Holy moly thank you so much for taking the suggestion Michael!!! I enjoy watching your awesome videos and I promise I will let you know if any good ideas for a math vid comes to mind! Keep up the great work!! By the way I knew this would make a nice thumbnail 😂
Yes! I thank you too for sharing a solution for the initial problem. By the way, as I asked in the last comment, I would like to know if you could provide any source on proving g(x)=lnx. Also if you remember my other suggestion, any thoughts on that one too?
10:00 I'm pretty sure you don't actually have to use the Axiom of Choice since we can fairly easily define such a function as follows: c([x])=min{x, 1/x}. Am I missing something or is it enough?
@@ricardocavalcanti3343 I also suspect that you don't actually need it, since you can defined them using the half open intervals defined at 13:20: take the first and third intervals to U, and the second and forth intervals to V.
I think you need the axiom of choice because you can create uncountably many interval defining U and V. Without the axiom of choice, you are restricted to a countable distribution in the ensemble [0;1)
Do you still need the axiom of choice if you define c([x]) = sup(phi([x]))? (The supremum of each equivalence class as a set.) Isn't the supremum uniquely defined for each set of real numbers?
I was thinking the same, but you don’t even need the supremum, maximum is enough as the size of the equivalence classes is at most 2. This guarantees the existence of at least one choice function, which is enough for the argument.
@@weker01 I believe the AoC comes into play because you need to make an infinite number of choices... the number of equivalence classes is infinite even though each one is finitely large.
@@jb31842 No I am pretty sure the function always taking the max is well defined as the union of the sets {({a,b}, min {a,b})} for all these {a,b} equivalance classes. The exisitance of this set follows from ZF and does not need ZFC as far as I can see. It may vary well be that the AC is used somewhere else in this proof but not here. I am pretty certain as I just passed a mathmatical logic course focusing on set theory. edit: We could also construct a choice function with transfinite induction easily
I remember that those functions are quite interesting and solutions to certain equations in thermodynamics. Have to try to find it - I was thinking about this some months ago :D
ya I got this at the end like "oh f(f(x)) = x^2 is easy that's just f(x) = x^(sqrt(2))... wait that also works for f(f(x)) = x^(-1) = x^i" so that's both tht hw questions answered
@@justinwebb7990 No for the seond question, a solution is f(x) = x^2 without a square root. You can easily find it (and a solution to the first question) if you assume the solution is of the form x^p for some real (or complex) number p.
New lighting is objectively better from a video standpoint, but I did love the yellower lights you had before, much more cozy and calm, easier on the eyes
I remember seeing a discussion of a similar functional equality somewhere last year, this time involving derivatives. The equation, if I'm not mistaken, was inv(f(x)) = f'(x), or simply a function whose derivative is its inverse. Can you do a video on this one as well, please?
16:14 i notice the exception -1 is outside of the bounds (0,inf), but also 0 itself maps to the number 1 which then stays at 1 forever, so again because of the bounds (but only so) it still holds true as f(0) is undefined
It is an old problem and the calculations were shown to me and others at the physics department when I was attending it ca. 1985, along with the Dirac's special functions. Apart from what had been known, I wonder where the theorems are here that either have been just proven or still need to be proven? In the margin conditions?
For complex numbers z I would take: f(z) = (z+i)/(1+i*z); z ≠ i or also f(z) = (z-i)/(1-i*z); z ≠ -i A similar function for real numbers with f(f(x)) = 1/x is not known to me. Best regards and Greetings from Germany, MartinN (Feles) PS: I like your videos and math problems.
I thought so too but plugging (x^i)^i into Wolfram Alpha gives a very strange result around 0 😢 (and I don't mean _at_ zero, of course that's undefined)
Michael’s chalks seem very pleasant to work with - soft on the board yet stable and sturdy. Where do you find that kind of chalk? The ones I have make too much unpleasant noise. 😂😄
Sometimes, I understand everything. Often, I have to watch again some parts of the video. But here ... I was lost after 3 minutes ! Maybe tomorrow, after a good night 😀.
