Overcomplicated indeed. If we notice that f(2xy)/y does not depend on y at all, we may argue that f(2xy) = C(x)y, pluggin in x=1 and substituting 2y -> t we'd have f(t) = At with A = C(1)/2 (of course it's trivial to show that C is a linear function of x too). Putting that back to original eq gives A=1
He did do one such problem, it's just that they're generally more gnarly. Also, the challenge is in proving that no other solution exists, a very, very hard proposition generally.
If you cheat a little and set y=0, you get f(x+1)f(0)=0. So either f(0) is zero or f(x+1) is zero. So either we have f(0)=0 or constant function that returns 0. Testing f(x)=x works. But f(x)=ax does not work for a !=1. So that gives a bit of intuition about what is going on.
@@raidingthealphasforthebobs4323 That is why I said cheat. I used the very powerful problem solving technique called "wishful thinking" :-). You solve a simpler version of the problem and see if it helps. In this case it works great because the identity function works for all types of numbers. The restriction to natural numbers was quite artificial and does not change the answer.
@@raidingthealphasforthebobs4323 oh that makes sense, does this channel use naturals are 1,... I was wondering why the 0 function wasn't a solution at the end. (I solved separately and used linearity to solve this problem, and I got both the 0 and identity functions as the only solutions.
Naturals are {1, 2, 3, …}. Solution depends on this too. Michael divides through by f(2) at about 2:20. it's only OK to do this because f maps o to {1,2,3,…} so f(2) can't be zero.
The whole time, it was obvious, that a = f(2), but was only used later. If |N had (as I learned it, well I've learned both styles) 0 in it, f(2) = 0 would imply another solution, but I know, you excempt 0 from IN.
I'm telling you, the same N→N strategy works 80% of the time. Remove the fs to see if LHS=RHS, a.k.a. f=id works, and when it does, prove it's unique, which is quick because you have the goal in mind.
I took basically the same approach except at around 2:10 I divided the 2nd equation by the first which gives the somewhat simpler t=f(2t)/f(2), or f(2t)=f(2) t
I think f being a polynomial is a good starting guess. Letf say it is of rank k then the rank of expressions on left and right depending on x are 2k^2=2k so either a constant or a first degree If we consider in y the rank is 1=k. If y=0 then 0=f(0)*f(x+1) so f(0)=0 so no free term and we have something like f(x)=ax 2y*a(ax^2+x)=a(x+1)*a*2xy axy(ax+1)=2a^2xy(x+1) for all x,y So ax+1=a(x+1) thus a=1
This was such a fun problem, I got f(x) = 0 or f(x) = x. (both of these definitely satisfy the problem) The way I did it was I noticed that y seemed a bit odd, like if we took a constant x, it looked like f would be linear over y. Because of this I started by considering a constant x, Then we notice that f(f(x^2)+x) is another constant and f(x + 1) is also a constant, we call these k_1 and k_2. Then we have that 2yk_1 = k_2f(2xy) which is simplified to 2y k_1/k_2 = f(2xy). Now we prove this is linear. To do that, consider f(a * 2xy) = f(2x(ay)) = 2(ay) k_1 / k_2 = a(2y k_1/k_2) = a f(2xy). Consider now f(2x(y + z)) where z is some number [I think real number, but as problem is defined over naturals, instead consider if k is some multiple of 1/2x, so this is adding 1, 2, 3 etc.] Then f(2x(y+z)) = 2(y+z) k_1/k_2 = 2y k_1/k_2 + 2z k_1/k_2 = f(2xy) + f(2xz). Thus we have proven f is a linear function. Because of this, we can apply our understanding of linear functions to make f significantly easier, the first one is that f(x) = ax, as f is a 1-dimensional linear function over the reals. Thus we have 2yf(f(x^2) + x) = 2y * a(a(x^2) + x) = 2y(a^2x^2 + ax) = 2axy(ax + 1). And we have that f(x+1)f(2xy) = a(x+1)a(2xy) = a2xy(ax + a) = 2a^2xy(x+1) So we have that 2axy(ax+1) = 2a^2xy(x+1) now consider if a = 0. Then this statement is true, so this is possible. Consider now if none of a,x,y are 0. Then ax+1 = a(x + 1) or ax + 1 = ax + a subtracting ax gives us a = 1. Thus we have shown either a = 1 or a = 0. We do need to prove a = 1 is valid if x or y are 0, but this is trivial as the above holds whenever x or y are 0.
