Multiplicative Functional Equation
Вставка
- Опубліковано 7 лис 2024
- f(xy) = f(x)f(y)
In this video, I solve the following neat question: Find all continuous functions f such that f(xy) = f(x) f(y). One solution would be f(x) = x^2, but can you think of other ones? I solve this with a clever transformation, by reducing this equation to one that is more familiar. I also discuss the case where x and y are 0 and/or negative, and even mention the non-continuous case. Enjoy!
Other functional equations:
f(x+y) = f(x) + f(y): • fx+y = fx + fy (Cauchy's functional equation)
f(x+y) = f(x)f(y): • f(x+y) = f(x)f(y)
f(f(f(x))) = x: • Press fff to pay respects
f'(x) = f(f(x)): • A new hard differentia...
Subscribe to my channel: / drpeyam
Check out my Teespring merch: teespring.com/...
How about det( )?
Whoa, fancy! 😍
Now that's class
omg I literally went to the comment section to write "I am dissapointed he didnt get the determinant function here"
I guess you were already ahead of me by 6 months :(
This guy is so pr0 that he says thanks for watching at the beginning of the video
Bon raisonnement
Your videos make math even more interesting. Thank you Dr. Peyam!! 👊🏼
if we did it for functions from ℝ\{0} to itself, we couldve used group theory since its basically finding the endomorphisms of group ℝ\{0}
🤮🤮🤮
@@drpeyam 😂😂😂
Nope, f(x)=1 also works but that isnt a bijection, is it? Also, |x|, |x^3| and many more. So yeah, while Aut(R\0) would give us many solutions, it still wouldn't cover all of them.
What you probably ment was the set of all endomorphisms
@@filippozar8424 oops sorry
The function is 0 at 0 and 1 everywhere else. Has to be f(x)=sgn(x)^2.
I was able to solve this by pretty much doing exactly what you did in the video
Haven't finished the video yet, but sqrt(xy) does not equal sqrt(x)sqrt(y) for x=y=-1.
Edit: I see where Peyam makes a more thorough statement at the end.
Your solutions are so good.
I have a question. If f is differentiable,
f is the following?
c>1=>f(x)=x^c or x^c(x=>0), -x^c(xf(x)=x,
f(x)=0 or 1
Thank you so much sir
Best explaination
This was a very interesting video! I am looking forward to more of your content.
what a wonderful video, it is very helpful for me thanks a lot
Dr Peyam I have a doubt whether x times abs x is odd or even and if it's graph is continuous at 0
I'd like to see a totally discontinuous solution that requires the axiom of choice.
Check out the video on f(x+y) =
f(x) + f(y) there I define one
About f(0), can't we say f(0)=f(0*0*0)=f((0*0)*0)=f(0*0)f(0)=f(0)f(0)f(0)=(f(0))³, then f(0) could be either 1, - 1 or 0? And another question: if f(x) = x^c, and c = ln(f(e)), how can I calculate c if I don't know f? I mean, it's like f(x) = x^(ln(f(e)))
Please do my favourite functional equation f(x+y) = g(x) + h(y). The assumptions on the functions f,g,h are that they are all real valued functions and f is continuous at 0.
Generalisation of Cauchy?
f(x)^2 can be also negative or imaginary
f(sqrt(x)) can be expressed by matrix A with integers, no square roots needed, then A^2 = xI
You should show why it can only be positive. It is not in video.
Show that cyclic group has only three elements, nothing, 1 and x.
Interesting fact is that those three group elements are providing solutions of all 3 characters: trivial, constant, bijective.
Cyclic group of four elements is having linear inverse dual that can be represented by -1 and f(x) = -1 or f(x) = -x doesn't hold the bonding condition of endomorphism. So the cyclic group representing character of solutions contains less than four elements.
Very cool! Thanks!
An amazing explication, thanks guy
Interesting video. One question though. At 10:53, you write it as f(00). What happens if I write it as f(01) or f(02)? Anything multiplied by 0 is 0.
Still f(0) but I don’t think you can conclude anything from that
Really cool Peyam
Maybe unpopular opinion: Discontinuous functions are way cooler
I feel like the space of discontinuous functions are just too too big to understand
Unpopular opinion. I hate those
But, they have much way more practical applications then continuous unfortunately
@@IoT_ ???? W hat?
@@thedoublehelix5661 all real systems are not continuous
I dont understand the f(x)=f(sqrt(x)*sqrt(x)) argument. Couldn't any f(x) function be written that way implying that no f(x) could ever have a domain with negative values, which isn't true.
so is absolute value function still continuous, despite not being differentiable at 0.
Yes
Woah that was pretty cool!
Hi
Thank you.
note +-(-x)^c is equal to +-(x)^c for integer c. otherwise it is complex
Thank
Nice video ! Thank you
Plugging in y=0, f(0)=f(0)f(x) for all x, so f(0)=0 or f(x) is identically 1. Also, f(x)=f(1)f(x), so f(1)=1 or f(x) is identically 0.
@@drpeyam lol rip, i believe you
Yeah sorry, you can’t define f(0) = 1, so f(0) has to be 0. But I think the function f(0) = 1 but f = 0 otherwise works, that’s why I included it
@@drpeyam That'd be a cool example, but if you take x=0 and y=/=0, f(xy)=f(0)=1, but f(x)f(y)=0, because f(y)=0.
Sorry I meant to say f is 0 at 0 and identically 1 otherwise
How about lim if each lim exsist
What if c = 1/2 when x f= sqrt(-x) = isqrt(x).
Very good!
17:22 1-del(x)
i think, that in case for c
you mean u cant define f(0)=1
what is the application case in real life from this math theory ?
Quantum mechanics
Group homomorphisms between the the reals under multiplication onto itself
Interesting vid
Revenge of the c's though...
You missed the case where f(x) is constant zero.
I mentioned the case 0
The function 1/x is continuous because it's defined on R \ {0}.
Here though we're looking at functions on R, and there is no way of defining 1/x at x=0 that is continuous.
@@dmhouse1024 But you can't define 1/x on R. The natural domain of it is R\{0}, you can't define it on R.
@@patryslawfrackowiak6690 what I mean is that there is no function f on R such that f(x) = 1/x for all x!=0 and f is continuous. In other words there is no way of "extending" 1/x to include a value at x=0 while preserving continuity.
@@dmhouse1024 Yes, and we're both right. There's no such function, as you've just said, but the function 1/x itself (without any extending) is continuous on all possible domain.
👍👍👍👍
can you please show the solution to the homework I'm only 16 i can't understand
It's a conditional identity, ain't true eternal, known as Cauchy's formula.
In the math Olympiad you'd not get the full points for the problem as you missed f(x) = 0 in the conclusion section, doctor.
Lmao
It's no one going to comment on that Tina Fey joke?
E
I don't think you can define c using f , it's like a circular reasoning..but still a great video :)
Impredicative? Yes. Circular? Not so much.
You forgot f(x)=0
Pintagili