Awesome content! I'm a student from germany who is actively participating in Math-Olympiade type contests here and this is my favourite channel showcasing these types of problems.
13:00 Wherever you are I hope you’re having a great day/night. Love you all. Other than that, I was thinking about an AMA. Maybe I’ll do that some day for anyone caring about me lol
I love your content. No abuses, no trivialities, just beautiful math on board. Finally, I found a channel that does not hold back punches. Will you be doing expositionary content as well?
I dont really comment on youtube videos, but I just couldnt resist saying that you are way underrated. Your content is gold and i really wish you had way more subscribers and i hope you are happy.
@@henrikolsson3626 "Homogeneous" means (broadly) "all the same". In this case, it means that the function f always appears to the same power: there are only linear terms in f. Given that, and the fact that f is a linear function, we know that changing a will simply scale each of the terms by the same amount and the equation will still be valid. Hope that helps!
@Adam Romanov Okay, first off, f(x)=ax is not the solution of your functional equation. And if you are setting a=0 anyway, then f(x) = a(sinx) will also satisfy :/ Second, I only made the homogeneous argument once I saw that the solution of the equation is a polynomial.
@Adam Romanov We can make the argument that if f(x) is a solution, then g(x) = cf(x) is also a solution for any constant c. This is really easy to see: take the equation and multiply through by c, and distribute as appropriate. The fact that this distribution works cleanly is what is meant by homogeneous
@@sahilbaori9052 But the point stands that f(x) = ax containing all possible solutions, and the equation being homogeneous, don't necessarily give you that a can be any real number. You could also find that a can only be 0, and you could also find that there are no solutions whatsoever. Those would both be consistent with the homogeneity. The homogeneity only tells you that if any non-zero real number works, then they all work, but A=>B is not equivalent to just B.
I solved it too! 1) x=y=0 - > f(0)=0 2) y=0 - > f(x^3)=x*f(x^2) 3) y=x - > f(x)=-f(-x) 4) y=1 and y=-1 give me two identities, adding it: f(x^3)=x*f(x^2)+x*f(1)-f(x) 5) from 2) and 4), f(x)=x*f(1) So, f(x)=ax with a=f(1) (it's an independent value). Very nice!
An alternative way to finish this solution is, from yf(x^2)+xf(y^2)=(x+y)f(xy), take y=x^3 and use f(x^6)=x^2f(x^4), and you pretty straightforwardly deduce f(x^2)=xf(x). Use this to further simplify the functional equation to xy(f(x)+f(y))=(x+y)f(xy), then take y=1.
Beautiful problem Michael thanks a lot! I appreciate that you always include exercises and theory of different levels of difficulty for the broad mathematical fans audice. Thanks!
As I read the problem it just reminded me the polynomial decomposition of the sum of two cubes. So if working with f(x)=x, would have worked for f(x)=ax as well. But of course that wouldn't be the proof for a generic solution. Great content!
There are problems in advanced calculus books (e.g., Rudin's _Principles of Mathematical Analysis_) with functional equations for logarithms and exponentials. All of the hints you see here apply to those problems as well.
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120, what is the value of s(3m)? Can someone please help me with this? It came in India. PRMO 2019 25th August Paper
In the end you didn't have to plug anything to prove that a can be anything because notice that for any linear function passing through origin, i.e. f(x) = ax, we have f(1)=a.
let x=y to see that f(x³)=xf(x²). this means f(x^n)=x^{n/3}*f(x^{2n/3}). When n=1, then f(x)=x^{1/3}*f(x^{2/3}). Continuing this with n=2/3 gives f(x)=x^{1/3+2/9}*f(x^{4/9}). Continuing this with n=4/9 gives f(x)=x^{1/3+2/9+4/27}*f(x^{8/27}). We can repeat the process and see that f(x)=x*f(1).
