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

@BrianGriffin83

Жыл бұрын

It is immediately evident that this is a solution, the point is to prove it's the only one...

@tomasstride9590

Жыл бұрын

To be honest I have to confess that some type of problems I find more difficult than others. These problem I do find difficult. Knowing which path to follow is often for me very difficult and I think it would have taken me a very long time indeed to find this particular route.

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

@GiornoYoshikage

Жыл бұрын

YEP

@yahav897

Жыл бұрын

I'm glad I'm not the only one that thought that!

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) 1/2 implies B 2/3 implies B 3/4 implies B f'(x) = 1 leaving f(x) = x + c as the only solution.

@Notthatkindofdr

Жыл бұрын

It was not given that the function is differentiable, so assuming that might have potentially lost some solutions.

@Anonymous-zp4hb

Жыл бұрын

@@Notthatkindofdr Valid point. I took 'strictly increasing' to mean f'(x) > 0 for all x in the reals. When what was meant, was probably that f(a) > f(b) for all a > b. If that's the case, then yes. My argument doesn't work.

@zsoltbihary3347

Жыл бұрын

I also immediately took the derivative and worked from there, getting f'(x) = 1. This illustrates me being a physicist, and not a mathematician. It did not even cross my mind that f(x) may not be differentiable :) As such, my solution is not valid, I only found a subset of the functions.

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.

@swenji9113

Жыл бұрын

I'm extremely curious about this problem. The first condition is redondant with the second one so you might as well remove it. Condition 2 is equivalent to saying that f restricted to exp(a*Q) is a logarithm for each real number a, but those logarithms might be different for different values of a, depending on the axiom of choice. If you choose to use the axiom of choice, you can construct solutions f while choosing exactly the images of a set of positive real numbers A provided that ln(A) is a Q-linearly independent set. Setting A ={2,3}, if you accept that ln(2) and ln(3) are linearly independent over Q, that makes a looooot of solutions to your problem. For example for any pair of integers (m,n), there is a solution f with f(2)=m, f(3)=n. I find the problem quite fun like you stated it but I don't think a math contest would ever give a problem that cannot be decided in ZF (there are models where there are no solutions). Are you sure that was exactly the phrasing of the problem?

@dneary

Жыл бұрын

@@swenji9113 You might be right about f(1)=0 not being in the original question, it is redundant. I also missed a very important ordering condition: if a\leq b then f(a) \leq f(b). Otherwise that is the problem. What bases have you found for a log function that make both f(2) and f(3) integers?

@swenji9113

Жыл бұрын

@@dneary Oh I see ! If the function is increasing then so is f(exp). Since f(exp) is Q-linear, it must be R-linear, and therefore f is a logarithm (or the zero function), say f(x) = k*ln(x). But if k is not equal to 0, then ln(3)/ln(2) would be rational. It know it isn't, but I don't know how to prove it... therefore k=0 and f=0. Is that it ?

@dneary

Жыл бұрын

@@swenji9113 it is! And proving it is straightforward with the fundamental theorem of arithmetic. If log_2(3) = p/q then 2^p=3^q for integers p,q

@swenji9113

Жыл бұрын

@@dneary Right! It was much simpler than I thought! Nice problem, thanks :)

Amazing problems I love these type of things

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

@user-en5vj6vr2u

Жыл бұрын

I was thinking this

@schweinmachtbree1013

Жыл бұрын

Yes but sadly only _quite_ consistent because the standard (and silly) notations sin²(x), cosⁿ(x), etc. mess it up :/ (there is no reason why we can't write trig identities as sin(x)² + cos(x)² = 1, sec(x)² - tan(x)² = 1 and so on though)

@byronwatkins2565

Жыл бұрын

@@schweinmachtbree1013 Many already do, but it is still uncommon.

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

@mantisbog

Жыл бұрын

Me? No.

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 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) 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 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 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!

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?

@jimschneider799

Жыл бұрын

Both differential and difference equations are pretty common throughout the sciences, the former relating functions and their derivatives, and the later relating the difference between function values to the difference between their arguments.

