Take x. Write it in base 3. Chop off the expansion after the first 1, or let it run forever if it’s in the Cantor set. Replace all 2’s with 1’s. Read in binary. This is f(x).
@@sergiokorochinsky49 Not exactly. I have very strong intuition for this, but if I were to prove it formally it would be a bit tricky. Here's a sketch of how you could try to do it. You can prove it for numbers that have a 1 somewhere, by induction in the number of digits before the 1. That determines what the function is for a dense subset of [0,1], and then the condition of not decreasing means that the formula also works for numbers with no 1s.
@@sergiokorochinsky49 Proof first the following If 0 < x < 1 / 3 then f ( x + 2 / 3 ) = 1 / 2 + f ( x ) From there and the fact that f ( x / 3 ) = f ( x ) / 2, the statement is "almost" obvious
This one was AWESOME. I'm totally stoked that I came close to solving it yesterday, put it on pause, slept on it, and then got it tonight. Thank you, Michael.
From the thumbnail before seeing the extra conditions, the solution to f(1/13) is clearly available without finding nice values or knowing the function is non-decreasing. That's because the only mappings available on f(p/q) are p -> 3p and p -> q-p and you can see there is a chain from p=1 to p=12 and back to p=1 via p1=1; p2=3x1=3; p3=3x3=9; p4=13-9=4; p5=3x4=12; p6=13-12=1. So, setting f(1/13) = u and following the chain: f(3/13) = 2u f(9/13) = 2*2u = 4u f(4/13) = 1 - 4u f(12/13) = 2(1-4u) = 2 - 8u f(1/13) = 1 - (2-8u) = 8u - 1 but that is equal to u; so 8u - 1 = u and so 7u= 1 and f(1/13) = u = 1/7 This method should be generic for solving these sort of functional equations as long as a chain exists. I don't yet know whether such a chain will exist for any f(1/n) or for arbitrary pairs of mappings.
We have to find f(145/2020) , i.e. f(29/404) Now 29/404 = 0.07 (Approx.) Then (29/404)*3*3 = 0.63 which lies in [1/3,2/3] Therefore we can just use the first rule twice f(145/2020) = f(29/404) = (1/2)*{f(87/404)} = (1/2)*(1/2)*{(f(261/404)} = (1/4)*(1/2) = 1/8
Plot of Cantor function. en.m.wikipedia.org/wiki/Cantor_function Your explanations are on point; this is truly one of the top math channels on YT. More power to you!
Bam! After watching like 25 of these Michael Penn videos I finally solved one myself before watching the video! Thanks Michael! Now I finally feel like I can buy a shirt because I earned it.
I calculated f(1/13) by a rotation of fractions. Called f(1/13)=x, obtaining f(3/13)=2x and f(9/13)=4x. Then f(4/13)= 1-4x and f(12/13) = 2-8x. Finally, f(1/13)=8x-1 but also x,then, f(1/13)=x=1/7.
do you (or anyone else) think that we have to prove that a function defined as in the statement can exist? What i mean by this is, what if by applying those 2 rules in the statement [F(x)=1/2F(3x); F(1-x)=1-F(x)] and the trick involved in finding F(1/13), we find out that for some particular values x1>x2, F(x1)
OH boy we're talking about the cantor function here. To solve this problem it would be better if you just knew more about math. Cause if you don't know about the cantor set and the cantor function beforehand (I didn't, i tried to *draw* the function, imagine that) it's gonna be pretty hard to get an instant idea of what you wanna do.
@@jongyon7192p You definitely did. Just look up "cantor funtion" in pbs infinite series (it's a youtube channel) and you'll see right away why your first 5 examples don't make any sense. (Clue - Look at the domain and range of the function).
