A Nice Functional Equation from India | Math Olympiads

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[coreyyanofsky]

without watching
set x = y = 0
f(0) = f(0)²
∴ f(0) = 0 or f(0) = 1
set y = 0, x free
f(x) = f(x)f(0)
∴ f(x) = 0 or f(0) = 1 but f(x) = 0 doesn't satisfy the original equation
set y = -x, x free
f(x)f(-x) = (1 + x)(1 - x)
∴ f(x) = (1 + x) or f(x) = (1 - x) and both satisfy the original equation

[rayl.y.]

Hey I found a vid where u solved this question a year ago. It's a vid titled British math Olympiads 2009.

[MathOrient]

I love your solution. Keep posting videos about functional equations :)

[TedHopp]

Nice. However, to be completely rigorous, we should go back and check that each solution actually satisfies the original functional equation.

[thbb]

for the first part f(0)=1, you don't need to explore the branch f(x)=0, if you simply start with x=0, y#0, you immediately get f(0)=1 as the only possibility, saving 1 minute.

[MrGeorge]

Solved it some days ago but my proof for f(1) either equal to zero or to two was more complicated. Yours is way easier. 😄

[chaosredefined]

Set y = 0
f(x) = f(x)f(0)
This is true if f(x)=0 or f(0) = 1.

If f(x) = 0, then we end up with xy = 0, which is not true for any situation where x=/=0 and y =/= 0. So, we have f(0) = 1.

Let y = -x
1 - x^2 = f(x)f(-x)

Note that, if we consider x=1, we get f(1)f(-1) = 0. Therefore, either f(1) = 0 or f(-1) = 0. So, there is at least one value k such that f(k) = 0.

Consider x = t - k, and y = k (such that f(k) = 0)
f(t) + (t-k)k = f(t-k)f(k)
f(t) + kt - k^2 = 0
f(t) = k^2 - tk

f(x+y) + xy = f(x)f(y)
(k^2 - kx - ky) + xy = (k^2 - kx)(k^2 - ky)
k^2 - kx - ky + xy = k^4 - k^3 x - k^3 y + k^2 xy
Put in x=y=0, we get
k^2 = k^4
This means that k = 0, k=1 or k=-1. But we already know that f(k) = 0, and f(0) = 1, so k =/= 0. So, k = 1 or -1.
Put in k=1
1 - x - y + xy = 1 - x - y + xy
That works

Put in k=-1
1 + x + y + xy = 1 + x + y + xy
That works.

So, f(x) = 1 - x or f(x) = 1 + x.

[stvcia]

I searched the solution as a linear form f(x)=ax+b. Аfter substitution into equation and comparison of both sides, got a²=1 (a=±1), b=1.

[Mewhenthewhenthe-xj]

This is the same video as the British functional equation 2009 video

[rickdesper]

You're not quite done. You've concluded that f(x) = x + 1 and f(x) = -x + 1 are the only possibilities that remain possible after looking at what's implied by using y = 0 and y = 1. But there remains a final step of showing that these definitions actually work in the original equation for all values of x. (They do, but you've only shown that no other solution is possible, not that these definitions are possible.)

[cmilkau]

f'(x + y) + y = f'(x) f(y)
f"(x + y) + 1 = f'(x) f'(y)
1 = a f'(x) - f"(x) where a = f'(0)

case a≠0:
particular solution: f’(x) = 1/a
homogeneous solution: f'(x) - 1/a = e^(ax + b)
f(x) = (1/a)(be^(ax) + x + c)

now by original equation, comparing coefficients of ye^(ax):
0 = be^(ax)y
rhs can only be constant when b = 0.
comparing coefficients of xy:
xy = xy/a², so a² = 1
comparing coefficient of x:
x/a = (x/a)(c/a)
so c/a = 1.

Indeed f(x) = ±x + 1 are both solutions.

case a=0:
f"(x) = -1
f'(x) = -x + p
f(x) = -x²/2 + px + q

0 = x²y²/4 by original equation, comparing coefficients of x²y². again rhs can't be constant, contradiction.

[violet_broregarde]

Wow I checked and yeah

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

if f(x) = x+1:
f(x+y)=x+y+1
x+y+1+xy=(x+1)(y+1)

if f(x) = 1-x:
f(x+y) = 1-x-y
1-x-y+xy = (1-x)(1-y)

[shacharh]

f(x) = f(x+0) + x*0 = f(x)*f(0)
therefore f(0) = 1
1 - x^2 = f(0) - x^2 = f(x +(-x)) +x*(-x) = f(x) * f(-x)
so f(x) * f(-1) = 1 - x^2 = (1-x) * (1+x)

if we assume that f is a polynomial then the only option is f(x) = 1 + x

[georgesbv]

Bad idea to replace both, x and y with 0.
f(x)=f(x)*f(0) gives a single solution for f(0)

[alesnecas]

What are you trying to find, i.e., what is the question?

[noctnightcr]

Sub y=0
f(x) = f(x)f(0)
so f(x)=0 for all real x or f(0)=1.

We consider the f(0)=1 case:
Sub (x, y)=(1, -1)
0=f(1)f(-1)
so f(1)=0 or f(-1)=0

Case 1: f(1)=0
Sub y=1
f(x+1)+x=0
so f(x)=1-x for all x. Checking gets
which works

Case 2: f(-1)=0
Sub y=-1
f(x-1)-x=0
so f(x)=x+1 for all x. Checking gets
which works

Thus the solutions are f(x)=1-x for all x or f(x)=x+1 for all x.

[diegobotto]

Pretty good video but I think you have an error in the first answer it should be -x-1 not -x+1

[mcwulf]

Subst y = 0 and we get f(x) = f(x)f(0) so f(0) = 1
Subst y = -x and we get f(0) + x.(-x) = 1 - x^2 = f(x)f(-x)
(1+x)(1-x) = f(x)f(-x)
So f(x) = (1+x) or (1-x)

[Dae-Ying-Kim]

* * So
the two only real number solution that
y
{ can be f(x)-1 } ROOT I. *
and
y
{ can be -f(x)+1 } ROOT II. *
and 2 of them = 2 or them = = = together of them, contain all real number cases of [ f(x+y)+xy ] = [ f(x)+f(y) ] = = the original question ! *
big thanks ! ! 004 : 041 010 : 041 ( Wedn ) 17th May 0002023 * *