The idea is to define the function piecewise, such that you have both a formula and its inverse in the same function, but at different places so that you don't immediately get the identity if you apply the function twice. He gives a nice example at the end with the mix of x↦1/x and -x.
If f(f(x) = 1/x, apply f on both sides two times, f(f(f(f(x) = f(f(1/x) but we know f(f(x)=1/x, substitute x=1/y, f(f(1/y)=y. And so f°f°f°f(x) = x (obviously the variable name doesn't matter. I used y just for it to be clear)
First homework: for z=e^(a+b*i) let f(z)=e^(a*i-b). So this is f:C-{0}->C-{0}. How I got this: easy searched for f(z)=z^alpha forms, we needed f(f(z))=1/z=z^(-1), from this z^(alpha^2)=z^(-1), hence alpha^2=-1, so alpha=i is a good solution. Ouch, this is not good [since b is not unique in z=e^(a+b*i)=e^a*(cos(b)+i*sin(b))], need to restrict abs(a),abs(b)
@@tac0cat14 I disagree. Both f(z)=z^i and f(z)=z^(-i) work. Look: f(z)=z^i=(re^(iθ))^i=(r^i)(e^(-θ))=(e^(-θ))e^(ilnr) => f^(-1)(z)=(e^θ)e^(-ilnr) and (1/f(z))=(e^θ)e^(-ilnr) f(z)=z^(-i)=(re^(iθ))^(-i)=(r^(-i))(e^θ)=(e^θ)e^(-ilnr) => f^(-1)(z)=(e^(-θ))e^(ilnr) and (1/f(z))=(e^(-θ))e^(ilnr) Also, it's not true that f°f(z)=z. Look: f(f(z))=((z^i)^i)=z^(-1) and, for f(z)=z^(-i), f(f(z))=((z^(-i))^(-i))=z^(-1)
Yes, but you have to specify the branch (since it is multivalued) and restrict the domain (since it is non injective, hence non invertible, if the domain is the whole complex plane).
This problem isn't any well-defined subject in mathematics. It is more a simply stated problem that requires more than expected to describe. That being said, my favorite resource for learning enough to jump into more abstract mathematics is The Book of Proof (www.people.vcu.edu/~rhammack/BookOfProof/BookOfProof.pdf)
Is it possible to have an equation where Derivative(f(Ln(x))) = 1/Squareroot(Derivative(Inverse(f(x/2)))^2+1/2+Sin(x pi)/(2x))? Not looking for the solution, just asking if it is possible. Thanks for the video.
Let n be a positive integer. Show that any number greater than n⁴/16 can be written in at most one way as the product of two of its divisors having difference not exceeding n. (1998 St. Petersburg City Mathematical Olympiad) Please solve this one. Thanks.
It turns out that we don't need the full axiom of choice here since the sets we are choosing from have two elements - we only need what is called the "axiom of binary choice" (which states that any family of two-element sets has a choice function). one might think that this is equivalent to the "axiom of finite choice" (any family of finite sets has a choice function), but it is not; it is strictly weaker. even more surprisingly, the axiom of binary choice does not imply the "axiom of ternary choice" (3-element sets) but it *does* imply the "axiom of quaternary choice" (4-element sets)! see here if interested math.stackexchange.com/questions/2293418/axiom-of-binary-choice-vs-axiom-of-finite-choice
I should remark that most of the time when one says "we don't actually need the full axiom of choice here; we only need a weakened version", this is not actually helpful in any way, because models of set theory in which the full axiom of choice doesn't hold but a weakened version does tend to be very contrived (indeed, most of the time they are constructed so that they have that property). perhaps an exception to this rule would be the fact that most, if not all, of "basic analysis" (which is not really well-defined, but you get the idea) can be done with the axiom of dependent choice as opposed to the full axiom of choice (because the axiom of dependent choice is what is needed to make certain types of arguments using sequences) [but maybe this is not an exception because I don't know how contrived the models of set theory in which countable choice holds but full choice doesn't are]
Why do you need the axiom of choice? Of any couple (x, 1/x), one must be greater than 1, the other lower than 1. Can't we just choose, say, the one that's lower than 1?
