a fun functional equation with an inverse twist 

I eyeballed the solution f(x) = x + C immediately at the beginning but I would never have come up with that proof lol.

8:17 I loved your change from weak induction to strong induction on the fly and the explanation of the difference. You made it understandable.

@4:05 2 times 2 is 4 - 1 that's 3 QUICK MATHS

This dude has taught me more math than my own teacher👌🏽

I solved it this way: Let g(x) be f(x)-x. Then the equation f(x) + f^(-1)(x) = 2x can be rewritten as g(x) =g(f^(-1)(x)). Since x is arbitrary we can plug in "x=f(x)" and get g(f(x)) = g(x). Now take x and y real numbers. Let a=g(x) and b= g(y). We have g(x) =a -> f(x) = x+a -> g(x+a) = g(f(x)) = a -> f(x+a) = x+ 2a -> ... After n steps (this is formalized by an induction argument) we deduce that f(x+na) = x+(n+1)a. Similarly f(y+nb) = y+ (n+1)b. I claim that a=b. Indeed, if, arguing by contradiction, it were not the case, then either a

Wow, lots of great tricks. I just guessed the solution after figuring out a few relationships between f(0), f(x0)=0 and f(2x0). My hunch was that it was a linear equation mx+n and I could prove quickly that m=1.

Nice! About the proof by induction: *it would've been fun to write down explicitly the base case (n=0), which boils down to x=x (assuming, as usual, that f_0 is the identity). *I think it's easier to replace x by f(x) in the induction hypothesis formula to get to the same result more easily, and with no need of strong induction.

I solved it in a slightly different way: Rather than substitute x for f(x), Subtract f(x), Apply f, then take the derivative to yield: 1 = ( 2 - f'(x) ) f'( 2x - f(x) ) If we call the left factor A = 2 - f'(x) and the right factor B = f'( 2x - f(x) ) Then because of the 'strictly increasing' rule, B > 0 and so A > 0 f'(x) < 2 implies B < 2 implies A > 1/2 implies B < 3/2 implies A > 2/3 implies B < 4/3 implies A > 3/4 implies B < 5/4 ... implies f'(x) = 1 f'(x) = 1 leaving f(x) = x + c as the only solution.

I hope your never stop using colors. I thoroughly enjoy how it's much more aesthetic; as math should be

Indeed it's amazing and more technical solution bravo .

@Michael Penn: I came across this nice math contest problem I thought you might like - it mixes at least 2 aspects you like a lot. The question is: "Find all functions f:R+ to R such that: 1) f(1) = 0 2) f(ab) = f(a) + f(b) 3) f(2) and f(3) are integers." I liked it a lot at the time, and still fish it out from time to time with students.

Amazing problems I love these type of things

Nice! The equation f(x) + f^-1(x) = a x has a solution if and only if a \geq 2. Solution for a > 2: Let r, r^-1, with r > 1, be the roots of the quadratic equation x^2 - a x + 1 = 0. Case 1: f(0) = 0. In this case there are 4 solutions: 1) f1(x) = r x 2) f2(x) = r x when x > 0 and f(x) = r^-1 x when x < 0 3) The inverse of f1 4) The inverse of f2. Case 2) f(x) ≠ 0. We may without loss of generality assume f(0) > 0 (because otherwise f^-1(0) < 0 and the equation is symmetric in f and f^-1). Let x1 be any positive number. Then f can be defined in the interval [0, x1] as any function such that f(0) = x1, f(x1) = a x1, and (*) For any x, y such that 0 \leq x < y \leq x1, we have r^-1 \leq (f(y) - f(x))/(y-x) \leq r. Any such function can be uniquely extended to a complete solution of the equation (i.e. defined in the real line R). Observations: 1) The condition (*) is equivalent to: (**) For any x, y such that 0 \leq x < y \leq x1, we have r^-1 \leq lower derivative of f(x) and upper derivative of f(x) \leq r. 2) The proof is pretty straightforward having in mind Michael's solution for a = 2. The only tricky part is perhaps showing there are no more solutions in case 1 above. Cheers!

