Every member in the equivalence class set has a single member on the domain: D := (-2, -1] U {0} U [1, 2) And so: Given absolutely any function on the domain D, you can extend it to create a function f that matches the given condition. Given any matching f, you can also extract such a g.
Curiously enough, if g is continuous on (-2, -1] and on [1, 2), as well as meeting the criteria that as n→2: g(n)→g(1), as well as n→-2: g(n)→g(-1); than f will be continuous everywhere EXCEPT on x=0 (for f to be continuous at x=0, g and f would need to be a constant)
Please consider the function f(x)=sin(2pi*log_2(x)). It fulfills the property and is not constant, but unfortunately it is only defined on ℝ⁺, which is also why your proof does not apply to it as neither f(0) nor the limit of f(1/2^n) as n goes to infinity are defined.
@@Noname-67 multiplying the input by 2 adds 2pi to the angle which doesnt change the value of the function so it does actually work, try it in wolfram alpha
@@justinbagci5656 that's only work if x is an integer power of 2. For example f(3) is something around 1 while f(6) is something around -1 Edit: I think I miscalculated something
wow thats a good one. yes it would be considered discontinuous on the whole on ℝ so thats why it isnt mentioned there. but indeed thats one of the equivalence relations examples discussed later in the videos
You can think of this equivalence relation as just two numbers being equivalent if they're shifts of each other in binary. If we were using 10 instead of 2, it would be clearer: 123, 0.123, and 12300000 would be all equivalent.
If I'm not mistaken, this means that any sequence of binary digits produces a unique equivalence class, and since there are uncoutably many sequences of binary digits, there are also uncountably many equivalence classes. Therefore, we can actually find a bijection between the real numbers and the set of equivalence classes, which in turn means that we can find a bijection between the set of all real functions and the set of all real functions satisfying f(2x)=f(x), which seems crazy. Correct me if I'm wrong!
You can also do some stuff with a linear combination of sines and cosines where the argument for both the sine and the cosine is (2*pi*log_2(x)). It works pretty nicely for x greater than 0.
That's what I got, too. It works for negatives as well if you change x to |x|. I didn't simplify by changing to base-2. That took me a while to figure out what you did there. Changing the base was more interesting than the video!
Sorry, watched without sound, will do with tomorrow. Curious to check other versions than continuous on R. Found next solution. Lets do replacement. x=2^t. So we have f(t)=f(t+1). In this case any periodic function with period 1 satisfied our condition. F.e. f(t) = {t}. Or better to put sin(2pi*t). Moving back, t=ln2(x). sin(2pi*ln2(x)). {ln2(x)}. Basically any periodic P(x) = P(x+1) have solution P(ln2(x)).But only on R+/0
nice! I also think f(any rational) = 1, f(any irrational) = 2 (and similar equivalence classes like algebraic/transcendental and infinitely more) is a discontinuous one that he missed
Nice video!! There is a nice way of stating the general case: Let p: R -----> R/~ Be the natural projection function (where ~ is the equivalent relation described in the video). Then a function satisfies f(x) = f(2x) iff f = h o p Where h is any function h: R/~ -----> R
Absolutely. I like to think of this universal property as the analog for the category of sets what the "Fundamental Homomorphism Theorem" is for the category of groups. www.wikiwand.com/en/Fundamental_theorem_on_homomorphisms
How well would f(x)=g(ln(x) mod ln(2)) fullfill the f(x)=f(2x) condition for x>0 ? edit: - g(x) can be an arbitrary function defined at least for some values 0
i like this idea. if you plug lnx/ln2 - floor(lnx/ln2) into wolframalpha it will actually draw a graph of the function when g(x) = x and it does seem to fullfill f(x)=f(2x)
@@drpeyam Ah yes, you're correct. I was assuming that the function was differentiable. If it were, then you end up with 2f' = f', and therefore f' must be zero.
@@mikeburns6603 Strange.. that seems to be correct, but the function proposed by Smiley1000, i.e. sin(2pi*log_2(x)), seems to be differentiable as well as fulfilling the requisite f(x)=f(2x) but it is not constant: 1) f(2x)=sin(2pi*log_2(2x))= sin(2pi(1+log_2(x)))= sin(2pi+2pi*log_2(x))= sin(2pi*log_2(x))=f(x) 2) it's derivative:cos(2pi*log_2(x))*2pi/(xln(2)) is only 0 for x=2^((2n+1)/4) n being any integer, because cos(y)=0 implies y=pi*(2n+1)/2 So what's going on here? I verified what you stated, but did not come to the same conclusion. The derivative of f(x)=f(2x) is f'(x)=2f'(2x) and that latter equation does not imply f'(x)=0 and Smiley1000's function does fulfill it: i.e. cos(2pi*log_2(x))*2pi/(xlnl(2))=2*cos(2pi*log_2(2x))*2pi/(2xln(2))=(2/2)*cos(2pi+2pi*log_2(x))*2pi/(xln(2))
@@mikeburns6603 I know that, it also isn't defined for negative x. But where is it stated that the domain of f needs to be R, or that it needs to be continuous at x=0?
