9:44 "i don't think there's other solutions to this differential equation " The omnipresent 0 function cries in the corner as it mourns on its acknowledgement
Another way to look at the problem is putting f(x) in the place of x. Then you have f'(f(x))=f(f(f(x))) with the RHS being f(f'(x)). Then f'=f^-1 which is a problem I believe you have already solved on the channel.
I think f’(x)=f^-1(x) => f’(f(x))=f(f’(x))=x, but it may not be the other way around, i.e., f’(f(x))=f(f’(x))x unless we can show f’(x)=f^-1(x) f’(f(x))=f(f’(x)).
@@maxgoldman8903 you are correct, in general fog=gof doesn't mean g=f^-1, e.g. f=x, g=x+1 then gof=fog. This means that f'=f^-1 is A solution to the problem but not the only one. I don't know if f'of=fof'=>f'=f^-1 can be shown.
This is a cool problem! You have proved that if f(x) is of the form alpha x^beta, these, plus the trivial solution alpha=0, are the solutions. But are we sure there aren't any other solutions?
This is very interesting, my first thought was “there must be a one parameter family of solutions, though probably without a closed form, just set f(0) to some value, and newtons method approximates a solution”, but because of the composition that doesn’t work because you need to know the behavior of f near f(0), too (when f(0) = 0 you do, which just gives you f(x) = 0)
Question for you (from someone uneducated in functional analysis) about "polynomial" functional equations, but the "exponent" is actually iteration. e.g. f(f(f(x))) + f(f(x)) + x = 0, and where '+' is not necessary integer addition -- what are these things and are they "interesting"? I guess, another way to phrase it might be: are there ring-like objects where instead of 'multiplication', we have function application?
What a neat diff eqⁿ and all, but could you attempt the integral from 0 to 1 of x^(1/x) dx? A solution has evaded me for a bit too long. Maybe you know a special function I don't, would be cool to see mr integral man.
There will definately be more functions, just not in the polynomial form.. let the degree of f(x) by a then derivation makes the degree a-1 and f(f(x)) always has degree 2a so a=-1. thus the polynomial CAN BE IN THE FORM OF- a/x + b/x^2 + c/x^3........ till whatever term. this can be tough to find but by case bashing (say its of the form a/x, then a/x + b/x2 and so on) we can find functions and if one is lucky there could by a possibility for a parameter which would imply there are finitely such functions... nevertheless when talking about functions there are endless possibilities.. and the possibility of there being a function with maybe factorials and other stuff is def possible
Multi-valued functions are continuous, you just don’t need to look at continuity in weird ways not compatible with Riemann surfaces! 😶 Well at least those multi-valued functions that one can get in complex analysis.
Then what is wrong with the following argument? f (prime) = f of f mens, (d/dx) f(x) = f(f(x)). Put y = f(x). Then, dy / dx = f(y). So, we are free to choose for f(y). For example, if we choose f(y) = 1 / y, then y dy = dx, so y^2 / 2. = x + c, or y = f(x) = sqr rt ( 2x + c1).
If f is invertible, with inverse g, then f' = f○f implies 1/g' = f'○g = f. That problem is equivalent to the fg' = 1 problem (derivative times the inverse equals 1).
Here'sanother way to look at this, by integrating both sides we get-> f(x)= int( f(f(x)) dx ....... (1) so in (1) put x--> f(x) so we get f(f(x))= int (f(f(f(x)))dx ....... (2) now substituting f(f(x)) from (2) in (1) we get f(x)= int(int(f(f(f(x))) dx) dx now the pattern maybe a little tough but continuing the same process we see that f(x)= in(int(int.......( f(f(f(......(x)))))) )dx) dx) dx..... where the number of integrals is n-1 and the number of f's in f(f(f....(x)))) is n so by algebrec anaylitic continuation we can put n=0 so when the number of integral becomes -1 and f(f(...x)))) becomes 0, thus we get f(x)= d/dx(1)= 0 this is one such function then putting n=-1 we get f(x)= d2/d2x ( f-(x)) this may look a little tough buut we get the same function mentioned in the video. Both ideas work its just the method shown in the video is a little incomplete whereas the method used here is basically impossible to solve....
