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.
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
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
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.
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.
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?
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.
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?
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.
@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 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?)
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 ...
😡
H u h
he never stopped :(
@guyy4792 yeah, but was there a good place for that?
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.
It bothered me the whole time you didn't say a = f(2).
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 still can be dealt with by considering the two cases. It just adds that f could be the 0 function
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
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
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.
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.
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?
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.
You can get a = f(a), which makes the last part easier to solve
After first two special cases you can get: a = f(2)
please can you do some more bmo problems : )
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.
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
The function is not defined for numbers which are not natural. So either x > 1 or x=1
@shivanandvp ohhhhhh damn ok ty
Please always remind us that your natural numbers excludes zero for every video related to natural numbers.
3:56 CRY! "y" is a strange kind of "t"!
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
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
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.
No good place to stop?
Cheerio my dearest!
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 C in |N
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
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?)