I found a way to compute f(x) for any x: Write x in ternary (base 3) and remove all digits after the first instance of the number 1 in the number. Then, replace all instances of the number 2 with a one. The resulting number interpreted in base 2 is f(x)
[1/3^n;2/3^n] -> 1/2^n | [7/3^(n+1);8/3^(n+1)] -> 3/2^(n+1) We see that the function is constant in other intervals as well. 1/13 is this a special limit point? What's going on around this point? 1/14 in [1/27;2/27] -> 1/8
I originally noticed that the function was some kind of fractal, with thirds of thirds of thirds of thirds... and I thought that I'd need to do some infinite product or something on 1/3 to get 1/13
As others have done using the first equation, f(145/2020)=f(435/2020)...=(1/4)*(1/2)=1/8. Interesting problem. I didn't know f(x) is called Cantor function.
Please help me in this question If each of the algebraic expressions lx^2 + mx + n mx^2 + nx + l nx^2 + lx + m Are perfect squares, then proove that (l+m)/n=-4 Plz help😢😢
(I'm not a math person so correct me where I'm wrong) If lx^2+mx+n is a square then it should be of the form (A+x)^2. Then it has a single root. Then the delta term (b^2-4ac, which shows up when solving quadratics) has to be zero. That means m^2-4ln=0 as well as l^2-4mn=0. Subtract these 2 equations: m^2-4ln-l^2+4mn=0. Collect the terms with n on the right side: m^2-l^2=4n(l-m). Rewrite left side: (m-l)(m+l)=4n(l-m). Simplify (m-l): m+l=-4n. And it's done: (m+l)/n=-4
Absolutely, just look at the question / remark of noahtaul above. Michael Penn does'nt proove the existence of such a function but it does'nt matter now.
@@reynaldopanji2066 it is left to the reader to understand that. Well if the input is the same and the function yields two or more different values for the same input, then obviously the given equation is not a function because it goes against the definition of a function and if this is the case we don't care about that because we're talking about functions here.
Yes, it is. See here: mathworld.wolfram.com/CantorFunction.html "Chalice (1991) showed that any real-valued function F(x) on [0,1] which is monotone increasing and satisfies 1. F(0)=0, 2. F(x/3)=F(x)/2, 3. F(1-x)=1-F(x) is the Cantor function (Chalice 1991; Wagon 2000, p. 132)."
f(145/2020) = (1/2) * f(435/2020) = (1/4) * f(1305/2020) = (1/4) * (1/2) = 1/8 This reminds me of the Cantor's middle third set. But how to show that this function is differentiable nowhere on (0,1)?
It is not true that the function is differentiable nowhere in ( 0 , 1 ) For example, the function is differentiable in the open interval ( 1 / 3 , 2 / 3 ) with f ' ( x ) = 0
The function in this puzzle is fascinating. I worked out how it acts on all of [0,1] before discovering it has a name, the Cantor function. I should have guessed based on its action on the open middle third Cantor set C. Letting Q be the rationals, I be the irrationals, D be the Dyadic rationals and T be the 3-adic rationals, then f maps lots of interesting sets to other interesting sets: f(Q ∩ [0,1]) = Q ∩ [0,1], f(T ∩ [0,1]) = D ∩ [0,1], f([0,1]\C) = D ∩ (0,1), f((C ∩ Q)\T) = (Q ∩ [0,1])\D, f(C ∩ I) = I ∩ [0,1].
One nice fact that highlights the symmetry of this equation: We have two expressions for f(x): The one you found: f(x) = (1/2)f(3x) and if we multiply the original by 2 we get: f(x) = 2f(x/3)
But have we even proven, that a solution to the functional equation exists? We only have shown, that if a solution exists, it must have the found values at the given points. We haven't even proven, that one could find different values by similar methods, wich would show, that no such solution exists.
Hi, When a non-increasing (or non-decreasing) function, defined on [a,b], takes the same value in a and b, whe can say that it is constant only if it is continue, and, here, we don't know if it is.
You don't even need the notion of continuity for that to happen. It's pretty much trivial but it was also proved in the video. Here is another proof: by absurd suppose f(c) is not equal to f(a), if it's less than f(a) then the function decreased: f(c)a. If it's greater than f(a) you have f(c)>f(b) with c
"Non-decreasing" IS increasing - there is no ambiguity. For a function to be (monotonic) increasing the condition is that for all 'a' greater 'b' the result f(a) must be greater OR EQUAL to f(b). You are confusing "increasing" and "strictly increasing" there.
