A Very Powerful Differential Equation 😊
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- Опубликовано: 26 мар 2024
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Tanx. One of my favorite channels.❤
Wow, thank you ❤️
Golden Ratio to the rescue once again!
but how do you prove that only polynomial solutions satisfy the given functional equation?
We can't. There may be more functions out there like this, but noone knows for sure.
This question has been asked on MathOverFlow and answers were given for how to properly solve the equation.
@@RUclips_username_not_foundI don't see how?? Can you share the link? Besides I found a way to short circuit the problem..if tou replace x with y which you know you can because x is presumadefined for all realmvalues you get f inverse y equals dy/dy..meaning f^-1(y) =1 but f^-1(y) equals x so x equals 1..that means x can only be 1so NO SOLUTIONS ..see.what zi mean? There's no flaws in my method so I don't see how power funccan work?
@@leif1075 @leif1075 Hi! I hope we have a good conversation, Mr Leif.
As you know, YT gets funky when it comes to links so that's why I avoided including it. I suppose you can use the info I gave you to find the post on your own.
>> "if tou replace x with y which you know you can because x is presumadefined for all realmvalues you get f inverse y equals dy/dy"
Are you expecting from me to just tell you that you can't simply change variables like that? 🤔
I think you can clearly see that your method is flawed as it gives you an incorect answer and it is easy to check that your result doesn't solve the equation.
>> "so I don't see how power funccan work?"
What does that have to do with whether your method works? These 2 are unrelated, as far as I can see.
@@leif1075 I thought you want to discuss the eqaution but it seems you lost interest. Well, have a nice day.
🎉
9:56 -b/(b+1) = -b/b^2 = -1/b
I thought of the theorem about deriving the inverse function.
To avoid any confusion of the inverse with the reciprocal 1/f, I prefer f_inv_ over f^(-1).
Then
(f_inv_(x))' = 1 / f'(f_inv_(x))
In this case, since f_inv_(x) = f'(x),
(f'(x))' = 1 / f'(f'(x))
f''(x) = 1 / f'(f'(x))
Is that correct? But even if, that doesn't help, does it?
f^(-1)(x) is always the inverse, (1/f)(x) is (f(x))^(-1). Also sin^(-1)(x) = arcsin(x). The confusion comes with sin²(x), but that's a horse of another color!
With g(y) = f^(-1)(y), we can say, that f'(x) = 1 / g'(f(x)). Doesn't really help, that's correct...
impressive , i thought of this problem when i was first introduced to functional equation but i couldn't solve it.
Thank you!
Nice video. What about f(f(x))=f`(x) ? I guess is also of the form x^a
The magic portion of the Golden Ratio.
In fact, it can further ve simplified. If we denote the Golden Ratio with a (phi), then the solution would look like:
f(x) = x ^ (phi) * (phi) ^ ( 1 - (phi) )
OR even further as
f(x) = (phi) * ( x / (phi) ) ^ (phi)
WOW, I"M ❤ING THIS!!!
😍
Indeed, f(x) = ϕ * (x/ϕ)^ϕ
is my favorite form.
f^(-1)(x) = ϕ * (x/ϕ)^(1/ϕ) = f'(x).
This really looks like transforming into some ϕ room, calculating the inverse, and then transforming back!
Want the equation at 7:19 implies that x raised to b-1- 1/b equals a constant ..the ab term bitnthat CANT BE THATS A FLAE..x can't LWAYS equal the same term because the it's not a function see what I mean..so isn't this totally wrong?? Did no one else notice this??
@@leif1075 I suspect, that x ^ (b - 1 - 1/b) can be quite constant, if b = ϕ, because ( ϕ - 1 - 1/ ϕ) = 0. I don't know, if that matters, but I'm pretty sure! BTW: b = ϕ is one of only and exactly two possible values, that allow this statement. The other one is b = - 1/ ϕ.
@@rainerzufall42but that's my point tbis implies xcan o ly equal that particular value..that I plies x can't take any value..which means NO SOLUTIONS..Surely I am not the only one who sees this? See what I mean?
you forgot to show the wolfram alpha result
He gives you f(x) = (G.R.)^(-(1 + √5)/(3 + √5)) times x^(G.R.).
Using Wolfram Alpha on your own: You can reflect this over the y=x line to get the inverse function. Then use Wolfram Alpha to look at f(x) taking the derivative and plot that separate. See if the two plots match (probably in a comparison or visual).
Remember the G.R. = (1 + √5)/2 are constants in the derivative of f(x) he states at the end of the problem. You can define -( 1 + √5)/(3 + √5) by omega rather than in terms of gamma but differentiating with respect to x brings down the golden ratio in terms of gamma and subtracts 1 from the golden ratio power of x (all given in his final boxed off answer at the end of the video leaving us to find the -gamma for b solution also)!
Thank you! Here's the result:
postimg.cc/dDhs67NZ
Here's the result:
postimg.cc/dDhs67NZ
WolframAlpha broadly ignores the properties of ϕ (GoldenRatio). The results are disappointing!
You mean f(x) = ϕ^(-1/ϕ) * x^ϕ ?
Because ϕ = Golden Ratio = (1 + sqrt(5))/2.
ϕ^2 = 1 + ϕ = (3 + sqrt(5))/2. Or ϕ - 1 = 1/ϕ.
Hmm, that's also even more simplified: f(x) = ϕ * (x/ϕ)^ϕ.
This question is to be answered until 09:59 am, for those who have the last digit of their DNI 3: What is the letter that you like the most in A, B, C, I, N , The m?
Math is magic, such a mysterious and odd equation[ f·¹(x)=f'(x)] has one of the most mysterious and beautiful maths constants contaned in the solutions, just wonderful, I hope their is a way to solve this without guessing I'd be very happy to know it.
This question has been asked on MathOverFlow and answers were given for how to properly solve the equation.
@@RUclips_username_not_found I searched but I couldn't find it, if you can please give me the link of the page where the answer is.
@@faruqatumceddal92dr peyam has made video about this and he said you can't solve this without guessing, its seem right to me given this differential equation is so weird i don't think you can solve it without guessing
@@faruqatumceddal92he also don't know if this is the only solution, maybe there is other solution but we don't know yet
@@faruqatumceddal92 @faruqatumceddal92 RUclips shadowbanned my reply because of the link, but here is how you find it: Search the question "ODE with inverse function. Solve f^{-1}(x)=f'(x)." on MathExchange and go to comments.
How to solve it without guessing
First, integrate both sides
U will get f(x) on the first side
The other side will take some time, but it's solvable
How do you know it's solvable. Please show.
@@cosmosapien597
Search that :
Integral f inverse
Very nice and powerful😊💯
Thank you 🙌❤️