Great video. These are truly good weak spots! I think that symmetry is the greatest weak spot in this kind of situations. I recommend that you use the P(x,y) notation when substituting since it is clear and short and it is the most common way in IMO solutions. I am using it too on my channel functional equations playlist. Good luck my friend and keep going!
That would probably be a long video because functions come in many different types. There is also a world of difference between functions from reals, from rationals, from naturals, functions that are equalities, inequalities, and divisibilities.
I've never tried this kind of problems before and now i figure out is not as complicated as i used to believe, you only need to understand the methods and a lot of practice. Thank you for the video, i hope you keep doing this.
Geometry Beautiful journey through Olympiad geometry Combi Olympiad combi Number theory MONT Inequalities Secretes in inequalities Functional eq. functional eq. By titu andereescu
Hello Letsthinkcritically! My name is Swetha, founder of Melodies for Math. I recently found your channel and subscribed, as I love your content. Your example in this video easily helped me understand how to work with functional equations. Math is cumulative, so building understanding across concepts is critical. As a group of high school students that explains various math concepts through song, we are so glad to support other channels with similar missions. I hope you have a great day! ~Swetha from Melodies for Math
@@alganpokemon905 sure, _post factum_ each of the subs is simple and you can even come with each on its own problem is that either one of them (or even 2 at a time) do not bring new info and would get discarded the only way I see how to stumble onto this trio together is to write down all possible substitutions and brute force all the tripples even IF there's some kind of intuition that you know, the video does not explain it by the video they just appear out of thin air
the functional equations problems posed by various "mathematical problem posing authorities" almost always phrased in a form which seems unbelieveble for a function to satisfy and it ALWAYS(almost always) turns out that the function was constant, linear or maybe even quadratic, it is a simple sleight of hand taking advantage of the fact that the most trivial functions (constant, linear, etc) satisfy unbeliveable constraints, why not show some NON-BORING functions satisfyieng extreme functional conditions
I think you made the solution more complex , while the solution is pretty straightforward put y = x + h where h is a very small quantity then you will get the derivative relation between f and g
Nekoliko bolj preprosto: i) g(f(x+y))=f(x)+(2x+y)g(y) {∂/∂x,∂/∂y }g(f(x+y))=g'(f(x+y))f'(x+y) ... oba odvoda sta enaka f'(x)+2g(y) = g(y)+(2x+y)g'(y) za y=0: f'(x) = - g(0)+2xg'(0) = 2Ax+B in po integraciji: f(x)=Ax^2+Bx+C za y=0 in x=0 dobimo iz i) g(f(0))=f(0) in za y=-x: f(0)=f(x)+xg(-x) ter g(x)=(f(-x)-f(0))/x = Ax-B v i) vstavimo y=0: g(f(x))=f(x)+2xg(0); sledi: A(Ax^2+Bx+C)-B = Ax^2+Bx+C-2xB in (A^2-A)x^2+(AB+B)x+AC-B-C=0 : A(A-1)=0; B(A+1)=0; C(A-1)=B 1) A=0, B=0, C=0: f(x)=0, g(x)=0 ... trivialna rešitev 2) A=1, B=0, C poljuben: f(x) = x^2+C, g(x) = x preizkus pa je že narejen v zgornjem videu.
suprise suprise, another esoteric complicated looking functional equation has a simple polynomial soulution, as always, why not cover some functional equations with more interesting solutions than g(x)=x and f(x)=x^2+C
Great video. These are truly good weak spots!
I think that symmetry is the greatest weak spot in this kind of situations.
I recommend that you use the P(x,y) notation when substituting since it is clear and short and it is the most common way in IMO solutions.
I am using it too on my channel functional equations playlist.
Good luck my friend and keep going!
Hello!!🍀
Can you upload another video which explains all tricks we can use in functional equations.
That would probably be a long video because functions come in many different types. There is also a world of difference between functions from reals, from rationals, from naturals, functions that are equalities, inequalities, and divisibilities.
God, functional equations are hard
There are some great functional equations related books you can read
Please never stop these :)
I've never tried this kind of problems before and now i figure out is not as complicated as i used to believe, you only need to understand the methods and a lot of practice. Thank you for the video, i hope you keep doing this.
