Strictly increasing actually doesn't imply that f is bijective. Try: f(x)=x-1 when x= 0. However if you assume that f is continuous, then f must be bijective. (Conversely, every strictly increasing bijective function is continuous)
Even with continuity we can't be sure that f is bijective simply by expanding the codomain to be bigger than the image. For example, f:R->R with f(x)=arctan(x) is continuous and strictly increasing but not bijective. But strictly increasing implies one to one, so we only need to demand f to be onto R
This example clearly highlights the difference of mind between pure and applied maths: for instance, an engineer, or a phisicist would spend at most 10 seconds in guessing as possibile solution the identity plus constant function, i.e. f(x)=x+k, and then he would classify such question as elementary, even banal; on the other hand, a pure maths theorist is interested in checking the unicity of the immediately guessed solution and this task will result a clever challenging work.
A geometric approach of this problem is the following. We know f and f^-1 are symmetric to y=x. We also get that f(x)-x = x- f^-1(x). That means that for every x, f and f^-1 should maintain equal distances from the point (x,x) on the line y=x. This intuitively leads to f being parallel to y=x , as it's the only way to preserve both of these symmetries, which is easily proved by contradiction.
HOMEWORK : Four integers a, b, c, and d with a ≤ b ≤ c ≤ d satisfy the property that the product of any two of them is equal to the sum of the other two. Given that the four numbers are not all equal, determine (if they exist) all the 4-tuples (a, b, c, d). SOURCE : Exeter Math Club Competition 2016 BONUS OPEN PROBLEM : What happens if we take 2n instead of 4 ? Product of n of them is equal to the other n.
Trivial solution is a = b = c = d = 0, but that is excluded. Another excluded solution is a = b = c = d = 2. The only accepted solution is then a = b = c = -1 and d = 2. For 2n-tuples, a similar pattern works for even n: x₁…x₂ₙ₋₁ = -1, x₂ₙ = n If x₂ₙ is in the product, the product becomes n∙(-1)ⁿ⁻¹, and the sum becomes n∙(-1); They are equal iif n is even. If x₂ₙ is in the sum, the product becomes (-1)ⁿ, and the sum becomes n+(-1)*(n-1) = 1; Again they are equal iif n is even.
@@nedbowlas913 Basically I have a dozen of bookmarks to math competitions websites, math magazines and miscellaneous stuff. I also have some PDF documents of books. Then I pick a problem whenever I feel it's original and hard enough.
SOLUTION *(-1, -1, -1, 2)* We can in fact solve the problem in real numbers. Let S = a + b + c + d. From the problem condition we see that ab + cd = (c+d) + (a+b) = S. Similarly, ac + bd = ad + bc = S. Notice that (a + b)(c + d) = (ac + bd) + (ad + bc) = 2S, and similarly (a + c)(b + d) = (a + d)(b + c) = 2S. By Vieta's Theorem we get that all three pairs (a + b, c + d), (a + c, b + d), (a + d, b + c) are solutions of the equation x² − Sx + 2S = 0, so all three pairs must be equal to (u, v) or (v, u) for some real numbers u, v. Since there are at most two distinct values in {a + b, a + c, a + d}, by Pigeonhole Principle at least two of b, c, d are the same. WLOG let b = c. Then by making the same observation on a, b, d gives that either there are two pairs of equal numbers among the four three of the four numbers are equal. If it's the first possibility then a, b, c, d is a permutation of p, p, q, q for some distinct reals p, q, but it's not difficult to see that {p + p, q + q} ≠ {p + q, p + q}. This leaves the second possibility (where the numbers is a permutation of p, p, p, q for p ≠ q), which satisfy the desired property. Now it remains to solve the following system of equations : p + p = pq, p + q = p². The first equation gives p = 0 or q = 2. If p = 0 then q = 0, which contradicts the requirement that p ≠ q. If q = 2, then p + 2 = p² has two solutions −1 or 2, but since p ≠ q= 2 we are forced to have p = −1, and by putting the numbers in non decreasing order we get (−1, −1, −1, 2) as the one and only possible quadruple.
@16:25. Correct me if I'm wrong, but I think the procedure here is not correct. If t+(m-1)a < x < t+ma, then t+ma < f(x) < t+(m+1)a, since f is increasing. This implies f(x)-x=b
He notes that f(x)=x satisfies the functional equation, so then proceeds to find other functions. Thus the assumption that f(x)!=x for some x. Pick such an x and call this t. Then either f(t)>t or f(t)t. And along the way he shows that if f(t)>t for _some_ t, then f(x)>x for _all_ x. Here t is a constant and x is a variable. Now t is a constant, so f(t)-t is also a constant. So a is a constant (and b also).
@@muratcan-k6x Hahagahaha, sorma. 6. sınıftan falan kalma. Reis, bu arada sen daha kolay bir çözüm görüyor musun şu soruya? Sâhiden bundan daha basit bir yöntem olmalıymış gibi geliyor bana.
I finally found a quick solution that's pretty good. This writeup relies on differentiability, but I haven't seen any correct solutions using it anyway (mostly incorrect uses of the derivative of function inverses). You can show the same thing with some sort of finite difference operator, it's just a bit less legible. Say you found that solutions of the form f(x) = x + c work by inspection. Assuming differentiability: f'(x) + 1/f'(f-1(x)) = 2 f'(f-1(x)) = 1/(2 - f'(x)) Now we know f'(x)>0 everywhere. We can also see f'(x) < 2 because otherwise it would be negative at f-1(x) from above. Here's the tricky part. Let's say f'(x) has bounds 0
Nice! It took me some thought to understand how you deduced the inequalities so I put it here for the next person who might be interested. Set y:=f-1(x) 1/U
At first, I was confused. However, I have realized that t is one fixed number during my second viewing and understood the flow of logic. The margin from g(x)=x at x=t, i.e. a, becomes margins at arbitrary x.
I was confused because I had a smudge on my screen that made an equal sign look like a minus and kept trying to figure out why he was subtracting this second similar equation from what looked like the correct answer.
