Nice try but piecewise without specifying what pieces is a bit of a scam lol. All functions are piecewise constant if i take my pieces to be singletons 😂
An example of f would be sin((2pi/ln(2))lnav(lnav(x))), where lnav is natural log of absolute value, which is of course not continuous at 1 so your point still stands
great answer. I think it could be any infinite sum of such a function and all its harmonics: SUM[k=1->inf] {A-k * Sin( (2 * PI * k / ln|2| ) * ln| ln|x| | + THETA-k) where A-k are scaling factors and THETA-k phase offset factors
This is dangerously close to the Vitali set construction. You might make a non-measurable set by accident! Don't do it, Dr. Peyam! You have so much to live for!
This is somewhat similar to a question I got for my Cambridge interview, which was "find a continuous function f such that f(x) = f(2x)". In that vein, I propose: f(x) = sin(2pi * log2(ln(x))).
f(x)=f(x*2^z) for all z in \mathbb{Z} using the continuity, we get f(0)=f(lim_{n to infinity} x*2^(-n))=lim_{n to infinity} f(x*2^(-n))=lim_{n to infinity} f(x)=f(x) for any x in \mathbb{R} so only the constant functions satisfy the condition f(x)=f(2x)
Once it's proven an even function and can limit the analysis to positive values, since the "repeated square root" construction sends every positive number to 1 in the limit, you don't need to do the "repeated squaring" construction to send (-1,1) to 0
You can assign a value only to a countable number of points of an everywhere discontinuous function. This means there will still be an uncountable number of functions that satisfy this equation, but have not been described.
6:40 actually it works for the set (0,infinity) and by continuity you can fill in the hole of 0 as simply being "the common limit as x goes to zero from either side"
I'm a little sad that you didn't explore this equivalence relation a little further. Especially, what is a good complete set of representatives to define f on. This is rather simple: First the equivalence class {0}, which is a singleton, then we have the class {-1,1}, so let's add 0 and 1 to our set. Next, we would like representatives for all classes of numbers > 1. This can be achieved by choosing any number x>1. The interval [x,x^2) contains representatives for all classes for numbers bigger than 1 (answering why is left as an exercise). A similar thing can be done for the classes between 0 and 1 by picking an interval (y^2,y] for some y between 0 and 1. In particular, the set {0,1} union [x,x^2) union (y^2,y] is a complete set of representatives which is sufficient to define f completely.
To add to this. You can generalize this of course to any even power k instead of 2. A set of representatives is then given by {0,1} union [x,x^k) union (y^k,y] You can also replace the 2 with any odd power j. Then you have to handle the negative numbers separetely so a set of representatives becomes {-1,0,1} union (-x^j,-x] union [-y,-y^j) union (y^j,y] union [x,x^j). And finally, you can extend this to any real power by again replacing the 2 in the interval definitions by this real number and restricting yourself to the positive real axis. Alternatively, you could move this problem to the complex plane in that case and see what happens there.
@@christophdietrich4240 I agree exploring the equivalence relation is pretty interesting and wish Peyam had gone down that rabbit hole a bit further. Nice breakdown of the findings though Chris! What the equivalence relation looks like on the complex plane would be tricky. I'm guessing if you looked at the polar coordinates the domain of r would be the same as the real case but I'm not too sure what the domain of the angle would be. Unlike for the real case there's also the mod 2*pi equivalence which complicates things. I'm pretty sure the answer will require axiom of choice since for most intervals I've tried out there seems to always be a close pair of angles within the set that are equivalent for e.g. 8(0.286 pi) = 0.288 pi mod 2 pi 32(0.3226 pi) = 0.3232 pi mod 2 pi 128(1.6063 pi) = 1.6064 pi mod 2 pi I think there is a result about the orbit of an irrational number under the map T(x) = 2x mod 1 being dense in the interval [0,1] which can be used here to show that every interval no matter how small contains a pair of equivalent angles.
@@christophdietrich4240 great job! What about the topolgy of a such set? It would be evdn connected? I tried, withou any sucess, to find some group action to establish the equivale relation and apply some results.
@@diegohcsantos Well, such a set is just the union of a couple of half open intervals not containing 0 or 1 with the singletons 0 and 1. So no, it would not be connected.