I would say that f^n --> n applications of f() is quite consistent with math notation... especially since f^{-1} is already present.

This got to be one of the most delicious proof I have ever seen. This is math artistic craft in action.

Hey Michael will you ever make a video about galois theory?

I did it by comparing the two expressions from taking x->f(x), and from applying f to both sides of the equation after rearranging. This gives: a f(x) - f(f(x)) = x f( ax - f(x)) = x Both of those being simultaneously true means that f(x) must be linear! So we can just set f(x) = mx + b, the inverse to x/m - b/m, and solve for m. You get a quadratic, and can find m = a/2 +- ( (a/2)^2 -1 )^(1/2) . b is forced to be zero generically, unless a=2. We require a >= 2 to avoid any problems with the square root, and can easily verify that 1/m + m = a.

I came from it at a completely different angle to what was done in the video. I basically used calculus which does use the assumption that f must be differentiable which people have already stated in the comments but I think it's still worth considering. I see that others have used this approach already for example by differentiating both sides which involves differentiating the inverse which isn't too bad to do. However, thinking that it wasn't possible to take the derivative of the inverse, I went for a different approach, it goes as follows: First eliminate the inverse, by doing the substitution y = f^-1(x), f(y) = x f(x) = f(f(y)). Sub in to equation, we get f(f(y)) + y = 2f(y), differentiate both sides to get f'(f(y))*f'(y) + 1 = 2f'(x) For clarity Let Z = f(y) f'(Z)*Z' + 1 = 2Z', now write out the derivatives in a slightly different way to get (df/dz)(dz/dy) + 1 = 2(dz/dy) df/dy + 1 = 2(dz/dy), Z' = f'(y) f'(y) + 1 = 2f'(y), f'(y) = 1, f(y) = y + C Now f(f^-1(x)) = f^-1(x) + C, x = f^-1(x) + C x = 2x - f(x) + C, f(x) = x + C

That chalk looks really cool!

Great video as always! I am curious where functional equations occur in math outside simply solving them? Are there any fields of math that deal with functional equations and have other ways of solving them?

First, note that f(x)=x is a trivial solution. Now suppose f(a) = b for some constants a != b. It follows that f^-1(b) = a. f(b) + f^-1(b) = 2b ---> f(b) + a = 2b ---> f(b) = 2b - a. f(b) - f(a) = (2b-a) - b = b - a. [f(b) - f(a)]/[b-a] = 1. This means that the average slope of f is 1 over any finite nonzero interval, which means all solutions are of the form f(x) = x + c for a constant c. I don't know if this is a rigorous solution but I think the general argument is sound.

I tried to solve this from the thumbnail alone... Wish I'd known about the strictly increasing

Could you expand on why taking the limit as n tends to infinity makes the strict inequality into a non-strict inequality? Does this fact have a name that is searchable?

Could you also solve this problem by first showing that f must be differentiable and using this fact to take the derivative of both sides of the equation?

I think it's interesting to consider the similar problem f(x) + f^(-1)(x) = x. It's easy to see that over R there are no solutions: Because f is increasing and invertible (EDIT: with an inverse defined everywhere on R) it must be continuous, so f(x) can only switch between being x at a fixed point. f cannot have any fixed points other than 0. Suppose wlog that f(x) > x for x > 0 (otherwise we could just switch its role with the inverse). But then f(x)>f(0)≥0 for x > 0, so the inverse g must satisfy g(y)>g(f(0))≥0 for y > f(0). So taking x = y > f(0) we end up with f(x) > x and g(x) > 0 so f(x)+g(x) > x. There are however some solutions if we allow f to be a partial function. For example f: [-a,a/2] -> [-a/2, a], f(x) = 1/2(x + √(3a²-2x²)) is a solution for those intervals.

for the generalized problem with ax, assuming the solution is of the form mx+b, we quickly find that if b is non zero, then m must be 1 and its only a solution for the case with a = 2. For b = 0, we get a quadratic equation m^2 -a*m + 1 = 0. From the solution we get that a >= 2 so that a solution exists with m being a positive real number.