So I found another solution to this problem: I got the equation f(x^3)=xf(x^2) like it was in the video, than you can see that this equation defines function f iteratively: f(x^3)=x^(1+2/3+4/9+8/27+...+(2/3)^N) * f(x^(2*(2/3)^N)) You can prove that by induction easily. Than what you have in exponent of the first x on the rhs is geometric series 1+2/3+4/9+8/27+...+(2/3)^N and it approaches 1/(1-2/3)=3 when N goes to inf And what you got is f(x^3)=x^3*f(1). And the rest of the solution is similar to what is shown in the video One thing with that solution that cannot be fully right is we made an assumption that f is continuous in 1 to make f(x^(2*(2/3)^N)) approach f(1)
Yeah, you've made an assumption of continuity here that we don't necessarily know to be true. You actually have to be very careful about continuity arguments in functional equations, because there might be pathological non-continuous solutions. For example, the very simple "for all reals x and y, f(x+y)=f(x)+f(y)" might seem to only be solved by f(x) = ax for a real. But that's only true for continuous f; a Hamel basis of the reals over the rationals gives rise to many other solutions, because there are actually many linear functions of R when we relax continuity
@@pitreason I don't believe you'll get anything faster than resolving the whole problem. Continuity has absolutely no reason to hold here, nor in any algebraic functional equation. Actually it's quite often that functional equations have very few continuous solutions but many monstruously complicated non-continuous ones. Take f(x+y) = f(x)+f(y) for example. This might look simple but it's not even possible to determine if it has non-continuous solutions or not ...
Did you do it in a way that proves that there cannot _also_ be non-polynomial solutions? For comparison, assuming the axiom of choice, Cauchy's functional equation f(x + y) = f(x) + f(y) has non-continuous solutions over the reals.
wow ! really ? I apologize on behalf of entire mathematical community that your precious concentration was hampered. Maybe Michael just become transparent so your view is not blocked when he is working on lower part of the board. SMH
idk, Prof articulates his answers very well and if he would write on the board in such a way it would just take longer for him, and in my opinion it doesnt really affect the quality of the video. Blackpenredpen is just too proficient with his writing skills lol
your habit of stating the answer before showing it really tunes me out You said you want to plug f(x) = ax to see what a can be and before you even do that, you say that a can be anything so there's no point in watching you plug f(x) = ax in because we already know. How about just plug in f(x) = ax and solve for a and find out what a can be after you solve it
Hello I'm Japanese collage student. I have a favorite TH-camr, but it's a Japanese TH-camr called Yobinori Takumi. I do a video called (Integral of the Week) once a week, so please take a look.
It was a quite *_EASY_* question. I have done by differentiating partially with respect to x and then replacing x with 0 and y with x in the final equation. We get f(x) = x*f'(0). Where we can't find value of f'(0) bcz for any value of f'(0) all equations are being satisfied. Thereby f(x) = k*x (bcz f'(0) is a constant , so i have written k in place of f'(0) where k belongs to R). ~ ♥️ from INDIA
@@jeffreyhellrung I have assumed that the function is differentiable and when i have solved the function comes out to be a linear function, therefore my assumption is correct. If I would have assumed that the function is non differnetiable then there will be no way to solve it! I'am gratefull that u have took interest in my solution. ❤
@@Anonymous-ov1vh In a sense, you are correct. However, fundamentally, if you add additional assumptions to the original problem, you end up solving a different problem :/ For example, if you make a different assumption, say that the function is differentiable *and* f(1) = 1, then now you would conclude that the solution is f(x) = x, but this doesn't end up capturing the totality of all solutions to the original problem!
@@jeffreyhellrung We can't assume that f(1) is a particular number but we can assume that f(1) is finite . We have to assume a broader possibility, and should prove the counter statement contradictory. (•‿•) By the way you are from which country ?
