Here's a quicker way: We already showed f is increasing, so if f(a) > a, then f(f(a)) > f(a) [by applying f on both sides], which in turn implies f(f(a)) > a. Again applying f on both sides would yield f(f(f(a))) > f(a) > a, which is a contradiction. For the other case, if f(a) < a, then by similar argument, you can show that f(f(f(a))) < f(f(a)) < f(a) < a, which is again a contradiction.
Note that if you replace the real numbers R by the complex numbers C, you do get continuous solutions which are not the identity. Namely, for each of the three solutions z of the cubic equation z^3 = 1, one can define f(x) = x*z with the property f(f(f(x))) = x * z^3 = x.
Are there any others? If we are using functions from Riemann sphere to itself, I suppose we have the ones he mentioned with the (x+A)/(x+B) ... And I guess, there would be similar such functions for any 2x2 matrix which cubes to the identity matrix?
@@drdca8263 If we're on the Riemann sphere, yeah. Then the only holomorphic functions that solve our problem are the Möbius transforms of order 1 and 3. Order 1 is just the identity, and order 3 corresponds to 2x2 matrices with characteristic polynomial x^2+x+1 (cause they need to solve x^3-1=0), meaning matrices with trace -1 and determinant 1.
5:30 You accidentally multiplied the numerator by the numerator, instead of the denominator on this board, the next board does have the correct equations however.
@@holliebuchanan6064 It's good for instructors to admit mistakes, I agree. But also students shouldn't hastily judge the subject of mathematics (or judge themselves in relation to the subject) because of some wrong algebra done on a board, which is usually just done to fill up time. If you care about learning the subject you should be able to fix the calculation, that's what really matters.
Before watching: 1) f³x=x => f is invertible (the inverse is f²) 2) An invertible continuous function from IR to IR is either strictly increasing or strictly decreasing. 3) If f is decreasing then f³ is decreasing, but f³=x is increasing so f is increasing. 4) a f(a)
Really nice and clean arguments, thank you. It is really shocking to me though that there exists this magical f(x) = x / (x - 1) which works almost everywhere.
I think it was said that because the function is both invertible (with inverse f∘f) and continuous then it has to be bijective (and thus one to one). So it either has to either strictly increase or strictly decrease otherwise at least one value would map to the same number and then could not be sent back to both of them with two more applications of f.
Take an interval I and the function Φ: T -> R, where T is the "lower triangle" of I^2, Φ(s,t)=f(s)-f(t) for (s,t) in T. T is convex, thus path connected, and Φ is continuous, and non-zero if f is "1-1", so it preserves its sign. So f is either strictly increasing or strictly decreasing.
The example for non-continuous solutions is overly complicated - and as stated not even a function! How to define f(-B)? And how to define f((B^2+A)/(B+1))? A much simpler infinite family of non-continuous solution is the following: Let a, b, c be three distinct real numbers. Define f(a)=b, f(b)=c, f(c)=a, and f(x)=x otherwise.
And in fact the function is constructed to be undefined at -B because then it is (vacuously) discontinuous there. It's one of the most natural ways to get a discontinuous function
The rational functions that satisfy this are (ax+b)/(cx+d) where a^2+ad+d^2+bc=0. For example, if a=1 and d=4, we can take b=-3 and c=7 to get (x-3)/(7x+4). Then we can just send 1/7 to 1/7 and -4/7 to -4/7. It turns out that if we only use rational coefficients, then a function like this can only have “order” equal to 1 (for the identity function), 2 (if a+d=0), 3 (the condition above), 4 (if a^2+d^2+2bc=0), or 6 (if a^2-ad+d^2+3bc=0). Then you can just reassign the values where the denominator is 0.
@@19divide53 The following was my thought process, but now that i think about it again, this argument is very similar to 2 above. As added context: this part was mostly proven by "look at it, it probably holds" (and having the example of the map S^1->(0,1] as a typical discontinuous one in my mind), and thus the argument might have to be refined a bit for the general case, but not by too much. The inverse image of d has to be one of the points b or c. If we take an open set U around d, the inverse is also open, and thus intersects both the intervals A and B (in general the connected components) around f^-1(d). Since we know that f is bijective (by the concotonation property) and maps connected components to connected components (by continuity), we know that f restricted to a connected component maps bijectively to another connected component. Now either A or B is mapped into J, but neither of them is in the original interval (-inf, a).
12:34 Don't we have a contradiction here already? Above we have f(a) < f(f(a)), and below that we have f(f(a)) < a < f(a). If we leave the a out of the second inequality, which we can do as it's a strict inequality, we have f(a) < f(f(a)) and f(f(a)) < f(a). They can't both be true.
for free we get the corollary that this is true on any linearly ordered set (or maybe it's enough to say a partially ordered set whose order is connected? idk) equipped with the order topology; that it's R in particular doesn't matter and that's cool.
