Solving the Functional Equation f(x-y)=f(x)f(y)

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[SyberMath]

To clarify things, I should've mentioned that f is continuous and we do not want any piecewise functions!

[David_W_Wood]

Possibly easier: f(x) = f(2x-x) = f(2x).f(x)

So either f(x) = 0 or f(2x) = 1

[armacham]

3:52 I disagree with something. You said: "It implies two things, either f(x) = 1 or f(x) = -1"

Another possibility exists:

Suppose you have a function where f(1) = 1 and f(2) = -1

That function would still (at least at f(1) and f(2)) satisfy the condition of 1 = f(x)^2

You have to do more work to establish that f(x) can never be equal to -1

[michaelz]

You can't have other solutions... if f(0) = 1, f(x)^2 = 1 leads to f(x) = 1 or - 1 for all x. Letting y = x/2 in this situation gives f(x/2) = f(x)f(x/2), so one can divide by f(x/2) to get f(x) = 1 for all x. So f(x) = 1 and f(x) = 0 are the only solutions.

[thierryvieville]

A playful and intersting problem, thanks : it is so nice to have this every day share.

Yes as pointed out by others we just need a bit more :
Ok, from the video, either f(x) = 0 for all x OR f(0) = 1 and f(x) = +- 1 but not necessarily constantly, but you are right.

Because, with x = 0 we obtain f(y) = f(-y) thus f is even, and thanks to this, writing x - y = z and y = -x thus z = x/2 = -y/2 we obtain
f(z) = f(x) f(y) = f(x) f(-x) = f(x)^2 >= 0
so that yes all values being positive we are left with f(x) = 1 constantly.

This is less elegant than Michael Z proposal by an alternative derivation.

[swenji]

I'd say :
For all x, f(x-0)=f(x)f(0)=f(0-x).
Then for all x, f(x)=f(x/2+x/2)=f(x/2)f(-x/2)=f(x/2)f(x/2)=f(x/2-x/2)=f(0), so f is constant without any computation on f.
Therefore, f est constant equal to c and the functional equation rewrites as c=c^2, thus the solutions are the constants 0 and 1

[marklarsen]

Thank you for the videos. I enjoy them.

I solved it from the initial screen shot, before watching.
While watching I found that you added the requirement that the domain is all real numbers.
With this domain, I agree: there are the two solutions stated.
But with other domains (that have closure under subtraction) there are more solutions.
If we choose the domain to be the integers, then there is the additional function of f(x) = {1 when x is even; -1 when x is odd}.
If we choose the domain to be the integers times some number b (where b is a non-zero real number), then the additional function is f(b*x) = {1 when x is even; -1 when x is odd}.
modulus
It works because the domain is not closed under "divide by 2".

I assert that there is a third function that solves the initial equation, if and only if the domain is closed under addition, closed under subtraction, but not close under "divide by 2". And the function is f(x) = {1, if x/2 is in the domain; -1 if x/2 is not in the domain}

[kprdqvg]

f=0 is a trivial solution. Let us look at cases where f is not zero. Note that partial_xf(x-y) = - partial_yf(x-y). Therefore, ( df(x)/dx ) f(y) =- f(x) (df(y)/dy). Thus (1/f(x))df(x)/dx = - (1/f(y))df(y)/dy = k, a constant, as the LHS is a function of x alone and the RHS a function of y alone. But x and y are really dummy variables here. Therefore the derivative of f is equal to the negative of itself, which implies that k = 0. This means, since f is not equal to 0, df(x)/dx = 0. Hence, f(x) = c, a constant.

[shoryaprakash]

f(x) ^2=1 does not implies that
f(x) =1 Or f(x) =-1 only
There is a case like a belong to set S and b belonging to R-S such as f(a) =1 and f(b) =-1 which we need to show that it doesnot work here. You need to add this one to complete your proof or try easier proof by taking y=x/2.

[pizzarickk]

Here's how I thought about 1^x being a solution;
I started with a simpler question, let's find f(x) such that;
f(y+x)= f(y)*f(x)
obviously, functions that satisfy this are the exponential functions, so
f(x) = a^x
so now, we can see that the fucntion f(x) that satisfies ; f(y-x)=f(y)*f(x) can be a^x but a^(-x) = a^(x), because then we'll have; f(y-x)=f(y)*f(-x) =f(y)*f(x). so now our goal is to find a function f(x) such that ;
f(x) = a^x AND f(x) = f(-x)
we can easily guess that the function that satisfies this is 1^x, because it's in the form of a^x and it has the same value for opposite inputs x and -x, which is equal to 1.

[brandoncastro]

This one is an equivalent solution: f(x)=[f(y)]^(2-x/y) if we take that f(y) is a constant. This solution satisfies the first ecuation! :D

[davidseed]

if x=y=0 then f(0)=f(0)^2 so f(0)=0 or 1.
put x=y, f(0)=f(x)^2 so f(x)=sqrt( 0 or 1) = 0 or 1

[robertgerbicz]

Put y=0:
f(x)=f(x)*f(0)
f(x)*(f(0)-1)=0 Hence f==0 or f(0)=1. So we can now assume that f(0)=1

Put x=y:
1=f(0)=f(x)^2 so f(x)=+-1 for all x, but we could still choose + or - sign differently.

Put x=0:
f(-y)=f(0)*f(y)=f(y) so f is an even function.

Put x/2 and y=-x/2:


so f(x)=1 for every x value. Hence we have only two solutions: f==0 or f==1.

[dmtri]

I think (without assuming the continouity of f) the solution is: f (x) =0 for every x, or Abs(f(x))=1 for every x....

[rodro]

Even easier, because you implied that f(0)=1, f(x)≠-1 because then 1=-1

[phuctranhuu]

f(x+y) = f(x).f(y)
We take logarith at both side.
Lnf(x+y) = lnf(x) + lnf(y)
Let's call g(x) = lnf(x) then we get
g(x+y) = g(x) + g(y)
And at this equation, we already know g(x) = a.x
=> lnf(x) = ax
=> f(x) = e^(ax) = (e^a)^x = b^x
(b=e^a)

🤤

[matematika]

It is posible to have f(x)=1 for some x and f(x)=-1 for other values of x. You should check that case.

[JohnRandomness]

f(x - 0) = f(x) f(0), so either f(x) = 0 or f(0) = 1.
Now take the latter case: 1 = f(x - x) = f(x) f(x) means that f(x) = ±1. If f(x) is continuous, then f(x) must be 1. If f(x) can be discontinuous, f(x) can be -1, 0, or 1 anywhere except that f(0) is 1 if f(x) isn't zero everywhere.

Therefore, requiring continuity, f(x) = 0 or 1, the constant.

[AsfuNtendstoINFINITY]

Nice one, Sybermath : problem and the solution:)

[Vladimir_Pavlov]

An amazing task! A challenge unique in its complexity and originality!