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.
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'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.
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 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.
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.
How to solve British functional equatiojs:
Put t for 2 and 2 for t ...
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.
😡
H u h
he never stopped :(
@guyy4792 yeah, but was there a good place for that?
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'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.
It bothered me the whole time you didn't say a = f(2).
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.
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)
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 can you do some more bmo problems : )
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"!
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)
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.
No good place to stop?
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
Cheerio my dearest!
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.
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?)