A Functional Equation from British Math Olympiads 2009

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #FunctionalEquations

PLAYLISTS 🎵 :

[paveljay]

After y=0 and discarding f(x)=0 we also could approach it by doing y=-x which yields f(x)f(-x) = 1-x^2 (1), then y=x gives f(x)^2=f(2x)+x^2 (2) and x->2x, y->-x yields f(2x)f(-x)=f(x)-2x^2 (3). Substituting the values of f(-x) and f(2x) from (1), (2) into (3) it follows the quadratic equation on f(x),
x^2f(x)^2 -2x^2f(x) +(x^2-x^4) = 0
Which after applying the quadratic formula finishes that f(x)=1+-x.

[FF-mswq]

Nice solution! And you stated the type of f very clearly at the beginning.

When you want to say that f is the zero function you can also use the terminology “f is *identically* zero” to make it clear.

Nice work.

[rssl]

Fire 🔥
I saw a similar problem from the 1993 polish Olympiad in another video
f(x+y)-f(x-y)=f(x) f(y)
It had the same approach of using 0 except you didn’t have to do the x=z-1 thing kinda but still eerily similar

[victorabaderamos]

Suggestion for next video: floor(log(x))=ceil(sin(x))

[wojtekk]

That's the kind of youtubers we need.
Not crappy vloggers, beauty products reviewers, etc. ;)

[angelmendez-rivera]

f(0)·f(t) = f(t) ==> f(t) = 0 for all t or f(0) = 1. However, f(t) = 0 for all t does not satisfy the equation. Therefore f(0) = 1.

The equation implies f(t)·f(–t) = f(0) – t^2 = 1 – t^2 = (1 + t)·(1 – t). Therefore, f(1)·f(–1) = 0. Therefore, f(1) = 0 or f(–1) = 0.

f(1)·f(t) = f(t + 1) + t = 0, or f(–1)·f(t) = f(t – 1) – t = 0. Therefore, f(t + 1) = –t or f(t – 1) = t. Therefore, f(t) = –(t – 1) = 1 – t, or f(t) = t + 1 = 1 + t. These do indeed satisfy the functional equation. Therefore, f(t) = 1 + (–1)^n·t for n in Z.

[prashantgujar]

Graph would be two perpendicular lines @45 deg to each X and Y axes, interacting at (0, 1)?

[lilac]

Solved it in my head just from the thumbnail.

First approach, plug in y = 0 to get f(0) = 1.

Then, plug y = -x,
And you get f(x)f(-x) = f(0) - x^2
--> f(x)f(-x) = (1+x)(1-x)
We can see that now we have two functions in the same form as left hand side, so either f(x) = 1+ x or f(-x) = 1+x, which gives us the solutions f(x) = 1+x or 1-x

Quite proud of this one, since I'm only 16 and we started functional equations only a week ago.

[jonorgames]

A similar, more advanced, problem, is:
Show that f(f(x))=x^2+x+1 has no solutions.

[Aditya_]

Ok so first find some value for which it is 0 and try to switch it up in a way the product of the two functions things cancel out and you get something simpler

[yufeng]

Another way to do it. After f(0)=1, replace y with dx. f(dx)=f(0)+f'(0)dx. plug it int and the whole thing become a very simple differential equation: f'=x/(f-1) solved it get the same result

[davidseed]

If you set f(x)=1+ax
Then the original equation becomes
(1+ax)(1+ay)=(1+ax+ay)+xy
a²xy=xy
a=±1
Not a proof, just reassuring.
And if you set f(x)=1+ax+bx^c
Then i think b musy be zero.
Would that be a proof?

[hussinmreih_]

hello, dear Sybermath.
I've been struggling with a question I found online and I'm there's no answer to check my solutions

k = (2+i sqrt (12))^n
find all real solutions where n is an integer that satisfy K to be a real, positive expression

[walterufsc]

f(x)f(y) = f(x+y) + xy

Setting y = 0, we have: f(x) f(0) = f(x) ==> f(0) = 1. [f(x) cannot be identically zero]

Setting y = -x: f(x) f(-x) = f(0) - x² = 1 - x²

Then f(x) f(-x) = (1 + x)(1 - x) = (1 - x)(1 + x)

Therefore f(x) = 1 + x or f(x) = 1 - x

[krstev]

Why was there such an easy problem in BMO

[applealvin]

Well only linear function satisfy the question, so we can try
f(x)=ax+b

[emanuellandeholm]

I found f(0) by setting x to 0 and then I found f(x) by setting y = -x.

[sorooshnazem]

After f(0)=1, you just need to put y:=-x