Don't solutions exist in pairs except for f(x)=x? f(t) is a solution iff its inverse is a solution, so you should be able to assume the first case, f(t) > t, without loss of generality: if f(t) < t then f^-1(t) > t (since they sum to 2t). Probably simpler to show that fact (f(t) is a solution exactly when its inverse is) than going through the whole argument again with f(t) < t. And that just follows from the defining property pretty easily.
log2(x) is the inverse of f(x) = 2^x. It's true that the domain and co-domain are not all of the real numbers, but that is not necessarily what surjective means. In your case, f(x) can be surjective because it goes from the domain (all real numbers) to the co-domain (all positive real numbers). And f^-1(x), which is a logarithm, goes from its domain (all positive reals) to its co-domain (all reals).
Maybe it's a regional terminology difference, but I was taught that surjectivity is required, and if you define your function's codomain to include values that aren't in its image, then it's not surjective, not bijective, and not invertible. You can't just tacitly assume the codomain is so restricted. It's academic in this case as the question uses the inverse for all x in R, so it must be taken to imply that the inverse has to exist, which becomes another constraint on our choice of function. It doesn't change the solution, I just wouldn't claim that strict monotonicity is enough for a function to be bijective.
Differentiate both sides wrt x (assuming a differentiable function). f(x) goes to f'(x) and f-1(x) goes to 1/f'(x) because it is a reflection of f(x) in y=x, and there is no problem with dividing by zero. Multiply the resulting expression and factorise to give (f'(x)-1)^2=0. So f(x) = x+a.
This isn't quite right. f-1(x)' is 1/f'(f-1(x)). You're evaluating the reflection over y=x at the wrong point. It's a coincidence that the result is correct.
You can solve this problem entirely geometrically! Rephrasing the equation as midpoints and applying cartesian geometry this reduces to showing that the trapezoid formed by (x, f(x)), (f(x), x), (x, f^-1(x)), (f^-1(x), x) is actually a square, and it follows that the only f satisfying this property for all real x must be f(x) = x - a.
Starting at 11:20, we take x not in {t+na: n in Z}, and show that f(x)>x. This implies that b = f(x)-x > 0. Basically, once we have f(x)>x for _one_ value of x, we get that f(x)>x for _all_ values of x. So b>0 if and only if a>0.
@@Happy_Abe I agree. I sometimes do not write clearly enough. I'll need to rewatch the video, though. It's possible his analysis shows that t+na must be in the range of f for all integers n, and thus the range actually must be R. Unspoken assumptions can be very difficult to spot.
@@bulgeo09 because thats not a function from R to R. a function from R to R needs to be defined at all R. at least for -1/x there is no way to fill the hole x=0 in a way that makes the function strictly increasing.
@@bulgeo09 that example doesn't quite work because it isn't defined at x=0 (so is not a function from R to R), but something like arctan does work: it is continuous and strictly increasing, but not bijective (because it is not surjective; it's image is (-pi/2, pi/2).)
@@schweinmachtbree1013 the refined statement should be a strictly increasing function does not imply bijectivtivity unless it is continuous and f: R->R. Otherwise, you’re arctan example would of course be a counterexample, disproving it.
@@tomatrix7525 that refinement is still not true - arctan is strictly increasing, continuous, and R->R. we can only conclude bijectivity onto the image, not complete bijectivity (to be able to conclude complete bijectivity all we can do is add surjectivity as an extra hypothesis)
I used (what I think is) an easier and faster method to solve this one. Since f is strictly increasing, the function has a positive slope (ignoring points at which there's no derivative). If it makes an angle of π/4+(theta), it's inverse makes an angle of π/4-(theta) (as inverse is reflection along y=x). So both these angles sum to π/2 and their tangents sum to 2 (by differentiating both sides of the given functional equation). Solving we get tan(theta)=1 so theta=π/4 or slope=1 is the only option for points where the function is differentiable. So if there exists a point of non-differentiability, it will be a point of discontinuity. But since f is mapped from R to R, an inverse can only exist if range of f=codomain of f so there is no possibility of a discontinuity for f. Also the functional equation holds for all real x so the domain of f inverse must be the set of all reals as well. This too implies that f cannot have any discontinuities. So f(x)=x+a is the only option.
hi Rayan. I am sorry but your proof is not correct, since the given equation is f(x)+f^-1(x)=2x and not what you suppose f(x)+f^-1(f(x))=2x. This imediately gives f(x) + x = 2x ie f(x)=x
I don't think you can assume f'(x) exists by monotonicity, but that the question involves f'(x) on all x in R should be a good enough stipulation that f'(x) exists
Interesting use of the sandwich theorem. ---- My first guess was proving on a set isomorphic to Z, proving on that to Q, and then using monotonicity to prove on R.
Definition: A monic linear polynomial function is a strictly monotone continuous bijection f from *R* to *R* satisfying the functional equation f(x) + f^{-1}(x) = 2x
@@gergonemes88 Good point, I guess it would have to be “translation of the real line” instead of “linear polynomial function” (although it was meant to be silly anyway so whatever xD) Edit: I guess “monic linear polynomial function” works too
I didn't have the patience for this sort of in depth analysis so I just instinctively chose f(x) = mx + b and deduced that m =1 and b could be any real number. ;-)
This is a good demonstration. Another way is to derive the equation in regard to x, noticing that f'(x)>0 as f(x) is strictly increasing, and f'(f^-1(x))=1/(f'(f(x)). Thus, f'(x)+1/(f'(f(x))=2. All the terms being strictly positive, 00. Hence f(x)=cx+d, w/ c,d real numbers. f^-1(x)= (x-d)/c and finally with the initial equation, c=1, d is a real number.
Actually I also thought about this way. But I thought that we will get this equation: f'(x) + 1/(f'(x)) = 2 where we get that: f'(x) = 1 hence f(x) = x +C - is the unique solution.
we can simply differentiate both side (knowing that the derivative of a function's inverse is one over the derivative of that function). We get the equation c+1/c=2 in terms of dy/dx. Solving for c we get dy/dx=+-1 but since strictly increasing it is just 1. Integrating we get f(x)=x+c
@@PETAphile Maybe? But I don't see any easy demonstration of this. However, Lebesgue's Theorem for the Differentiability of Monotone Functions tells us the function is differentiable almost everywhere. We can probably use John Steven's demonstration to show the function must be f(x)=x+c on each domain it is differentiable on, and show the c must be the same for all those domains because of continuity (a monotonous, bijective function must be continuous). John Steven's demonstration idea basically works, he just glossed over some major difficulties we need to address to make sure we are allowed to use this demonstration.