That chalk looks really cool!

@BridgeBum

Жыл бұрын

The chalk does "pop" with whatever they are doing to the saturation, I personally find what it does to his skin (especially his hands) distracting.

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

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

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?

@thomashoffmann8857

Жыл бұрын

Just an example : 1/n is always > 0 but at the limit it is equal zero. Maybe it's about closed sets: "A closed set contains all its limit points" As [0, infinity) is closed on the left boundary, it must contain all limit points there.

@stewartzayat7526

Жыл бұрын

I don't know if it has a name, but it's a well known property of limits that is proved in a standard calculus course.

@MH-sf6jz

Жыл бұрын

Closed sets have the property that if a sequence is in a closed set, and it converges, then the limit of the sequence is also in the closed set. Notice that 1/n is in the closed set [0,1], since the sequence converges, and [0,1] is closed, so the limit is also in the closed set. This property can be easily shown with proof by contradiction.

@TheEternalVortex42

Жыл бұрын

It's just an application of the "squeeze theorem".

@Hyakurin_

Жыл бұрын

If x_n>a for all n (or at least definitely) and x_n converges to x, we can't have x0 in the definition and get a-x>x_n-x for n sufficiently large. This gives a0=a but 1/n->x=0=a, so in general we have to consider the non strict inequality. In fact this follows from the fact that [a,+infty[ is closed in euclidean topology.

Hi Dr. Penn!

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 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.

@ffggddss

Жыл бұрын

Actually, the inverse fn is (x - b)/m = x/m - b/m I think the rest of your method will still arrive at the same conclusion... Except I think you want 1/m = m, not -m, which would make m = ±i. Fred

@RobbieRosati

Жыл бұрын

@@ffggddss thanks for the correction, I updated my comment. Actually I get that b must be zero unless a=2, and then I get the same two solutions for m.

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)

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

@pavanato

Жыл бұрын

Great solution! I believe you just made one typo on "f'(f(y))*f'(y) + 1 = 2f'(x)" It should have been 2'f(y) on the right hand side

@kiyagiles655

Жыл бұрын

@@pavanatoyour right, it should be 2f'(y) not 2f'(x). Of course not making any typos was way too good to be true

@kiyagiles655

Жыл бұрын

I also tried out differentiating both sides straight off, firstly I showed that (f^-1(x))' = 1/f'(f^-1(x)). f^-1(x) = y, x = f(y), 1 = f'(y)*dy/dx, dy/dx = 1/f'(y) = 1/f'(f^-1(x)) f'(x) + 1/f'(f^-1(x)) = 2 f'(x)f'(2x - f(x)) + 1 = 2f'(2x - f(x)) Let y = 2x - f(x), dy/dx = 2 - f'(x) (2 - dy/dx)f'(y) + 1 = 2f'(y) f'(y)*dy/dx = 1, (df/dy)(dy/dx) = 1 df/dx = 1, f(x) = x + C. However if I integrate both sides of f'(y)*dy/dx = 1, We get f(y) = x + C, which doesn't make sense, what am I missing?

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)

@snared_

9 ай бұрын

there was no difference of cubes in this video, what are you even on about?

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

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.

4:05 quick maffs

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.

@douglasfeather3745

Жыл бұрын

Your reasoning is not correct. You got [f(b) - f(a)]/[b-a] = 1 but not for all a and b, you have only established this for those a and b for which f(a) = b.

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?

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

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

@orenfivel6247

Жыл бұрын

kzread.info/dash/bejne/q52rq7usZNbWmrQ.html

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.

@swenji9113

Жыл бұрын

True but in general the hard part for this type of problems is to prove that solutions must be of that form

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

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.

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

4:05 Quick Maths

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.

@Noct1um

Жыл бұрын

Well, if f:R -> R and f is strictly increasing then f is also invertible and does NOT has to be continuoues. Short example: f(x)=x, for -x>0 and f(x)=e^x, for x>0 or x=0.