you cannot substitute the x in the integral by f(x). because in reality you have (1): f(x) = int (from somewhere to x) f(f(t)) dt then substituting x->f(x) changes the integration bounds and not the integrand
I'm sure you've actually solved this exactly differential equation 10 months ago... I think the video was called "A very interesting differential equation: derivative equals composition"
@@maths_505 I think it would because many people who are interested in mathematics and are viewers of your channel can communicate and discuss about mathematics and get help for mathematics.
f^(n-1)(x)=f°^n(x) a•ß!/(ß+1-n)!•x^(ß+1-n)=a^sum[k=0,n-1](ß^k)•x^ß^n ß^n=ß+1-n a•ß!/(ß+1-n)!=a^((1-ß^n)/(1-ß)) ß!/(ß+1-n)!=a^(n/(1-ß)) a=(ß!/(ß+1-n)!)^((1-ß)/n) let n=2 ß^2=ß-1 ß^2-ß+1=0 ß=(1+-isqrt(3))/2 ß=e^(+-ipi/3) a=ß^((1-ß)/2) a=e^+-(ipi/6•e^(-+ipi/3)) if n=3, f"(x)=f(f(f(x))) aß(ß-1)x^(ß-2)=a^(1+ß+ß^2)•x^ß^3 ß^3=ß-2 aß(ß-1)=a^((3-ß)/(1-ß)) ß(ß-1)=a^(2/(1-ß)) a=(ß^2-ß)^((1-ß)/2) this is solvable with the cubic formula, but I'd rather not
Awesome content as always. I would like to share that I was precisely thinking about this kind of Differential Equations yesterday after TH-cam recommended me this video: th-cam.com/video/rNUfiQgj6ZI/w-d-xo.html from Michael Penn (pretty similar approach, not sure if that was the source of inspiration of this video as well). However, I was thinking more in the equation f'(x) = f_n(x) (being f_n the nth composited function f). Doing the same trick you can lead to interesting conclusions about the parameters of the power functions that make the job. I haven't reach any marvelous result yet, but probably you could take a look if interested. It was nice to share results (and thoughts) with you today. Always a pleasure to see your videos, definitely they had become a recurrent highlight of the day. Kind regards from Mexico!
this looks as if it gives 0 for all x. but maybe not 😅 without the z in the denominator the sum converges to the integral of log(1-t x)/t for t from 0 to 1. Ok this is a divergent integral, and the 1/z seems to counteract this divergence. Have you plotted the result?
Insane maths gameplay
Cranking derivatives
360 no-scoping integrals
9:44 "i don't think there's other solutions to this differential equation "
The omnipresent 0 function cries in the corner as it mourns on its acknowledgement
alpha = 0
1:18 He rules it out since it's boring.
@@tomkerruish2982 That's the funny part.
i cannot think of a non-trivial differential equation that has only 0 as a solution
@@opensocietyenjoyer non homogenous ODEs, differenial equations with y in the denominator
Cool that the preposition "of" adds an extra "of" when you read fof(x) like fofofex
Another way to look at the problem is putting f(x) in the place of x. Then you have f'(f(x))=f(f(f(x))) with the RHS being f(f'(x)). Then f'=f^-1 which is a problem I believe you have already solved on the channel.
I think f’(x)=f^-1(x) => f’(f(x))=f(f’(x))=x, but it may not be the other way around, i.e., f’(f(x))=f(f’(x))x unless we can show f’(x)=f^-1(x) f’(f(x))=f(f’(x)).
@@maxgoldman8903 you are correct, in general fog=gof doesn't mean g=f^-1, e.g. f=x, g=x+1 then gof=fog. This means that f'=f^-1 is A solution to the problem but not the only one. I don't know if f'of=fof'=>f'=f^-1 can be shown.
How about something like f'(x) = f*f, where * is some sort of convolution. Solve by integral transforms!
Excellent idea
This is a cool problem! You have proved that if f(x) is of the form alpha x^beta, these, plus the trivial solution alpha=0, are the solutions. But are we sure there aren't any other solutions?
@@emanuellandeholm5657 someone commented a couple of cool solutions. If you scroll down you should find them.