Am I missing something, isn't the derivative of an inverse function is one over that function, composed with the inverse of the function? Shouldn't this invalidate this calculation?
I have tried in vain to prove the result if we suppose f is differentiable. It seems manageable, but I can’t find the way. Does anybody have such a proof?
Why are we assuming f(x) =/= x? Are we making it part of the problem statement or what? I mean, are we changing the problem statement after reading it the first time? Why would we do that?
I have a better method Define f(x_n)=x_(n+1) and x_1=x Thus we get the recurrence x_(n+1)+x_(n-1)=2x_n Thus by telescoping we get: x_(n+1)-x_n=f(x)-x Again by telescopic summation we get: x_(n+1)=nf(x)-(n-1)x Now using that f is strictly we get f(x)=x
We're not trying to say that x is of the form t+na. We've found a particular t and the corresponding a, and we're only letting n vary arbitrarily. Consider f(x)=x+sin(a(x-t)/2π)+a. It's equal to x+a at x=t+na, but not at other values, and it doesn't fit the functional equation, so we must have more to prove.
I have a problem suggestion: if you have fractions a/b and c/d (all integers, b not 0, d not 0), then: a/b + c/d = (ad+bc)/(bd). In general, the sum is *not* just (a+c)/(b+d), a mistake many students make when learning about fractions. However... sometimes (a+c)/(b+d) *does* happen to give you the right answer. So, what are all the types of {a, b, c, d} (as above) solutions that work?
@@angelmendez-rivera351 I haven't had the chance to look this over in detail yet, but I'm delighted you tried it out :) did you like it? i personally thought it was a really fun question :-D
I'm not 100% sure, but I think it's saying that any real number a will satisfy the requirements. As f(x)=x+a, f^-1(x)=x-a. This means that f(x)+f^-1(x)=(x+a)+(x-a)=2a
f is bijective (because it increases strictly on R and is continuous on R) thus for any number a in f(R) (we may suppose it is R), there exists one and only one real number x such that f(x) = a. In particular, any number a can be written as f of someone. But in fact, I don’t think you need all that reasoning because you can actually name anything you want by anything else.
Isn't there any way to proceed by assuming x = f(y) for some y? That way we get rid of the f^-1 and we get f(f(y)) + y= 2f(y), which seems to be a better place to begin with.
This is an intuitive way to _suggest_ the result. Our back of a napkin calculation proceeds as follows, writing function applications as fx rather than f(x). ffy + y = 2fy ffy - 2fy + y=0 (move everything to LHS) (f^2 - 2f + 1)y = 0 (factor out the y) (f-1)^2 = 0 (for all y!=0 we can divide by the y, then factorise the quadratic) Hence f=1. So fx = x. Obviously this misses the f(x)=x+a case, and is a rough exploration of an idea, but suggests that f(x) must be linear. Then one has to proceed to find a rigorous proof, which is what Michael gives. It is quite simple in nature, stating from one point where f(x)>x, then proceeding to show that f(x)=x+a, leaving the case f(x)
The proof he gives shows, among other things, that if f(x)>x for _one_ value of x, then f(x)>x for _all_ values of x. This is the result of step 3 in the summary below. The structure of the proof is: 1.) to assume that f(t)>t for one value of t. 2.) Then show that f(x)=x+a for all x in {t+na for n in N},then for all x in {t+na for n in Z}. 3.) Then for any x not in {t+na for n in Z}, he shows that f(x)>x, 4.) so we repeat the argument to see that f(x)=x+b for x in {t+nb for n in Z}. 5.) Finally an inequality is used to show that a=b.
Yo okay so I did something here that no one else seems to have picked up on. (f^-1)'(x) = 1/f'(x) So given f(x) + (f^-1)(x) = 2x Take the derivative, we'll say y = f'(x) y + 1/y = 2 y² - 2y + 1 = 0 (y-1)² = 0 y - 1 = 0 y = 1 This means f'(x) = 1 Now integrating, f(x) = x + C Wow!
I may have missed it but why bother assuming a doesn’t equal zero in the very beginning? Was there a step where something was divided by a at some point? It seems like you could ignore that assumption and get the same result.
We observe that f(x) satisfies the functional equation right at the start. Then to look for any other f(x), we assume that f(x)!=x for some x. This splits into two cases: f(x)x. This gives a=f(x)-x>0 as one case, and a=f(x)-x0 case, since the a
I actually don't know how to go about solving this, other than that f(x) = x is one obvious solution. This will be interesting to watch. Edit after watching: Wow, yeah, I wouldn't have come up with basically any of that.
if f(t) < t then t < f^{-1}(t) a = f^{-1}(t) - t We have if f solution then f^{-1} inverse of f is also solution. Because if f strictly increasing we have f^{-1} strictly increasing and if for all x in R f(x) + f^{-1}(x) = 2x Then for all x in R f^{-1}(x) +( f^{-1})^{-1} (x) = 2x and f^{-1}(t) > t Then for all x in R f^{-1}(x)=x+a Then for all x in R f(x)=x-a
Hello Michael Can u solve this problem This problem was in my math assignment sheet The problem is : Lim x------>0 (f(x)) Where f (x) is : (xsin(sin(x) ) - sin^2(x) ) / x^6 And we have to solve it *WITHOUT* L'Hopitals rule.. The ans is 1/18 This is a very interesting problem that's why I wanted share this with you.