Pretty typical solution for a functional equation of this type! Would be far more interested in seeing a functional equation that wasn't from some competition so the solution can be a lot more complex.
2:53 Overcomplicated! From the first formula, you can see that a = f(2). That's also obvious because of f(2t) = f(2*1) = f(2) = f(2)*1 for t=1. For IN = { 0, 1, ... } even more dangerous: For f(2) = 0, you can't divide by it. In this case f(2t) can be anything, unless for t=1 it's 0.
@@coolbikerdad Yes, if you exclude 0 from IN that's true. I had already edited that assumption in. Thanks. My main point stands in both cases: Obviously a = f(2).
@@coolbikerdadFor reference though zero is often included in the Natural Numbers, there’s no universal consensus on that. Michael makes clear as the video progresses he’s excluding zero but really that should be made explicit in the original statement of the problem for clarity.
1. Let x = 1, y is free 2yf(f(1)+1) = f(2)f(2y). because f(2) is finite then f(2y) must divide 2y. Using this fact. For all even numbers a: f(a) = a*g(a) 2. Replace f(2xy) for 2xy*g(2xy) in the original eq: 2yf(f(x^2)+x) = f(x+1)*2xy*g(2xy) then f(f(x^2)+x) = f(x+1)*x*g(2xy). LHS does not depend on y so g(2xy) is constant. therefore f(a) = k*a for some k 3. x is event, y is free 2yk(kx^2 + x)= f(x+1)*2xyk then kx^2 + x = f(x+1)x then kx + 1 = f(x + 1). so k =1 and if a is odd then f(a) = a. 4. profit?
I think all you had to do was to set x = 1 => 2y f(f(1)+1) = f(2) f(2y) => f(2y) = C*y => f(x) = ax => (put in original equation) => a = 1 => f(x) = x Or do I miss something?
To take it at the least amount of seriousness, there's this question: Why call nature "god" and thus attach all the baggage that comes along with the word, "god"? We already have a word for nature: "Nature." It works just fine to describe the workings of energy and matter. What is the point of using the word "god" for this? What is gained?
Of course not everybody excludes 0 from the Natural Numbers, that really should have been made explicit in the original problem description. If you do include 0 as a Natural Number, you’ll get the constant zero function as another (trivial) solution and then have to contend with solving for the case where f(0) = 0 and there exist a c>0 where f(c) > 0 and solve that. The solution f(x) = x for all x in the video is still one such solution, but determining whether or not it’s the only such solution when 0 is allowed in the range is a bit tricky (i.e. can you rule out all other functions where f(x) = either 0 or x depending on the value of x for all x?)
@possiblybottomlesspit "Mathematics" is maths by convention, now we have an adjective here, not a noun, I don't mind! I haven't participated, but I can smell BS.
Of course a British equation would have a variable called t.
How to solve British functional equatiojs:
Put t for 2 and 2 for t ...
It bothered me the whole time you didn't say a = f(2).
Overcomplicated indeed. If we notice that f(2xy)/y does not depend on y at all, we may argue that f(2xy) = C(x)y, pluggin in x=1 and substituting 2y -> t we'd have f(t) = At with A = C(1)/2 (of course it's trivial to show that C is a linear function of x too). Putting that back to original eq gives A=1
These functional equations always seem to end up as f(x) = x or f(x) = k for some constant k.
Yeah i think some quadratic or sinusoidal functionals would spice things up pretty nicely.
He did do one such problem, it's just that they're generally more gnarly. Also, the challenge is in proving that no other solution exists, a very, very hard proposition generally.
@@MrRyanroberson1for sure
If you cheat a little and set y=0, you get f(x+1)f(0)=0. So either f(0) is zero or f(x+1) is zero. So either we have f(0)=0 or constant function that returns 0. Testing f(x)=x works. But f(x)=ax does not work for a !=1. So that gives a bit of intuition about what is going on.
U can't plug 0 since it isn't a natural
@@raidingthealphasforthebobs4323 That is why I said cheat. I used the very powerful problem solving technique called "wishful thinking" :-). You solve a simpler version of the problem and see if it helps. In this case it works great because the identity function works for all types of numbers. The restriction to natural numbers was quite artificial and does not change the answer.
@@raidingthealphasforthebobs4323 oh that makes sense, does this channel use naturals are 1,... I was wondering why the 0 function wasn't a solution at the end. (I solved separately and used linearity to solve this problem, and I got both the 0 and identity functions as the only solutions.