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120, what is the value of s(3m)? Can someone please help me with this? It came in India. PRMO 2019 25th August Paper
3 years late but here you go,s(33m) can be written as S(11*3m) which is equal to s(3m*10+3m*1) since multiplying by 10 just adds a zero at the end, the sum of the digits does not change. hences(33m) is equal to s(6m), now since we know that s(33m)=120, s(6m)=120 so, s(3m)=s(6m)/2=60!!, see you at ioqm 2025
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120, what is the value of s(3m)? Can someone please help me with this? It came in India. PRMO 2019 25th August Paper
@@TechToppers the first property is trivial as we are working in base 10. Multiplying any number by 10 adds a zero-digit at the end of the number. E.g. s(2) = 2 = 2 + 0 = s(20) = s(10*2). The second property is not immediate, but for simplicity consider integers 1
Awesome content! I'm a student from germany who is actively participating in Math-Olympiade type contests here and this is my favourite channel showcasing these types of problems.
Welcher Jahrgang bist du? Schon mal in die Bundesrunde geschafft?
13:00
Wherever you are I hope you’re having a great day/night. Love you all.
Other than that, I was thinking about an AMA. Maybe I’ll do that some day for anyone caring about me lol
Count me in!
what's an AMA?
DJ VALENTE DO CHP AMA = Ask Me Anything. I have people asking questions about me from time to time, so yeah
@@goodplacetostop2973 It would be great to watch it 😂
Ya sure just mention us the time
I love your content. No abuses, no trivialities, just beautiful math on board. Finally, I found a channel that does not hold back punches. Will you be doing expositionary content as well?
I dont really comment on youtube videos, but I just couldnt resist saying that you are way underrated. Your content is gold and i really wish you had way more subscribers and i hope you are happy.
Since the equation is homogenous, we can say that it'll work for any "a" without actually plugging and checking it.
What does homogenous mean? I might know it as I don't learn math in English, is it that I can swap x and y around and it won't affect the equation?
@@henrikolsson3626 "Homogeneous" means (broadly) "all the same". In this case, it means that the function f always appears to the same power: there are only linear terms in f. Given that, and the fact that f is a linear function, we know that changing a will simply scale each of the terms by the same amount and the equation will still be valid. Hope that helps!
@Adam Romanov Okay, first off, f(x)=ax is not the solution of your functional equation.
And if you are setting a=0 anyway, then f(x) = a(sinx) will also satisfy :/
Second, I only made the homogeneous argument once I saw that the solution of the equation is a polynomial.
@Adam Romanov We can make the argument that if f(x) is a solution, then g(x) = cf(x) is also a solution for any constant c. This is really easy to see: take the equation and multiply through by c, and distribute as appropriate. The fact that this distribution works cleanly is what is meant by homogeneous
@@sahilbaori9052 But the point stands that f(x) = ax containing all possible solutions, and the equation being homogeneous, don't necessarily give you that a can be any real number. You could also find that a can only be 0, and you could also find that there are no solutions whatsoever. Those would both be consistent with the homogeneity. The homogeneity only tells you that if any non-zero real number works, then they all work, but A=>B is not equivalent to just B.
I solved it too!
1) x=y=0 - > f(0)=0
2) y=0 - > f(x^3)=x*f(x^2)
3) y=x - > f(x)=-f(-x)
4) y=1 and y=-1 give me two identities, adding it: f(x^3)=x*f(x^2)+x*f(1)-f(x)
5) from 2) and 4), f(x)=x*f(1)
So, f(x)=ax with a=f(1) (it's an independent value).
Very nice!
It's mathS in Britain mate! MathS!
Hello fellow bri ish
mafs
An alternative way to finish this solution is, from yf(x^2)+xf(y^2)=(x+y)f(xy), take y=x^3 and use f(x^6)=x^2f(x^4), and you pretty straightforwardly deduce f(x^2)=xf(x). Use this to further simplify the functional equation to xy(f(x)+f(y))=(x+y)f(xy), then take y=1.
nice
Beautiful problem Michael thanks a lot! I appreciate that you always include exercises and theory of different levels of difficulty for the broad mathematical fans audice. Thanks!
As I read the problem it just reminded me the polynomial decomposition of the sum of two cubes. So if working with f(x)=x, would have worked for f(x)=ax as well. But of course that wouldn't be the proof for a generic solution. Great content!
in 4:23 it is true only if y=0 so why we change whole equation
thank you bro ❤❤
12:36 "I've got a little itch back there, would you mind?”
