a fun functional equation

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[Hyakurin_]

I think one also should check a function of the form f=0 for x≠0 and f(0)≠0 after the first observation, and is easy to check it's a solution

[High_Priest_Jonko]

I like how you walk through the problem very naturally like somebody would if they were trying to solve it themselves. Doesn't feel like you're just copying a solution on the board

[henrik]

1:50 but now you are missing the solution that f(x) = 0 for all x != 0 and f(0) = a for any real a. Or more generally: For any solution you can change the value of f(0) to be arbitrary.

[carstenmeyer]

1:22 Maybe a (very) small mistake -- there is an entire set of trivial solutions dependent on a parameter *c ∈ ℝ:*

*f: ℝ -> ℝ, f(x) := c * 𝛿(x) // 𝛿(x) = 1 if x = 0 and zero everywhere else*

You can also prove any solution with (at least) one zero *f(y_0) = 0* must be a trivial solution of the form above.

[alexey_burkov]

I think that the assumption at the start when x=x and y=0, is not right a bit.
For example the function f(x) that is 0 everywhere except x=0, and f(0)=c< ∞, then everything is satisfied.

[benjaminvatovez]

Thank you for the solution.
1) xf(x)=0 thus f(x)=0 for all x/=0, f(0) is free even after we check in the general equation.
2) If we put the rational solution in the genereal equation, we find out that a must equal 1 thus f(x)=1/(x²-x+1) which is positive.

[CraigNull]

Since each side of the equation reverses sign when x and y are swapped, you can prove xf(x) is continuous everywhere (and f continuous at nonzero domain). Then you're set up for a differential equation for xf(x) by dividing both sides by x-y and taking limits. This is another way to get to a functional form for f

[adrianarnaezsanchez]

Functional equations videos are awesome Michael

[AntonioLasoGonzalez]

1:49 it is not an ok assumption since f(x)=a (for any a) if x=0 or 0 if x is not equal to 0 satisfies, and in this case you can't divide by xf(x) (since it is equal to 0 for all x)

[jimallysonnevado]

The function f(0)= a and f(x) = 0 if x is not equal to 0 also works. I think when it was in the part where xf(x) f(0) = xf(x), he should be onsidering two cases. Since f is not identically 0 it must be non-zero somewhere. This splits into two cases, it is non-zero only on x=0 or there is another point x not equal to 0 such that f(x) is not 0. In the first case we get a family of solutions. In the second case pick x element of R/{0} and then we get the solutions he got from above.

[manucitomx]

That was fun and educational.
Thank you, professor.

[arvindsrinivasan]

After seeing the trivial solution of f(x)=0, if we look at when this isn’t true, you can solve this naturally without plugging in special values.

We start with


Dividing both sides by xyf(x)f(y) you get


Multiplying the right side out gives


We can rearrange this and separate all xs from ys and get


This is true for all x and y, so
1/(xf(x))-x-1/x=a for some constant a.

Therefore, we get the desired solution
f(x)=1/(x^2+ax+1)

The last part to make it work for all reals is the same as in the video.

[khoozu]

I think the accurate function which satisfy the eqn f(1/x)=x^2f(x) should be f(x)=1/(Ax^2+Bx+A)
Then apply f(0)=1, u will get A=1,
And thus f(x)

[ronaldjorgensen]

clear presentation thank you for your time and persistence

[cicik]

also f(1) = 1/A+B+C; where you know that C = 1 so you can colclude that coefficient at square is A = 1 and the function has form f(x) = 1/x²+bx+1 where b in (-2..2), however are you sure that we can not restrict b further?

[kikilolo]

1:27 I think you missed all the solutions where f(x)=0 for x not 0 and f(0)=cte.

[laprankster]

When substituted (x, y)=(x, 1) I set f(1) equal to a constant c and then substituted the function into the original equation to find what f(1) is. Turns out it didn’t matter what f(1) was.

[christianaustin]

1:50 why is assuming x != 0 an okay assumption?

[wesleydeng]

You can just start from 4:15 and reach the same conclusion. (noting that f(1)=0 is a special case). Also, should be xf(x)=0, not f(x)=0 as mentioned by other comments.

[BJ-sqsi]

Granted that the problem is posed with the constraint mapping from f R -> R for the sake of the video,
in practice would applying a constraint that excludes such limits of -+infinity at a point ever be useful for generalizing a theorem or maybe some practical application?