@@maths_505: You are "maths" which tells me you’re not from the U.S. but probably from Britain. Yet you don’t have an accent!
@@roberttelarket4934 I'm actually from Pakistan and we call it both maths and math here but I grew up calling it maths.
This is very interesting, my first thought was “there must be a one parameter family of solutions, though probably without a closed form, just set f(0) to some value, and newtons method approximates a solution”, but because of the composition that doesn’t work because you need to know the behavior of f near f(0), too (when f(0) = 0 you do, which just gives you f(x) = 0)
Question for you (from someone uneducated in functional analysis) about "polynomial" functional equations, but the "exponent" is actually iteration. e.g. f(f(f(x))) + f(f(x)) + x = 0, and where '+' is not necessary integer addition -- what are these things and are they "interesting"? I guess, another way to phrase it might be: are there ring-like objects where instead of 'multiplication', we have function application?
Parental advisory
Nice Analysis. Thanks.
What a neat diff eqⁿ and all, but could you attempt the integral from 0 to 1 of x^(1/x) dx? A solution has evaded me for a bit too long. Maybe you know a special function I don't, would be cool to see mr integral man.
Aight
0x is also a valid solution too ,it's the first thing came to my mind it's nice and simple but But it kind of has no fun.😅
There will definately be more functions, just not in the polynomial form.. let the degree of f(x) by a then derivation makes the degree a-1 and f(f(x)) always has degree 2a so a=-1. thus the polynomial CAN BE IN THE FORM OF- a/x + b/x^2 + c/x^3........ till whatever term. this can be tough to find but by case bashing (say its of the form a/x, then a/x + b/x2 and so on) we can find functions and if one is lucky there could by a possibility for a parameter which would imply there are finitely such functions... nevertheless when talking about functions there are endless possibilities.. and the possibility of there being a function with maybe factorials and other stuff is def possible
Now this is the type of comment I was hoping for
hm... if f(x) = 1/x
then f(f(x)) = x
somewhere your logic is flawed, since the degree of f(f(x)) is not -2 😉
One can try with power series...
This is why i quit calculus, too many sweats
Multi-valued functions are continuous, you just don’t need to look at continuity in weird ways not compatible with Riemann surfaces! 😶
Well at least those multi-valued functions that one can get in complex analysis.
I will memorize these equations so I can have a shortcut that is simpler to calculate than the power rule for this specific case.
Then what is wrong with the following argument? f (prime) = f of f mens, (d/dx) f(x) = f(f(x)). Put y = f(x). Then, dy / dx = f(y). So, we are free to choose for f(y). For example, if we choose f(y) = 1 / y, then y dy = dx, so y^2 / 2. = x + c, or y = f(x) = sqr rt ( 2x + c1).
By choosing f(y), you also choose f(x) (in your case f(x) = 1/x). Yet, in your final solutiin you get a different f(x) which is is a contradiction
Exactly
If f is invertible, with inverse g, then f' = f○f implies 1/g' = f'○g = f. That problem is equivalent to the fg' = 1 problem (derivative times the inverse equals 1).
What program do u use?
just curious how can we prove that power functions are the only solutions to this problem?
Alpha_1 can be simplified to e^(π(√3+i)/6) alpha_2 can also be similarly simplified
Já enjoei essas questões.
LET HIM COOK 🗣️🗣️🗣️🔥🔥🔥!!!
Did you change the thumbnail?
thumbnail goes hard
Here'sanother way to look at this, by integrating both sides we get-> f(x)= int( f(f(x)) dx ....... (1) so in (1) put x--> f(x) so we get f(f(x))= int (f(f(f(x)))dx ....... (2) now substituting f(f(x)) from (2) in (1) we get f(x)= int(int(f(f(f(x))) dx) dx now the pattern maybe a little tough but continuing the same process we see that f(x)= in(int(int.......( f(f(f(......(x)))))) )dx) dx) dx..... where the number of integrals is n-1 and the number of f's in f(f(f....(x)))) is n so by algebrec anaylitic continuation we can put n=0 so when the number of integral becomes -1 and f(f(...x)))) becomes 0, thus we get f(x)= d/dx(1)= 0 this is one such function then putting n=-1 we get f(x)= d2/d2x ( f-(x)) this may look a little tough buut we get the same function mentioned in the video. Both ideas work its just the method shown in the video is a little incomplete whereas the method used here is basically impossible to solve....