@@diegohcsantosThe group action is the Z-action induced by the bijection x -> x² on [0, + infinity). In general "solving" a functional equation of the form f•g = f, for a bijection g, is the same thing as describing the orbits of g
Obviously, the quotient has at most the cardinality of R. If we also assume CH holds then I think it's enough to observe that if the cardinality of the quotient were less than aleph(1) then it is at most countable. Since each class is itself countable, it has zero Lebesgue measure. Also, given that the classes form a partition of the reals and our assumption that there is a countable number of them, countable additivity implies that R has zero Lebesgue measure, which is a contradiction. PS: I know proofs by contradiction are a bit frowned upon, not very insightful and a somewhat lazy way to prove things. But hey, it's still a proof. 😂 I also realize that the challenge was to find an explicit bijection, but I couldn't resist.
Intuitively, I would say that making such a bijection would require the axiom of choice, and you're not going to be able to specify such a bijection explicitly.
@@modestorosado1338 Hahaha. I know what you mean xD There's a bijection from the equivalence classes to the interval [0,1] so in particular the set has the same cardinality as R. First we'll show there is a bijection h from the equivalence classes excluding [0] to the interval [1,4], namely the function h([1]) = 4 and h([x]) = x^{2^m} where m is chosen such that the output value is in the interval [2,4) for x>1 and h([x]) = 4x^{2^m} where m is chosen such that x^{2^m} is in then interval [1/4,1/2) when 0
@@elephantdinosaur2284 Very clever. I didn't think it was possible to describe the bijection so explicitly. I had posted another comment pointing out that the way you had defined h didn't work for [x] when 0
It turns out that while such a function can't be nonconstant and continuous when defined on all the reals, if we restrict our domain to reals strictly greater than 1 it is not difficult to find nonconstant continuous examples.
@@adityadwivedi4412 I don't think that the domain being complete obliges the function to be constant. Assuming that such a non-constant function can be found on x>1, as asserted by Mitch Kovacs, then when restricted to x≥2 it will have the same property.
Such a function is sin(2πlog_2(log_2(x))). This is because sin(2πx) has a period of 1, and if you square x, you double log_2(x) and increase log_2(log_2(x)) by 1, so the function remains the same. Note that the domain of log_2(log_2(x)) is x>1. One can easily adapt this argument to show that such functions are precisely the continuous periodic functions of period 1 applied to log_2(log_2(x)): If f(x) is continuous function on x>1 such that f(x^2)=f(x) for all x>1, let g(x)=f(2^2^x). Then g is continuous and g(x+1)=f(2^2^(x+1))=f(2^(2×2^x))=f((2^2^x)^2)=f(2^2^x)=g(x), so g is periodic with period 1. The converse argument is as I gave for the specific example above.
For the domain [2,infty) I think we can extend any continuous function h defined on the interval [2,4] with h(2)=h(4) to the entire domain by setting f(x) = h([x]). Let (x_n) be a sequence converging to L. For some M we have 2
Could this also be done by a proof by contradiction? Assume a non constant function f exists such that f(x) = f(x^2) Therefore f inverse (f(x)) = f inverse (f(x^2)) therefore x=x^2 for all x? Which is a contradiction Okay I just realised this only works for one is to one functions but thought I might as well comment anyways to see what you all think
I think in this type of exercise occurs the confussion of one-one functions and some of injective functions. With this we can prove that a-a = b-b , a=b so this is false and with this type of functions one-one and a little bit of code we can say that a=b in this case is a simple confussions and lack of understanding of this terms
In your very first line you have to be careful that you are making an extra assumption that your function is one to one and therefore has an inverse. I actually would have just been stuck at x=0, x=1. Good thing you know your ontology - can tell apart apples from oranges.
@@theproofessayist8441 thanks for reply. Yeah, a lot of assumptions are made! :D For example x [as in f(x) ] is a variable but then I find x value (like a constant) must be equal to 0 and 1, and back to variable. This could easily confuse me when I was a teenager, lol.
f(x)= x^(1/Logb(x)) f(x^n)=(x^n)^(1/Logb(x^n)) =(x^n)^(1/(n•Logb(x))) =x^(n/(n•Logb(x))) =x^(1/Logb(x)) =f(x) →f(x^n)=f(x) →f(x^2)=f(x) b and n can be every thing,but not zero. I have a question. What is f(0) and f(1)?