That side trip through the IVT was totally unnecessary. When you start it, you already have everything written on the board to show a contradiction. Specifically, you have that a < f(a) implies f(a) < f(f(a)) which further implies f(f(a)) < a. Chain those together and you end up with a < a. Boom!
maybe it's useful to know that the set of the functions (ax+b)/(cx+d) with the operation of composition is isomorphic to the set of 2x2 matrices (a,b, c,d) with ordinary matrix multiplication, so what was done in the beginning was just computing matrix multiplication
I guess there is no need to use the intermediate value theorem, as the inequality swaps for the values f(a) and a, which contradict the fact that the function is increasing.
For all odd number of compositions f(x) = x is the only solution. Funnily for even compositions there can be more (not only in the 2 case as seen at the beginning)
3:58 little did he realize A=B is f(x) = 1 and if (a+1)x + a(b+1) = (b+1)x^2 + (b^2 + a)x then b+1 = 0; b = -1; (a+1)x = (a+1)x works for all a which means the solution is (x+a)/(x-1) = (a-x)/(1-x) so i wonder if (a-x)/(1-x)^n is always valid
I tried a linear algebra approach, not sure if it is valid though. The minimal polynomial gives three eigenvalues: 1 and two complex: -120 and +120 degree rotations, so that the only real eigenvector is the identity function ...
Hello Michael : I found f(f) = 1x ; for real number a generalized, but f(f(f)) = 1x is hard, i try before 13 years go. I still think about. For f(f)) = 1x, it s for every symmetric function g(x,y) = g(y,x) ; so f is solution of equations g(f,x) = 1;
One could more generally consider continuous functions f where f^n = f o f o ... o f (n times) = id. The argument for the case n = 3 could be easily generalized to show that there are no nontrivial solutions for any odd n. In the case of even n, any continuous involution (i.e., f o f = id) would be a solution for n = 2 (and of course also for all other even n). Is there a solution for even n > 2 that is not an involution?
I think there is no solution for even n>2 that is not an involution. I'd argue, that if we split the reals into intervals (like i did in my comment), for even numbers we can argue the the outermost intervals have to be mapped to each other, thus giving a pair that is mapped to each other. Thus it has to be an involution, since now all of the points have to be mapped to one other and back again, lest they be mapped to the outermost and "stuck in a loop".
I would like to see videos on the "snake oil method" for combinatorial identities. The derivation of sum of inverse squares = pi^2/6 using 1/(1-xy) integrated over x and y from 0 to 1 would also be nice. I would in general like to see some proofs from "proofs from the book" by aigner and ziegler. They are some hard to understand, but extremely interesting proofs for sure!
Am I the only one shocked by the "increasing" part of the proof? I didn't find even one comment on how this part is flawed. f not being decreasing on R doesn't make it increasing on R... The proof is a bit more complicated than that.
The "fact" at 8:02 is not really a fact, because it's false. It's indeed true that if f is continuous and invertible on an interval I, then f is also strictly increasing or strictly decreasing; the problem is that the contrary doesn't hold completely. If f is strictly monotone, f is invertible and f is continuous almost everywhere. In particular, If f is strictly monotone f can only have a countable set of discontinuous points, and every discontinuity must be a jump discontinuity. In this problem, we have a continuous f from the start, but it's never stated that f is invertible by the hypothesis: thus, either we add the invertibility hypothesis, to imply that f is strictly monotone, or we assume that f is strictly monotone by hypothesis, without any other information. Indeed, as noticed by others, the continuity of f is unnecessary to prove the fact that f(x) must be equal to x; the only fact that we really use is that f is strictly increasing.
Nice excercise!! In oder to avoid Mobious transforms and tedious handling, I would recommend the no continous function f that maps each k . a_1a_2a_3 ........ to the number k.a_3a_1a_2 ........ taking in consideration the 3 cycle (1,2,3).
Here's a neat way to see that f^n(x)=x has no increasing, continuous solutions for n>1, besides f(x)=x: Suppose such an f exists. Note that f is a bijection from R to R. Let G be the set of all increasing, continuous, bijective functions from R to R. Note that x belongs to G as well, and this set forms a group under function composition, with identity element x. Fix an enumeration of the rationals q_1, q_2, q_3, .... Let g and h be distinct elements of G. Note that knowing the value of g on each q_i determines g by continuity. Define "
The functional cube root of the identity function, nice. Composition is such an under appreciated operation in math outreach IMHO. You can find some seriously hard problems just by trying to solve functions like f^x = g(y). Like the functional square root of sin(x), or even solutions to fractional application problems. This is largely untreaded ground in mathematics.
