a fun functional equation with an inverse twist
HTML-код
- Опубликовано: 7 фев 2025
- 🌟Support the channel🌟
Patreon: / michaelpennmath
Merch: teespring.com/...
My amazon shop: www.amazon.com...
🟢 Discord: / discord
🌟my other channels🌟
mathmajor: / @mathmajor
pennpav podcast: / @thepennpavpodcast7878
🌟My Links🌟
Personal Website: www.michael-pen...
Instagram: / melp2718
Twitter: / michaelpennmath
Randolph College Math: www.randolphcol...
Research Gate profile: www.researchga...
Google Scholar profile: scholar.google...
🌟How I make Thumbnails🌟
Canva: partner.canva....
Color Pallet: coolors.co/?re...
🌟Suggest a problem🌟
forms.gle/ea7P...
I eyeballed the solution f(x) = x + C immediately at the beginning but I would never have come up with that proof lol.
It is immediately evident that this is a solution, the point is to prove it's the only one...
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
YEP
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
y+nb < x+ma < x+(m+1)a < y+(n+1)b does not hold in many cases, for example
i) b = 0, then y + nb = y + (n+1)b so they can't be strict inequalities
ii) b = 1, then y + nb and y (n+1)b just differs by one, and not by at least 3 as the chain of strict inequalities implies
iii) a < 0, then x+ma < x+(m+1)a is false
etc.
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.
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.
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
Indeed it's amazing and more technical solution bravo .
I hope your never stop using colors. I thoroughly enjoy how it's much more aesthetic; as math should be
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.
It was not given that the function is differentiable, so assuming that might have potentially lost some solutions.
@@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.
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.
@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.
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?
@@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?
@@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 ?
@@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
@@dneary Right! It was much simpler than I thought! Nice problem, thanks :)
This got to be one of the most delicious proof I have ever seen. This is math artistic craft in action.
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.
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
@@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.
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!
Hey Michael will you ever make a video about galois theory?
Me? No.
If I assume f to be differentiable, it is obvious: just differentiate both sides which gives f’(x)+(1/f’(x))=2 so with z= f’(x), z^2-2z+1=0 ie (z-1)^2=0 so z=f’(x)=1 ie f(x)=x+a.
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?
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?
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.
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.
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.
It's just an application of the "squeeze theorem".
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.
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.
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.
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
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
@@pavanatoyour right, it should be 2f'(y) not 2f'(x). Of course not making any typos was way too good to be true
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?
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?
4:05 quick maffs
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?
upd: (f'(x)-1)²=0, missed
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.
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.
@@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.
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?
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.
True but in general the hard part for this type of problems is to prove that solutions must be of that form
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?
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.
Amazing problems I love these type of things
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 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)
there was no difference of cubes in this video, what are you even on about?
I would say that f^n --> n applications of f() is quite consistent with math notation... especially since f^{-1} is already present.
I was thinking this
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)
@@schweinmachtbree1013 Many already do, but it is still uncommon.
I really don't like the negative power notation for inverse functions. Isn't there a less confusing notation?
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.
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?
you got a solution where a=0
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.
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]
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.!
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
4:05 Quick Maths
That chalk looks really cool!
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.
I tried to solve this from the thumbnail alone... Wish I'd known about the strictly increasing
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
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).
we see this problem before in this channel but with different approach: "a FUNctional equation..."
ruclips.net/video/yiwIYs4spjQ/видео.html
Didn't you make a video on this problem about a year ago?
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.
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
we don't know if f(x) is differentiable or not, but yeah i came up with this solution too
4:05
Quick math
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.
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.
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)
You actually used the fact that the function is strictly increasing right at the beginning where you did f^-1(f(x))=x.
04:05
two times two is four,
minus one is three.
QUICK MATHS !!!
4:04 2+2 is 4, minus 1 that's 3 (QUICK MAFS)
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?
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 ]
It was not given that the function is differentiable, so assuming that might have potentially lost some solutions.
@@Notthatkindofdr "f(x) increasing on x" so definitely has inverse, and (if continuous) must be differentiable!
@@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.
@@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.
The generalized solution is found by scaling:
g(x) = a^-1 * f(a x)
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)
@@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.
Hi Dr. Penn!
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.
4:06
2 times 2 is 4, minus 1 is 3
Quick maths
Is it the identity function
For the homework question f(x)+f⁻(x)=ax the solution is f(x)=½(a±✓(a²-4))x
ok, my brain is fried. I could not follow.
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.
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.
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?
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.
@@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
@@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.
@@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)
@@alexsandersouzadasilva1291 I haven't played with it, but since the reals are a subset of the complex I'm guessing there are still solutions. :)
It's really such a shame so many of these functional equations are just linear
Good Place to Stop hasn't posted yet, so I had to watch the video all the way to the end 🤷🏾♂️
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.
Thank you for that feedback. I'll try to adhere to that in future edits.
Have a good day!
-Stephanie
MP Editor
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?
Weak Induction is a bit simpler / more elegant. That's probably the only reason
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.
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
What's with the filter on the video?
That's not a filter. That's a light and what looks like fluorescent chalk.
@@0xTJ I mean, I think a saturation increase is more likely than Michael buying new chalk and a new light
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!
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.
You have nice colours, Michael Penn
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
You're actually assuming that the solution is a polynomial (or at least an analytical function) which is pretty hard to prove
1... 2sec eyeballing
looks like a bad "red nose" cold ...
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.
-its a strictly increasing function from R->R, so it is bijective- (if it weren't surjective, it wouldn't have an inverse).
@@samueljele exp: *R* → *R* is strictly increasing but not bijective - it does have an inverse, just not one from *R* to *R*
The formulation of the question has f inversed hence we already know were only looking for bijections. I dont think Michael missed anything here
@@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!
"The solution needs to be of the form f(x) = ax + b"
Why?
f(x)=ax+b doesn't even satisfy the original equation.. unless a=1 of course, which is the solution in the video
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
How do you prove that f is differentiable?
The derivative of f¯¹(x) is not 1/f'(x). It is 1/f'(f¯¹(x)).
i did exactly the same!
f(x)=1.
Definitely not invertible
it is probably equal to zero. am I right?
Here first
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.