a nice functional equation
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If f(x) = ax, then f( x^3 ) = a x^3.
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
@@monzurrahman8307 That's not necessary, the check starting at 10:00 also works fine for a = f(1), if done properly.
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.
The check should be checked :)
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.
functional equation techniques can be a useful way to look at differential equations in some cases.
Yes, often the form is a giveaway, but doesn't offer any insight to the method of proof, just gives a guess solution.
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
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.
11:23
Are there any functional equation problems where the solution turns out to be something more interesting than a constant or linear function?
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
Would have it been wrong to say that, since f(x^2) is even, then x*f(x^2) is odd?
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
Can't assume f continuous at x=1.
@@aadfg0 i think he just did
@@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.
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.
@@aadfg0 ok, I see it now. Thank you for pointing this out.
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 🤔
Everytime you upload any video made me realise you are really genius and I really wait everyday specially for your video
A way to prove that it’s an odd function without first proving f(0) = 0: substitute y = -x.
Then you would need to also prove f is continuous
@@hiimgood I don't recall michael ever proving f is continuous, so why would you need to for subbing in y = -x?
Do you use Hagoromo chalk?
By inspection, the original problem is the formula for the sum of cubes. This immediately leads to f(x) = x
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)
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)
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.
You did this problem before right?
Do we need to prove this is the only solution?
Since it was shown that if the functional equation is true, the form f(x) = ax follows automatically, there can be no other solutions.
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.
totally confusing; not math as I know it
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.
If you set a³ = f(1), then the solution is correct. This can be done as x³ is bijective over the real numbers
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.
@@monzurrahman8307 Not necessary, the check can also easily be done with a = f(1).
@@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.
Where are my shqipez at 🇦🇱🇦🇱