Cool, I solved the non-continuous case, but not formally. Assign any outputs you want for all x values in the ranges: (-2,-1], 0, [1,2). Then all the remaining x values have determined outputs.
So how about f(x) = sin(2π log₂x) ? This can be further extended to any periodic function: f(x) = f(ax), f(x) = g(P logₐx) where g(x) is periodic and P is its period.
@@drpeyam Easily extended to R though. Just use absolute value of x and use whatever value you want at 0. With the added benefit that with this method you can produce functions with the desired property that can be easily computed, given a computable periodic function g.
The real line quotient the equivalence relation x ~ 2^N x for N integer, as a set is just [-1, -1/2) U {0} U (1/2, 1]. As a topological space, however, one needs to notice that R/~ must be connected, and in the quotient, the class of {0} is an accumulation point of all other classes. It should not be hard to formally prove that R/~ is homeomorphic to a compact interval [0,1], and the set of all such functions is in one-one correspondence with functions [0,1] ---> R.
I think another way to show the continuous case is that f(x) = f(2x) implies that every point (x,y) = (0.5x,y) so x = 0.5x which means x = 0 and so f(x) = f(2x) = f(0) The discontinuous case can also be shown by saying that for every point (x,y) there exists a point (1/2x,y) (which sounds very similar to the previous case) so x has to be 2^m in order for every x to be divisible by 2 leading to f(x) being 1 for x = 2^m for all m in Z and 0 o/w, you could also generalise this to f(x) is a if x = b/2^m and 0 o/w for all a, b and m in Z
You could also use the epsilon delta definition of continuity to say f must be a constant function f is continuous at 0 let f(0) = c let f(k) =/= c for some k then, for ε < |f(k) - c| for any δ we choose there exists λ=x/2^n < δ meaning that there will always be an intrusive bit, f(x) is not constant f(x) is not continuous at x=0 (the converse is left as an exercise to the reader) However, if f(x) is not required to be continuous at x=0, then f(x) = g(log_2 |x|) for any continuous periodic function g(x) with period 1 (take g(x) = sin(2pi x) for example) This is because at 0, there will always be an intrusive bit which is arbitarily close to 0, but for any other value, you can always filter those out by choosing a sufficiently small interval. As a final statement, if f(x) does not have to be continuous at x=0, it can be defined piecewisely for -ve and +ve numbers f(x) = g(log_2 x) if x>0 h(log_2 -x) if x
Well, the set of the equivalence classes introduced in the video is certainly uncountably infinite, for if it were countable, the real numbers would be a countable union of countable sets (and therefore a countable set by Cantor's diagonal proof). It is clear that its cardinality is less than or equal to the cardinality of the reals, and if we choose to accept CH, this implies that there are exactly as many equivalence classes as there are real numbers. Of course, picking any bijection between the two sets allows us to define a function satisfying the functional equation f(2x) = f(x) with f[R] = R. I don't know how this would work out without CH, though ...
@@beatoriche7301 Of course the set of equivalence classes has the same cardinality as R as it should be simple to construct a bijection between (-2,-1] u {0} u [1,2) and R
In fact we didn't need to talk about the continuous case at all, we could have just jumped straight into the "most general case" which includes the continuous case. However the solution to the "most general case", doesn't give us any insight into what happens with interesting constraints like continuity, where f(x) = constant: we had to reason this from scratch. So the question is, are there any other interesting constraints apart from continuity, that give another interesting class of solutions, that are not obvious from the "most general case" solution?
Hi Dr. Peyman! Out of curiosity for those watching, is there a particular course where math majors could encounter an abundance of functional equations outside of ODEs or PDEs?
The general formulation I came up with is that your f can be any function that can be defined in terms of some g as below: f(x) = g(log2(x) - floor(log2(x))) or, if you want to use “frac” to mean “the fractional part”, it is more concisely f(x) = g(frac(log2(x))) Taking the logarithm makes the problem equivalent to g(y) = g(y+1), where y = log2(x). Then, the frac function transforms y into a period-1 sawtooth wave in the logarithmic space. Thus, we can plug the result into any function we want, and that function gets to map [0,1) to anything at all, with no restrictions.
@@tarsofelix4414 Since (for positive x) log2(-x) = iπ/log(2) + log2(x), my original proposal already works for negative numbers! However, you did give me the idea to make it more general. Since the real part of a logarithm is the only thing that changes when you multiply by two (i.e., double the distance away from the origin, or the absolute value), we really only need to take the fractional part of the real component of the logarithm, and can leave the imaginary part alone. So if we use a function reFrac(x) = frac(re(x)) + im(x)×i (where re and im extract the real and imaginary components, respectively), this function will do just that. The final result is: f(x) = g(reFrac(log2(x))) If you put "plot frac(re(log2(x+yi))) + im(log2(x+yi))×i" into Wolfram Alpha (or your graphing system of choice) you can see how this most general solution turns out!