In love with this problem
Great work. May you pls take a look at PUTNAM 2001 A6
Can you give use your routine to have a body like that ?
Good teaching makes this kind of questions look so easy.
There are problems in advanced calculus books (e.g., Rudin's _Principles of Mathematical Analysis_) with functional equations for logarithms and exponentials. All of the hints you see here apply to those problems as well.
At 9 :37 you could also let x equal 1 and it works out..
great content as always
so this is equivalent to saying f is linear. I wonder if we can generalize this to 2 or even n dimensions?
How do we know that don't exist non linear functions that satisfy the condition?
But can you do Euler’s identity?
What's your idea bro??
If I find f(x²)=0 can I then say f(1)=0 because f(1²)=f(x²)=0?
by which criteria is a competition math question worthy of being featured in a video?
It has to have a good place to stop
@@malawigw god damnit
The criteria of fun math that most TH-camrs don't like to touch in the fear that it does not reach masses.
Thanks sir
Sir can you do some Philippine Mathematical Olympiad problems?
Sir Great Explanation😊👍👍❤
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120,
what is the value of s(3m)?
Can someone please help me with this?
It came in India. PRMO 2019 25th August Paper
What is a good book for Coordinate Geometry with tough exercises?
In the end you didn't have to plug anything to prove that a can be anything because notice that for any linear function passing through origin, i.e. f(x) = ax, we have f(1)=a.
Thank you, it is very interesting
Русский
I like the Dune t shirt
Trinomial (three termed object ) lol 😂
True
100k subs let's go!
let x=y to see that f(x³)=xf(x²). this means f(x^n)=x^{n/3}*f(x^{2n/3}). When n=1, then f(x)=x^{1/3}*f(x^{2/3}). Continuing this with n=2/3 gives f(x)=x^{1/3+2/9}*f(x^{4/9}). Continuing this with n=4/9 gives f(x)=x^{1/3+2/9+4/27}*f(x^{8/27}). We can repeat the process and see that f(x)=x*f(1).
It's awesome
So I found another solution to this problem:
I got the equation f(x^3)=xf(x^2) like it was in the video, than you can see that this equation defines function f iteratively:
f(x^3)=x^(1+2/3+4/9+8/27+...+(2/3)^N) * f(x^(2*(2/3)^N))
You can prove that by induction easily. Than what you have in exponent of the first x on the rhs is geometric series
1+2/3+4/9+8/27+...+(2/3)^N and it approaches 1/(1-2/3)=3 when N goes to inf
And what you got is f(x^3)=x^3*f(1). And the rest of the solution is similar to what is shown in the video
One thing with that solution that cannot be fully right is we made an assumption that f is continuous in 1 to make f(x^(2*(2/3)^N)) approach f(1)
Yeah, you've made an assumption of continuity here that we don't necessarily know to be true. You actually have to be very careful about continuity arguments in functional equations, because there might be pathological non-continuous solutions. For example, the very simple "for all reals x and y, f(x+y)=f(x)+f(y)" might seem to only be solved by f(x) = ax for a real. But that's only true for continuous f; a Hamel basis of the reals over the rationals gives rise to many other solutions, because there are actually many linear functions of R when we relax continuity
I guess it can be proven somehow that this is continuous. Maybe I'll do it some other time;)
The fastest way to prove that this is continuous is to solve the problem a different way, and then note that all solutions are continuous
@@pitreason I don't believe you'll get anything faster than resolving the whole problem. Continuity has absolutely no reason to hold here, nor in any algebraic functional equation. Actually it's quite often that functional equations have very few continuous solutions but many monstruously complicated non-continuous ones. Take f(x+y) = f(x)+f(y) for example. This might look simple but it's not even possible to determine if it has non-continuous solutions or not ...
Swenji look guys I’ve just suggested another way to solve this, I don’t insist that this one is better than showed in the video
It's 'a' cubed at the end.