I don't like these kind of exercises because it's very very difficult to come up with the solution, u have to know previous results and on top of that (the worst) u can't generalize it ,that means that in the best scenario u could remember how to solve it , it only deals with exactly this problem , u shift a little bit to other scenario and bad luck u are done, I'm not attracted at all Most difficult dosent mean most interesting (look at the amount of suscribers) Look the amount the subscribers of mind your decisions ,terrible simple visual exercises hiding big puzzles ,a huuuge amount of subscribers Complexity is inverse to the amount of subscribers at some point A)mind your decisions is more visual and has simple exercises 2M suscribers B) a more visual channel that dosent even solve problems less mental challenge is 3 brown 1 blue result? 3.6M suscribers .... Less complexity more suscribers which is a result that amazes me because I would think the most difficult more attractive that's why we all love maths but facts are facts
Since this exam was rigged in 1993, the problem was also faked out likewise. Notice the denominators, they must be mocking the nitpick nerdies taking the exam...
I looked up some physics olympiad problems and they are interesting. I would like to post one from time to time if there is interest. Like this comment if you are interested.
Take x. Write it in base 3. Chop off the expansion after the first 1, or let it run forever if it’s in the Cantor set. Replace all 2’s with 1’s. Read in binary. This is f(x).
That's how I did it!
Wow, that rather elegant. I thought about using iterations. How do plot looks like?
Is the proof of this affirmation supposed to be "obvious"?
@@sergiokorochinsky49 Not exactly. I have very strong intuition for this, but if I were to prove it formally it would be a bit tricky. Here's a sketch of how you could try to do it. You can prove it for numbers that have a 1 somewhere, by induction in the number of digits before the 1. That determines what the function is for a dense subset of [0,1], and then the condition of not decreasing means that the formula also works for numbers with no 1s.
@@sergiokorochinsky49 Proof first the following
If 0 < x < 1 / 3 then f ( x + 2 / 3 ) = 1 / 2 + f ( x )
From there and the fact that f ( x / 3 ) = f ( x ) / 2, the statement is "almost" obvious
This one was AWESOME. I'm totally stoked that I came close to solving it yesterday, put it on pause, slept on it, and then got it tonight. Thank you, Michael.
From the thumbnail before seeing the extra conditions, the solution to f(1/13) is clearly available without finding nice values or knowing the function is non-decreasing. That's because the only mappings available on f(p/q) are p -> 3p and p -> q-p and you can see there is a chain from p=1 to p=12 and back to p=1 via p1=1; p2=3x1=3; p3=3x3=9; p4=13-9=4; p5=3x4=12; p6=13-12=1.
So, setting f(1/13) = u and following the chain:
f(3/13) = 2u
f(9/13) = 2*2u = 4u
f(4/13) = 1 - 4u
f(12/13) = 2(1-4u) = 2 - 8u
f(1/13) = 1 - (2-8u) = 8u - 1 but that is equal to u; so 8u - 1 = u and so 7u= 1 and f(1/13) = u = 1/7
This method should be generic for solving these sort of functional equations as long as a chain exists. I don't yet know whether such a chain will exist for any f(1/n) or for arbitrary pairs of mappings.
7:51 Small correction but it should be f(436/1993) not f(440/1993).
8:27 This would also change f(1320/1993) to f(1308/1993).
The thumbnails are getting really interesting lately...
Yes agreed and the problems as well.
We have to find f(145/2020) , i.e. f(29/404)
Now 29/404 = 0.07 (Approx.)
Then (29/404)*3*3 = 0.63 which lies in [1/3,2/3]
Therefore we can just use the first rule twice
f(145/2020)
= f(29/404)
= (1/2)*{f(87/404)}
= (1/2)*(1/2)*{(f(261/404)}
= (1/4)*(1/2)
= 1/8
Small correction at 6:47. The value of 519/1993 is slightly BIGGER than a quarter (about 0.26).