you can't invert a constant function because it isn't injective (excluding the trivial case where the domain is a single point, in which case if f(a)=b, f^-1(b)=a)
For complex functions with f(f(z))=1/z, I have a solution which would be a lot easier to explain geometrically so here's just my result: f: C\{0,i,-i}-->C\{0,i,-i}, f(r, phi)=... ...= (1/r, phi) if either (r, phi) is in (0;1)x(0,pi) or in (1;inf)x(pi;2pi) ...=(r, 2pi-phi) if (r, phi) is either in (1;inf)x(0;pi) or in (0;1)x(pi;2pi) ...=(r,pi) if phi=0 ...=(1/r,0) if phi=pi ...=(1 pi-phi) if r=1 and phi is in (0,pi/2) ...=(1,pi+phi) if r=1 and phi in (pi/2 pi) ...=(1,phi-pi) if r=1 and phi in (pi;3pi/2) ...=(1,3pi-phi) if r=1 and phi in (3pi/2;2pi) I see no easy way to include i and -i, however...
f(f(x))=f(x)^2 is just equivalent to f(y)=y^2 for all y in the image of f. Examples: f(x)=x^2 f(x)=0 f(x)=0 if x is rational but not equal to 1, f(x)=1 otherwise f(x)= 2^(2n) if x=2^n for some integer n, f(x)=0 otherwise f(x)=x^2 if x is not negative, f(x)=-12345x+23pi^e otherwise ;)
Homeworks pretty obvious. Both are to find the "half function" of xⁿ, where n=-1 in the first case and n=2 in the second. In each case, it's just x^√n. : First one was my "easy" solution to the original problem before I noticed it was f(x) rather than f(z). f(z) = zⁱ. This means f(f(z)) = zⁱⁱ = z⁻¹ = 1/z. [Second solution is z⁻ⁱ, so a general solution is presumably a strange combination of the two]. Second, f(x) = x^√2. Then f(f(x)) = (x^√2)^√2) = x^(√2*√2) = x².
While the examples were written as an infinite list, I think the point there was to show that they could be any arbitrary values. They should both be uncountable.
I had a go at this before watching the more-complicated-than-I-expected explanation. I managed to confuse myself at one point, and I'm clearly wrong, but could someone please explain to me why this doesn't work? (Hoping I'm not being too dumb... :) If f(f(x)) = 1/x, then f(-1)( f(f(x)) ) = f(-1)(1/x). The LHS = f(x) and the RHS = 1/f(1/x) by the original condition. So f(x) = 1/f(1/x), or f(1/x) f(x) = 1. That's obviously not right, because (for example) f(x) = x^2 satisfies that condition but not f(f(x)) = 1/x.
I'm not sure whether you actually use the axiom of choice for your choice function. The axiom of choice only yields existence, but this should be clear, since you can choose e.g. C([x]) := max [x] = max {x, 1/x}. Maybe one needs the AOC to prove that the max-function exists on this subdomain of finite subsets of P(R). However, as I understood your reasoning, You only state that any such choice function yields a desired function f. But this means that you show "C choice function ==> f has property", meaning at this point you don't even need existence.
The problem is if y=f(x) And suppose that f is of class (at least) c¹ and apply f to both sides and differentiate u get the weird differential equation : 1= -y'²/y² which is impossible for a real valued function
“That’s a good place to [start rock climbing]”
Big fan of the new lighting! It looks great!!
Perfect, was wondering about this recently.
Great video, Michael! Also, I love the new lighting, definitely adds to the production quality
Holy moly thank you so much for taking the suggestion Michael!!! I enjoy watching your awesome videos and I promise I will let you know if any good ideas for a math vid comes to mind! Keep up the great work!!
By the way I knew this would make a nice thumbnail 😂
Yes! I thank you too for sharing a solution for the initial problem.
By the way, as I asked in the last comment, I would like to know if you could provide any source on proving g(x)=lnx. Also if you remember my other suggestion, any thoughts on that one too?
On which video did you duscussed the question with a continuity constraint? I'm really curious and would love to read.