Not sure if this proves f(x)=f(0) for the continous case, because it only says that the limit as N->inf is f(0) for each starting f(x), but that limit never gets reached
Nope the values of sequence are all equal to each other, since it is not increases or decreases then the limit it approaches must equal to all those values
a function that satisfies this equation is ‘f(x) = the note perceived by humans when listening to an x Hz tone’ =P this is because multiplying the frequency of a sound wave by 2 shifts it up by one octave, which we perceive as the same note. similarly, dividing by 2 shifts it down an octave for example, 440 Hz is defined to be an A note, and because of this property, an 880 Hz tone also sounds like an A. similarly, so does 220 Hz. and we get the following equivalence class of all frequencies that sound like an A note, {…, 55, 110, 220, 440, 880, 1760, 3520, …} :D
I'm sorry but I don't undertand what property (theorem, justification....) permits to write the equation visible at 0:39 ? For me, it's litteraly coming out of nowhere... :D
@@drpeyam errrrr... if y = x /2, by taking f(y) = f(2y), shouldn't we have by substituion f(x/2) = f(2(x/2)) instead of f(x) = f(2(x/2)) ? I understand that it's the last part of the equation, but how then can this be equal to f(x), since we are talking about x/2 and not x ?
y = 1 if x is rational, y = 0 if x is irrational. There are lots of answers if they don't have to be continuous. Trig/ln2 functions that everybody below found are well worth noting.
Are you sure you found all of them? what about something like f(any rational) = 1, f(any irrational) = 2. This is also a sort of equivalence class -based function (bc multiplication by 2 preserves rationality/irrationality), but it's not in your "most general case", I think
Challenge: Consider the quotient topology induced by ℝ → ℝ/~ as well its restriction on subspaces. What familiar space is (0, ∞)/~ homeomorphic to? How about (−∞, 0)/~? What happens when you add in the origin (the only singleton equivalence class)? What famous (but trivial) topology does this space ℝ/~ have?
What about f(x)=A*x^(i*2*pi*k/ln2), with k = {0, 1, -1, 2, -2,...}, A is any real number? This function satisfies the condition in the complex space and it includes all the constant solutions you mentioned when k=0 by just varying A. You can verify that f(2x)=f(x) for all real x.
@@BCQM_BCQM Dr. Peyam requires f to be continuous on R in his statement of the problem, so you will have to define f(0). A function can only be called continuous if it is defined everywhere, the limit is defined everywhere and both are equal everywhere.
One question is left: how many are these equivalence classes? SPOILER: . . . . . . . . . . . . This is quite trivial once you realize that the elements in the left-closed, right-open interval [1,2[ all belong to a different class. But still, it's interesting to notice. Also, there are 2^(2^aleph_0) functions solving that equation
One interesting example is the function f(x)=sin(2 π ln(x)/ln(2)). In general, if g(x) is periodic with period ln(2), then f(x)=g(ln(x)) will fit this criterion.
why can't we define continuous function on [1, 2) (such that derivative at 1 and 2 would be 0) so that it would be continuous on the whole interval [0, oo). isn't it?
I attempt to solve it using calculus, assuming f is double differentiable on R s.t. f(2x)=f(x), then 2f'(2x)=f'(x) => 2^k f'(2^k x)=...=2f'(2x)=f'(x)=f'(x/2)/2=f'(x/4)/4=...=f'(x/2^n)/2^n by continuity of f'(x), f'(x) = lim n->inf f'(x/2^n)/2^n=f'(0)*0=0 =>f(x)= int f'(x) dx = 0+C=C where C is a constant
The function that relates frequency to the notes of the musical scale (as positions on a circle) is an example. (You can invert a logarithmic spiral and apply a sawtooth/mod function, so it is piecewise continuous.)
Okay also der trick mit der äquivalenzrelation für die allgemeine lösung funktioniert klarerweise für probleme der form f(c*x) =f(x) bzw f(x^a) =f(x) für geeignete a respektive c in R, sodass man eine Äquivalenzrelation bilden kann. Großartig
I think we can generalize the first part of this video to: if X and Y are complete metric spaces, f : X → Y is continuous, g : X → X is a contraction mapping in the sense of the Banach fixed point theorem, and for all x we have f(g(x)) = f(x), then f is constant. The video has X = ℝ, Y = ℝ, and g(x) = x/2.
@@drpeyam No, only the ones with period equal to 1/n, n any (positive) integer; e.g. all the periodic trigonometric functions do not fulfill that condition because their period is 2pi or pi.
For a bit of a stronger statement, the function does not have to be continuous everywhere, but only continuous at x=0 for the first case to hold
An even stronger statement, the first case is the only solution if and only if f(x) is continuous at x=0
I love how Dr. Peyam always thanks us for watching before we even watch the video! Is Dr. Peyam really from the future? 🤔
Ayh! Cool to see you here Dr. Weselcouch! Love your irrational number approximation videos!