If we knew that f is differentiable then it is very easy to prove that f(x) = x + c. Perhaps one can prove that a function f that satisfies our hypotheses must be differentiable
Didn't we define 'a' as "f(t) - t" (from f(t) = t + (f(t) - t) ). So a isn't a constant? I don't see how this is a valid solution - it doesn't shout to me as a valid answer to "Find all strictly increasing f:." This isn't a complaint that I think you're wrong - I'm always blown away by your amazing proofs! But this is a stumbling block for me that I don't understand :(
t is just some fixed value such that a=f(t)-t>0 is also just some fixed value. he then looked at all x of the form t+n*a and the values of x between those points. so he proved that for every fixed t, you can find such a relationship which implies that you can write f(x)=x+a for an arbitrary real a (including the case a
you said bijective implies investable but arent they the same? I might be wrong but its a minor point what I wanted to say is that while teaching math I made up some functions that are fun to draw the graphs of qualitatively I started with 1/(x^2+1) or e^-x^2 which is similar visually then 1/((x-4)^2+1) or. e^-(x-4)^2 and. 1/(x^2+1) + 1/((x-4)^2+1) ... sin(x). sin(pi x) sin(pi x^2) sin(pi x)/(x^2+1) sin^3(pi x) sin^3(pi x^2) / (x^2+1). and stuff like that ... love your videos maybe your viewers would also like to see some function graphs also
I must be missing something because I solved this problem very easily in 2 minutes. Simply differentiate the equation wrt x. Then observe that the derivative of f-inverse is the reciprocal of the derivative of f. So, using that you will get an equation of the form f'(x)+1/(f'(x)) = 2. This has only one solution f'(x) =1 by AM-GM inequality. So, you get f(x)=x+c for a constant c in R. Done. What's wrong with this?
@@rinsim That's a good point. Let's think a bit more about it. In the equation, the right hand side is clearly differentiable, so the left hand side must be too. If the sum of a function and its inverse is differentiable, can we something about f? If f is differentiable, its inverse will be too. If it is not, its inverse won't be either, but their sum has to be differentiable. I wonder if we can say that it is the former and not the latter.
@@divyanshgupta7042 yeah I know, the problem my teacher suggested asked to prove that the only strictly increasing function f such that ff(x) =x is f(x)=x
Does strictly increasing imply bijective without having continuity as well? Consider f(x)=x for x more than or equal to 1 and x-1 for x less than 1. This is increasing and yet not surjective as .5 isn't in the codomain
Is my solution correct? We take the derivative f'(x) + f^-1 ´ (x) = 2 f'(x) + 1/f´(x) = 2 (f´(x))^2 + 1 = 2f´(x) (f´(x))^2 - 2f´(x) +1 = 0 f´(x) = 1 for all x in R f(x) = x+c
Please Michael Uncle, solve problem 3 from 1993 IMO in your next video. It is an interesting game problem. Please Mike Sir, please solve the problem. Yours sincerely/ obediently, Riju Bhatt
My first thought was f(x) = ax + b. Solving this eqn. gives a = 1 and b is free. Doesn't prove there aren't any other solutions tho. Setting x = f(x) gives f^2(x) + f^⁻1(f(x)) = 2x => f^2(x) + x = 2x => f^2(x) = x. This is Babbage's functional equation and it has a solution set with two parameters. Not sure why this is wrong, x = f(x) covers the entire domain of x. Edit: wait, I see the problem now... RHS becomes f(2x). :D
Strictly increasing actually doesn't imply that f is bijective. Try: f(x)=x-1 when x= 0. However if you assume that f is continuous, then f must be bijective. (Conversely, every strictly increasing bijective function is continuous)
Continuity of f on R is indeed important to suppose at the beginning
err, I don't get how that function isn't bijective?
well, arctan(x) is strictly increasing and continuous and isn't bijective (cause it isn't subjective onto R)
@@phee4174 It is, but the domain of the inverse is not all of the reals.
Even with continuity we can't be sure that f is bijective simply by expanding the codomain to be bigger than the image. For example, f:R->R with f(x)=arctan(x) is continuous and strictly increasing but not bijective. But strictly increasing implies one to one, so we only need to demand f to be onto R
This example clearly highlights the difference of mind between pure and applied maths: for instance, an engineer, or a phisicist would spend at most 10 seconds in guessing as possibile solution the identity plus constant function, i.e. f(x)=x+k, and then he would classify such question as elementary, even banal; on the other hand, a pure maths theorist is interested in checking the unicity of the immediately guessed solution and this task will result a clever challenging work.
A geometric approach of this problem is the following. We know f and f^-1 are symmetric to y=x. We also get that f(x)-x = x- f^-1(x). That means that for every x, f and f^-1 should maintain equal distances from the point (x,x) on the line y=x. This intuitively leads to f being parallel to y=x , as it's the only way to preserve both of these symmetries, which is easily proved by contradiction.
HOMEWORK : Four integers a, b, c, and d with a ≤ b ≤ c ≤ d satisfy the property that the product of any two of them is equal to the sum of the other two. Given that the four numbers are not all equal, determine (if they exist) all the 4-tuples (a, b, c, d).
SOURCE : Exeter Math Club Competition 2016
BONUS OPEN PROBLEM : What happens if we take 2n instead of 4 ? Product of n of them is equal to the other n.
Trivial solution is a = b = c = d = 0, but that is excluded.
Another excluded solution is a = b = c = d = 2.
The only accepted solution is then a = b = c = -1 and d = 2.
For 2n-tuples, a similar pattern works for even n: x₁…x₂ₙ₋₁ = -1, x₂ₙ = n
If x₂ₙ is in the product, the product becomes n∙(-1)ⁿ⁻¹, and the sum becomes n∙(-1); They are equal iif n is even.
If x₂ₙ is in the sum, the product becomes (-1)ⁿ, and the sum becomes n+(-1)*(n-1) = 1; Again they are equal iif n is even.
@@nedbowlas913 Basically I have a dozen of bookmarks to math competitions websites, math magazines and miscellaneous stuff. I also have some PDF documents of books. Then I pick a problem whenever I feel it's original and hard enough.
SOLUTION
*(-1, -1, -1, 2)*
We can in fact solve the problem in real numbers. Let S = a + b + c + d. From the problem condition we see that ab + cd = (c+d) + (a+b) = S. Similarly, ac + bd = ad + bc = S. Notice that (a + b)(c + d) = (ac + bd) + (ad + bc) = 2S, and similarly (a + c)(b + d) = (a + d)(b + c) = 2S.
By Vieta's Theorem we get that all three pairs (a + b, c + d), (a + c, b + d), (a + d, b + c) are solutions of the equation x² − Sx + 2S = 0, so all three pairs must be equal to (u, v) or (v, u) for some real numbers u, v. Since there are at most two distinct values in {a + b, a + c, a + d}, by Pigeonhole Principle at least two of b, c, d are the same. WLOG let b = c. Then by making the same observation on a, b, d gives that either there are two pairs of equal numbers among the four three of the four numbers are equal.
