a nice functional equation

[@elephantdinosaur2284]

If f(x) = ax, then f( x^3 ) = a x^3.

[@monzurrahman8307]

Yeah, he should have set a = (f(1))⅓, which you can do as x³ is bijective over the reals, and then f(x³) = a³x³ and everything works from there

[@bjornfeuerbacher5514]

@@monzurrahman8307 That's not necessary, the check starting at 10:00 also works fine for a = f(1), if done properly.

[@Brollyy349]

The check at the end is incorrect - "a" term should be outside the cubing brackets on the LHS, you should pull it out first as a shared factor, then apply sum of cubes formula and distribute it in the second bracket to get the conclusion.

[@jardozouille1677]

The check should be checked :)

[@LouisEmery]

The second term on the RHS (f(x^2)+f(y^2) - f(xy)) was suggestive right off the bat. Looked like the factor of x^3+y^3 to begin with. Functional equations are found in math contest not in math courses (that I took). They are kind of fun since they are mostly procedure with some looking ahead.

[@bwcbiz]

functional equation techniques can be a useful way to look at differential equations in some cases.

[@fahrenheit2101]

Yes, often the form is a giveaway, but doesn't offer any insight to the method of proof, just gives a guess solution.

[@spiderjerusalem4009]

Just by looking at the form, it reminisces one of the factorization of x³+y³ whence, by letting f(x)=xg(x) for some function g, it suffices to prove that g is a constant function Upon arriving at f(x³)=xf(x²) (implying g is an even function), move the terms of the original eq yf(x²)+xf(y²)=(x+y)f(xy) substituting f(x)=xg(x), then subsequently divide both sides by xy xg(x²)+yg(y²)=(x+y)g(xy) (but now i'm frankly confused whether or not it is allowed to substitute values for x,y such that xy=0. The eq is true if x=y=0, but case in which precisely one of them is 0 is uncertain) subs y=0, xg(x²)=xg(0) g(x²)=g(0) g(x)=g(0), for x≥0 since g is an even function g(-x)=g(x)=g(0), for x≥0 => g(x)=g(0) for any x g≡1

[@goodplacetostop2973]

11:23

[@danielbranscombe6662]

I used f(x)=x^(1/3)f(x^(2/3)) if you keep applying it n times you end up with f(x)=x^(1/3+2/9+4/27+....+2^n/3^(n+1))f(x^(2^n/3^n)) taking the limit as n goes to infinity we end up with f(x)=xf(1) thus if f(1)=c f(x)=cx and we can check that this works for any value of c

[@aadfg0]

Can't assume f continuous at x=1.

[@jacksonsmith2955]

@@aadfg0 i think he just did

[@danielbranscombe6662]

@@aadfg0 where do I assume continuity? I establish the recursive identity f(x)=x^(1/3)f(x^(2/3)) the same way as in the video if you apply this identity to itself N times you get the closed form I showed above taking the limit results in f(x)=xf(1). I don't believe I need continuity for the limit. There are two parts to the limit. The exponent of the external x is a geometric series, so known limiting value of 1. Then there is the exponent of x inside the function, which tends to 0 thus giving x^0 is 1.

[@aadfg0]

a_n := x^(2^n/3^n) -> 1 I completely agree with. But in order to say that a_n -> 1 implies f(a_n)-> f(1), you must have f is continuous.

[@danielbranscombe6662]

@@aadfg0 ok, I see it now. Thank you for pointing this out.

[@julienbernu1661]

I solved that in a completely different way: - start like you did to establish, f(x^3) = x f(x^2), and thus f is odd. - then plug y -> -y in the main equation and divide by x-y on both sides, using f odd and taking the limit as y approaches x, find that f is differentiable on |R* with d/dx f(x^3) = 3 f(x^2) - now differentiate f(x^3) = x f(x^2) to also find d/dx f(x^3) = f(x^2) + 2x^2 f '(x^2) - equating the 2 expressions for d/dx f(x^3), find that 2 f(x^2) = 2 x^2 f '(x^2), or f(x) = x f '(x) on |R*+ - above differential equation is easily solved as dx/x = df/f => ln(x) + C = ln(f) => f(x) = C x on |R*+, and with f odd, f(x) = C x on |R.