@@guccisavage2739 Thanks I'm glad you like them!
he was the only one that showed up to Dr. Hawking’s party for time travelers.
It is a simple observation that in the future, we have in fact watched the video in it's entirety.
i’d like to think that he thanks us for tuning in his vids because it seems he appreciates that hehe
Wow, I realized intuitively that it must be the constant function but had no idea how to prove it. This was so nicely done, too. Really elegant.
Thank you!!!
What if my teacher were half as excited as him ?
Then half your class would go on to win Field medals
Every member in the equivalence class set has a single member on the domain:
D := (-2, -1] U {0} U [1, 2)
And so: Given absolutely any function on the domain D, you can extend it to create a function f that matches the given condition. Given any matching f, you can also extract such a g.
Curiously enough, if g is continuous on (-2, -1] and on [1, 2), as well as meeting the criteria that as n→2: g(n)→g(1), as well as n→-2: g(n)→g(-1); than f will be continuous everywhere EXCEPT on x=0
(for f to be continuous at x=0, g and f would need to be a constant)
7:53 was great 😂 "But also negative one, negative two... Oh hmm-hmm-hmm... *Nah*... But also negative one, negative two!"
Yes, …, -1/2, -1, -2, -4, …. Is a different equivalent class than …, 1/2, 1, 2, 4... . Dr Peyam realized this as he was talking. 😊
Please consider the function f(x)=sin(2pi*log_2(x)). It fulfills the property and is not constant, but unfortunately it is only defined on ℝ⁺, which is also why your proof does not apply to it as neither f(0) nor the limit of f(1/2^n) as n goes to infinity are defined.
Your function is wrong because it only work if x is a integer power of 2, it doesn't work when x=3 or something else, it doesn't work
@@Noname-67 multiplying the input by 2 adds 2pi to the angle which doesnt change the value of the function so it does actually work, try it in wolfram alpha
@@justinbagci5656 that's only work if x is an integer power of 2. For example f(3) is something around 1 while f(6) is something around -1
Edit: I think I miscalculated something
@@Noname-67 You really did miscalculate something because
sin(2pi*log_2(6))
= sin(2pi*log_2(2*3))
= sin(2pi*(log_2(2)+log_2(3)))
= sin(2pi*(1+log_2(3)))
= sin(2pi+2pi*log_2(3)))
= sin(2pi*log_2(3)))
wow thats a good one. yes it would be considered discontinuous on the whole on ℝ so thats why it isnt mentioned there. but indeed thats one of the equivalence relations examples discussed later in the videos
You can think of this equivalence relation as just two numbers being equivalent if they're shifts of each other in binary.
If we were using 10 instead of 2, it would be clearer: 123, 0.123, and 12300000 would be all equivalent.
There is no 2 and 3 in binary
@@adityadwivedi4412 I was talking about base 10 when I said "if we were using 10 instead of 2".
A very interesting observation!
If I'm not mistaken, this means that any sequence of binary digits produces a unique equivalence class, and since there are uncoutably many sequences of binary digits, there are also uncountably many equivalence classes. Therefore, we can actually find a bijection between the real numbers and the set of equivalence classes, which in turn means that we can find a bijection between the set of all real functions and the set of all real functions satisfying f(2x)=f(x), which seems crazy. Correct me if I'm wrong!
f(2x) = f(x)
Assume that f(x + y) = ( f(x) + f(y) )/2
f(x+x) = ( f(x) + f(x) )/2 -> f(2x) = f(x)
f(x) = f(x + 0) = ( f(x) + f(0) ) /2
2f(x) - f(x) = f(0)
f(x) = f(0)
since f(0) is a constant therefore f(x) = c, c € R
You can also do some stuff with a linear combination of sines and cosines where the argument for both the sine and the cosine is (2*pi*log_2(x)). It works pretty nicely for x greater than 0.
Oh that's pretty cool!
That's what I got, too. It works for negatives as well if you change x to |x|. I didn't simplify by changing to base-2. That took me a while to figure out what you did there. Changing the base was more interesting than the video!
Sorry, watched without sound, will do with tomorrow. Curious to check other versions than continuous on R. Found next solution.
Lets do replacement. x=2^t. So we have
f(t)=f(t+1). In this case any periodic function with period 1 satisfied our condition. F.e. f(t) = {t}. Or better to put sin(2pi*t). Moving back, t=ln2(x).
sin(2pi*ln2(x)). {ln2(x)}. Basically any periodic P(x) = P(x+1) have solution P(ln2(x)).But only on R+/0
9:12 that's me every time I want to talk about a really precise theorem 😂
@AKS I was in such a situation many years back, but I was not so souvereign like the Dr here...
f(x) = cos(2pi(log_2(x)) is an example of a non-trivial function that satisfies this and is continuous on all positive numbers.
Only defined on x>0 though
@@drpeyam "f(x) = cos(2π log₂ |x|)" ? heh...