If it's the first possibility then a, b, c, d is a permutation of p, p, q, q for some distinct reals p, q, but it's not difficult to see that {p + p, q + q} ≠ {p + q, p + q}. This leaves the second possibility (where the numbers is a permutation of p, p, p, q for p ≠ q), which satisfy the desired property.
Now it remains to solve the following system of equations : p + p = pq, p + q = p². The first equation gives p = 0 or q = 2. If p = 0 then q = 0, which contradicts the requirement that p ≠ q. If q = 2, then p + 2 = p² has two solutions −1 or 2, but since p ≠ q= 2 we are forced to have p = −1, and by putting the numbers in non decreasing order we get (−1, −1, −1, 2) as the one and only possible quadruple.
love this one
@@velian9133 Oh really? I have ELMO solutions bookmarked somewhere but I did miss that 😅
The thumbnail😁
Lit
@16:25. Correct me if I'm wrong, but I think the procedure here is not correct. If
t+(m-1)a < x < t+ma, then t+ma < f(x) < t+(m+1)a, since f is increasing. This implies f(x)-x=b
Ooh I love functional equations because of this channel
This is a great problem
It's easy to show that the function is differentiable a.e. by taking x
Why are there only two cases f(t)>t and f(t)
Yup, and I'm also really wondering why he chose this really convoluted way. I think there has to be an easier way to prove this result.
He notes that f(x)=x satisfies the functional equation, so then proceeds to find other functions. Thus the assumption that f(x)!=x for some x. Pick such an x and call this t. Then either f(t)>t or f(t)t. And along the way he shows that if f(t)>t for _some_ t, then f(x)>x for _all_ x. Here t is a constant and x is a variable.
Now t is a constant, so f(t)-t is also a constant. So a is a constant (and b also).
@@muratcan-k6x Hahagahaha, sorma. 6. sınıftan falan kalma.
Reis, bu arada sen daha kolay bir çözüm görüyor musun şu soruya? Sâhiden bundan daha basit bir yöntem olmalıymış gibi geliyor bana.
I finally found a quick solution that's pretty good. This writeup relies on differentiability, but I haven't seen any correct solutions using it anyway (mostly incorrect uses of the derivative of function inverses). You can show the same thing with some sort of finite difference operator, it's just a bit less legible.
Say you found that solutions of the form f(x) = x + c work by inspection. Assuming differentiability:
f'(x) + 1/f'(f-1(x)) = 2
f'(f-1(x)) = 1/(2 - f'(x))
Now we know f'(x)>0 everywhere. We can also see f'(x) < 2 because otherwise it would be negative at f-1(x) from above. Here's the tricky part. Let's say f'(x) has bounds 0
Nice! It took me some thought to understand how you deduced the inequalities so I put it here for the next person who might be interested.
Set y:=f-1(x)
1/U
At first, I was confused. However, I have realized that t is one fixed number during my second viewing and understood the flow of logic. The margin from g(x)=x at x=t, i.e. a, becomes margins at arbitrary x.
I was confused because I had a smudge on my screen that made an equal sign look like a minus and kept trying to figure out why he was subtracting this second similar equation from what looked like the correct answer.
Don't solutions exist in pairs except for f(x)=x? f(t) is a solution iff its inverse is a solution, so you should be able to assume the first case, f(t) > t, without loss of generality: if f(t) < t then f^-1(t) > t (since they sum to 2t). Probably simpler to show that fact (f(t) is a solution exactly when its inverse is) than going through the whole argument again with f(t) < t. And that just follows from the defining property pretty easily.
very elegant!
Being strictly increasing doesn't imply surjectivity, so it can't imply bijectivity. Consider f(x) = 2^x for example.
yes, but injectivity is sufficient for existence of an inverse function
log2(x) is the inverse of f(x) = 2^x. It's true that the domain and co-domain are not all of the real numbers, but that is not necessarily what surjective means. In your case, f(x) can be surjective because it goes from the domain (all real numbers) to the co-domain (all positive real numbers). And f^-1(x), which is a logarithm, goes from its domain (all positive reals) to its co-domain (all reals).
Maybe it's a regional terminology difference, but I was taught that surjectivity is required, and if you define your function's codomain to include values that aren't in its image, then it's not surjective, not bijective, and not invertible. You can't just tacitly assume the codomain is so restricted.
It's academic in this case as the question uses the inverse for all x in R, so it must be taken to imply that the inverse has to exist, which becomes another constraint on our choice of function. It doesn't change the solution, I just wouldn't claim that strict monotonicity is enough for a function to be bijective.
Another example if f(x) = arctan(x)
Differentiate both sides wrt x (assuming a differentiable function). f(x) goes to f'(x) and f-1(x) goes to 1/f'(x) because it is a reflection of f(x) in y=x, and there is no problem with dividing by zero. Multiply the resulting expression and factorise to give (f'(x)-1)^2=0. So f(x) = x+a.
This isn't quite right. f-1(x)' is 1/f'(f-1(x)). You're evaluating the reflection over y=x at the wrong point. It's a coincidence that the result is correct.
You can solve this problem entirely geometrically! Rephrasing the equation as midpoints and applying cartesian geometry this reduces to showing that the trapezoid formed by (x, f(x)), (f(x), x), (x, f^-1(x)), (f^-1(x), x) is actually a square, and it follows that the only f satisfying this property for all real x must be f(x) = x - a.
At 11:29, does one not again need to differentiate between the two cases b > 0 and b < 0?
Starting at 11:20, we take x not in {t+na: n in Z}, and show that f(x)>x. This implies that b = f(x)-x > 0. Basically, once we have f(x)>x for _one_ value of x, we get that f(x)>x for _all_ values of x. So b>0 if and only if a>0.
Sir you work is very appreciable
Strictly increasing implies injective but not surjective so we don’t know the function is bijective and thus we don’t know it’s invertible
Fair enough. But it is invertible on its range, which, yes, should have been specified to be all of R.