I have solved this question in 2 minute in my head without copy and pen by comparing polynomial degrees rules
Did you do it in a way that proves that there cannot _also_ be non-polynomial solutions?
For comparison, assuming the axiom of choice, Cauchy's functional equation f(x + y) = f(x) + f(y) has non-continuous solutions over the reals.
4:10 try not to cover the board like this, it hampers concentration and disturbs the view
wow ! really ? I apologize on behalf of entire mathematical community that your precious concentration was hampered.
Maybe Michael just become transparent so your view is not blocked when he is working on lower part of the board.
SMH
@@ancientwisdom7993 see how blackpenredpen writes on the lower part of board. He never blocks vision completely
idk, Prof articulates his answers very well and if he would write on the board in such a way it would just take longer for him, and in my opinion it doesnt really affect the quality of the video. Blackpenredpen is just too proficient with his writing skills lol
wow
Hi,
This was before hair cutting, so I understand my request about the camera is not taken into account 😄
your habit of stating the answer before showing it really tunes me out
You said you want to plug f(x) = ax to see what a can be and before you even do that, you say that a can be anything so there's no point in watching you plug f(x) = ax in because we already know. How about just plug in f(x) = ax and solve for a and find out what a can be after you solve it
im a 14 year old kid who isn't good at math why am i here
Hello I'm Japanese collage student.
I have a favorite TH-camr, but it's a Japanese TH-camr called Yobinori Takumi. I do a video called (Integral of the Week) once a week, so please take a look.
Who wants to win a pack of chocolate ?
Am I the only one finding this video kinda slow and aimed for kids like in 4:45 , I prefer the harder ones
Kids needs to learn math too
Directly BMO??😅
It was a quite *_EASY_* question.
I have done by differentiating partially with respect to x and then replacing x with 0 and y with x in the final equation.
We get f(x) = x*f'(0).
Where we can't find value of f'(0) bcz for any value of f'(0) all equations are being satisfied.
Thereby f(x) = k*x (bcz f'(0) is a constant , so i have written k in place of f'(0) where k belongs to R).
~ ♥️ from INDIA
I think the issue with your solution is you have to first justify that f is differentiable.
@@jeffreyhellrung
I have assumed that the function is differentiable and when i have solved the function comes out to be a linear function, therefore my assumption is correct.
If I would have assumed that the function is non differnetiable then there will be no way to solve it!
I'am gratefull that u have took interest in my solution. ❤
@@Anonymous-ov1vh In a sense, you are correct. However, fundamentally, if you add additional assumptions to the original problem, you end up solving a different problem :/ For example, if you make a different assumption, say that the function is differentiable *and* f(1) = 1, then now you would conclude that the solution is f(x) = x, but this doesn't end up capturing the totality of all solutions to the original problem!
@@jeffreyhellrung We can't assume that f(1) is a particular number but we can assume that f(1) is finite .
We have to assume a broader possibility, and should prove the counter statement contradictory.
(•‿•)
By the way you are from which country ?
@@Anonymous-ov1vh USA; you?
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120,
what is the value of s(3m)?
Can someone please help me with this?
It came in India. PRMO 2019 25th August Paper
3 years late but here you go,s(33m) can be written as S(11*3m) which is equal to s(3m*10+3m*1) since multiplying by 10 just adds a zero at the end, the sum of the digits does not change. hences(33m) is equal to s(6m), now since we know that s(33m)=120, s(6m)=120 so, s(3m)=s(6m)/2=60!!, see you at ioqm 2025
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120,
what is the value of s(3m)?
Can someone please help me with this?
It came in India. PRMO 2019 25th August Paper
For the digit sum (base 10), we have the following two properties:
s(10*a) = s(a), for all integers a
s(a + b)
From where did you get to know these properties??
@@TechToppers the first property is trivial as we are working in base 10. Multiplying any number by 10 adds a zero-digit at the end of the number. E.g. s(2) = 2 = 2 + 0 = s(20) = s(10*2).
The second property is not immediate, but for simplicity consider integers 1