Still undefined on x = 0@@gastonsolaril.237
seems to break at x=0
nice! I also think f(any rational) = 1, f(any irrational) = 2 (and similar equivalence classes like algebraic/transcendental and infinitely more) is a discontinuous one that he missed
Nice video!! There is a nice way of stating the general case:
Let p: R -----> R/~
Be the natural projection function (where ~ is the equivalent relation described in the video).
Then a function satisfies f(x) = f(2x) iff f = h o p
Where h is any function
h: R/~ -----> R
Absolutely. I like to think of this universal property as the analog for the category of sets what the "Fundamental Homomorphism Theorem" is for the category of groups.
www.wikiwand.com/en/Fundamental_theorem_on_homomorphisms
Dr. Peyam: Uses Red Pen.
Me: "Wait, that's illegal! You're not bprp..."
2+2=4 is also forbidden ;)
How well would f(x)=g(ln(x) mod ln(2)) fullfill the f(x)=f(2x) condition for x>0 ?
edit:
- g(x) can be an arbitrary function defined at least for some values 0
i like this idea. if you plug lnx/ln2 - floor(lnx/ln2) into wolframalpha it will actually draw a graph of the function when g(x) = x and it does seem to fullfill f(x)=f(2x)
@@justinbagci5656 Well, for g(x)=sin(2pi*x/ln(2)) the function f(x) becomes effectively the same as the one proposed (in the comment) by Smiley1000.
I'm a common Indian man , I see my country's flag in the thumbnail , I click it
cool! i love this idea of a video finding all functions that fullfill some property! keep it up
日本人でおすすめにでてきた同士はいますか？笑
いるで。でも元々英語の数学や物理の解説見てたから、ノーカンかも。
@@suou7938 すごい！英語のもの見てるんですね！僕は大雑把な理解しか出来ませんでした…笑
For the continuous case, you can also just take the derivative using the chen lu and show that f'(x)=0 everywhere, so it's a constant.
Care to elaborate? :) Also continuous doesn’t imply differentiable
@@drpeyam Ah yes, you're correct. I was assuming that the function was differentiable. If it were, then you end up with 2f' = f', and therefore f' must be zero.
@@mikeburns6603 Strange.. that seems to be correct, but the function proposed by Smiley1000, i.e. sin(2pi*log_2(x)), seems to be differentiable as well as fulfilling the requisite f(x)=f(2x) but it is not constant:
1) f(2x)=sin(2pi*log_2(2x))=
sin(2pi(1+log_2(x)))=
sin(2pi+2pi*log_2(x))=
sin(2pi*log_2(x))=f(x)
2) it's derivative:cos(2pi*log_2(x))*2pi/(xln(2))
is only 0 for x=2^((2n+1)/4) n being any integer, because cos(y)=0 implies y=pi*(2n+1)/2
So what's going on here?
I verified what you stated, but did not come to the same conclusion.
The derivative of f(x)=f(2x) is f'(x)=2f'(2x) and that latter equation does not imply f'(x)=0 and Smiley1000's function does fulfill it:
i.e.
cos(2pi*log_2(x))*2pi/(xlnl(2))=2*cos(2pi*log_2(2x))*2pi/(2xln(2))=(2/2)*cos(2pi+2pi*log_2(x))*2pi/(xln(2))
@@Apollorion That function is not differentiable at zero (or even defined) but everywhere else is good.
@@mikeburns6603 I know that, it also isn't defined for negative x. But where is it stated that the domain of f needs to be R, or that it needs to be continuous at x=0?
Do you say "All right, thanks for watching" when you start your actual in-person classes?!
Yeah I actually say “Thanks for coming”
I love how excited you are while explaining
Cool, I solved the non-continuous case, but not formally.
Assign any outputs you want for all x values in the ranges: (-2,-1], 0, [1,2). Then all the remaining x values have determined outputs.
So how about f(x) = sin(2π log₂x) ? This can be further extended to any periodic function:
f(x) = f(ax), f(x) = g(P logₐx) where g(x) is periodic and P is its period.
Not defined on R
@@drpeyam Easily extended to R though. Just use absolute value of x and use whatever value you want at 0. With the added benefit that with this method you can produce functions with the desired property that can be easily computed, given a computable periodic function g.
7:57 a wild thought appears!!! Hahaha. Loved it. That's genuine.
take a shot every time he says “however”
You also could change the condition of continuity for the existence of lim f(x) when x tends to +-inf. You obtain something different :)
The real line quotient the equivalence relation x ~ 2^N x for N integer, as a set is just [-1, -1/2) U {0} U (1/2, 1]. As a topological space, however, one needs to notice that R/~ must be connected, and in the quotient, the class of {0} is an accumulation point of all other classes. It should not be hard to formally prove that R/~ is homeomorphic to a compact interval [0,1], and the set of all such functions is in one-one correspondence with functions [0,1] ---> R.