Love the introduction of climbing to the channel. I’m also do some climbing over in Europe alot so it is something I like
6:06 Wouldn't that be { {1} } instead of {1}? Because it's a set of sets.
No, it would technically be { [1] } but that's annoying to write out.
@@RandomBurfness [1] = {1}.
That video made think if is it possible to define fractional compositions of real/complex valued functions, to generalize that problem.
10:00 I'm pretty sure you don't actually have to use the Axiom of Choice since we can fairly easily define such a function as follows:
c([x])=min{x, 1/x}.
Am I missing something or is it enough?
You are right
The Axiom of Choice is needed in a previous step, where he failed to mention it: to define the sets U and V.
@@ricardocavalcanti3343 I also suspect that you don't actually need it, since you can defined them using the half open intervals defined at 13:20: take the first and third intervals to U, and the second and forth intervals to V.
@@putinrulz I agree with you. In that example you don't need the Axiom of Choice. Only in the more general case, discussed earlier.
I think you need the axiom of choice because you can create uncountably many interval defining U and V. Without the axiom of choice, you are restricted to a countable distribution in the ensemble [0;1)
Is the chalk on your sweater from writing math, or is it from rock climbing? Do all your hobbies involve chalk?
The people need to know.
He was even a gymnast. He must have liked bars lets just say.
Its from chalk climbing!
Do you still need the axiom of choice if you define c([x]) = sup(phi([x]))? (The supremum of each equivalence class as a set.) Isn't the supremum uniquely defined for each set of real numbers?
I was thinking the same, but you don’t even need the supremum, maximum is enough as the size of the equivalence classes is at most 2. This guarantees the existence of at least one choice function, which is enough for the argument.
yea I also don't see why the AC is needed, as all the equivilance classes are finite.
@@weker01 I believe the AoC comes into play because you need to make an infinite number of choices... the number of equivalence classes is infinite even though each one is finitely large.
@@jb31842 No I am pretty sure the function always taking the max is well defined as the union of the sets {({a,b}, min {a,b})} for all these {a,b} equivalance classes. The exisitance of this set follows from ZF and does not need ZFC as far as I can see. It may vary well be that the AC is used somewhere else in this proof but not here.
I am pretty certain as I just passed a mathmatical logic course focusing on set theory.
edit: We could also construct a choice function with transfinite induction easily
Didn't he already use AoC when defining the partition ~?
I remember that those functions are quite interesting and solutions to certain equations in thermodynamics. Have to try to find it - I was thinking about this some months ago :D
I believe f(z)=z^i works for the first HW question. Ya sorta gave that away in the beginning.
Yep, that's right.
ya I got this at the end like "oh f(f(x)) = x^2 is easy that's just f(x) = x^(sqrt(2))... wait that also works for f(f(x)) = x^(-1) = x^i" so that's both tht hw questions answered
oh I just realized I misread the second question lol
@@justinwebb7990 No for the seond question, a solution is f(x) = x^2 without a square root. You can easily find it (and a solution to the first question) if you assume the solution is of the form x^p for some real (or complex) number p.
more generally c*x^(+-i)
New lighting is objectively better from a video standpoint, but I did love the yellower lights you had before, much more cozy and calm, easier on the eyes
Lejos, el mejor video hasta el momento. Muchas gracias
Creo que tú y yo seremos los únicos que comentemos en español este video
@@zeravam jajaja, siiii
I remember seeing a discussion of a similar functional equality somewhere last year, this time involving derivatives. The equation, if I'm not mistaken, was inv(f(x)) = f'(x), or simply a function whose derivative is its inverse. Can you do a video on this one as well, please?
He already did: ua-cam.com/video/rNUfiQgj6ZI/v-deo.html
@@ricardocavalcanti3343 oh, nice! Didn't know this one, thanks!
new lighting looks great mate
f(x) = x² satisfies f(f(x)) = f(x)².
f(x) = 1 too
and so functions like
f(x) = x² if x in [-1,1]
= 1 otherwise
satisfies the pb as well
The objective of this homework is to see the procedure to get that solution, very good solution by the way
All functions f(x) = x^(2ⁿ) satisfy this
(with n being a non-zero positive integer)
It's kinda like the argument in 2:34 I believe.