[@whiteboar3232]

Would have it been wrong to say that, since f(x^2) is even, then x*f(x^2) is odd?

[@gaufqwi]

Are there any functional equation problems where the solution turns out to be something more interesting than a constant or linear function?

[@Didymus888]

Sure, what kind a solution is there to f(x+p) = f(x)? What about f(xy) = f(x) + f(y)? Or f(x + y) = f(x)f(y)? Answer: sine/cosine, logarithm, and exponential

[@zyklos229]

Wonder, if you take a step back, that not only f, but also x and y can be considered searchable objects. So for any fixated f one could find the x and y that fulfil the equation. Or for given x, y find f. So "the" solution is not only "f regardless of x and y", but is sort of extendable to a triple (f, set for x, set for y) -> true or false 🤔

[@ShaunakDesaiPiano]

A way to prove that it’s an odd function without first proving f(0) = 0: substitute y = -x.

[@JordHaj]

Then you would need to also prove f is continuous

[@assassin01620]

​​@@JordHaj I don't recall michael ever proving f is continuous, so why would you need to for subbing in y = -x?

[@dawkinsfan660]

Do you use Hagoromo chalk?

[@adamwho9801]

By inspection, the original problem is the formula for the sum of cubes. This immediately leads to f(x) = x

[@charleyhoward4594]

how does he get the result at 4:11 ??? x=x^(1/3) doesn't yield -f(x) when u subs, in f(x^3)=xf(x^2)

[@williamwarren5234]

Subbing x=x^(1/3): f(x^(1/3)) = x^(1/3)f(x^(2/3)) = -f(-x) With the last equality coming from subbing x = -x^(1/3) like he did on the board.

[@mirokajevskivelevski8895]

Can you solve this problem one from the 2023 MMO (macedonian math olympiad)? Find f (R to R) s.t.x*f(x+y)+y*f(y-x)=f(x²+y²)

[@yuseifudo6075]

You did this problem before right?

[@yeech]

Do we need to prove this is the only solution?

[@bjornfeuerbacher5514]

Since it was shown that if the functional equation is true, the form f(x) = ax follows automatically, there can be no other solutions.

[@fahrenheit2101]

Yes, that's always what's being asked, and in this case was shown. You plug in values, and things reduce to very nice forms.

[@kishanpaswan9615]

Everytime you upload any video made me realise you are really genius and I really wait everyday specially for your video

[@JordHaj]

Such a good, methodical procedure, and such a big flop at the end. f(x^3) = ax^3, not (ax)^3 so the only values of a that satisfy the equation are 0 and ±1. Edit: a similar flop on my part resulted in my methodical procedure being incorrect, as I also had (ax + ay) as the first factor.

[@monzurrahman8307]

If you set a³ = f(1), then the solution is correct. This can be done as x³ is bijective over the real numbers

[@bjornfeuerbacher5514]

The check at the end works for every value of a, if done properly. Left hand side: a x³ + a y³ Right hand side: (x + y) ( ax² + ay² - axy ) That's the same, regardless of what a actually is.

[@bjornfeuerbacher5514]

@@monzurrahman8307 Not necessary, the check can also easily be done with a = f(1).

[@JordHaj]

​@@bjornfeuerbacher5514 yes, you're correct. I made the same mistake as Michael did in the end, where I had (ax + ay) as the first factor on the left side instead of (x + y). I will edit my comment to reflect that.

[@tzonic8655]

Where are my shqipez at 🇦🇱🇦🇱

[@charleyhoward4594]

totally confusing; not math as I know it