I think another way to show the continuous case is that f(x) = f(2x) implies that every point (x,y) = (0.5x,y) so x = 0.5x which means x = 0 and so f(x) = f(2x) = f(0)
The discontinuous case can also be shown by saying that for every point (x,y) there exists a point (1/2x,y) (which sounds very similar to the previous case) so x has to be 2^m in order for every x to be divisible by 2 leading to f(x) being 1 for x = 2^m for all m in Z and 0 o/w, you could also generalise this to f(x) is a if x = b/2^m and 0 o/w for all a, b and m in Z
You could also use the epsilon delta definition of continuity to say f must be a constant function f is continuous at 0
let f(0) = c
let f(k) =/= c for some k
then, for ε < |f(k) - c|
for any δ we choose
there exists λ=x/2^n < δ
meaning that there will always be an intrusive bit,
f(x) is not constant f(x) is not continuous at x=0
(the converse is left as an exercise to the reader)
However, if f(x) is not required to be continuous at x=0, then f(x) = g(log_2 |x|) for any continuous periodic function g(x) with period 1 (take g(x) = sin(2pi x) for example)
This is because at 0, there will always be an intrusive bit which is arbitarily close to 0, but for any other value, you can always filter those out by choosing a sufficiently small interval.
As a final statement, if f(x) does not have to be continuous at x=0, it can be defined piecewisely for -ve and +ve numbers
f(x) = g(log_2 x) if x>0
h(log_2 -x) if x
For the general solution, if we give a different value for each equivalence class, what kind of image can we obtain by those functions ?
Well, the set of the equivalence classes introduced in the video is certainly uncountably infinite, for if it were countable, the real numbers would be a countable union of countable sets (and therefore a countable set by Cantor's diagonal proof). It is clear that its cardinality is less than or equal to the cardinality of the reals, and if we choose to accept CH, this implies that there are exactly as many equivalence classes as there are real numbers. Of course, picking any bijection between the two sets allows us to define a function satisfying the functional equation f(2x) = f(x) with f[R] = R. I don't know how this would work out without CH, though ...
@@beatoriche7301 Of course the set of equivalence classes has the same cardinality as R as it should be simple to construct a bijection between (-2,-1] u {0} u [1,2) and R
@@smiley_1000 You're right! That's way more elegant than what I came up with.
f(x)=a
Where a is a constant
It can't have any more answers
Yes it can!
In beginning, is it not enough to assume f is continuous at 0 ?
Is there any reason for the flag-like colors in the thumbnail? I couldn't find a plausible explanation
maybe this was a question asked in an indian contest?
No just random
In fact we didn't need to talk about the continuous case at all, we could have just jumped straight into the "most general case" which includes the continuous case. However the solution to the "most general case", doesn't give us any insight into what happens with interesting constraints like continuity, where f(x) = constant: we had to reason this from scratch. So the question is, are there any other interesting constraints apart from continuity, that give another interesting class of solutions, that are not obvious from the "most general case" solution?
8:03 You think about extending positive equivalence classes to negative numbers?
Hi Dr. Peyman! Out of curiosity for those watching, is there a particular course where math majors could encounter an abundance of functional equations outside of ODEs or PDEs?
Yes and course is named functional analysis
@@Theloverguy 0 have inverse in R
Sin(log(x)), you just have to choose the base such that doubling x results in adding 2pi (not continuous though)
Hello. At 2:23 I thing it should be written x->0 instead of N->0 line 2
The general formulation I came up with is that your f can be any function that can be defined in terms of some g as below:
f(x) = g(log2(x) - floor(log2(x)))
or, if you want to use “frac” to mean “the fractional part”, it is more concisely
f(x) = g(frac(log2(x)))
Taking the logarithm makes the problem equivalent to g(y) = g(y+1), where y = log2(x). Then, the frac function transforms y into a period-1 sawtooth wave in the logarithmic space. Thus, we can plug the result into any function we want, and that function gets to map [0,1) to anything at all, with no restrictions.
You can include negative numbers with a small tweak:
f(x) = g(frac(log2(|x|)))
This function will be defined for all real numbers except 0.
@@tarsofelix4414 Since (for positive x) log2(-x) = iπ/log(2) + log2(x), my original proposal already works for negative numbers! However, you did give me the idea to make it more general. Since the real part of a logarithm is the only thing that changes when you multiply by two (i.e., double the distance away from the origin, or the absolute value), we really only need to take the fractional part of the real component of the logarithm, and can leave the imaginary part alone.
So if we use a function reFrac(x) = frac(re(x)) + im(x)×i (where re and im extract the real and imaginary components, respectively), this function will do just that.
The final result is:
f(x) = g(reFrac(log2(x)))
If you put "plot frac(re(log2(x+yi))) + im(log2(x+yi))×i" into Wolfram Alpha (or your graphing system of choice) you can see how this most general solution turns out!