That's funny, a couple of weeks ago I realised that all examples of continuous one-to-one functions who aren't strictly monotone are defined on disjoint unions of intervals. So I proved that all continuous one-to-one functions on an interval are either strictly increasing or strictly decreasing. And now I saw it being used here. The proof is as follows for anyone interested: Let f: I -> R continuous and one-to-one, where I is an interval of R. Define T={(s,t) ε I^2 : s R the *continuous* function Φ(s,t) = f(s)-f(t). Since T is convex, thus path-connected, a version of intermediate value theorem holds; if the sign of Φ changes between two pairs (s',t') , (s'',t'') there is a (s,t) "between" them such that Φ=0. This contradicts that f is one-to-one, thus Φ is either positive or negative for all points in T, so f is either strictly increasing or strictly decreasing.
Re increasing claim for f( ), didn't explicitly suppose or prove f invertable. Which yes is required for fact (lemma), and excludes most continuous functions seen in Calculus
You can't use decreasing as the opposite of increasing in your contradiction. Some functions are neither. Your proof is still essentially valid though, if you just remove the first line.
A slightly different approach. Assume there is a point x s.t. x ≠ f(x). Then also f(f(x)) ≠ f(x), so x, f(x) and f(f(x)) are three distinct points. Now take the supremum of all r such that intervals B_r = (x-r, x+r), f(B_r), f(f(B_r)) are disjoint. Then some of these three sets should share a boundary point (otherwise you could increase r). Let's say that X and Y=f(X) are the ones that have intersecting boundaries and let s be a point in the intersection. Since s lies in both closures of X and f(X), we have that f(s) lies in cl f(X) and cl f(f(X)), and f(f(s)) lies in cl f(f(X)) and cl f(f(f(X))) = cl X. Since s, f(s) and f(f(s)) are three distinct points, then X, f(X) and f(f(X)) should be three intervals on a line with each two intersecting on the boundaries, which is impossible on a real line.
any single-valued function whose graph is symmetric through the x=y line also satisfies f(f(x))=x, since we know (x,f(x)) is a point in the graph we know (f(x),x) is too because it's symmetric, and (f(x),f(f(x))) is also a point on the graph and since it's a single-valued function there can only be one y for each x so we know f(f(x))=x
f(x) is increasing so its invertible so if G is the inverse of f(x) (im on a phone so i cant type the -1 power) then x = G(G(G(x))) by taking G of both sides three times. B/c f(x) = x is the only function with the property f(x) = g(x) = x, we can see that G(G(x)) = x. By taking F of both sides we get G(x) = F(x) which is again only satisfied by f(x) = x
A visual way of interpreting this: the condition (f○f○f)(x) = x is equivalent to f○f being the inverse of f. This implies that f is a homeomorphism from R to itself (an automorphism in Top), visually this can be interpreted as a function which internally "deforms" R without overlaps. Since R is a line this would visually correspond to "stretching", "shrinking" parts of R as well as possibly shifting and reflecting it. If f reflects the line, then applying it three times would still reflect it, so f can't reflect it in this case. It seems intuitive that applying f a few times won't return R to the original state. More precisely, if you pick and point, repeatedly applying f will always gravitate it towards some fixed point of f, so if after 3 applications of f it must return to its place of origin, it must have never moved.
To whom it may concern: the Mathematica solution in 07:50 is A -> -1 - B - B^2. , and yields True for the three folded substitution, which means Nest[(-1 - B - B^2 + #)/(B + #) &, x, 3] == x // Simplify yields True.
Doesn't need to be. The point is that there are multiple discontinuous solutions and he is giving u a family of them. These are locally continuous so define a piecewise solution from all of them. If u really want a continuous and invertible solution, then f(x)=x is the only solution.
If you define a piecewise function from multiple functions satisfying this property, the piecewise function won’t necessarily satisfy the property itself.
@@comexk Hmm not sure i agree. The requirement is the function satisfy f(f(f(x)))=x. Lets say i find two solutions g(x) and h(x) that satisfy this. Then g(g(g(x)))=x and h(h(h(X)))=X. Now lets define a new function s(x)=g(x) for xa. Then in general s is not continuous. To check if s(s(s(x)))=x, we need to consider two case, xa. For the former, s(x)=g(x) and we know g satisfies this property. For the latter s(X)=h(X) and this too satisfies this property. Thus s also satisfies this property. Infact, since we don't require g in the region x>a and h in the region x
If f(x) = 1/(1-x), then f(f(f(x))) = x. f is continuous everywhere except for x = 1, which causes division by zero. So this proposition requires that f be continuous everywhere.