To get a solution very quickly, we just differentiate on both sides. We get f’+1/f’=2 (the derivative of the inverse is one over the derivative of f. Just solve for f’ by solving the quadratic equation x^2-2x+1=0. f’ is equal to 1. f is therefore x+C, where C is any real number. C is any number, since f^-1=x-C, and f+f^-1 cancel each other out. Now, we just have to show that this is the only solution to this functional equation, which means f is differentiable.

instead of induction, letting aₙ=f iterated n times aₙ₊₁-2aₙ+aₙ₋₁=0 aₙ₊₁-aₙ=aₙ-aₙ₋₁ i.e. aₙ is an arithmetic sequence aₙ-aₙ₋₁=Δa₀ aₙ-a₀=nΔa₀ fₙ(x)=n(f(x)-x)+x

f(f(x))=2f(x)-x We can find the characteristic equation of the recurrence relation (a_n+2)=2(a_n+1)-(a_n) with (a_0)=x and (a_1)=f(x) which gives the exact solution. Am I correct?

Didn't you make a video on this problem about a year ago?

Hi Dr. Penn!

4:05 quick maffs

Just a little poking around reveals what is probably only a part of the answer: f(x) = x + a, where a is any constant. Then f⁻¹(x) = x - a, and f(x) + f⁻¹(x) = 2x + a - a = 2x Multiplying by a constant doesn't work. [f(x) = ax; f⁻¹(x) = x/a] Let's see what else can work . . . Fred Son of a gun! I hit it on the nose without all those intermediate steps! Of course, what I didn't achieve was a proof that my solution was the entire solution. Thanks, Prof.!

For the homework question f(x)+f⁻(x)=ax the solution is f(x)=½(a±✓(a²-4))x

4:05 Quick Maths

The difference of cubes directly translates to a hexagonal numbers: Consider the set of lattice points in the cube of size (n+1) NOT in the cube of size n. Viewed from the line x=y=z, we see one point in the center (n+1, n+1, n+1) with six edges of length (n+1) around the edge. This flattened image is in fact a hexagonal number. Not much of a proof but it will give a good intuition on why they match up. (If you can visualize it)

i did myself the part with induction, but i didn't have idea how to move forward with this so i kinda gave up on that and tried some other obscure way to prove only possible solution is in form f(x)=x+c. First i proved strictly increasing bijection from R->R must be continuous, then i proved that for infinitely many sequences x_n f(x_n)=x_n+c but i had some troubles proving that constant "c" is same for every sequence and some other 'technical' problems so eventualy i gave up with whole problem and skipped to your video

we see this problem before in this channel but with different approach: "a FUNctional equation..."

f(x)+f^-1(x)=2x y=f(x)と置き、両辺を微分する。 dy/dx+dx/dy=2 左辺を通分して整理する。 dx^2+dy^2=2dxdy 2dxdyを移項して因数分解する。 (dx-dy)^2=0 dx=dy 両辺にい∫をつけて左右入れ替える。 ∫dy=∫dx y=x+C [終]

I really don't like the negative power notation for inverse functions. Isn't there a less confusing notation?

Shouldn't the induction hypothesis made for "some k>=2" to avoid the case f0 that is not defined.

My first thought was that since the inverse of a function is basically mirrored along x = y, x+y = c and x - y = c could be f(x) and f-1(x)

The same result can be obtained considering the curves of a function and its inverse are each other's mirrors to the y=x line.

The generalized solution is found by scaling: g(x) = a^-1 * f(a x)

4:05 Quick math

but we can simply take the derivative, so: f'(x)+1/f'(x) equal to 2 => (f'(x)-1)²=2=>f'(x)=1, then y=x+C is immediately the answer? is it the wrong way?