Naturals are {1, 2, 3, …}. Solution depends on this too. Michael divides through by f(2) at about 2:20. it's only OK to do this because f maps o to {1,2,3,…} so f(2) can't be zero.
@@theartisticactuary if N was {0,1,...} it still can be dealt with by considering the two cases. It just adds that f could be the 0 function
😡
H u h
he never stopped :(
@guyy4792 yeah, but was there a good place for that?
Maybe he said "Good place to stop", but it was left out in the video edit. :)
The whole time, it was obvious, that a = f(2), but was only used later.
If |N had (as I learned it, well I've learned both styles) 0 in it, f(2) = 0 would imply another solution, but I know, you excempt 0 from IN.
After first two special cases you can get: a = f(2)
I'm telling you, the same N→N strategy works 80% of the time. Remove the fs to see if LHS=RHS, a.k.a. f=id works, and when it does, prove it's unique, which is quick because you have the goal in mind.
I took basically the same approach except at around 2:10 I divided the 2nd equation by the first which gives the somewhat simpler t=f(2t)/f(2), or f(2t)=f(2) t
after doing x=2t, y=1 i did x=2t+1, y=1 and got some pretty good results too!
I think f being a polynomial is a good starting guess. Letf say it is of rank k then the rank of expressions on left and right depending on x are
2k^2=2k so either a constant or a first degree
If we consider in y the rank is 1=k.
If y=0 then 0=f(0)*f(x+1) so f(0)=0 so no free term and we have something like f(x)=ax
2y*a(ax^2+x)=a(x+1)*a*2xy
axy(ax+1)=2a^2xy(x+1) for all x,y
So
ax+1=a(x+1) thus a=1
Let degree of f(x) is n then by the given equation 2n^2 =2n
n=0,1 so we can see f(x) may be linear or constant.
this assumes f is a polynomial
@@tashquantum Pretty safe assumption after you reach f(2y)=C*y. You know C is a natural number.
@@gregoryknapen9133which doesn’t immediately rout much restrictions on f(odd)
You can get a = f(a), which makes the last part easier to solve
This was such a fun problem, I got f(x) = 0 or f(x) = x. (both of these definitely satisfy the problem)
The way I did it was I noticed that y seemed a bit odd, like if we took a constant x, it looked like f would be linear over y.
Because of this I started by considering a constant x,
Then we notice that f(f(x^2)+x) is another constant and f(x + 1) is also a constant, we call these k_1 and k_2.
Then we have that 2yk_1 = k_2f(2xy) which is simplified to 2y k_1/k_2 = f(2xy).
Now we prove this is linear.
To do that, consider f(a * 2xy) = f(2x(ay)) = 2(ay) k_1 / k_2 = a(2y k_1/k_2) = a f(2xy).
Consider now f(2x(y + z)) where z is some number [I think real number, but as problem is defined over naturals, instead consider if k is some multiple of 1/2x, so this is adding 1, 2, 3 etc.]
Then f(2x(y+z)) = 2(y+z) k_1/k_2 = 2y k_1/k_2 + 2z k_1/k_2 = f(2xy) + f(2xz).
Thus we have proven f is a linear function.
Because of this, we can apply our understanding of linear functions to make f significantly easier, the first one is that f(x) = ax, as f is a 1-dimensional linear function over the reals.
Thus we have 2yf(f(x^2) + x) = 2y * a(a(x^2) + x) = 2y(a^2x^2 + ax) = 2axy(ax + 1).
And we have that f(x+1)f(2xy) = a(x+1)a(2xy) = a2xy(ax + a) = 2a^2xy(x+1)
So we have that 2axy(ax+1) = 2a^2xy(x+1)
now consider if a = 0.
Then this statement is true, so this is possible.
Consider now if none of a,x,y are 0.
Then ax+1 = a(x + 1) or ax + 1 = ax + a subtracting ax gives us a = 1.
Thus we have shown either a = 1 or a = 0. We do need to prove a = 1 is valid if x or y are 0, but this is trivial as the above holds whenever x or y are 0.
Please always remind us that your natural numbers excludes zero for every video related to natural numbers.
please can you do some more bmo problems : )
I solved this before I knew that f(x) could only take natural numbers and I found another solution: f(x) = 0 (for x-even), 1 (x-odd)
So basically, a number theorist and a set theorist would find two different solutions. Because they differ in whether 0 is in |N. 😆
Pretty typical solution for a functional equation of this type! Would be far more interested in seeing a functional equation that wasn't from some competition so the solution can be a lot more complex.