So it's a comparison between the incomplete and the impossible to solve😂😂😂
you cannot substitute the x in the integral by f(x). because in reality you have (1):
f(x) = int (from somewhere to x) f(f(t)) dt
then substituting x->f(x) changes the integration bounds and not the integrand
f(x)=0 is a solution
i honestly thought f(x) = 0 would be a solution that someone would care to put in a video
I like to think you record your videos first try with no notes or prep
Thanks for the explicit content warning lmao
"(complex number)^(complex number)" - very bad idea.
isn't surreal^surreal worse ?
I'm sure you've actually solved this exactly differential equation 10 months ago... I think the video was called "A very interesting differential equation: derivative equals composition"
Now that I've released 400 videos it's become hard to keep track of the problems I've solved😭
@@maths_505 th-cam.com/video/C6fZVwqhbnE/w-d-xo.html&ab_channel=Maths505 😀
Could you make a discord server?
Would that be useful to the community?
@@maths_505 I think it would because many people who are interested in mathematics and are viewers of your channel can communicate and discuss about mathematics and get help for mathematics.
Yes! I want a math discord server that isnt swarming with users
You didn’t have enough room in the margin? Is your name Fermat?!
Very Fast thinking and solving problem. One question is make simple when we put there any appropriate constant?
f(x) = 0
@@DerMathematicker indeed
Graph it
f^(n-1)(x)=f°^n(x)
a•ß!/(ß+1-n)!•x^(ß+1-n)=a^sum[k=0,n-1](ß^k)•x^ß^n
ß^n=ß+1-n
a•ß!/(ß+1-n)!=a^((1-ß^n)/(1-ß))
ß!/(ß+1-n)!=a^(n/(1-ß))
a=(ß!/(ß+1-n)!)^((1-ß)/n)
let n=2
ß^2=ß-1
ß^2-ß+1=0
ß=(1+-isqrt(3))/2
ß=e^(+-ipi/3)
a=ß^((1-ß)/2)
a=e^+-(ipi/6•e^(-+ipi/3))
if n=3,
f"(x)=f(f(f(x)))
aß(ß-1)x^(ß-2)=a^(1+ß+ß^2)•x^ß^3
ß^3=ß-2
aß(ß-1)=a^((3-ß)/(1-ß))
ß(ß-1)=a^(2/(1-ß))
a=(ß^2-ß)^((1-ß)/2)
this is solvable with the cubic formula, but I'd rather not
Fascinating
f(x) = 0 😂
allo
:))
Awesome content as always. I would like to share that I was precisely thinking about this kind of Differential Equations yesterday after TH-cam recommended me this video: th-cam.com/video/rNUfiQgj6ZI/w-d-xo.html from Michael Penn (pretty similar approach, not sure if that was the source of inspiration of this video as well).
However, I was thinking more in the equation f'(x) = f_n(x) (being f_n the nth composited function f). Doing the same trick you can lead to interesting conclusions about the parameters of the power functions that make the job. I haven't reach any marvelous result yet, but probably you could take a look if interested.
It was nice to share results (and thoughts) with you today. Always a pleasure to see your videos, definitely they had become a recurrent highlight of the day. Kind regards from Mexico!
Math was fun and interesting up until differential equations. I then realized how most of it is fake and arbitrary. Sad
balls
1sties
Please stop saying "OK cool". It's extremely annoying.
"Ok cool!" Became the catchphrase of the channel.
Stopping would be like changing π for some random letter
OK cool
Nuh uh
The ohhkayy coool is the reason why people have subscribed in the first place!
@@maths_505 Interesting. Could you add time codes for OK cool so I can press / unpress m on time? :)
AsymptoticSum[(-Log[1 - t x]/t)/z/. t -> n/z, {n, 1, z}, z -> Infinity]
this looks as if it gives 0 for all x. but maybe not 😅
without the z in the denominator the sum converges to the integral of log(1-t x)/t for t from 0 to 1. Ok this is a divergent integral, and the 1/z seems to counteract this divergence. Have you plotted the result?