Watched this video but didn't understand the following: -why does continuously substituting x for f(x) yield ALL solutions to the functional equation he defined - at 15:38 why does going from the inequality less than 0 imply that the two parts are equal?

Find all strictly increasing functions, such that f(f(x))+f(x)=2x-3.How do I do this? Can anyone help to demonstrate that f(x)=x+constant?

Sir, I had a doubt that how can we assume x=f(y) for some y and rename it as x. What if x is not the range of f. Like for ex. for f(x)=e^x then it can never be -1 for real x. Please resolve my query

May be easy by differenziation for dx: f(x)+f*(x)=2x (the inverse function has the same tangent in x but inverse sign) dy/dx+dx/dy=2dx dy^2+dx^2=2*dx*dy dy^2+dx^2-2*dx*dy=0 (dy-dx)^2=0 dy=dx so y=x+c

Is it the identity function

ok, my brain is fried. I could not follow.

Is the "strictly increasing" condition really necessary? If you assume f is linear then f(x) = x + a is the only solution with or without the increasing condition.

4:04 2+2 is 4, minus 1 that's 3 (QUICK MAFS)

Hi, I have a doubt. For the question, As f(x) gives real, I put x=0 and assumed f(x) must be a polynomial. I got f(x) = 2x+c, And then I substituted into f(x)+f-¹(x)=2x and got c = (-x) That gave me f(x)=x instead of f(x)=x+a What went wrong?

A very elegant solution to the problem. If you remove the requirement that the solution is strictly increasing then there are other solutions. A simple example is f(x) = x+1 if x is an integer, otherwise f(x)=x. However, this, and similar solutions are not continuous. If the requirement is that the solution is continuous rather than strictly increasing then is f(x)=x+c the only solution?

04:05 two times two is four, minus one is three. QUICK MATHS !!!

From f_2(x) = 2f(x) - x we can get f_2(x) - f(x) = f(x) - x Since this is true for all x in R and n in N, f_n+2(x) - f_n+1(x) = f_n+1(x) - f_n(x) = f(x) - x By using the same argument this is also true for all n in Z. Be k_x = f(x) - x and A_x = {..., f^-1(x), x, f(x), f_2(x), ...} the orbit of x, from the previous equation we have A_x = {x + n . k_x | n in Z}. Pick any x < y, if k_x < k_y then be g_n = f_n(y) - f_n(x) = y + n . k_y - x - n . k_x = (y-x) + n.(k_y - k_x) g_0 > 0 and lim g_n = -inf There exists n such that g_n < 0 f_n(y) - f_n(x) < 0 f_n(y) < f_n(x) which is impossible since x < y and f san increasing function, so f_n s also an increasing function. Similar argument if k_x > k_y. So for all x, we have f(x) - x is a constant.

You actually used the fact that the function is strictly increasing right at the beginning where you did f^-1(f(x))=x.

We can use calculus to solve this: Let g=f⁻¹(x) the inverse function (i.e. f(g)≡ x) You need to know something about the differential of the inverse function, dg/dx = 1/(f '(g)) so starting from functional f(x) + f⁻¹(x)= 2x then differentiating w.r.t x get f '(x) + df⁻¹(x)/dx = 2 ⇒ f '(x) + dg/dx = 2 ⇒ f '(x) + 1/(f '(g)) = 2 _ eqn(A) the above still holds if we replace x with g since x is just a place holder! f '(g) + df⁻¹(g)/dg = 2 _ eqn(B) but df⁻¹(g)/dg =1/(f '(x)) we can let h=f '(x) so eqn(B) becomes, f '(g) +1/h = 2 or f '(g) = 2 -1/h substitute for f '(g) in eqn(A) get; h + 1/(2 -1/h) = 2 solving for (h -1)² = 0 or f '(x) =1 integrating gives the answer *f(x) = x +const.* [ btw: same solution for the generalized form he suggested h + 1/(a -1/h)= a ⇒ a(h -1)² = 0 so f '(x) =1 ]

4:06 2 times 2 is 4, minus 1 is 3 Quick maths

Great work. I have only one question: Isn't it simpler to apply f on the starting equation instead of setting x=f(x)? But probably I just found my mistake, my suggestion affords f to be linear. Nevertheless I do not know why x=f(x) is allowed.