@lock_ray

Жыл бұрын

@@Noct1um yeah it's invertible, but its inverse is not defined on [0,1), I should have made that clear. We need the inverse of f to be defined everywhere on R in order for the functional equation to make sense on all of R.

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.!

4:05 Quick math

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

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?

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?

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

@rgqwerty63

Жыл бұрын

Yes you got 99% the right idea, just remember the initial problem had a factor of 2 on the RHS so the substitution that reduces the 2nd problem to the initial one exactly is g(x) = 2/a * f(ax/2)

@Notthatkindofdr

Жыл бұрын

@@rgqwerty63 I disagree. Your scaled function g(x) satisfies the same functional equation (with the same a) as f(x) does. It doesn't reduce to the initial problem.

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

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 g_0 > 0 and lim g_n = -inf There exists n such that g_n Similar argument if k_x > k_y. So for all x, we have f(x) - x is a constant.

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]

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

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?

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

@AlcyonEldara

Жыл бұрын

The equation is true for all x, so it is also true for f(x). You are taking it backward: he doesn't pick x then find y, he picks x then apply the equation to f(x).

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?

@_legoved_

4 ай бұрын

upd: (f'(x)-1)²=0, missed

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

Is it the identity function

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.

@AlcyonEldara

Жыл бұрын

f(x) = x + a isn't a linear function. And yes, the condition is necessary: f(x) = x for x in Q, f(x) = x+1 for x in R\Q would work.

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

@sdchy5042

Жыл бұрын

we don't know if f(x) is differentiable or not, but yeah i came up with this solution too

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?

@danielsandoval9478

Жыл бұрын

you got a solution where a=0

1... 2sec eyeballing

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

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 ]

@Notthatkindofdr

Жыл бұрын

It was not given that the function is differentiable, so assuming that might have potentially lost some solutions.

@tomctutor

Жыл бұрын

@@Notthatkindofdr "f(x) increasing on x" so definitely has inverse, and (if continuous) must be differentiable!

@Notthatkindofdr

Жыл бұрын

@@tomctutor The function given by f(x)=x (when x0) is increasing and invertible, but not differentiable at 0. You could add in infinitely many non-differentiable points by making f(x) consist of line segments with ever-increasing slopes.

@tomctutor

Жыл бұрын

@@Notthatkindofdr Ok like the |x-1| not diff'ble at x=1 but cont's. By my calculus method I have found "a" solution to the functional mentioned. Your are claiming there could be other functions (not diff'ble) that possibly solve the problem! Since I did differentiate my trial solution f(x) then we may assume that it is indeed diff'ble a priori. Maybe I am being pedantic here, but this leads to a solution. Inverse functionals are notoriously difficult to solve in any event since there are not necessarily unique: e.g. f(x) = f⁻¹(x) which is same as f(x) - f⁻¹(x)= 0x is it not, would lead to f(x) = x but we all know that f(x) = 1/x also works, as does f(x) = (1-x)/(1+x) etc. The latter two examples aren't differentiable over ∀ℝ though.

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

You have nice colours, Michael Penn

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.

@ChefSalad

5 ай бұрын

You could set x=f(z) and get the same results. The x or the z are just dummy variables. This kind of replacement of x with f(x) is always fine as long as you replace all of the x's with f(x)'s. He tends to reuse variables a lot, which can be a bit confusing at first, but you eventually get used to it. Usually when he reuses a variable like this, he changes the equal sign to an arrow, which helps avoid confusion. I don't know why he didn't do that in this 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

@kemkyrk8029

Жыл бұрын

You're actually assuming that the solution is a polynomial (or at least an analytical function) which is pretty hard to prove

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

What's with the filter on the video?

@0xTJ

Жыл бұрын

That's not a filter. That's a light and what looks like fluorescent chalk.

@skylardeslypere9909

Жыл бұрын

@@0xTJ I mean, I think a saturation increase is more likely than Michael buying new chalk and a new light

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.