Not sure if this proves f(x)=f(0) for the continous case, because it only says that the limit as N->inf is f(0) for each starting f(x), but that limit never gets reached
Nope the values of sequence are all equal to each other, since it is not increases or decreases then the limit it approaches must equal to all those values
A*sin(2pi*log_2|x|)+B*cos(2pi*log_2|x|) works for any non-zero x
A nice nonconstant but continous example is sin(2pi *log2(x)) i believe.
a function that satisfies this equation is ‘f(x) = the note perceived by humans when listening to an x Hz tone’ =P
this is because multiplying the frequency of a sound wave by 2 shifts it up by one octave, which we perceive as the same note. similarly, dividing by 2 shifts it down an octave
for example, 440 Hz is defined to be an A note, and because of this property, an 880 Hz tone also sounds like an A. similarly, so does 220 Hz. and we get the following equivalence class of all frequencies that sound like an A note, {…, 55, 110, 220, 440, 880, 1760, 3520, …} :D
I'm sorry but I don't undertand what property (theorem, justification....) permits to write the equation visible at 0:39 ? For me, it's litteraly coming out of nowhere... :D
It’s just iteration. If f(x) = f(2x) for all x, then f(y) = f(2y) for all y and just let y = x/2
@@drpeyam errrrr... if y = x /2, by taking f(y) = f(2y), shouldn't we have by substituion f(x/2) = f(2(x/2)) instead of f(x) = f(2(x/2)) ? I understand that it's the last part of the equation, but how then can this be equal to f(x), since we are talking about x/2 and not x ?
f(x) = f(2(x/2)) is always true but f(2(x/2)) = f(x/2) is true because f(2*anything) = f(anything), that’s why we get f(x) = f(x/2)
@@drpeyam Thank you for taking the time to respond ! ;)
Amazing, elegant and to the point, I loved it !! Subscribed immediatly !
I love your videos man! You’re very underrated.
So is maths (unfortunately)
By whom, please?
Awwww thanks!!!
seems like "all functions" should come with a clarification, all f : A --> B for what sets A, B
Best example of f(x)=f(2x) is Signum function.
That is particularly boring
f=0 is even simpler
@@Wurfenkopf but particularly uninteresting
y = 1 if x is rational, y = 0 if x is irrational. There are lots of answers if they don't have to be continuous. Trig/ln2 functions that everybody below found are well worth noting.
Are you sure you found all of them? what about something like f(any rational) = 1, f(any irrational) = 2.
This is also a sort of equivalence class -based function (bc multiplication by 2 preserves rationality/irrationality), but it's not in your "most general case", I think
i dont even know what kind of fields of math you post videos of but every video has such good explanations that I always end up learning something new
Along exactly similar lines, if f is a periodic function and lim_{x --> infinity} f(x) exists as a number L, then f is constant with value L.
Challenge: Consider the quotient topology induced by ℝ → ℝ/~ as well its restriction on subspaces. What familiar space is (0, ∞)/~ homeomorphic to? How about (−∞, 0)/~? What happens when you add in the origin (the only singleton equivalence class)? What famous (but trivial) topology does this space ℝ/~ have?
that’s too easy
ℝ/~ has weird topology, are you sure it's something famous?
What about f(x)=A*x^(i*2*pi*k/ln2), with k = {0, 1, -1, 2, -2,...}, A is any real number? This function satisfies the condition in the complex space and it includes all the constant solutions you mentioned when k=0 by just varying A. You can verify that f(2x)=f(x) for all real x.
gracias Doctor Peyam... me gustan estos videos de funciones de variable real y todas esas cosas! genial 👏
Maybe f(x)=sin(2πln(abs(x))/ln2) is a solution
I had the same idea, but your function is not defined for x = 0. But since 0*2 = 0, this should be easy to fix.
@@smiley_1000 since the limit as x approaches 0 doesn't exist so it will never be continuous; then I decided not to define any f(0).
@@BCQM_BCQM Dr. Peyam requires f to be continuous on R in his statement of the problem, so you will have to define f(0). A function can only be called continuous if it is defined everywhere, the limit is defined everywhere and both are equal everywhere.
The true most-general-case would be to find all functions, f, with f(2x)=f(x) where the domain is any (unital) ring-and no other restrictions!
Any function on the fractional part of log2(x) fits this definition, I believe.
One question is left: how many are these equivalence classes?
SPOILER:
.
.
.
.
.
.
.
.
.
.
.
.
This is quite trivial once you realize that the elements in the left-closed, right-open interval [1,2[ all belong to a different class. But still, it's interesting to notice.
Also, there are 2^(2^aleph_0) functions solving that equation
One interesting example is the function f(x)=sin(2 π ln(x)/ln(2)). In general, if g(x) is periodic with period ln(2), then f(x)=g(ln(x)) will fit this criterion.
Definitely! Except your function is not defined on all of R
@@drpeyam In order to define the function on R, you could use f(x)=g(ln|x|), and let f(0)=0.
Yes so a particular instance of the general formula in the video :)
If f is a continuous function on [0, inf] and f(2x) is less equal to f(x) for all x, does this imply f is deceasing?