What's with the filter on the video?

My solution was first proving the function must be linear (the coefficients of higher order polynomials won't cancel out because of the strictly increasing condition). Then I just solved for f^-1 using a generic linear polynomial ax +b and then plugged them into the equation and solved for the coefficients and got a=1 and b is arbitrary

It's really such a shame so many of these functional equations are just linear

You have nice colours, Michael Penn

Good Place to Stop hasn't posted yet, so I had to watch the video all the way to the end 🤷🏾‍♂️

1... 2sec eyeballing

Hi, . ¿can be this a proof?: By definition we have f + f ' = 2x. we know that 2x = x + x and general 2x = ax + (2-a)x for any `a` belong to the reals. Without loss of generality, the potential solution is: f + f ' = ax + (2-a)x with f = ax and f ' = (2-a)x. By definition of an inverse function, we need that: f (f '(x)) = x so a((2-a)x) = x iff (2a-a^2)x = x iff (a^2-2a+1)x = 0. So we need resolve a^2 - 2a + 1 = 0. Factoring this we have that (a-1)^2 = 0 with the solution when a = 1. The solution to the system is f = ax, f ' = (2-a)x, a = 1.

f(x) + 1/f(x)=2x f(x)=a a^2 - 2xa +1=0 If we make ∆>= 0 x^2 -1 >=0 (x+1)(x-1)>=0; ∆=0, x=1 a=2x/2, a=1, what answers the equality. ∆>0 a=( 2x+(x^2 - 1)^(1/2))/2 Teacher generally I try to solve your problems before watch it and this was my solution, and is to late here in my country and I couldn't verify with details if this solution is correct. If it is wrong could you explain where is my mistake?

I think I read once that proving with induction is equivalent to proving with strong induction. If this is true, why do we ever use induction? Why do we not always use strong induction?

To the editor, please place the attention grabbing animations and sounds asking to subscribe at better times in the videos. We are trying to focus on the math, and any distraction during nonobvious steps is annoying and hurts our focus. Please place the sub and channel adverts at times of low congitive load for example when an already derived line is being copied.

looks like a bad "red nose" cold ...

Dear Michael, I really love your videos but often these immediate "guessing" steps or proposals like let's use f(x) instead of x seem completely out of nowhere to me. But to you it seems like a totally normal thing that you've already done 100x in other problems. To me it would be very helpful to maybe take a bit more time and explain what the more general idea behind those substitutions is, why you're allowed to do that and how one can spot them. Thanks!

I think that the proof is not totally correct and you need to prove before that f is a surjection before doing some substitutions.

Just came from seeing the thumbnail to say "x=1" ok bye

The solution needs to be of the form f(x) = ax + b, so we would have (a+1/a)x + (1-1/a)b = cx in general, but this means (1-1/a) = 0 which means a=1, which means that the only possible value of c is (a+1/a)=(1+1/1)=2… and that’s a good place to stop!

it is probably equal to zero. am I right?

Take derivative of both sides: f'(x) + 1/f'(x) = 2 Multiply by f'(x) and rearrange f'(x)^2 -2 f'(x) + 1 = 0 Solve quadratic for f' f'(x) = 1 Integrate f(x) = x + c

f(x)=1.

Here first

x+c+x-c=2x✓ cx^n+(x/c)^(1/n)=2x=>x=1 cx+x/c=2x c²+1=2c c=1 y=x y=x+c см.выше. ч.т.д.