Early Math authors used Q.E.D. at the end of a Proof. Paul Halmos introduced the Box, ☐ for article conclusions, and Michael, "A Good Place to Stop". Some viewers (like me) have difficulty sustaining the intensity of enjoyment, needing breaks!☺
I did it differently: After setting x=y=0 and getting f(0)=1/2, if you set y=0 you get f(xz)>=f(x), meaning for all real numbers r=xz except 0, f(r) is an upper bound of f over the reals except 0, so f is constant over the reals except 0, and doing the same thing as before to find f(0), you get f=1/2.
I really wish you did more non-trivial functional equations. I get almost frustratingly bored by the constant or identity function being the answer every time or RARELY a simple example of a logarithm or something. I really enjoy the concept and I hope you find a chance to make non-trivial functional equations.
It's not so much about the answer itself as it is about the process of getting there. Try coming up with the solution yourself and I doubt you'll get frustratingly bored every time!
@@LIVIU2003S I do, that's why I'm saying this. I know the method is illustrative of harder problems but I would like to actually see the harder problems it illustrates, ya know? Functional equations are so cool and it's miserable to realize every problem you get is near-trivial. I love the vast majority of Dr. Penn's work, this is one minor critique from a longtime fan, but it feels worth saying
@@jplikesmaths true. And some viewers (like me) can't stand the short form content and prefer longer videos. Dr. Penn has made plenty of them in the past, my comment is only that the functional equation videos are frustratingly trivial and I would like to see more
@Alex G. The reality is that even though this channel is targeting a niche of a niche, UA-cam is still a numbers game. It's also time consuming to make a lot of videos on every topic, so the more representative problems are usually chosen. Video length is a factor as mentioned, and so is the perceived difficulty of a problem. If you're interested in the topic there are some books about Functional Equations you can buy. You can also find quite a few problems from past IMO shortlists, you should find the official page if you google it
I did something sightly different on one of the inequalities for some reason. y=1 z=0 yields (1/2 -f(0))f(x)>= 1/2(1/2-f(0)). If you suppose f(0)=1/2. But that's impossible for f(0). Similar stuff happens to assuming f(0)>1/2. That leaves the only possible relation to be f(0)=1/2. But I like the symmetry of your solution much, much better.
Great problem! I would like to point out that this year the IMO P2 was a functional inequality, so maybe they made it so that this way both functional equations lovers and inequality lovers will be satisfied 😂
Hey michal, i would really be intrested in a video where you show how question like there are thought of... Like what is problem writers process to create a problem like this
Hey Michael in your second channel do you plan to cover lots of maths content in your tutorials cause I'd love to learn from you as you explain stuff so seemlessly
Since x is a free variable, it can take any real value. So since f(x) >= 1/2 for y = z = 1, and we have no constraints on x, f(x) >= 1/2 for all values of x, regardless of what y and z are.
4:27 That was a really nice set of steps.
8:57
Early Math authors used Q.E.D. at the end of a Proof. Paul Halmos introduced the Box, ☐ for article conclusions, and Michael, "A Good Place to Stop". Some viewers (like me) have difficulty sustaining the intensity of enjoyment, needing breaks!☺
I did it differently: After setting x=y=0 and getting f(0)=1/2, if you set y=0 you get f(xz)>=f(x), meaning for all real numbers r=xz except 0, f(r) is an upper bound of f over the reals except 0, so f is constant over the reals except 0, and doing the same thing as before to find f(0), you get f=1/2.
I really wish you did more non-trivial functional equations. I get almost frustratingly bored by the constant or identity function being the answer every time or RARELY a simple example of a logarithm or something. I really enjoy the concept and I hope you find a chance to make non-trivial functional equations.
It's not so much about the answer itself as it is about the process of getting there. Try coming up with the solution yourself and I doubt you'll get frustratingly bored every time!
Problem is some viewers won’t be interested in a 30-35 min video
@@LIVIU2003S I do, that's why I'm saying this. I know the method is illustrative of harder problems but I would like to actually see the harder problems it illustrates, ya know? Functional equations are so cool and it's miserable to realize every problem you get is near-trivial. I love the vast majority of Dr. Penn's work, this is one minor critique from a longtime fan, but it feels worth saying
@@jplikesmaths true. And some viewers (like me) can't stand the short form content and prefer longer videos. Dr. Penn has made plenty of them in the past, my comment is only that the functional equation videos are frustratingly trivial and I would like to see more
@Alex G. The reality is that even though this channel is targeting a niche of a niche, UA-cam is still a numbers game. It's also time consuming to make a lot of videos on every topic, so the more representative problems are usually chosen. Video length is a factor as mentioned, and so is the perceived difficulty of a problem. If you're interested in the topic there are some books about Functional Equations you can buy. You can also find quite a few problems from past IMO shortlists, you should find the official page if you google it
I saw this problem in the morning and thought about it on and off all day today. Thanks for the exposure to inequalities in functional equations.
I have learned so much from this channel. Keep up the content! 👍
Michael, brilliant; just brilliant
A more direct approach:
x=y=0 yields f(0)=½.
y=0 yields f(xz)≥f(x).
This easily yields that f must be constant, and thus f(x)=½.
(f(x)=0 for all x except 0, f(0)=½) satisfies f(xz)≥f(x) as well, so your proof is incomplete.
I did something sightly different on one of the inequalities for some reason. y=1 z=0 yields (1/2 -f(0))f(x)>= 1/2(1/2-f(0)). If you suppose f(0)=1/2. But that's impossible for f(0). Similar stuff happens to assuming f(0)>1/2. That leaves the only possible relation to be f(0)=1/2. But I like the symmetry of your solution much, much better.
Wow you crushed this one!
Thanks a million, Michael! Nuts!
As a JEE aspirant loved it!!
Great problem!
I would like to point out that this year the IMO P2 was a functional inequality, so maybe they made it so that this way both functional equations lovers and inequality lovers will be satisfied 😂
Ayy good to see you man!! IMO P2 was pretty easy this year. Everyone in the Indian team solved it I believe. But cool question!
Hey michal, i would really be intrested in a video where you show how question like there are thought of...
Like what is problem writers process to create a problem like this
Hey Michael in your second channel do you plan to cover lots of maths content in your tutorials cause I'd love to learn from you as you explain stuff so seemlessly
thank you
Fun, thanks.
What is the painting at the heading of this video? It's gorgeous.
Thanks.
Please consider accepting donations from "Buy me a coffee" (even if you don't drink coffee).
dont see this one ...
Why did UA-cam recommend me this?
"a nice functional inequality" sounds like something Jordan Peterson would say.
asnwer= 1 isit ha ri nang ha
just because when y and z =1 then x>=1/2 doesn't mean it's always >= 1/2?
Since x is a free variable, it can take any real value.
So since f(x) >= 1/2 for y = z = 1, and we have no constraints on x, f(x) >= 1/2 for all values of x, regardless of what y and z are.
@@DavidSavinainen x is free but I still don't see how that means that the same result will hold for all values of y and z
Too obvious, there are a lot of functional equations/inequalities with much more interesting idea
👍☐
Op
Try imo 2022 p2