A Functional Equation from Putnam and Beyond

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#ChallengingMathProblems #FunctionalEquations

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[ishaanlakhera]

It was so easy, tho so difficult to think about that idea

[マツ-js]

I'm Japanese. this video is very helpful because I can improve my English listening skills as well as learn math.

[fredthelegend]

I really liked this question. It was the perfect sort of difficulty for me, as I was able to solve it, but at the same time, it was still difficult enough that I had to think about it for a while in order to work out how to answer it, as I am still very new to functional equations. I referred back to other functional equation videos in order to work this one out. Great video anyway!

[manojsurya]

Adding the 2 equations was a brilliant idea, I did not think that it wud simplify to x, great work

[paulortega]

Nice problem. Same kind of idea. Let A=(x-3)/(x+1) and B=(x+3)/(1-x). If you repeatably substitute x in A with A you cycle through A ---> B, B ---> x, x ---> A, A---> B, ... So, doing the same repeated substitution for this problem you get the 3 equations f(A)+f(B) = X, f(B)+f(X)=A, and f(X)+f(A)=B which quickly reduces to f(x)=(A+B-x)/2=( 8x/(1-x^2) - x)/2

[diogenissiganos]

Woah, that's a really interesting one!

[satyapalsingh]

Wow ! My heart is filled with joy .Marvellous . Your method of solving is unique .God bless you !

[pvector]

Here's a slight reformulation of the solution. If you set g(x) = (x-3) / (x +1), then the original equation can be rewritten as

f(g(x)) + f(g(g(x)) = x

By substituting x = g(x) and x = g(g(x)), and using the curious fact that g(g(g(x))) = x we also have

f(g(g(x)) + f(x) = g(x)
f(x) + f(g(x)) = g(g(x))

So we have three linear equations with three unknowns: f(x), f(g(x)), and f(g(g(x)). For example by adding the second and the third, and subtracting the first we get:

2f(x) = g(x) + g(g(x)) - x

EDIT: I just realized that several other posters proposed the same idea before. Well, I'm keeping the post in case someone finds this mini-writeup useful.

[احسانملکی-فث]

This video made my day. Every good math video makes my day. Thank you for bringing this nice problem. Good luck.

[ED-iqmv]

Finally i found technique for equations like this…..thanks i have been looking for this for along time…..this is really really useful for someone like me

[snejpu]

Wow. That's the first functional equation I have ever solved and it's correct. What can I say, I did exactly what you do in the video. I even used the same letters for a substitution. : )

[esteger]

I've only started watching recently, and this is my favorite so far. Before viewing, I was able to solve the problem after much trial and error, with a similar but less efficient method. In other words, it took more algebra. Anyway, I've learned a lot about solving functional equations. Thanks!

[yoav]

So nice problem! I solved it by my self and i am happy i got the same answer:)

[mahajankeshav]

This channel is really very interesting and informative.

[juniorcandelachillcce]

Bonito ejercicio, me hizo recordar a mi primer año en la universidad. Saludos desde Perú.

[babitamishra]

You really make a good recap and recreational stuff, really enjoy solving and learning.

[jasonleelawlight]

I spent a good amount of time figuring out how to plug in the right values to get enough equations to cancel out the noisy parts. It’s essentially the same as what you did.

[miloradtomic]

One of wonderful Functional Equation. Congratulations on being methodical.
Thanks a lot.

[christopherrice]

SyberMath you never fail to impress me and astound me with the math problem videos you come up with. Keep up the awesome content👏👏👏👏!!!!

[txikitofandango]

Holy crap I found the answer by trial and error. I reasoned through and evaluated f at 0 and plus or minus 1/3, 2, 3, and 5. Can prove that f is odd. Plotted the points. Noted that there's probably vertical asymptotes at plus and minus 1, so put an (x-1)^2 in the denominator. Fiddled around with cubic equations in the numerator. Presto. Okay, now to figure out how to actually do it.