@@tomkerruish2982 that’s all I’m trying to say
@@Happy_Abe I agree. I sometimes do not write clearly enough. I'll need to rewatch the video, though. It's possible his analysis shows that t+na must be in the range of f for all integers n, and thus the range actually must be R. Unspoken assumptions can be very difficult to spot.
@@tomkerruish2982 yeah, clearer to just speak them out from the beginning
Michael probably assumed that f was continuous, in which case it's bijective since it's strictly monotone
"strictly increasing" does not imply bijectivity unless the function is continuous.
Even so why can’t it be bounded above and converge ie -1/x it’s cts and strictly increasing and not bijective
@@bulgeo09 because thats not a function from R to R. a function from R to R needs to be defined at all R. at least for -1/x there is no way to fill the hole x=0 in a way that makes the function strictly increasing.
@@bulgeo09 that example doesn't quite work because it isn't defined at x=0 (so is not a function from R to R), but something like arctan does work: it is continuous and strictly increasing, but not bijective (because it is not surjective; it's image is (-pi/2, pi/2).)
@@schweinmachtbree1013 the refined statement should be a strictly increasing function does not imply bijectivtivity unless it is continuous and f: R->R. Otherwise, you’re arctan example would of course be a counterexample, disproving it.
@@tomatrix7525 that refinement is still not true - arctan is strictly increasing, continuous, and R->R. we can only conclude bijectivity onto the image, not complete bijectivity (to be able to conclude complete bijectivity all we can do is add surjectivity as an extra hypothesis)
16:38 Enjoying a sunny 28 degrees (Celsius obviously) 😎 Have a good day everyone!
Though, it’s been a while Michael hasn’t been a thumbnail.
Sir you are hard worrking
@@deepakgoswami7882 Yeah. Not gonna lie, it's getting harder and harder to come every day with original problems.
@@goodplacetostop2973 No sir but I even appriciate your work and dedication. thanks for all that . By heart
This was absolutely brilliant! I wish I hadn’t given up, but I’m not sure I would have thought to set up inequalities (which was the first step).
I used (what I think is) an easier and faster method to solve this one. Since f is strictly increasing, the function has a positive slope (ignoring points at which there's no derivative). If it makes an angle of π/4+(theta), it's inverse makes an angle of π/4-(theta) (as inverse is reflection along y=x). So both these angles sum to π/2 and their tangents sum to 2 (by differentiating both sides of the given functional equation). Solving we get tan(theta)=1 so theta=π/4 or slope=1 is the only option for points where the function is differentiable. So if there exists a point of non-differentiability, it will be a point of discontinuity. But since f is mapped from R to R, an inverse can only exist if range of f=codomain of f so there is no possibility of a discontinuity for f. Also the functional equation holds for all real x so the domain of f inverse must be the set of all reals as well. This too implies that f cannot have any discontinuities. So f(x)=x+a is the only option.
Michael's proof does not assume differentiability (though its result implies that f is everywhere differentiable).
hi Rayan. I am sorry but your proof is not correct, since the given equation is f(x)+f^-1(x)=2x and not what you suppose f(x)+f^-1(f(x))=2x. This imediately gives
f(x) + x = 2x ie f(x)=x
Every day, Michael Penn gets up and chooses violence 😅
can you please also do some combinatorics as well??
loved this video
Much simpler version, assuming that f is differentiable (I think one could prove that):
f(x) + f^-1(x) = 2x | differentiate
f'(x) + (f^-1(x))' = 2 | inversion rule
f'(x) + 1/f'(x) = 2
(f'(x))^2 - 2 f'(x) + 1 = 0
=> f'(x) = 1 +- Sqrt[1 - 1] = 1
=> f(x) = x + C
I did not understand the last argument, wouldn’t a=b only for n tending to infinity? How does this generalize?
Would a function defined in steps work? For example, f(x)=x+2 for x0.
Yes, function inverses are not affected by the fact that a function is piecewise.
I don't think you can assume f'(x) exists by monotonicity, but that the question involves f'(x) on all x in R should be a good enough stipulation that f'(x) exists
Interesting use of the sandwich theorem. ---- My first guess was proving on a set isomorphic to Z, proving on that to Q, and then using monotonicity to prove on R.
Definition: A monic linear polynomial function is a strictly monotone continuous bijection f from *R* to *R* satisfying the functional equation f(x) + f^{-1}(x) = 2x
f(x) = 2x is a linear polynomial function and it does not satisfy that functional equation.
@@gergonemes88 Good point, I guess it would have to be “translation of the real line” instead of “linear polynomial function” (although it was meant to be silly anyway so whatever xD)
Edit: I guess “monic linear polynomial function” works too
I didn't have the patience for this sort of in depth analysis so I just instinctively chose f(x) = mx + b and deduced that m =1 and b could be any real number. ;-)
And it is of interest to consider what real solutions apply for f(x) + f-1(x) = nx where n is any real number!
I think I am smart now. Excellent.
This is a good demonstration. Another way is to derive the equation in regard to x, noticing that f'(x)>0 as f(x) is strictly increasing, and f'(f^-1(x))=1/(f'(f(x)). Thus, f'(x)+1/(f'(f(x))=2. All the terms being strictly positive, 00. Hence f(x)=cx+d, w/ c,d real numbers. f^-1(x)= (x-d)/c and finally with the initial equation, c=1, d is a real number.
Actually I also thought about this way. But I thought that we will get this equation:
f'(x) + 1/(f'(x)) = 2
where we get that:
f'(x) = 1
hence f(x) = x +C - is the unique solution.
Reminds me of the proofs in college lab that took up 6 chalkboards
we can simply differentiate both side (knowing that the derivative of a function's inverse is one over the derivative of that function). We get the equation c+1/c=2 in terms of dy/dx. Solving for c we get dy/dx=+-1 but since strictly increasing it is just 1. Integrating we get f(x)=x+c
You would need to prove the function and its inverse are differentiable first.
@@Falanwe Isn’t the differentiability of the function and its inverse implied by the fact that their sum is a differentiable function?
@@PETAphile Maybe? But I don't see any easy demonstration of this.
However, Lebesgue's Theorem for the Differentiability of Monotone Functions tells us the function is differentiable almost everywhere. We can probably use John Steven's demonstration to show the function must be f(x)=x+c on each domain it is differentiable on, and show the c must be the same for all those domains because of continuity (a monotonous, bijective function must be continuous).