That's ok, when he multiplied it by 3 que got approximately 0.78, so it didn't interfere in anything
@@pedroribeiro1536 Luckily :) It was just a slip of a tongue
12:36
Good place to start at 0:00
Lol you are so lucky ... Michael gives you a shoutout to you in every single one of his videos unintentionally!
Semper Fidelis
This is the best moment
Did you create an account just for this?
7:47 I think that 440 should be 436
Correct. "Luckily" it doesn't change the result.
Yeah its so lucky, imagine he wrote 443, so it will become 1329, but 1329/1993 is greater than 2/3 lol
Plot of Cantor function.
en.m.wikipedia.org/wiki/Cantor_function
Your explanations are on point; this is truly one of the top math channels on YT. More power to you!
Bam! After watching like 25 of these Michael Penn videos I finally solved one myself before watching the video! Thanks Michael! Now I finally feel like I can buy a shirt because I earned it.
wait, what about Sweden?
I calculated f(1/13) by a rotation of fractions. Called f(1/13)=x, obtaining f(3/13)=2x and f(9/13)=4x. Then f(4/13)= 1-4x and f(12/13) = 2-8x.
Finally, f(1/13)=8x-1 but also x,then, f(1/13)=x=1/7.
One can deduce what the function is for any real x in [0,1] by using ternary expansions
do you (or anyone else) think that we have to prove that a function defined as in the statement can exist? What i mean by this is, what if by applying those 2 rules in the statement [F(x)=1/2F(3x); F(1-x)=1-F(x)] and the trick involved in finding F(1/13), we find out that for some particular values x1>x2, F(x1)
OH boy we're talking about the cantor function here.
To solve this problem it would be better if you just knew more about math. Cause if you don't know about the cantor set and the cantor function beforehand (I didn't, i tried to *draw* the function, imagine that) it's gonna be pretty hard to get an instant idea of what you wanna do.
This question is already answered down in the comments
@@g0z3 did i make a mistake here?
f(3)=2
f(-2)=-1
f(-2/3)=-1/2
f(5/3)=3/2
f(5/9)=3/4
but 1/3
@@jongyon7192p The domain of f is set to [0,1], so most of those values are wrong. E.g., f(3) is undefined, not 2.
@@jongyon7192p You definitely did. Just look up "cantor funtion" in pbs infinite series (it's a youtube channel) and you'll see right away why your first 5 examples don't make any sense. (Clue - Look at the domain and range of the function).
This really brings me back to my high school math team days. Bergen County Academies (AAST back in the day)!
I found a way to compute f(x) for any x:
Write x in ternary (base 3) and remove all digits after the first instance of the number 1 in the number. Then, replace all instances of the number 2 with a one. The resulting number interpreted in base 2 is f(x)
This was an very interesting problem, I enjoyed it!
[1/3^n;2/3^n] -> 1/2^n | [7/3^(n+1);8/3^(n+1)] -> 3/2^(n+1)
We see that the function is constant in other intervals as well.
1/13 is this a special limit point?
What's going on around this point?
1/14 in [1/27;2/27] -> 1/8
(1/13)=0,(002) in 3 digit system. We enter a certain interval if there is a number one. => 1/13 special limit point.
Sweden also belongs to the Nordic countries.
4:48 is 440 just approximate?
I originally noticed that the function was some kind of fractal, with thirds of thirds of thirds of thirds... and I thought that I'd need to do some infinite product or something on 1/3 to get 1/13
As others have done using the first equation, f(145/2020)=f(435/2020)...=(1/4)*(1/2)=1/8.
Interesting problem. I didn't know f(x) is called Cantor function.
Please help me in this question
If each of the algebraic expressions
lx^2 + mx + n
mx^2 + nx + l
nx^2 + lx + m
Are perfect squares, then proove that (l+m)/n=-4
Plz help😢😢
(I'm not a math person so correct me where I'm wrong) If lx^2+mx+n is a square then it should be of the form (A+x)^2. Then it has a single root. Then the delta term (b^2-4ac, which shows up when solving quadratics) has to be zero. That means m^2-4ln=0 as well as l^2-4mn=0. Subtract these 2 equations: m^2-4ln-l^2+4mn=0.