Amazing you are helping us.. the imo aspirants so much by providing how to look for insights
When you have g(x) as linear or constant function, just set x=0 in the original equation and you have f(y)
10:41 How is B equal zero ? It is x=1/2 . Please reply how did you get B equal zero ?
can you suggest math olympiad books ?
Geometry
Beautiful journey through Olympiad geometry
Combi
Olympiad combi
Number theory
MONT
Inequalities
Secretes in inequalities
Functional eq.
functional eq. By titu andereescu
Looking forward for more such illustrations on fe :)
Let a,b,c be positive real such that a
Since a
use am hm inequality
Hello Letsthinkcritically! My name is Swetha, founder of Melodies for Math. I recently found your channel and subscribed, as I love your content. Your example in this video easily helped me understand how to work with functional equations.
Math is cumulative, so building understanding across concepts is critical. As a group of high school students that explains various math concepts through song, we are so glad to support other channels with similar missions.
I hope you have a great day!
~Swetha from Melodies for Math
10:18 if A*A = A then A = 1 or A=-1. What happens when A = -1? Does this give the same solutions, or are there more solutions?
A = -1 is not a solution to that equation. It implies A = 1 or 0, but we are looking at the case where A is non-zero.
WHICH BOOK SHOULD I PREFER FOR ALGEBRA I AM A BEGGINER IN ALGEBRA
Challenge and thrill of pre College mathematics will be best👍
how did you come up with the three substitutions to sum up?
they are quite simple subs.
just playing around with it and being observant.
@@alganpokemon905 sure, _post factum_ each of the subs is simple
and you can even come with each on its own
problem is that either one of them (or even 2 at a time) do not bring new info and would get discarded
the only way I see how to stumble onto this trio together is to write down all possible substitutions and brute force all the tripples
even IF there's some kind of intuition that you know, the video does not explain it
by the video they just appear out of thin air
great video :)
YES
Are you allowed to just swap x and y like that in a functional equation?
yes king
the functional equations problems posed by various "mathematical problem posing authorities" almost always phrased in a form which seems unbelieveble for a function to satisfy and it ALWAYS(almost always) turns out that the function was constant, linear or maybe even quadratic, it is a simple sleight of hand taking advantage of the fact that the most trivial functions (constant, linear, etc) satisfy unbeliveable constraints, why not show some NON-BORING functions satisfyieng extreme functional conditions
What e legend
I subbed
I think you made the solution more complex , while the solution is pretty straightforward put y = x + h where h is a very small quantity then you will get the derivative relation between f and g
It is not given that the functions are differentiable.
Please good writing
Nekoliko bolj preprosto:
i) g(f(x+y))=f(x)+(2x+y)g(y)
{∂/∂x,∂/∂y }g(f(x+y))=g'(f(x+y))f'(x+y) ... oba odvoda sta enaka
f'(x)+2g(y) = g(y)+(2x+y)g'(y)
za y=0: f'(x) = - g(0)+2xg'(0) = 2Ax+B in po integraciji: f(x)=Ax^2+Bx+C
za y=0 in x=0 dobimo iz i) g(f(0))=f(0) in za y=-x: f(0)=f(x)+xg(-x) ter g(x)=(f(-x)-f(0))/x = Ax-B
v i) vstavimo y=0: g(f(x))=f(x)+2xg(0); sledi:
A(Ax^2+Bx+C)-B = Ax^2+Bx+C-2xB in (A^2-A)x^2+(AB+B)x+AC-B-C=0 :
A(A-1)=0; B(A+1)=0; C(A-1)=B
1) A=0, B=0, C=0: f(x)=0, g(x)=0 ... trivialna rešitev
2) A=1, B=0, C poljuben: f(x) = x^2+C, g(x) = x
preizkus pa je že narejen v zgornjem videu.
The g looks like a s
Your reasoning doesn't seem to be correct.
suprise suprise, another esoteric complicated looking functional equation has a simple polynomial soulution, as always, why not cover some functional equations with more interesting solutions than g(x)=x and f(x)=x^2+C
Bad writing
Very poor explanation
I thought it was brilliant, what makes you think otherwise?