Another example: f(x) = 0.
Lighting is a 10 out of 10 improvement
Great to hear, don't get too excited though.... I have a big back catalogue from before I changed the lights!
A nicer way to write this: when is f^{-1}(x) = (f(x))^{-1}
nice lighting!
16:14 i notice the exception -1 is outside of the bounds (0,inf), but also 0 itself maps to the number 1 which then stays at 1 forever, so again because of the bounds (but only so) it still holds true as f(0) is undefined
Nice new lighting!
Wow this is Awesome!!
18:52 Homework
19:36 Good Place To Stop
Ty
People hate quarentine but humanity were needing it
It is an old problem and the calculations were shown to me and others at the physics department when I was attending it ca. 1985, along with the Dirac's special functions. Apart from what had been known, I wonder where the theorems are here that either have been just proven or still need to be proven? In the margin conditions?
The new lighting is great.
At the end, when looking for complex functions one option is f(x)=x^i
I like this exam so much , and you give the good solution.
This is brilliant
For complex numbers z I would take:
f(z) = (z+i)/(1+i*z); z ≠ i
or also
f(z) = (z-i)/(1-i*z); z ≠ -i
A similar function for real numbers with f(f(x)) = 1/x is not known to me.
Best regards and Greetings from Germany, MartinN (Feles)
PS: I like your videos and math problems.
if you're expanding to complex numbers, wouldn't f(x) = x^i do the job?
I thought so too but plugging (x^i)^i into Wolfram Alpha gives a very strange result around 0 😢 (and I don't mean _at_ zero, of course that's undefined)
Would it be a good idea to construct such a function with a laurent series?
I'm dept. of mathematics korean university student,
but this problem is too difficult to solve...
I respect you so much...
I was literally thinking about this the night before. Holy shit. I think my prayers were answered.
I haven't even watched this video and I figured it out: f(x)= x^±i
Michael’s chalks seem very pleasant to work with - soft on the board yet stable and sturdy. Where do you find that kind of chalk? The ones I have make too much unpleasant noise. 😂😄
Hagaromo chalk
Sometimes, I understand everything. Often, I have to watch again some parts of the video. But here ... I was lost after 3 minutes !
Maybe tomorrow, after a good night 😀.
The idea is to define the function piecewise, such that you have both a formula and its inverse in the same function, but at different places so that you don't immediately get the identity if you apply the function twice. He gives a nice example at the end with the mix of x↦1/x and -x.
Amazing! I just thought about this problem! 😃
Interestingly, it can be written as f^(-1)(x) = f^(-1)(x). Silly notation... 😁
f^(-1)(x)=(f(x))^(-1) more like this
*Hello from Turkey.*
Try this: Find all continuous f:S^1 \to S^1 such that f^{-1} and f^{-1} (z)=1/f(z). I think answer is very easy but ...but maybe not so elementary
i wonder if there are any analytic functions that solve this
For the original equation I let f(x) = (A+Bx)/(C+Dx) , one solution is A=B=C=D and another is A = +/- (BC)^1/2 and D = -/+ (BC)^1/2
New lighting is better, it looks like you have a wider shot too but I might be imagining it
what's an equivalence class?
If elements are books, equivalences classes are shelves. ;D
@@samucabrabo Great answer
@@zeravam haha. I took that from some book or article. But I don't remember which one.
Lighting is great
2:10 why???????
If f(f(x) = 1/x, apply f on both sides two times, f(f(f(f(x) = f(f(1/x) but we know f(f(x)=1/x, substitute x=1/y, f(f(1/y)=y. And so f°f°f°f(x) = x (obviously the variable name doesn't matter. I used y just for it to be clear)
@@justacutepotato2945 omg thks s2
First homework: for z=e^(a+b*i) let f(z)=e^(a*i-b). So this is f:C-{0}->C-{0}.
How I got this: easy searched for f(z)=z^alpha forms, we needed f(f(z))=1/z=z^(-1), from this
z^(alpha^2)=z^(-1), hence alpha^2=-1, so alpha=i is a good solution.