Nice number identities! There's also: 4^3 + 0^3 + 7^3 = 407 9^4 + 4^4 + 7^4 + 4^4 = 9474 8^4 + 2^4 + 0^4 + 8^4 = 8208 The below pattern stops with 2401: square root(81) = 8 + 1 cube root(512) = 5 + 1 + 2 fourth root(2401) = 2 + 4 + 0 + 1 The number 2401 is the largest number that satisfies "for an n-digit number the nth root gives the sum of the digits of the number".
Can “all real numbers” whatever that even means be defined as equivalence classes under such a function? EDIT: (less confrontationally) can all real numbers be defined in such an equivalence class?
Let A be the set of all algebraic numbers And define f to be characteristic function of A then if x is an element of A then x² is also an element of A since (A,•) is a group . And if x is not an element of A i.e a transcendental number then x² is definitely not an elementary of A. Coz if it does then x= √x² is algebraic. Which proves that for any real number x of IR the relation f(x) = f(x²) holds.
AGHH i am still in tenth grade, but i wish to apply to UC berkeley and get my engineering undergrad for aerospace. then i wll try to pursue astrophysics as it is my goal in life to become an astrophysicist and research!!!
Hey! We have quite the similar goals! I always wanted to be an astrophysicist or a theoretical physicist. I've adored space since forever, almost since i was a little kid. And I see the same enthusiasm I had, back when I was in my 10th grade in you! Its quite touching and makes me feel nostalgic, ahh the simpler times, ehm i digress but anyways. Make sure you work REALLY REALLY hard for these next two years, these years matter way more than the last 10 did, trust me on this. And yes as always I wish you the best of luck in your future endeavors. Ahaha it's always great to meet people with mutual interests especially when those interests are physics, math and space it really does strike a chord within me. I bet you must be a really fun person to talk to in person :) Well anyways thanks for reading my long rant. And most of all have fun, there's a lot of things yet to be unraveled in this vast cosmos, lots of things to be discovered, lots of possibilities!
I were also surprised xD Is this a common Germanism in English? 🤔 - like Kindergarten or entgegen/zusammen (in stereochemistry) xD I'm from Germany ^^ PS: Nice video 👍😉
@@martinnimczick839 I think a fair few english speakers use it. It's not super common, and not rarely used, somewhere in between but more on the uncommon side now a days :)
@@omkarnathgupta484 yes but f(x) = f(x²) should be true for every x not just the cube roots of unity. Also f is defined as a real function so x does belong to R.
This thumbnail made me think of f(x^2) = f(x)^2, which means f must be a function which preserves square-root-ness (for x, x^2, no matter how many times you apply f to both, the left is always sqrt right
Isn't this basically almost identical to your f(2x)=f(x) video? If g(x) = f(e^x), then we are led to g(2x) = g(x), etc. Combine this with the fact that f(x^2) = f(x) implies f is even, and you have the current video as a consequence.
Oh no. Continuous is not that funny. sin also too easy. I propose C+{ln2(ln2(x))}. But yep, [1;~]. Replacement x=2^2^t do the trick. f(2^2^t)=f((2^2^t)^2)=f(2^2^(t+1). f(t)=f(t+1). Any periodic gives solution. But f(ln2(ln2(x)). Also it can be shifted left-right and up-down, so solution will have 2 constants
Ah ha! Constant functions and piecewise functions!
Hello, blackpenredpen. Nice to see you here.
Nice try but piecewise without specifying what pieces is a bit of a scam lol. All functions are piecewise constant if i take my pieces to be singletons 😂
General solution: f(x)=g(ln(ln(|x|))), for |x|>1, f(x)=h(ln(-ln(|x|))), for |x|
An example of f would be sin((2pi/ln(2))lnav(lnav(x))), where lnav is natural log of absolute value, which is of course not continuous at 1 so your point still stands
great answer. I think it could be any infinite sum of such a function and all its harmonics:
SUM[k=1->inf] {A-k * Sin( (2 * PI * k / ln|2| ) * ln| ln|x| | + THETA-k)
where A-k are scaling factors and THETA-k phase offset factors
Note: this answer is mathematically equivalent to Dr. Peyam's solution
This is dangerously close to the Vitali set construction. You might make a non-measurable set by accident! Don't do it, Dr. Peyam! You have so much to live for!