Collect the terms with n on the right side: m^2-l^2=4n(l-m). Rewrite left side: (m-l)(m+l)=4n(l-m). Simplify (m-l): m+l=-4n. And it's done: (m+l)/n=-4
How could we have the graph of the function ?
Plot of Cantor function. en.m.wikipedia.org/wiki/Cantor_function
Awesome!🔥
Is there any book on problems on functional equations? If anybody knows?
Best thumbnail ever!
So does this give a continuous curve or steps?
Isn't this function a Cantor's Stairs?
Absolutely, just look at the question / remark of noahtaul above.
Michael Penn does'nt proove the existence of such a function but it does'nt matter now.
I don't understand one thing, in this video, If x
You're asking a question that doesn't make sense. How would a function assume a different value given the same input?
@@RandomBurfness Because it was written like that haha. I assume that it is possible, so not everything written is possible?
@@reynaldopanji2066 it is left to the reader to understand that. Well if the input is the same and the function yields two or more different values for the same input, then obviously the given equation is not a function because it goes against the definition of a function and if this is the case we don't care about that because we're talking about functions here.
@@jomama3465 Thank you for explaining sir, I do need some explanations for further learning not question for question 🙏
@@reynaldopanji2066 dont call me sir, I'm only 17😂
f(145/2020)=f(29/404)=f(7.25/101)=f(21.75/101)/2=f(65.25/101)/4=0.5/4=1/8=0.125, since 65.25/101 is a little less than 2/3
I really enjoyed this!
Has anyone thought to try and graph this function?
Yes, look up the cantor function
@@romajimamulo oh haha ofc... can’t remember wether I was trying to trick someone into doing this but yes that’s the one
7:50 it should be 436/1993.
In this case it doesn't have an impact on the answer.
Is this Cantor's function?
Yes, it is. See here: mathworld.wolfram.com/CantorFunction.html
"Chalice (1991) showed that any real-valued function F(x) on [0,1] which is monotone increasing and satisfies
1. F(0)=0,
2. F(x/3)=F(x)/2,
3. F(1-x)=1-F(x)
is the Cantor function (Chalice 1991; Wagon 2000, p. 132)."
@@alonamaloh no wonder he didnt try to obtain a closed form
Agh... you posted first.
I arrived at the same conclusion
@@angelmendez-rivera351 I noticed that, but I didn't write the Wolfram page I was quoting.
@@alonamaloh I'm guessing the proof of this shouldn't be hard. Might try it as an exercise.
Is it possible to find the expression of f(x)?
It makes me curious what the graph of this function would look like... But I think we'd need to get many more values to plot it approximately.
en.wikipedia.org/wiki/Cantor_function
Look up the Cantor function!
f(145/2020) = (1/2) * f(435/2020) = (1/4) * f(1305/2020) = (1/4) * (1/2) = 1/8
This reminds me of the Cantor's middle third set.
But how to show that this function is differentiable nowhere on (0,1)?
It is not true that the function is differentiable nowhere in ( 0 , 1 )
For example, the function is differentiable in the open interval ( 1 / 3 , 2 / 3 ) with f ' ( x ) = 0
This is the Cantor function!
oof, it's been a long time since I've done these and I didn't recognize the function right away. It just looked like functional whatever to me, lol.
Plot the graph of this function pls )
I want to know how to prove the function is well-defined. Maybe the function is somewhere decreasing or maybe some f(x) has different values.
The non decreasing property is part of the definition.
Am I correct? f(145/2020) = 1/8
I got the same.
Yes, just double the first rule
I think it would be great to see a kind of fractal zoom on this function.
The function in this puzzle is fascinating. I worked out how it acts on all of [0,1] before discovering it has a name, the Cantor function. I should have guessed based on its action on the open middle third Cantor set C. Letting Q be the rationals, I be the irrationals, D be the Dyadic rationals and T be the 3-adic rationals, then f maps lots of interesting sets to other interesting sets:
f(Q ∩ [0,1]) = Q ∩ [0,1], f(T ∩ [0,1]) = D ∩ [0,1], f([0,1]\C) = D ∩ (0,1), f((C ∩ Q)\T) = (Q ∩ [0,1])\D, f(C ∩ I) = I ∩ [0,1].
