geometry, oh sorry, forgot about no geo :D). It helps using this trick when you have guessed the functions (which is mostly the case in all FE's). Like, by now, I had guessed that the solutions are most probably $f(x)=x-1$ and $f(x)=1-x$. Then I noted that the second solution follows from the first, as if $f$ was a solution, then $-f$ will also be a solution. This brought me to a nice revelation, i.e. $f(0)=1$ can be assumed and $f(0)=-1$ will immediately give the other solution (as I had by now proved that $f(1)=0$ and $f(0)= \pm 1$). Now some manipulations with $g$ brought me to the necessary solution. Finally while posting the FE, I completely removed the mention of any such function $g$, thus making the solution in $f$ purely.

Before proceeding to the next FE, I'd like to mention another trick which I couldn't mention before. This is basically used to show that if $f(c)=k$, then $c=p$, where $k,c,p$ are constants. It has two main parts: First of all show that there exists some $x \in \mathbb{R}$ such that $f(x)=k$. Then assume that $f(c)=k$, and try to arrive at some sort of contradiction to prove that $c=p$. Remember that the first step is really important, although it is oft forgotten. You can see some examples which use this trick here and here.

Now, Let's go on to APMO 2016 Problem 5. This is an FE on positive reals, meaning that we no longer have zeroes at our service. I consider it to be one of the most difficult FE's to do on an exam, because of how time consuming it can be (If I remember correctly, it took me 4 hours to solve this one :oops:). We start with some obvious things, like proving injectivity and surjectivity, which any person having some basic knowledge in FE's should be able to do. Although there is a small room for mistake. One can easily prove that $f$ is surjective $\forall x> \frac{1}{2}$. But from this concluding that $f$ is surjective for all $x$ is a big mistake, something which I did while solving the problem, only to rectify it later. After this the problem actually flows on, slowing moving us to $f$ being involutive, multiplicative and finally additive, proving and using $f(1)=1$ and $f(2)=2$ in the way. In the end, all one needs to do is use the famous identity that if $f$ is multiplicative as well as additive, then either $f \equiv 0$ or $f(x)=x$. Plugging this into the original equation, we get the required solution.

Let's move on to the last FE of this topic: IMO 2008 Problem 4. This problem has a very short solution, but if not taken care of then one can easily make the huge mistake of ignoring the so-called Pointwise Value Trap (To learn what they are see this handout by v_Enhance). After getting that either $f(a)=a$ or $f(a)=\frac{1}{a} \text{ } \forall a \in \mathbb{R^+}$, which is quite easy to discover, one needs to check on the aforementioned Trap. I know of two methods to check: First is assuming that it is possible for the two values of $f(x)$ to hold for distinct numbers simultaneously, and then arriving at a contradiction. The second is to assume that if one of the solution holds for some number $t$, then it holds for all $x$ which lie in the domain of $f$. I personally prefer the first one, but the choice can depend from person to person and question to question. To understand it better, see the discussion in the beginning here. Also see my solution to IMOSL 2007 A4. It gives a nice insight on FE's in general.

Before concluding with this topic, I'll just post some tips.

1. Try making your solution presentable. Using lemmas while writing is a beautiful way to do this. See my solution to APMO 2016 Problem 5 for an example. (This can be used in geometry problems also)

2. After solving an FE, always remember to plug in the solution and check it, as this always carries some marks.

3. If an FE has more than one non-symmetric solutions, then always check for the Trap mentioned above.

4. Before starting an FE, always try to discover the solutions to it first. This helps a lot later on in how to move forward with the FE, and motivates what to plug into the FE. Basically check whether $f$ can be constant or not, and then try putting $f(x)=ax+b$ or $f(x)=ax^2+bx+c$ and try solving for $a,b,c$. Remember that mostly all Olympiad FE's have usually a max of 2 or 3 solutions, and which generally are of the above given forms only.

In the end, remember that FE's have room for a lot of creativity. So don't think too much before trying something out of the way, as it always helps while solving them.
This post has been edited 6 times. Last edited by math_pi_rate, Oct 16, 2018, 7:27 AM

Geo Geo everywhere, nor a point to see.

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  • duh i found this masterpiece after the owner went inactive noooooooooooo :sadge: :(

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  • Hi everyone! Sorry but I doubt if I'll be reviving this anytime soon :oops:

    I might post something if I find anything interesting, but they'll mostly be short posts, unlike the previous ones!

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  • @below be patient. He must be busy with some other work. :)

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  • Try this: Prove that a two colored cow has an odd number of Agis Phesis

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