Thank god I can change the speed of youtube-movies to say 0,5 because I cannot think so fast... haha
Choosing a representative out of every equivalence class for the general case? smells a lot like axiom of choice
wait for every (nonzero) equivalence class [x] the set {|y| in [x] : |y|>1} has a minimum so i think the axiom of choice is not needed right?
0:40 German outta nowhere? Is this some sort of running gag?
why can't we define continuous function on [1, 2) (such that derivative at 1 and 2 would be 0) so that it would be continuous on the whole interval [0, oo). isn't it?
Why do you complicate. 2*x=x gives 0 and infinity as the solution.
No cos(2pi*ln(x)/ln(2)) IS solution
No, not defined on all of R
I attempt to solve it using calculus, assuming f is double differentiable on R s.t. f(2x)=f(x),
then 2f'(2x)=f'(x)
=> 2^k f'(2^k x)=...=2f'(2x)=f'(x)=f'(x/2)/2=f'(x/4)/4=...=f'(x/2^n)/2^n
by continuity of f'(x), f'(x) = lim n->inf f'(x/2^n)/2^n=f'(0)*0=0
=>f(x)= int f'(x) dx = 0+C=C where C is a constant
But f is not differentiable
Also the differentiation is unnecessary, the approach in the video is much simpler
Would something like f(x)=sin(log(|x|)*2*pi/log(2)) work?
The function that relates frequency to the notes of the musical scale (as positions on a circle) is an example. (You can invert a logarithmic spiral and apply a sawtooth/mod function, so it is piecewise continuous.)
Gives me such David Cross vibes ("Mr. Show")
f(x) = the same constant for every prime number?
I’m not smart but isn’t it 0, 1 and -1 ? And maybe infinity ?
May I ask what is the name of this type of equation?
I’d call it a dilation equation
this would be a fun calculus assigment
Dr. Peyam is like mathematician version of mehdi
Dirichlet function?
Well I don't know any of these so i can't understand, can you please help me how i could find the conceptual lessons upon this.
See the playlist
You are amazing !
What about f(x)=d/dx (ln(x))
Solution:
f(x) = {the largest prime factor of x} works for all (integers) x > 1.
What about cos(2*pi*k) = 1?
sin(2*pi*k) = 0?
No here x is any real number, not just an integer
Okay also der trick mit der äquivalenzrelation für die allgemeine lösung funktioniert klarerweise für probleme der form f(c*x) =f(x) bzw f(x^a) =f(x) für geeignete a respektive c in R, sodass man eine Äquivalenzrelation bilden kann. Großartig
I think we can generalize the first part of this video to: if X and Y are complete metric spaces, f : X → Y is continuous, g : X → X is a contraction mapping in the sense of the Banach fixed point theorem, and for all x we have f(g(x)) = f(x), then f is constant.
The video has X = ℝ, Y = ℝ, and g(x) = x/2.
Very nice observation, thank you!
I loved your video ,by the way can u plz makr videos on laplace conversation or some tips on it ❤️
What if the condition becomes f(x)=f(x+1) for all real numbers?
But that’s just all the periodic functions
@@drpeyam No, only the ones with period equal to 1/n, n any (positive) integer; e.g. all the periodic trigonometric functions do not fulfill that condition because their period is 2pi or pi.
Beautiful one
If functions equal, derivatives also equal.
f’(x)=f’(2x)=2*f’(x)
f’(x)=0
f(x)=const
Yeah, but no one told that f should be derivable
@@Wurfenkopf f is not derivative. f’ is.
If f’(x)==2*f’(x) it means f’(x)=0
@@dmitrykatyshev9125
DerivaBle. A function such that it HAS a derivative.
It's not sufficient that the function is continuous to be derivable
What about f(x) = sin(2*pi*lnx/ln2),
Isn’t f(x)=f(2x) in this case for all x> 0?
We need all x in R
@@drpeyam oh yeah, right, and I guess even if you modded you still wouldn’t have a value for x=0...
F(x)= f(2x) … x = 2 x
X-2x=0
-x =0
X=o F(X)= F0)
Me: just do f(x)=f(2x)=constant :sunglasses:
I got a+bcos(c+2dπln(x)/ln(2)), where d is an integer.
y=mod(1,ln(x)/ln(2)) also works.
y=f(mod(1,ln(x)/ln(2))) again.
easy. Divide both side by f and we get x = 0
*Algebruh 100*
Wow, just intuitively I guessed f(x) = 0 will be the only function possible. But missed the continuous and non-continuous parts of the function.
Сool. I really want to know more about equations these type.
i saw my flag's color and came real quick
it should be continuous at least in x=0.
写像fに何が入っても定数になるってこと？
What about f(x)=1/2*f(2x)?
That would be interesting! I did the case f(cx) = cf(x) in another video, but that’s different
Oh wait, nvm. The video assumes that it works for *all* constants, not just one specific constant.