Ouch, this is not good [since b is not unique in z=e^(a+b*i)=e^a*(cos(b)+i*sin(b))], need to restrict abs(a),abs(b)
Same way I thought.😉
Another one is alpha=-i.
@@EngMorvan if that were the case then f ° f =x and f ° f ° f ° f = 1/x
@@tac0cat14 I disagree. Both f(z)=z^i and f(z)=z^(-i) work. Look:
f(z)=z^i=(re^(iθ))^i=(r^i)(e^(-θ))=(e^(-θ))e^(ilnr)
=>
f^(-1)(z)=(e^θ)e^(-ilnr)
and
(1/f(z))=(e^θ)e^(-ilnr)
f(z)=z^(-i)=(re^(iθ))^(-i)=(r^(-i))(e^θ)=(e^θ)e^(-ilnr)
=>
f^(-1)(z)=(e^(-θ))e^(ilnr)
and
(1/f(z))=(e^(-θ))e^(ilnr)
Also, it's not true that f°f(z)=z. Look:
f(f(z))=((z^i)^i)=z^(-1)
and, for f(z)=z^(-i),
f(f(z))=((z^(-i))^(-i))=z^(-1)
Actually, no restriction on a is needed, since e^a is unique for any real number a.
Brouwer would not like the proof. Is there a proof without the Axiom of Choice?
Doesn't f(x)=x^i just work?
Yes, but you have to specify the branch (since it is multivalued) and restrict the domain (since it is non injective, hence non invertible, if the domain is the whole complex plane).
Is not y= 1/x work
The inverse of f(x) = 1/x is f^(-1)(x) = 1/x , because f(f^(-1)(x)) = f^(-1)(f(x)) = 1/(1/x) = x, but 1/f(x) = 1/(1/x) = x
What are some good resources for learning this kind of maths?
This problem isn't any well-defined subject in mathematics. It is more a simply stated problem that requires more than expected to describe. That being said, my favorite resource for learning enough to jump into more abstract mathematics is The Book of Proof (www.people.vcu.edu/~rhammack/BookOfProof/BookOfProof.pdf)
@@MichaelPennMath thank you! I'll check it out. I'm surprised that this isn't a well defined field of its own though!
I dont get why ffff(x)=x at 2:19
I prefer the older lighting since with this setup, there is some glare on the Blackboard making it look a little bit white.
Interesting integral question, Integrate arctan(x)arcos(x) from 0 to 1, like so Michael sees this
f is a functional fourth root. Id^(1/4)
Is it possible to have an equation where Derivative(f(Ln(x))) = 1/Squareroot(Derivative(Inverse(f(x/2)))^2+1/2+Sin(x pi)/(2x))? Not looking for the solution, just asking if it is possible. Thanks for the video.
The choice principle is strong with this one.
For the last question:
f(x) = 0
f(x) = 1
f(x) = x^2
Yep, I thought those too.
x^n can be called as the complete ans for the last qn
@@gamingDivyaa No. Because if f(x)=x^3. Then f(f(x))=x^9 and f(x)^2=x^6
Let n be a positive integer. Show that any number greater than n⁴/16 can be written in at most one way as the product of two of its divisors having difference not exceeding n.
(1998 St. Petersburg City Mathematical Olympiad)
Please solve this one.
Thanks.
V10? That's serious stuff.
