A functional equation that didn't quite make the IMO.

We present a solution to a problem involving a functional equation from the 1985 International Mathematics Olympiad long list.

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[Utesfan]

So I started with the case y=0. This gives f(0)=1 or f(x)=0. If f(x)=0 we are done. Otherwise, considering x=y we now get 2f(x)^2 - f(2x) = 1. But the limit of the RHS must go to 0 as x goes to infinity. This contradiction excludes the case f(0)=1, leaving only f(x)=0.

[DarGViD]

Can't you just let x = y + a, and let y to inf? Then you immediately get the answer.

[איתןגרינזייד]

After you found out that f(0)=0 you could have substituted in the given equation y=0 and get 2f(X)=2f(X)*f(0)=0=>f(X)=0

[aydin]

Without that limit condition, f(x) = cos(ax) and f(x) = cosh(ax) work for all real a.

[kaloan]

After 2:20 you can just plug in y=0, so 2*f(x)=2*f(x)*f(0), and if f(x) is nonzero, then f(0) is 1, contradiction.

[blazedinfernape]

If the second constraint was not there then cosx would have been a solution

[martiniquevodka]

Thx to u i again like maths.We are told by professor in school that maths will be interesting the next year.First year of university still boring but having daily problems solving/ theory videos is cool

[parameshwarhazra]

Oh my god Michael penn you surprised me again with another awesome video . Nice . Keep it up. With love from India

[gagaoolala]

Much more straighforward solution: let a = x + y, b = x - y. Then f(a)+f(b) = 2f((a+b)/2)f((a-b)/2). Tend b to infinity to get f(a) = 0. QED.

[utsav]

Who disliked this quality video has no heart.

[friedrich]

f(0)=0. Let y=0, then f(x)+f(x)=2f(x)f(0)
2f(x)=2*0.
f(x)=0

[selimvirtanen]

I did it like this: Let y=0 to get f(x)=f(x)f(0), which gives us f=0 (constant) or f(0)=1. Assuming the latter let y=x to get f(2x)=2f(x)^2 -1. By the limit, there is some m>0 that for all x>m we have |f(x)|<1/2. However, now for such x we have f(2x)< -1/2, which is a contradiction.

[sugarfrosted]

6:14 I jumped to "it's odd and even, it's zero"

[richardsmith]

prove that function is odd by obtaining 2 equations, the first one as it is and for the 2nd equation- interchange x with y, ud obtain f(x-y) = f(y-x), then substitute in the original equation x = 0 and y =0 individually, and ud obtain f(x) = 0 for all x

[mathunt]

You had f(0)=0, so then just set y=0 in the original equation to see that 2f(x)=0 which implies f(x)=0 and we're done.

[praharmitra]

Another (I think, simpler) way of proceeding is after 2:20, you set $x=a$, $y=0$ which gives $2f(a) = 2f(a) f(0) = 0$. Thus, $f(a) = 0$.

[luckywssbi]

The funny thing about case 2, it would actually imply that f(x) is both an even and an odd function so you can reach the contradiction sooner.

[renpnal]

First thing I tried was the y = x substitution and after that the y = 0 substitution. Most of the time went to me trying to figure out what I missed, because I didn't think it could be that easy.

[HAL-ojjb]

I guess that the problem only made it to the longlist was because having only f(x)=0 as a solution was a little bit unsatisfying

[ach]

I think the limit is the fastest way to do this: take x=x+c, y=c-> f(2c+x) + f(x) = 2f(x+c)f(c), then you take the limit of c to infinity->f(inf) + f(x) = 2*f(inf)*f(inf)=0->f(x)=0 for all x.