If you're willing to imagine a special number '∞' with 1/0 = ∞ and 1/∞ = 0 then it even works for x=1. (I know this isn't a proper use of that symbol, but it works out here.)
This gives matrix multiplication vibes. take the column vectors (1,1) and (A,B) as a 2x2 matrix and square it, this will yield (1+A , 1+B) (A+AB, A+B^2). Multiply this matrix by the column vector (x,1) and divide the top entry by the bottom and you’ll get f(f(x))= ((A+1)x+ A+AB)/((B+1)x+A+B^2)
interesting…this sounds like relating the function operation of composition to a hyperoperation of repeated compositions. could we then ask about “composition roots”?
11:15 I think there’s a small mistake. In the video you proved by contradiction that f is increasing, but you didn’t prove it’s strictly increasing. That is, when you assumed f was strictly decreasing (i.e. a f(b) ) you got a contradiction, which means f is increasing but not necessarily strictly increasing (i.e. a>b implies f(a) ≥ f(b), meaning f could be constant in some parts) But you then assumed in the next portion that f is strictly increasing in the proof. It doesn’t kill the proof but it does mean a bunch of > signs should be ≥
A function from R to R which has an inverse must be strictly increasing or decreasing. If not strictly in/decreasing then f(a) = f(b) => f^-1(f(a)) = f*-1(f(b)) => a = b.
@@normanstevens4924 The problem didn’t assume the function is invertible though. It only assumed continuity and that composing the function three times returns x.
@@normanstevens4924 Sure, but again that’s something you derive, it’s not an assumption. (Again, I’m not saying this is a big problem, a couple of small steps wraps it up.)
11:53 I'm probably missing something, but doesn't 𝑓(𝑓(a)) < 𝑓(a) together with the fact that 𝑓 is increasing, already imply that 𝑓(a) < a? Since if 𝑓 is strictly increasing doesn't 𝑓(u) < 𝑓(v) ⇒ u < v?
12:14 - why is intermediate value theorem needed here? Don't we just have a < f(a) < f(f(a)) < a (contradiction) from 3 inequalities on the board?
True - and your way is more elegant - but his proof works even if that part of his argument can be done another way w/o using the IVT.
this is a more natural solution, but using IVT was cool
@@DeanCalhounUsing IVT means using continuity twice (it was already used to conclude monotonicity), so ... meh
Here's a quicker way:
We already showed f is increasing, so if f(a) > a, then f(f(a)) > f(a) [by applying f on both sides], which in turn implies f(f(a)) > a. Again applying f on both sides would yield f(f(f(a))) > f(a) > a, which is a contradiction.
For the other case, if f(a) < a, then by similar argument, you can show that f(f(f(a))) < f(f(a)) < f(a) < a, which is again a contradiction.
Yeah I'm surprised he seemingly missed that and used IVT instead
@@tomaszadamowski yeah, this answer seemed more natural to me.
Truee lol I was sos surprised lmao
Right on! I was thinking about exactly the same way to do it, which is both faster and more intuitive.
He is just lazy
Note that if you replace the real numbers R by the complex numbers C, you do get continuous solutions which are not the identity. Namely, for each of the three solutions z of the cubic equation z^3 = 1, one can define f(x) = x*z with the property f(f(f(x))) = x * z^3 = x.
Are there any others?
If we are using functions from Riemann sphere to itself, I suppose we have the ones he mentioned with the (x+A)/(x+B) ...
And I guess, there would be similar such functions for any 2x2 matrix which cubes to the identity matrix?
@@drdca8263 If we're on the Riemann sphere, yeah. Then the only holomorphic functions that solve our problem are the Möbius transforms of order 1 and 3. Order 1 is just the identity, and order 3 corresponds to 2x2 matrices with characteristic polynomial x^2+x+1 (cause they need to solve x^3-1=0), meaning matrices with trace -1 and determinant 1.
Also x^z, where z³=1
@@nikoladjuric9904 Can that be defined continuously?
Also f(x) = A+z*(x) , where z³=1 but z≠1 , and A is any constant.
5:30 You accidentally multiplied the numerator by the numerator, instead of the denominator on this board, the next board does have the correct equations however.
That's the moment, when he cleans the board and continues with the correct formula!
@@rainerzufall42 This is how students learn to hate Mathematics, alas. For the love of Mike, admit that a mistake was made and correct it.
@@holliebuchanan6064 It's good for instructors to admit mistakes, I agree. But also students shouldn't hastily judge the subject of mathematics (or judge themselves in relation to the subject) because of some wrong algebra done on a board, which is usually just done to fill up time. If you care about learning the subject you should be able to fix the calculation, that's what really matters.