This is somewhat similar to a question I got for my Cambridge interview, which was "find a continuous function f such that f(x) = f(2x)". In that vein, I propose: f(x) = sin(2pi * log2(ln(x))).
f(x)=f(x*2^z) for all z in \mathbb{Z}
using the continuity, we get f(0)=f(lim_{n to infinity} x*2^(-n))=lim_{n to infinity} f(x*2^(-n))=lim_{n to infinity} f(x)=f(x) for any x in \mathbb{R}
so only the constant functions satisfy the condition f(x)=f(2x)
@@Leidl.Michael Differentiability continuity ;)
@@Convergant? what do you mean? wheres my mistake
@@Leidl.Michael Differentiability continuity
@@Convergant Continuity doesn't imply differentiability
yay, you have all three on your channel now :)
f²(x) = f(x)
f(x²) = f(x)
f(x)² = f(x)
Oh really? Wow 🤩
Aren't the first and last the same?
@@vinlebo88 I think the first one is using iterated function notation: f(f(x)) = f(x)
@@vinlebo88 What Aaron said :)
@@AaronRotenberg thanks
Once it's proven an even function and can limit the analysis to positive values, since the "repeated square root" construction sends every positive number to 1 in the limit, you don't need to do the "repeated squaring" construction to send (-1,1) to 0
You can assign a value only to a countable number of points of an everywhere discontinuous function. This means there will still be an uncountable number of functions that satisfy this equation, but have not been described.
You should explore all the continuous functions that satisfy this equation on other subsets of R
I'm sure Peyam puts some math questions in German for bonus points. Anyone who listens to his German will be able to solve them 😂
It’s true though lol
6:40 actually it works for the set (0,infinity) and by continuity you can fill in the hole of 0 as simply being "the common limit as x goes to zero from either side"
The perfect set to use where all numbers are unlinked by f is: [0] union [1] union [2,4) union (1/4,1/2]
Whatever happens on Dr Peyam's channel, stays on Dr Peyam's channel.
What about colabs
Unless it goes to bprp
I dont understand a word, but I love your enthusiasm
du sollst deutsch lernen 😂😂😂😂
Closed Set is the Vegas of Math! 😜
I'm a little sad that you didn't explore this equivalence relation a little further. Especially, what is a good complete set of representatives to define f on. This is rather simple: First the equivalence class {0}, which is a singleton, then we have the class {-1,1}, so let's add 0 and 1 to our set. Next, we would like representatives for all classes of numbers > 1. This can be achieved by choosing any number x>1. The interval [x,x^2) contains representatives for all classes for numbers bigger than 1 (answering why is left as an exercise). A similar thing can be done for the classes between 0 and 1 by picking an interval (y^2,y] for some y between 0 and 1. In particular, the set {0,1} union [x,x^2) union (y^2,y] is a complete set of representatives which is sufficient to define f completely.
To add to this. You can generalize this of course to any even power k instead of 2. A set of representatives is then given by {0,1} union [x,x^k) union (y^k,y] You can also replace the 2 with any odd power j. Then you have to handle the negative numbers separetely so a set of representatives becomes {-1,0,1} union (-x^j,-x] union [-y,-y^j) union (y^j,y] union [x,x^j). And finally, you can extend this to any real power by again replacing the 2 in the interval definitions by this real number and restricting yourself to the positive real axis. Alternatively, you could move this problem to the complex plane in that case and see what happens there.
@@christophdietrich4240 I agree exploring the equivalence relation is pretty interesting and wish Peyam had gone down that rabbit hole a bit further. Nice breakdown of the findings though Chris!
What the equivalence relation looks like on the complex plane would be tricky. I'm guessing if you looked at the polar coordinates the domain of r would be the same as the real case but I'm not too sure what the domain of the angle would be. Unlike for the real case there's also the mod 2*pi equivalence which complicates things.
I'm pretty sure the answer will require axiom of choice since for most intervals I've tried out there seems to always be a close pair of angles within the set that are equivalent for e.g.
