Solving a Suggested Functional Equation

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#ChallengingMathProblems #FunctionalEquations

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[SyberMath]

Dear friends, here is the video that I'm talking about, a similar problem:

[GregWeidman]

I really appreciate when you go back and check the results.

[SeekingTheLoveThatGodMeans]

It looked from the start that f(x) = x + 1/3 but thank you for the more rigorous proof/demonstration.

[PunmasterSTP]

Functional? More like “fun for all”! Thanks for sharing all these neat problems!

[jungtaewon]

Substituting x, y, z which x+y+z=-1, f(x+f(y+f(z)))=0. So, there exists a real number t which f(t)=0. Substituting y=z=t, f(x)=x+2t+1. And t is constant. So, f(x) is linear, and we can know that f(x)=x-t because f(t)=0. f(x)=x+1/3(because -t=2t+1, t=-1/3.) is the only function that satisfies the equation.

[jonathanward]

This was a great problem. Out of the countless times I have watched Syber videos this is the first time I had trouble understanding. I had to watch one part of the video many many times to get it but I eventually did. Just a little hazy. Overall great job though.

[synesthesia]

This one was extremely easy, you can just look at it and see that it's x+1/3

[michaelz]

As a function of x, it's of the form f(x + a) = x + b + 1 for some a and b independent of x. So replacing x by x - a we get f(x) = x + c for c = -a + b +1. Plugging this form into the equation leads to c = 1/3. That's all you need to do here.

[salimbravo]

Great contribution. Thanks for everything. 👍👍👍

[Wurfenkopf]

We don't need the hypothesis that f is continuous.
Just substitute y = -f(z) and x = -f(0) and we're done!

[ahmadmazbouh]

You know that the problem is cool when it looks complex but while solving it it suddenly turns to be as simple as this

[wesleydeng]

Let x=y=0, z=-1 => f(f(f(-1))) = 0. Then let y=z=f(f(-1)), one gets f(x) = x + 2*f(f(-1)) + 1 = x + c. Put it back in the original equation, c=1/3. Being continuous is not needed.

[manojsurya]

That's a cool problem syber, I had to rewatch the video coz I couldn't follow up the first time, since ur steps were different n amazin

[אליסושנסקי]

Really like functional equations, and I hope you continue to upload them. My method is similar to yours, but I would still like to present it:
After we get f(x)=x-c+1
I would like to call 1-c=n, while n is a constant.
That means that f(x)=x+n
Then substitute in the original equation x=0 and y=0.
We get f(f(f(z)))=z+1
Substitute x instead of z and get f(f(f(x)))=x+1
But we already know f(x)=x+n and from there we can conclude that f(f(x))=x+2n, and if we repeat the process we get f(f(f(x)))=x+3n
Then we understand x+3n =x+1
And then n=⅓
In the end we can say that f(x)=x+n= x+⅓

[brinzanalexandru]

Another way to prove that f is linear is to set x=y=0 so we get f(f(f(z)))=z+1 now replace in the initial equation z with f(z) and x=y=0 and using the first equation we got obtain f(z+1)-f(z)=1 which is a constant so f must be a linear function.

[suntzu]

4:02
In x+y+f(z)-2c+2, put
f(z)=z+1-c.

Therefore, it will become:
(x+y+z+3-3c)=(x+y+z+1)
Solving this, c=2/3.

[skylardeslypere]

Without using f is continuous, you can set y=-1/3 and z = -1/3 + f(-1/3)
Then, plugging it in and simplifying*, you get f(x) = x - 2/3 + f(-1/3).
Then you can "guess" that f(x) = x+A for some constant A. Plugging it in the equation
f(x+f(x+f(x))) = 3x+1 (what you get when x=y=z), you can solve for A which gives you A = 1/3

* f(-1/3 + f(-1/3 + f(-1/3))) = 0 using the definition

[suntzu]

Putting x=0 and y=0, we get
f(f(f(z)))=(z+1)

Is there a general way to solve this?

[KeithKessler]

If we set x=y=z=0 we get f(f(f(0)))=1. It's obvious that adding 1/3 to the argument of the function satisfies the given equation.

[krishnanadityan]

Put x=-f(0) and y=-f(z) gives
f(z)=z+1-2f(0) from which we get by putting z=0, the following expression
f(0)=1-2f(0). This gives f(0)=1/3
Thus, we get f(z)=z-1/3 or
f(x)=x-1/3 !!!