Before watching:
1) f³x=x => f is invertible (the inverse is f²)
2) An invertible continuous function from IR to IR is either strictly increasing or strictly decreasing.
3) If f is decreasing then f³ is decreasing, but f³=x is increasing so f is increasing.
4) a f(a)
Really nice and clean arguments, thank you. It is really shocking to me though that there exists this magical f(x) = x / (x - 1) which works almost everywhere.
Good place to stop was actually at 11:48 since we have a
Hi,
At 12:00 you already have the contradiction because we have f(a) < f(f(a)) and f(f(a)) < f(a) .
You cannot prove that f is increasing supposing that f is decreasing. It can be neither increasing nor decreasing
yeah! why is no one seeing this
I think it was said that because the function is both invertible (with inverse f∘f) and continuous then it has to be bijective (and thus one to one).
So it either has to either strictly increase or strictly decrease otherwise at least one value would map to the same number and then could not be sent back to both of them with two more applications of f.
He literally wrote on the blackboard, and pointed it out: a continuous invertible function is either strictly increasing or strictly decreasing.
Take an interval I and the function Φ: T -> R, where T is the "lower triangle" of I^2, Φ(s,t)=f(s)-f(t) for (s,t) in T. T is convex, thus path connected, and Φ is continuous, and non-zero if f is "1-1", so it preserves its sign. So f is either strictly increasing or strictly decreasing.
@@ThomasBushnellBSG
Deleted/edited also. 👍🏻
The example for non-continuous solutions is overly complicated - and as stated not even a function! How to define f(-B)? And how to define f((B^2+A)/(B+1))?
A much simpler infinite family of non-continuous solution is the following: Let a, b, c be three distinct real numbers. Define f(a)=b, f(b)=c, f(c)=a, and f(x)=x otherwise.
Yes, f(x)=1/x is not a function.
And in fact the function is constructed to be undefined at -B because then it is (vacuously) discontinuous there. It's one of the most natural ways to get a discontinuous function
14:06 The identity function remains undefeated
The rational functions that satisfy this are (ax+b)/(cx+d) where a^2+ad+d^2+bc=0. For example, if a=1 and d=4, we can take b=-3 and c=7 to get (x-3)/(7x+4). Then we can just send 1/7 to 1/7 and -4/7 to -4/7.
It turns out that if we only use rational coefficients, then a function like this can only have “order” equal to 1 (for the identity function), 2 (if a+d=0), 3 (the condition above), 4 (if a^2+d^2+2bc=0), or 6 (if a^2-ad+d^2+3bc=0). Then you can just reassign the values where the denominator is 0.
Easier method would be to derive a = f(f(f(a))) > f(f(a)) > f(a) > a, which is a contradiction
11:38 the contradiction is right there on the board.
There was no need to continue for 2 more minutes.
One can also go about this a bit differently, with more of a look towards topology.
We can take the set {x,f(x),f(f(x))}={a,b,c} with a
Could you explain more why the inverse image of an open set around d is not open?
@@19divide53
The following was my thought process, but now that i think about it again, this argument is very similar to 2 above. As added context: this part was mostly proven by "look at it, it probably holds" (and having the example of the map S^1->(0,1] as a typical discontinuous one in my mind), and thus the argument might have to be refined a bit for the general case, but not by too much.
The inverse image of d has to be one of the points b or c. If we take an open set U around d, the inverse is also open, and thus intersects both the intervals A and B (in general the connected components) around f^-1(d). Since we know that f is bijective (by the concotonation property) and maps connected components to connected components (by continuity), we know that f restricted to a connected component maps bijectively to another connected component. Now either A or B is mapped into J, but neither of them is in the original interval (-inf, a).
Of course in Complex set, 120deg rotation with e^i*2pi/3 is one that works.
As well as 240 with 4π/3 or -2π/3 and 360 with 0 or 2π (the cube roots of unity).
12:34 Don't we have a contradiction here already? Above we have f(a) < f(f(a)), and below that we have f(f(a)) < a < f(a). If we leave the a out of the second inequality, which we can do as it's a strict inequality, we have f(a) < f(f(a)) and f(f(a)) < f(a). They can't both be true.
0:54 your two fold example could equally well apply to all complex numbers too, and most other domains over which subtraction makes sense.
I think it’s easier that a
Interestingly, in the case of the discontinuous version, setting A = -2 and solving for constant B gives us -(Golden ratio) and 1/Golden ratio.
for free we get the corollary that this is true on any linearly ordered set (or maybe it's enough to say a partially ordered set whose order is connected? idk) equipped with the order topology; that it's R in particular doesn't matter and that's cool.