8(0.286 pi) = 0.288 pi mod 2 pi
32(0.3226 pi) = 0.3232 pi mod 2 pi
128(1.6063 pi) = 1.6064 pi mod 2 pi
I think there is a result about the orbit of an irrational number under the map T(x) = 2x mod 1 being dense in the interval [0,1] which can be used here to show that every interval no matter how small contains a pair of equivalent angles.
@@christophdietrich4240 great job! What about the topolgy of a such set? It would be evdn connected? I tried, withou any sucess, to find some group action to establish the equivale relation and apply some results.
@@diegohcsantos Well, such a set is just the union of a couple of half open intervals not containing 0 or 1 with the singletons 0 and 1. So no, it would not be connected.
@@diegohcsantosThe group action is the Z-action induced by the bijection x -> x² on [0, + infinity). In general "solving" a functional equation of the form f•g = f, for a bijection g, is the same thing as describing the orbits of g
Thank you so much
What about the case:
f(x)=f(x²+¼) when f is cont. ?
Hoping you reply
So clear explanation ❤️❤️❤️❤️. Loved it😁
Nice video peyam. Finally on midterm so I can watch more of your videos
Please consider the function f(x)=sin(2pi*log_2(log_2(x))).
Challenge: Find the cardinality of this equivalence relation by establishing an explicit bijection between the classes and a more well-known set. 😉
Obviously, the quotient has at most the cardinality of R. If we also assume CH holds then I think it's enough to observe that if the cardinality of the quotient were less than aleph(1) then it is at most countable.
Since each class is itself countable, it has zero Lebesgue measure. Also, given that the classes form a partition of the reals and our assumption that there is a countable number of them, countable additivity implies that R has zero Lebesgue measure, which is a contradiction.
PS: I know proofs by contradiction are a bit frowned upon, not very insightful and a somewhat lazy way to prove things. But hey, it's still a proof. 😂
I also realize that the challenge was to find an explicit bijection, but I couldn't resist.
Intuitively, I would say that making such a bijection would require the axiom of choice, and you're not going to be able to specify such a bijection explicitly.
@@modestorosado1338 Hahaha. I know what you mean xD
There's a bijection from the equivalence classes to the interval [0,1] so in particular the set has the same cardinality as R.
First we'll show there is a bijection h from the equivalence classes excluding [0] to the interval [1,4], namely the function h([1]) = 4 and h([x]) = x^{2^m} where m is chosen such that the output value is in the interval [2,4) for x>1 and h([x]) = 4x^{2^m} where m is chosen such that x^{2^m} is in then interval [1/4,1/2) when 0
@@elephantdinosaur2284 Very clever. I didn't think it was possible to describe the bijection so explicitly. I had posted another comment pointing out that the way you had defined h didn't work for [x] when 0
@@modestorosado1338 Thanks Modesto! Yeah I only noticed my error after I posted. My bad xD
brilliant explanation
It turns out that while such a function can't be nonconstant and continuous when defined on all the reals, if we restrict our domain to reals strictly greater than 1 it is not difficult to find nonconstant continuous examples.
@@adityadwivedi4412 I don't think that the domain being complete obliges the function to be constant. Assuming that such a non-constant function can be found on x>1, as asserted by Mitch Kovacs, then when restricted to x≥2 it will have the same property.
Such a function is sin(2πlog_2(log_2(x))). This is because sin(2πx) has a period of 1, and if you square x, you double log_2(x) and increase log_2(log_2(x)) by 1, so the function remains the same. Note that the domain of log_2(log_2(x)) is x>1.
One can easily adapt this argument to show that such functions are precisely the continuous periodic functions of period 1 applied to log_2(log_2(x)):
If f(x) is continuous function on x>1 such that f(x^2)=f(x) for all x>1, let g(x)=f(2^2^x). Then g is continuous and g(x+1)=f(2^2^(x+1))=f(2^(2×2^x))=f((2^2^x)^2)=f(2^2^x)=g(x), so g is periodic with period 1. The converse argument is as I gave for the specific example above.
For the domain [2,infty) I think we can extend any continuous function h defined on the interval [2,4] with h(2)=h(4) to the entire domain by setting f(x) = h([x]).
Let (x_n) be a sequence converging to L. For some M we have 2
Sir, you're a gentleman.