John Steven's demonstration idea basically works, he just glossed over some major difficulties we need to address to make sure we are allowed to use this demonstration.
Am I missing something, isn't the derivative of an inverse function is one over that function, composed with the inverse of the function? Shouldn't this invalidate this calculation?
@@Balequalm Makes it a lot more complicated, but I think it's manageable. I don't have the courage to write it up though.
a be arbitrary from R
and f(a) = b
=> f⊣(b) = a
and f(b) + f⊣(b) = 2b => f(b) + a = b + f(a) => (f(b) - f(a))/(b - a) = 1
I have tried in vain to prove the result if we suppose f is differentiable. It seems manageable, but I can’t find the way. Does anybody have such a proof?
Why are we assuming f(x) =/= x? Are we making it part of the problem statement or what? I mean, are we changing the problem statement after reading it the first time? Why would we do that?
We do so to find other solutions than f(x) =x. So if f(x) x (as a function), then there must be a t\in\R s.t. f(t) t.
I think you should practice writing at 90°
It would improve your writing at the lower section of the board
I have a better method
Define f(x_n)=x_(n+1) and x_1=x
Thus we get the recurrence x_(n+1)+x_(n-1)=2x_n
Thus by telescoping we get:
x_(n+1)-x_n=f(x)-x
Again by telescopic summation we get:
x_(n+1)=nf(x)-(n-1)x
Now using that f is strictly we get f(x)=x
But the answer is wrong
Isn’t all x in t+na, since t and a are just Real numbers, n is an integer, so really we can generate every real number x through this combination?
We're not trying to say that x is of the form t+na. We've found a particular t and the corresponding a, and we're only letting n vary arbitrarily. Consider f(x)=x+sin(a(x-t)/2π)+a. It's equal to x+a at x=t+na, but not at other values, and it doesn't fit the functional equation, so we must have more to prove.
@@iabervon just rewatched, I understand now, basic mistake, thanks pal
How do you solve this functional equation: f(x^2)/f(x) = 1 - x with domain: -1 < x < 1
I have a problem suggestion:
if you have fractions a/b and c/d (all integers, b not 0, d not 0), then:
a/b + c/d = (ad+bc)/(bd).
In general, the sum is *not* just (a+c)/(b+d), a mistake many students make when learning about fractions. However... sometimes (a+c)/(b+d) *does* happen to give you the right answer.
So, what are all the types of {a, b, c, d} (as above) solutions that work?
Michael takes problem suggestions on Google Forms (see the description)
@@schweinmachtbree1013 thanks! (I'm new. My apologies!)
@@angelmendez-rivera351 I haven't had the chance to look this over in detail yet, but I'm delighted you tried it out :) did you like it? i personally thought it was a really fun question :-D
But "a' has a f(X) in it so I am confused how it becomes a number.
I'm not 100% sure, but I think it's saying that any real number a will satisfy the requirements.
As f(x)=x+a, f^-1(x)=x-a. This means that f(x)+f^-1(x)=(x+a)+(x-a)=2a
f is bijective (because it increases strictly on R and is continuous on R) thus for any number a in f(R) (we may suppose it is R), there exists one and only one real number x such that f(x) = a. In particular, any number a can be written as f of someone.
But in fact, I don’t think you need all that reasoning because you can actually name anything you want by anything else.
Remember t is fixed. f(x) is variable, but f(t) is simply a real number such that f(t) =/= t
Thanks all. I need to work through it again to see what's going on.
Isn't there any way to proceed by assuming x = f(y) for some y? That way we get rid of the f^-1 and we get
f(f(y)) + y= 2f(y), which seems to be a better place to begin with.
that's a good place to start.
This is an intuitive way to _suggest_ the result. Our back of a napkin calculation proceeds as follows, writing function applications as fx rather than f(x).
ffy + y = 2fy
ffy - 2fy + y=0 (move everything to LHS)
(f^2 - 2f + 1)y = 0 (factor out the y)
(f-1)^2 = 0 (for all y!=0 we can divide by the y, then factorise the quadratic)
Hence f=1. So fx = x.
Obviously this misses the f(x)=x+a case, and is a rough exploration of an idea, but suggests that f(x) must be linear. Then one has to proceed to find a rigorous proof, which is what Michael gives. It is quite simple in nature, stating from one point where f(x)>x, then proceeding to show that f(x)=x+a, leaving the case f(x)
Why does f(t) have to be greater than or less than t for all values of t? Couldn't it oscillate around f(t)=t?
The proof he gives shows, among other things, that if f(x)>x for _one_ value of x, then f(x)>x for _all_ values of x. This is the result of step 3 in the summary below.
The structure of the proof is:
1.) to assume that f(t)>t for one value of t.
2.) Then show that f(x)=x+a for all x in {t+na for n in N},then for all x in {t+na for n in Z}.
3.) Then for any x not in {t+na for n in Z}, he shows that f(x)>x,
4.) so we repeat the argument to see that f(x)=x+b for x in {t+nb for n in Z}.
5.) Finally an inequality is used to show that a=b.
Yo okay so I did something here that no one else seems to have picked up on.
(f^-1)'(x) = 1/f'(x)
So given f(x) + (f^-1)(x) = 2x
Take the derivative, we'll say y = f'(x)
y + 1/y = 2
y² - 2y + 1 = 0
(y-1)² = 0
y - 1 = 0
y = 1
This means
f'(x) = 1
Now integrating,
f(x) = x + C
Wow!
Very nice problem
I may have missed it but why bother assuming a doesn’t equal zero in the very beginning? Was there a step where something was divided by a at some point? It seems like you could ignore that assumption and get the same result.
We observe that f(x) satisfies the functional equation right at the start. Then to look for any other f(x), we assume that f(x)!=x for some x. This splits into two cases: f(x)x. This gives a=f(x)-x>0 as one case, and a=f(x)-x0 case, since the a
@@Chalisque That doesn’t really answer my question. Why not assume f(x) >= x instead of f(x) > x? (i.e. that a>=0 instead of a>0) ?
This doesn't allow one to cut up the real line into intervals like [t+na,t+(n+1)a]
@@MichaelPennMath Got it, thanks!