That side trip through the IVT was totally unnecessary. When you start it, you already have everything written on the board to show a contradiction. Specifically, you have that a < f(a) implies f(a) < f(f(a)) which further implies f(f(a)) < a. Chain those together and you end up with a < a. Boom!
3:53 why didn't you put f(x) into f•f, but instead put f•f(x) into f? this was much more convoluted this way
maybe it's useful to know that the set of the functions (ax+b)/(cx+d) with the operation of composition is isomorphic to the set of 2x2 matrices (a,b, c,d) with ordinary matrix multiplication, so what was done in the beginning was just computing matrix multiplication
I guess there is no need to use the intermediate value theorem, as the inequality swaps for the values f(a) and a, which contradict the fact that the function is increasing.
For all odd number of compositions f(x) = x is the only solution. Funnily for even compositions there can be more (not only in the 2 case as seen at the beginning)
Sarkovskii theorem (1964): if f:R->R is continous with x0 such that f(f(f(x0)))=x0, then f has periodics points of any period
3:58 little did he realize A=B is f(x) = 1 and if (a+1)x + a(b+1) = (b+1)x^2 + (b^2 + a)x then b+1 = 0; b = -1; (a+1)x = (a+1)x works for all a which means the solution is (x+a)/(x-1) = (a-x)/(1-x) so i wonder if (a-x)/(1-x)^n is always valid
I tried a linear algebra approach, not sure if it is valid though. The minimal polynomial gives three eigenvalues: 1 and two complex: -120 and +120 degree rotations, so that the only real eigenvector is the identity function ...
Hello Michael :
I found f(f) = 1x ; for real number a generalized, but f(f(f)) = 1x is hard, i try before 13 years go. I still think about.
For f(f)) = 1x, it s for every symmetric function g(x,y) = g(y,x) ; so f is solution of equations g(f,x) = 1;
One could more generally consider continuous functions f where f^n = f o f o ... o f (n times) = id. The argument for the case n = 3 could be easily generalized to show that there are no nontrivial solutions for any odd n. In the case of even n, any continuous involution (i.e., f o f = id) would be a solution for n = 2 (and of course also for all other even n). Is there a solution for even n > 2 that is not an involution?
I think there is no solution for even n>2 that is not an involution. I'd argue, that if we split the reals into intervals (like i did in my comment), for even numbers we can argue the the outermost intervals have to be mapped to each other, thus giving a pair that is mapped to each other. Thus it has to be an involution, since now all of the points have to be mapped to one other and back again, lest they be mapped to the outermost and "stuck in a loop".
I would like to see videos on the "snake oil method" for combinatorial identities.
The derivation of sum of inverse squares = pi^2/6 using 1/(1-xy) integrated over x and y from 0 to 1 would also be nice.
I would in general like to see some proofs from "proofs from the book" by aigner and ziegler. They are some hard to understand, but extremely interesting proofs for sure!
He did a video about the second suggestion a while ago but I don't remember the name of it. Other suggestions would be great to see
Am I the only one shocked by the "increasing" part of the proof? I didn't find even one comment on how this part is flawed. f not being decreasing on R doesn't make it increasing on R... The proof is a bit more complicated than that.
Yes, but it is continous and invertible, so must be increasing or decreasing. We eliminated decreasing case.
Consider the discrete domain {0,1,2}. Your f(x) could simply be f(x)=x+1 (mod 3). Slightly more interesting: {0..5} and f(x)=x+2 (mod 6).
The "fact" at 8:02 is not really a fact, because it's false. It's indeed true that if f is continuous and invertible on an interval I, then f is also strictly increasing or strictly decreasing; the problem is that the contrary doesn't hold completely. If f is strictly monotone, f is invertible and f is continuous almost everywhere. In particular, If f is strictly monotone f can only have a countable set of discontinuous points, and every discontinuity must be a jump discontinuity. In this problem, we have a continuous f from the start, but it's never stated that f is invertible by the hypothesis: thus, either we add the invertibility hypothesis, to imply that f is strictly monotone, or we assume that f is strictly monotone by hypothesis, without any other information. Indeed, as noticed by others, the continuity of f is unnecessary to prove the fact that f(x) must be equal to x; the only fact that we really use is that f is strictly increasing.
The invertibility actually follows just from f○f○f=id, as this implies f○f=f^-1 (and thus bijectivity follows)
@@AboutMoreGames You're right, the functional equation f○f○f=id implies that the inverse function of f exists.