Could this also be done by a proof by contradiction? Assume a non constant function f exists such that f(x) = f(x^2)
Therefore f inverse (f(x)) = f inverse (f(x^2)) therefore x=x^2 for all x? Which is a contradiction
Okay I just realised this only works for one is to one functions but thought I might as well comment anyways to see what you all think
The fly on the left side at 6:25 is the unsung hero of the video
I think in this type of exercise occurs the confussion of one-one functions and some of injective functions. With this we can prove that a-a = b-b , a=b so this is false and with this type of functions one-one and a little bit of code we can say that a=b in this case is a simple confussions and lack of understanding of this terms
Please can you do it for me : f(n pow 2 + m pow 2)=f(n) pow 2 + f(m) pow 2. f from N to N
I feel sad or bored I watch Dr Peyam
Love your videos!!
Dr Peyam, Please do a video on the properties of the function f(x)=sin(cos(tan(x)))
What properties? lol
This is the first time i actually understood this thank you very much
Me, has high-school level of maths:
f(x²) = f(x)
x² = x
x = 0, x = 1
f(x) = f(x²) = f(0) = f(1) means f is constant function
That's all :D
I realised I was stuck and that's not a solution, after watching your video.
Thanks!
In your very first line you have to be careful that you are making an extra assumption that your function is one to one and therefore has an inverse. I actually would have just been stuck at x=0, x=1. Good thing you know your ontology - can tell apart apples from oranges.
@@theproofessayist8441 thanks for reply. Yeah, a lot of assumptions are made! :D For example x [as in f(x) ] is a variable but then I find x value (like a constant) must be equal to 0 and 1, and back to variable. This could easily confuse me when I was a teenager, lol.
Given f(x), can you find g such that f(x)=g(g(x))?
4:00 as above, so below
f(x)= x^(1/Logb(x))
f(x^n)=(x^n)^(1/Logb(x^n))
=(x^n)^(1/(n•Logb(x)))
=x^(n/(n•Logb(x)))
=x^(1/Logb(x))
=f(x)
→f(x^n)=f(x)
→f(x^2)=f(x)
b and n can be every thing,but not zero.
I have a question. What is f(0) and f(1)?
I only want to show how a funktion f can look like,if f(x)=f(x²) and ask something.
Thanks sir
Respect from INDIA
Please make video on numbers of types
1^3 + 5^3 +3^3 = 153
3^3 + 7^3 +0^3 = 370
3^3 + 7^3 +1^3 = 371
Cube root(19683) =1+9+6+8+3 = 27
1^4 + 6^4 +3^4 + 4^4 = 1634
(6*9) + (6+9) = 69
Nice number identities! There's also:
4^3 + 0^3 + 7^3 = 407
9^4 + 4^4 + 7^4 + 4^4 = 9474
8^4 + 2^4 + 0^4 + 8^4 = 8208
The below pattern stops with 2401:
square root(81) = 8 + 1
cube root(512) = 5 + 1 + 2
fourth root(2401) = 2 + 4 + 0 + 1
The number 2401 is the largest number that satisfies "for an n-digit number the nth root gives the sum of the digits of the number".
Can “all real numbers” whatever that even means be defined as equivalence classes under such a function?
EDIT: (less confrontationally) can all real numbers be defined in such an equivalence class?
Yes indeed, it’s called the Cauchy construction of R, there’s some video on that on my channel
Without the continuous hypothesis, you could say the situation is not.... IDEAL !
Or is it ?
Let A be the set of all algebraic numbers And define f to be characteristic function of A then if x is an element of A then x² is also an element of A since (A,•) is a group . And if x is not an element of A i.e a transcendental number then x² is definitely not an elementary of A. Coz if it does then x= √x² is algebraic. Which proves that for any real number x of IR the relation f(x) = f(x²) holds.
Do you know how resolve (f(x))^2 = f(x)?
For each x, f(x) = 0 or 1
From what competition is that?
Came up with it myself
This question literally was on my Calc 1 exam today
AGHH i am still in tenth grade, but i wish to apply to UC berkeley and get my engineering undergrad for aerospace. then i wll try to pursue astrophysics as it is my goal in life to become an astrophysicist and research!!!