I actually don't know how to go about solving this, other than that f(x) = x is one obvious solution. This will be interesting to watch.
Edit after watching: Wow, yeah, I wouldn't have come up with basically any of that.
f(x) = x + root(x^2-1), x - root(x^2-1)
if f(t) < t then t < f^{-1}(t)
a = f^{-1}(t) - t
We have if f solution then f^{-1} inverse of f is also solution.
Because if f strictly increasing we have f^{-1} strictly increasing and
if for all x in R f(x) + f^{-1}(x) = 2x
Then for all x in R
f^{-1}(x) +( f^{-1})^{-1} (x) = 2x
and f^{-1}(t) > t
Then for all x in R f^{-1}(x)=x+a
Then for all x in R f(x)=x-a
Hello Michael
Can u solve this problem
This problem was in my math assignment sheet
The problem is :
Lim x------>0 (f(x))
Where f (x) is :
(xsin(sin(x) ) - sin^2(x) ) / x^6
And we have to solve it *WITHOUT* L'Hopitals rule..
The ans is 1/18
This is a very interesting problem that's why I wanted share this with you.
Good Place To Start at 0:05
If we knew that f is differentiable then it is very easy to prove that f(x) = x + c. Perhaps one can prove that a function f that satisfies our hypotheses must be differentiable
What a journey
Didn't we define 'a' as "f(t) - t" (from f(t) = t + (f(t) - t) ). So a isn't a constant? I don't see how this is a valid solution - it doesn't shout to me as a valid answer to "Find all strictly increasing f:."
This isn't a complaint that I think you're wrong - I'm always blown away by your amazing proofs! But this is a stumbling block for me that I don't understand :(
t is just some fixed value such that a=f(t)-t>0 is also just some fixed value. he then looked at all x of the form t+n*a and the values of x between those points. so he proved that for every fixed t, you can find such a relationship which implies that you can write f(x)=x+a for an arbitrary real a (including the case a
@@demenion3521 Ah I think I follow! That makes sense now, thanks! :)
The reason is that we showed there existed a certain constant t for which t
Homework help - what happens if you exchange f(x) with its inverse?
you said bijective implies investable but arent they the same? I might be wrong but its a minor point what I wanted to
say is that while teaching math I made up some functions that are fun to draw the graphs of qualitatively
I started with 1/(x^2+1) or e^-x^2 which is similar visually
then 1/((x-4)^2+1) or. e^-(x-4)^2 and. 1/(x^2+1) + 1/((x-4)^2+1) ...
sin(x). sin(pi x) sin(pi x^2) sin(pi x)/(x^2+1) sin^3(pi x) sin^3(pi x^2) / (x^2+1). and stuff like that ...
love your videos maybe your viewers would also like to see some function graphs also
Michael u just turned me onnnn lol
I must be missing something because I solved this problem very easily in 2 minutes. Simply differentiate the equation wrt x. Then observe that the derivative of f-inverse is the reciprocal of the derivative of f. So, using that you will get an equation of the form f'(x)+1/(f'(x)) = 2. This has only one solution f'(x) =1 by AM-GM inequality. So, you get f(x)=x+c for a constant c in R. Done. What's wrong with this?
You are assuming f differentiable.
@@rinsim That's a good point. Let's think a bit more about it. In the equation, the right hand side is clearly differentiable, so the left hand side must be too. If the sum of a function and its inverse is differentiable, can we something about f? If f is differentiable, its inverse will be too. If it is not, its inverse won't be either, but their sum has to be differentiable. I wonder if we can say that it is the former and not the latter.
Continue abstract algebra playlist and do advance abstract algebra
It's similar to an exercise my teacher gave me at school the other day only instead of the given equation it said that f(x) =f^-1(x)
CAN WE SOLVE THIS USING f(x)=x
@@divyanshgupta7042 actually we had to prove that f(x) =x
@@giratin5911 it is proven the graphs of f(x) and finvers(x) are mirror images of each other along the line y=x
this is the fact
@@divyanshgupta7042 yeah I know, the problem my teacher suggested asked to prove that the only strictly increasing function f such that ff(x) =x is f(x)=x
"I wanna notice"
Does strictly increasing imply bijective without having continuity as well? Consider f(x)=x for x more than or equal to 1 and x-1 for x less than 1. This is increasing and yet not surjective as .5 isn't in the codomain
Way to prove indirectly Z is dense in R
Is my solution correct?
We take the derivative
f'(x) + f^-1 ´ (x) = 2
f'(x) + 1/f´(x) = 2
(f´(x))^2 + 1 = 2f´(x)
(f´(x))^2 - 2f´(x) +1 = 0
f´(x) = 1 for all x in R
f(x) = x+c
bro lmao f inverse x does not mean 1/f of x
@@srishiridisai9294 the derivative of the inverse of a function is the inverse of the derivative of that function
en.wikipedia.org/wiki/Inverse_functions_and_differentiation?wprov=sfla1
@@TheMahri77 if you read that link, you'd see that you got the derivative wrong
Nice elementary solution.
ok boomer
Please Michael Uncle, solve problem 3 from 1993 IMO in your next video. It is an interesting game problem. Please Mike Sir, please solve the problem.
Yours sincerely/ obediently,
Riju Bhatt
Does these kinda functional equations show up in other fields of math or just in math contests? Invertibility also implies bijectivity.
Thumbnail : PJ sir style 😂
Incrrasing doesn't imply bijective
My first thought was f(x) = ax + b. Solving this eqn. gives a = 1 and b is free. Doesn't prove there aren't any other solutions tho.
Setting x = f(x) gives f^2(x) + f^⁻1(f(x)) = 2x => f^2(x) + x = 2x => f^2(x) = x. This is Babbage's functional equation and it has a solution set with two parameters. Not sure why this is wrong, x = f(x) covers the entire domain of x. Edit: wait, I see the problem now... RHS becomes f(2x). :D
I solved it yesterday)
Where did you find it plz?
I like the thumbnail
I love functional Equations but cannot solve them
Sir please teach me mathematics...please sir
Didn't watch, but f(x)=x should do the trick, right?
Soon to be shown video will be FUNNYctional equation.