Don't forget to include the solution f(x) = 0 for all x.
2:53 Overcomplicated! From the first formula, you can see that a = f(2). That's also obvious because of f(2t) = f(2*1) = f(2) = f(2)*1 for t=1. For IN = { 0, 1, ... } even more dangerous: For f(2) = 0, you can't divide by it. In this case f(2t) can be anything, unless for t=1 it's 0.
f(2) cannot be zero as f:N->N
@@coolbikerdad Yes, if you exclude 0 from IN that's true. I had already edited that assumption in. Thanks.
My main point stands in both cases: Obviously a = f(2).
@@coolbikerdadFor reference though zero is often included in the Natural Numbers, there’s no universal consensus on that. Michael makes clear as the video progresses he’s excluding zero but really that should be made explicit in the original statement of the problem for clarity.
Can't f(x) be 0 if x is even and 1 if x is odd. I think that works.
f(x) = (1 -(-1)^x)/2
@shivanandvp ohhhhhh damn ok ty
1. Let x = 1, y is free
2yf(f(1)+1) = f(2)f(2y). because f(2) is finite then f(2y) must divide 2y. Using this fact. For all even numbers a: f(a) = a*g(a)
2. Replace f(2xy) for 2xy*g(2xy) in the original eq:
2yf(f(x^2)+x) = f(x+1)*2xy*g(2xy) then f(f(x^2)+x) = f(x+1)*x*g(2xy). LHS does not depend on y so g(2xy) is constant. therefore f(a) = k*a for some k
3. x is event, y is free
2yk(kx^2 + x)= f(x+1)*2xyk then kx^2 + x = f(x+1)x then kx + 1 = f(x + 1). so k =1 and if a is odd then f(a) = a.
4. profit?
1: How do you prevent f(f(1)+1) =100000000 and f(2)=1 ? I guess you “only” obtain that f(2y) may jump between finitely many linear functions
@@HagenvonEitzen the point 3 proves that if f(a) is ka when x is even then f(b) = b when y is odd.
3:56 CRY! "y" is a strange kind of "t"!
and to think, i took the 2024 bmo less than a month ago
if you could, please have a look at it, it was absolute agony to do, the questions were mostly combinatorics and game theory (not my style)
Most of them are set by the same guy. You've got to learn his style, he's very big on game theory and - shocker - number theory
I think all you had to do was to set x = 1 => 2y f(f(1)+1) = f(2) f(2y) => f(2y) = C*y => f(x) = ax => (put in original equation) => a = 1 => f(x) = x
Or do I miss something?
f(2y)=... means you only know the values for even numbers, doesn't tell you anything about the odd ones
No good place to stop?
Cheerio my dearest!
To take it at the least amount of seriousness, there's this question:
Why call nature "god" and thus attach all the baggage that comes along with the word, "god"?
We already have a word for nature: "Nature." It works just fine to describe the workings of energy and matter. What is the point of using the word "god" for this? What is gained?
that's cool
where the and thats a good place to stop
*where's
f(x) could be linear or constant
f cannot be constant because if f=C then 2y * C=C^2 for all y in |N which is impossible since we do not include 0 in |N
Of course not everybody excludes 0 from the Natural Numbers, that really should have been made explicit in the original problem description. If you do include 0 as a Natural Number, you’ll get the constant zero function as another (trivial) solution and then have to contend with solving for the case where f(0) = 0 and there exist a c>0 where f(c) > 0 and solve that. The solution f(x) = x for all x in the video is still one such solution, but determining whether or not it’s the only such solution when 0 is allowed in the range is a bit tricky (i.e. can you rule out all other functions where f(x) = either 0 or x depending on the value of x for all x?)
It's the British MATHS Olympiad. Not 'math'. I have participated in it so I should know.
It's always maths in Britain!
Its full name is the British Mathematical Olympiad... so no 's'. I've also participated, so I should know!
@possiblybottomlesspit "Mathematics" is maths by convention, now we have an adjective here, not a noun, I don't mind! I haven't participated, but I can smell BS.
meth
@rainerzufall42 Oh yeah, I don't mind either! Well aware we say maths (I'm a brit), just thought the original comment was quite hoity-toity lol