Nice excercise!! In oder to avoid Mobious transforms and tedious handling, I would recommend the no continous function f that maps each k . a_1a_2a_3 ........ to the number k.a_3a_1a_2 ........ taking in consideration the 3 cycle (1,2,3).
11:31
Couldn't you conclude here that:
a < f(a) < f(f(a)) < a
Which means:
a < a
Which is a contradiction?
Here's a neat way to see that f^n(x)=x has no increasing, continuous solutions for n>1, besides f(x)=x:
Suppose such an f exists. Note that f is a bijection from R to R. Let G be the set of all increasing, continuous, bijective functions from R to R. Note that x belongs to G as well, and this set forms a group under function composition, with identity element x.
Fix an enumeration of the rationals q_1, q_2, q_3, ....
Let g and h be distinct elements of G. Note that knowing the value of g on each q_i determines g by continuity. Define "
The functional cube root of the identity function, nice. Composition is such an under appreciated operation in math outreach IMHO. You can find some seriously hard problems just by trying to solve functions like f^x = g(y). Like the functional square root of sin(x), or even solutions to fractional application problems. This is largely untreaded ground in mathematics.
That's funny, a couple of weeks ago I realised that all examples of continuous one-to-one functions who aren't strictly monotone are defined on disjoint unions of intervals. So I proved that all continuous one-to-one functions on an interval are either strictly increasing or strictly decreasing. And now I saw it being used here. The proof is as follows for anyone interested:
Let f: I -> R continuous and one-to-one, where I is an interval of R. Define
T={(s,t) ε I^2 : s R the *continuous* function Φ(s,t) = f(s)-f(t). Since T is convex, thus path-connected, a version of intermediate value theorem holds; if the sign of Φ changes between two pairs (s',t') , (s'',t'') there is a (s,t) "between" them such that Φ=0. This contradicts that f is one-to-one, thus Φ is either positive or negative for all points in T, so f is either strictly increasing or strictly decreasing.
Re increasing claim for f( ), didn't explicitly suppose or prove f invertable. Which yes is required for fact (lemma), and excludes most continuous functions seen in Calculus
You can't use decreasing as the opposite of increasing in your contradiction. Some functions are neither. Your proof is still essentially valid though, if you just remove the first line.
If fofof=i can we replicate this and get
fofofofo.........=i.
If it is true it implies f=i.
you talk about homework in the video. Is this supposed to be a companion video to a course you teach?
A slightly different approach. Assume there is a point x s.t. x ≠ f(x). Then also f(f(x)) ≠ f(x), so x, f(x) and f(f(x)) are three distinct points. Now take the supremum of all r such that intervals B_r = (x-r, x+r), f(B_r), f(f(B_r)) are disjoint. Then some of these three sets should share a boundary point (otherwise you could increase r). Let's say that X and Y=f(X) are the ones that have intersecting boundaries and let s be a point in the intersection. Since s lies in both closures of X and f(X), we have that f(s) lies in cl f(X) and cl f(f(X)), and f(f(s)) lies in cl f(f(X)) and cl f(f(f(X))) = cl X. Since s, f(s) and f(f(s)) are three distinct points, then X, f(X) and f(f(X)) should be three intervals on a line with each two intersecting on the boundaries, which is impossible on a real line.
That's a really nice argument
Not sure what the HW is supposed to be since you already proved f is increasing?
Do you really need IVT there? Seems like the inequality a
The negation of increasing is not decreasing 🤔
f(x)=A/x is my favorite of the the functions such that f ( f ( x )) = x
A, x 0
I thought of this a month ago. It took a discord walkthrough.
Wtf. He could use transitivity of
OMG, he didn’t produce the tightest proof possible? Oh, the horror!
@@bookert2373He si getting messier and messier, very disrespectful to do such a shabby job
@@pepefrogic3034 I don’t mean to be disrespectful, but perhaps you should be looking for a different channel, one that would properly respect you?
any single-valued function whose graph is symmetric through the x=y line also satisfies f(f(x))=x, since we know (x,f(x)) is a point in the graph we know (f(x),x) is too because it's symmetric, and (f(x),f(f(x))) is also a point on the graph and since it's a single-valued function there can only be one y for each x so we know f(f(x))=x
N fold composition please
f(x) is increasing so its invertible so if G is the inverse of f(x) (im on a phone so i cant type the -1 power) then x = G(G(G(x))) by taking G of both sides three times. B/c f(x) = x is the only function with the property f(x) = g(x) = x, we can see that G(G(x)) = x. By taking F of both sides we get G(x) = F(x) which is again only satisfied by f(x) = x
A visual way of interpreting this: the condition (f○f○f)(x) = x is equivalent to f○f being the inverse of f. This implies that f is a homeomorphism from R to itself (an automorphism in Top), visually this can be interpreted as a function which internally "deforms" R without overlaps. Since R is a line this would visually correspond to "stretching", "shrinking" parts of R as well as possibly shifting and reflecting it. If f reflects the line, then applying it three times would still reflect it, so f can't reflect it in this case. It seems intuitive that applying f a few times won't return R to the original state. More precisely, if you pick and point, repeatedly applying f will always gravitate it towards some fixed point of f, so if after 3 applications of f it must return to its place of origin, it must have never moved.