@oliverherskovits7927

Жыл бұрын

You've assumed without loss of generality f being a linear function. But this is loss of generality as we don't yet know that f must be linear. Hence this proof isn't valid. You need to prove that f must be linear, then your argument works.

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

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?

@BridgeBum

Жыл бұрын

F inverse isn't the same as 1/f. So step #1. The inverse function relation is if y is f(x), f inverse(y) = x. Classic examples are sin and arcsin or e^x and ln x.

@alexsandersouzadasilva1291

Жыл бұрын

@@BridgeBum I knew it, but if you put the values the result is correct. We shouldn't use math as its laws, but make it do whatever we want

@BridgeBum

Жыл бұрын

@@alexsandersouzadasilva1291 Sort of - this is more about math notation than it is math itself. Math notation is just a social construct, so going against the grain there is just going to lead to problems.

@alexsandersouzadasilva1291

Жыл бұрын

@@BridgeBum f(x)= x + a, solves the equation proposed very well, but f(x) goes from real to real, but if a permits solutions in complex numbers f(x) would be defined in complex too what is a contradiction with the domain of f(x)

@BridgeBum

Жыл бұрын

@@alexsandersouzadasilva1291 I haven't played with it, but since the reals are a subset of the complex I'm guessing there are still solutions. :)

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?

@fatman3762

Жыл бұрын

Weak Induction is a bit simpler / more elegant. That's probably the only reason

@backyard282

Жыл бұрын

We could but we often don't need it, that's the only reason. Often we only need to use the fact that it holds for just n, and not all numbers up to n. But yeah we could absolutely use strong induction every single time.

@self8ting

Жыл бұрын

You have to use only what you need. I think it's partly in order to be able to reuse parts of proofs to prove results with slightly different initial conditions. If you use strong tools that are not usable for the result you want you may have to find weaker tools to redo the original proof until the point the initial conditions diverge. So it's better to directly do proofs with the weakest arguments possible from the get go. I don't know if it's clear but it's the best I can do ahah

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.

@MichaelPennMath

Жыл бұрын

Thank you for that feedback. I'll try to adhere to that in future edits. Have a good day! -Stephanie MP Editor

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!

@timayovyk2036

Жыл бұрын

For this one, I imagine that the general idea behind the substitution was to get rid of the f^(-1) and then figure out what can happen from there. He mentioned that it gave a good notation for f_2(x), and had the thought that this could be a repeatable strategy. Although I agree it would be nice to get more explanation behind those aspects a lot of it likely just comes from experience and knowing what types of strategies work in certain scenarios and just trying them, it might not necessarily have a sophisticated reason.

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

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

@samueljele

Жыл бұрын

-its a strictly increasing function from R->R, so it is bijective- (if it weren't surjective, it wouldn't have an inverse).

@schweinmachtbree1013

Жыл бұрын

@@samueljele exp: *R* → *R* is strictly increasing but not bijective - it does have an inverse, just not one from *R* to *R*

@rgqwerty63

Жыл бұрын

The formulation of the question has f inversed hence we already know were only looking for bijections. I dont think Michael missed anything here

@samueljele

Жыл бұрын

@@schweinmachtbree1013 yes sorry, I mixed things up. That was stupid. But the thing with the inverse still holds.

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!

@bjornfeuerbacher5514

Жыл бұрын

"The solution needs to be of the form f(x) = ax + b" Why?

@backyard282

Жыл бұрын

f(x)=ax+b doesn't even satisfy the original equation.. unless a=1 of course, which is the solution in the video

it is probably equal to zero. am I right?

f(x)=1.

@BrianGriffin83

Жыл бұрын

Definitely not invertible

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

@SzanyiAtti

Жыл бұрын

How do you prove that f is differentiable?

@mrphlip

Жыл бұрын

The derivative of f¯¹(x) is not 1/f'(x). It is 1/f'(f¯¹(x)).

@abelferquiza1627

Жыл бұрын

i did exactly the same!

Here first

@ngc-fo5te

Жыл бұрын

Nope.

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.