Hey! We have quite the similar goals! I always wanted to be an astrophysicist or a theoretical physicist. I've adored space since forever, almost since i was a little kid. And I see the same enthusiasm I had, back when I was in my 10th grade in you! Its quite touching and makes me feel nostalgic, ahh the simpler times, ehm i digress but anyways. Make sure you work REALLY REALLY hard for these next two years, these years matter way more than the last 10 did, trust me on this. And yes as always I wish you the best of luck in your future endeavors. Ahaha it's always great to meet people with mutual interests especially when those interests are physics, math and space it really does strike a chord within me. I bet you must be a really fun person to talk to in person :) Well anyways thanks for reading my long rant. And most of all have fun, there's a lot of things yet to be unraveled in this vast cosmos, lots of things to be discovered, lots of possibilities!
@@manamritsingh969 Thanks a lot :)
Did he really say „you can play this Spiel again“?
I did!
I were also surprised xD
Is this a common Germanism in English? 🤔 - like Kindergarten or entgegen/zusammen (in stereochemistry) xD
I'm from Germany ^^
PS: Nice video 👍😉
@@martinnimczick839 I think a fair few english speakers use it. It's not super common, and not rarely used, somewhere in between but more on the uncommon side now a days :)
So then, isn't the solution to this just the constant function f(x) = c?
If we use f(x)=1+x²+x then f(w)=f(w²)
No f(w²) = 1 + w⁴ + w²
If w is cube root of unity
In qstn there is not mention that x belongs to real
@@omkarnathgupta484 yes but f(x) = f(x²) should be true for every x not just the cube roots of unity.
Also f is defined as a real function so x does belong to R.
Ok👍
There is a book to study calculus in this way?
I think spivak or apostol
@@drpeyam Thank you!
This thumbnail made me think of f(x^2) = f(x)^2, which means f must be a function which preserves square-root-ness (for x, x^2, no matter how many times you apply f to both, the left is always sqrt right
wonderful professor
Isn't this basically almost identical to your f(2x)=f(x) video? If g(x) = f(e^x), then we are led to g(2x) = g(x), etc. Combine this with the fact that f(x^2) = f(x) implies f is even, and you have the current video as a consequence.
It is!
The moment I read the title, I immediately thought "easy, constant function"
I feel like I am part of a closed but infinite set of your subscribers.
What if we did f(x^2)=f(f(x))?
Then f(x) = x^2. At least - 1 of the solutions
I wish everyone could speak English as you do...
nice tie and shirt
I'm a Japanese high school student. I couldn't understand what you said
私は日本人ですが、日本の方に視聴して頂けて嬉しいです
@@aashsyed1277 I think so too.
My favorite are function ,shape equation, vector and differential calculus．
Lets play the Spiel again 😅🙌🙌
Nice ❤
i have a Déjà vu = Déjà vu🟩
To extend to negative #'s (even continuation) ,we need absolute value i guess:
f(x)=f(|x|)=f(|x|^(2*1/2))=f(|x|^(1/2))=f(|x|^(1/4))=...
No entendí nada pero sos un genio
very nice
Very interesting
██ For x=0 & x=1 - f(x)=anything. You don't analyze these 2 cases.
Yeah I do, [0] and [1] are equivalence classes of their own
@@drpeyam f(0)=3 -> f(0^2)=3, f(1)=657 -> f(1^2)=657
😲 wow
So, not real answer.
how can x^2 = x, doesnt make sense.
SPIEL MIT MIR!
Oh no. Continuous is not that funny. sin also too easy. I propose C+{ln2(ln2(x))}. But yep, [1;~]. Replacement x=2^2^t do the trick. f(2^2^t)=f((2^2^t)^2)=f(2^2^(t+1).
f(t)=f(t+1). Any periodic gives solution. But f(ln2(ln2(x)). Also it can be shifted left-right and up-down, so solution will have 2 constants
advice: change notation. let g(t) = f(2^2^t) = f(x), so we may have our original condition be g(t) = g(t+1).
Its just 0
My guess: f is just a constant fumction
@@cmdlp4178 no, it has to be either o or 1. Not just any const. Number....
@@איתיסמואלוב f(x)=C, f(x^2)=C
f(x)=f(x^2)
@@איתיסמואלוב No. Any constant function works. Your misconception is that for for x>=1, f(x) = f(1), and same for x
Spiel mit mir 😂😂😂
314th like