Can you upload videos of problem solving in continuity and differentiability also sir??
One nice fact that highlights the symmetry of this equation:
We have two expressions for f(x):
The one you found:
f(x) = (1/2)f(3x)
and if we multiply the original by 2 we get:
f(x) = 2f(x/3)
Can you also upload some physics videos ?
Can you please try AIME I problem 14 of 2012??
Please... Kind request Sir...
Thanks! Please, solve a few functional equations.
I wanna see its graph but I doubt that I can make it in desmos 😅
You can look up the cantor function
What function satisfy these conditions? (Not an fluent english speaker, dont bash me)
But have we even proven, that a solution to the functional equation exists?
We only have shown, that if a solution exists, it must have the found values at the given points.
We haven't even proven, that one could find different values by similar methods, wich would show, that no such solution exists.
Doesn't Sweden participate?
I did the same mistake, you have Iceland, but not Sweden, so it is not "Scandinavian", but "Nordic".
@@CM63_France sweden is a nordic country
they do, he misspoke
forgetting sweden smh 😔
Sorry! I'll do a problem from the Swedish math olympiad to make it up to you.
Fint namn 😊
Awesome.....
I think you forgot Iceland in you list of Nordic Countries
Solution to every problem is i8♾3.14
Hi,
When a non-increasing (or non-decreasing) function, defined on [a,b], takes the same value in a and b, whe can say that it is constant only if it is continue, and, here, we don't know if it is.
You don't even need the notion of continuity for that to happen. It's pretty much trivial but it was also proved in the video. Here is another proof: by absurd suppose f(c) is not equal to f(a), if it's less than f(a) then the function decreased: f(c)a. If it's greater than f(a) you have f(c)>f(b) with c
f(145/2020)=0.5f(435/2020)=0.25f(1305/2020)=0.25*0.5=0.125
Since f(1305/2020)≈0.65, it is between 1/3 and 2/3 which means it equals 0.5.
0.5*0.25= 0.125 Check your calculations
nice
That's cool
But, is the devil's staircase slippery?
I'm so early I can't even see the "that's a good place to stop" guy...
7/32
100k!
May I see a smile on your face next video??? 🤔🤔🤔
1/8
"Non-decreasing" IS increasing - there is no ambiguity.
For a function to be (monotonic) increasing the condition is that for all 'a' greater 'b' the result f(a) must be greater OR EQUAL to f(b). You are confusing "increasing" and "strictly increasing" there.
Calculation ugh!!
Mike you're always scratching, in other videos also.
I don't like these kind of exercises because it's very very difficult to come up with the solution, u have to know previous results and on top of that (the worst) u can't generalize it ,that means that in the best scenario u could remember how to solve it , it only deals with exactly this problem , u shift a little bit to other scenario and bad luck u are done, I'm not attracted at all
Most difficult dosent mean most interesting (look at the amount of suscribers)
Look the amount the subscribers of mind your decisions ,terrible simple visual exercises hiding big puzzles ,a huuuge amount of subscribers
Complexity is inverse to the amount of subscribers at some point
A)mind your decisions is more visual and has simple exercises 2M suscribers
B) a more visual channel that dosent even solve problems less mental challenge is 3 brown 1 blue result? 3.6M suscribers ....
Less complexity more suscribers which is a result that amazes me because I would think the most difficult more attractive that's why we all love maths but facts are facts
Who comes up with this kind of bs problems ?!?!?!
Since this exam was rigged in 1993, the problem was also faked out likewise. Notice the denominators, they must be mocking the nitpick nerdies taking the exam...
Can you also upload some physics videos ?
I looked up some physics olympiad problems and they are interesting. I would like to post one from time to time if there is interest. Like this comment if you are interested.
@@MichaelPennMath Yeah would love to see them.