To whom it may concern: the Mathematica solution in 07:50 is A -> -1 - B - B^2. , and yields True for the three folded substitution, which means Nest[(-1 - B - B^2 + #)/(B + #) &, x, 3] == x // Simplify yields True.
Bro didn't spot a < a
But the discussed function is not defined for all real numbers.
Doesn't need to be. The point is that there are multiple discontinuous solutions and he is giving u a family of them. These are locally continuous so define a piecewise solution from all of them. If u really want a continuous and invertible solution, then f(x)=x is the only solution.
If you define a piecewise function from multiple functions satisfying this property, the piecewise function won’t necessarily satisfy the property itself.
@@comexk Hmm not sure i agree. The requirement is the function satisfy f(f(f(x)))=x. Lets say i find two solutions g(x) and h(x) that satisfy this. Then g(g(g(x)))=x and h(h(h(X)))=X. Now lets define a new function s(x)=g(x) for xa. Then in general s is not continuous. To check if s(s(s(x)))=x, we need to consider two case, xa. For the former, s(x)=g(x) and we know g satisfies this property. For the latter s(X)=h(X) and this too satisfies this property. Thus s also satisfies this property. Infact, since we don't require g in the region x>a and h in the region x
@@shohamsen8986x
@@joshuahanson9905Im not sure i understand your concern. Why does g(x)
Nice video. You're really digging these rainbow paint thumbnails aren't you?
Nice one.
Idempotent homomorphisms in abstract algebra can be really interesting and have some niche use.
If f(x) = 1/(1-x), then f(f(f(x))) = x. f is continuous everywhere except for x = 1, which causes division by zero. So this proposition requires that f be continuous everywhere.
If you're willing to imagine a special number '∞' with 1/0 = ∞ and 1/∞ = 0 then it even works for x=1. (I know this isn't a proper use of that symbol, but it works out here.)
This gives matrix multiplication vibes.
take the column vectors (1,1) and (A,B) as a 2x2 matrix and square it, this will yield (1+A , 1+B) (A+AB, A+B^2). Multiply this matrix by the column vector (x,1) and divide the top entry by the bottom and you’ll get f(f(x))= ((A+1)x+ A+AB)/((B+1)x+A+B^2)
Indeed, there is an homomorphism from the group of nonsingular 2×2 matrices to the fractional linear transformations!
Rotation by 2π/3.
Yes, if complex
interesting…this sounds like relating the function operation of composition to a hyperoperation of repeated compositions. could we then ask about “composition roots”?
Aren't those usually called _eigen functions_ ?
11:15 I think there’s a small mistake. In the video you proved by contradiction that f is increasing, but you didn’t prove it’s strictly increasing. That is, when you assumed f was strictly decreasing (i.e. a f(b) ) you got a contradiction, which means f is increasing but not necessarily strictly increasing (i.e. a>b implies f(a) ≥ f(b), meaning f could be constant in some parts)
But you then assumed in the next portion that f is strictly increasing in the proof. It doesn’t kill the proof but it does mean a bunch of > signs should be ≥
A function from R to R which has an inverse must be strictly increasing or decreasing. If not strictly in/decreasing then f(a) = f(b) => f^-1(f(a)) = f*-1(f(b)) => a = b.
@@normanstevens4924 The problem didn’t assume the function is invertible though. It only assumed continuity and that composing the function three times returns x.
@@Bodyknock f(f(x)) is the inverse of f(x)
@@normanstevens4924 Sure, but again that’s something you derive, it’s not an assumption. (Again, I’m not saying this is a big problem, a couple of small steps wraps it up.)
I often wonder apart from academic purposes what real value do such problems have?
This one seems very useful to me. To have proven knowledge that there's no strange continuous function that will return to the identity
Stability and chaotic systems are all about repeated applications of a function.
@@hybmnzz2658 that is a very useful piece of information. I will check it out for details.
11:53 I'm probably missing something, but doesn't 𝑓(𝑓(a)) < 𝑓(a) together with the fact that 𝑓 is increasing, already imply that 𝑓(a) < a? Since if 𝑓 is strictly increasing doesn't 𝑓(u) < 𝑓(v) ⇒ u < v?