It turns out that we don't need the full axiom of choice here since the sets we are choosing from have two elements - we only need what is called the "axiom of binary choice" (which states that any family of two-element sets has a choice function). one might think that this is equivalent to the "axiom of finite choice" (any family of finite sets has a choice function), but it is not; it is strictly weaker. even more surprisingly, the axiom of binary choice does not imply the "axiom of ternary choice" (3-element sets) but it *does* imply the "axiom of quaternary choice" (4-element sets)! see here if interested math.stackexchange.com/questions/2293418/axiom-of-binary-choice-vs-axiom-of-finite-choice
I should remark that most of the time when one says "we don't actually need the full axiom of choice here; we only need a weakened version", this is not actually helpful in any way, because models of set theory in which the full axiom of choice doesn't hold but a weakened version does tend to be very contrived (indeed, most of the time they are constructed so that they have that property). perhaps an exception to this rule would be the fact that most, if not all, of "basic analysis" (which is not really well-defined, but you get the idea) can be done with the axiom of dependent choice as opposed to the full axiom of choice (because the axiom of dependent choice is what is needed to make certain types of arguments using sequences) [but maybe this is not an exception because I don't know how contrived the models of set theory in which countable choice holds but full choice doesn't are]
As per the video we have f(f(x))=1/x. Handwavily it's clear x=0 must be excluded from the domain of f. For x>0, let f(x)=-1/x^2. For x
some heavy duty stuff after a long time.
Here is a pretty "heavy duty" unlisted video: ua-cam.com/video/n8AzVj_hocQ/v-deo.html
@@MichaelPennMath thanks a lot.
Why do you need the axiom of choice? Of any couple (x, 1/x), one must be greater than 1, the other lower than 1. Can't we just choose, say, the one that's lower than 1?
How about f’(x) = 1/f(x)
By the way, just out of curiosity. What is the inverse of constant function?
you can't invert a constant function because it isn't injective (excluding the trivial case where the domain is a single point, in which case if f(a)=b, f^-1(b)=a)
I think it will be a relation..?
@@moonlightcocktail Well, every function is a relation, so yes
A multivalued function where if f(x)=1 where f has the domain of (a,b), then f^(-1)(x) can take any value from (a,b). Basically a useless function.
Nice
oh i was just thinking ofusing x^i since f^-1(x) = x^-i = 1/f(x) but if you can do it in only real numbers that's cool too
For complex functions with f(f(z))=1/z, I have a solution which would be a lot easier to explain geometrically so here's just my result:
f: C\{0,i,-i}-->C\{0,i,-i},
f(r, phi)=...
...= (1/r, phi) if either
(r, phi) is in (0;1)x(0,pi)
or in (1;inf)x(pi;2pi)
...=(r, 2pi-phi) if (r, phi) is either in (1;inf)x(0;pi)
or in (0;1)x(pi;2pi)
...=(r,pi) if phi=0
...=(1/r,0) if phi=pi
...=(1 pi-phi) if r=1 and
phi is in (0,pi/2)
...=(1,pi+phi) if r=1 and
phi in (pi/2 pi)
...=(1,phi-pi) if r=1 and
phi in (pi;3pi/2)
...=(1,3pi-phi) if r=1 and
phi in (3pi/2;2pi)
I see no easy way to include i and -i, however...
f(f(x))=f(x)^2 is just equivalent to
f(y)=y^2 for all y in the image of f. Examples:
f(x)=x^2
f(x)=0
f(x)=0 if x is rational but not equal to 1, f(x)=1 otherwise
f(x)= 2^(2n) if x=2^n for some integer n, f(x)=0 otherwise
f(x)=x^2 if x is not negative, f(x)=-12345x+23pi^e otherwise ;)
I’m halfway through wondering when tf golden ratio gonna show up
Homeworks pretty obvious. Both are to find the "half function" of xⁿ, where n=-1 in the first case and n=2 in the second. In each case, it's just x^√n. : First one was my "easy" solution to the original problem before I noticed it was f(x) rather than f(z). f(z) = zⁱ. This means f(f(z)) = zⁱⁱ = z⁻¹ = 1/z. [Second solution is z⁻ⁱ, so a general solution is presumably a strange combination of the two]. Second, f(x) = x^√2. Then f(f(x)) = (x^√2)^√2) = x^(√2*√2) = x².
f(f(x)) should be equal to [f(x)]^2, not to x^2.
@@ricardocavalcanti3343 In the case given, f(x) = x^√2, so f(f(x)) = (x^√2)^√2 = x^(√2*√2) = x²
@@SlidellRobotics But the proposed problem was f(f(x)) = [f(x)]^2, not f(f(x)) = x^2. x^√2 is a solution to the latter, not to the former.
Wow!! How can I learn to solve these Cool stuff? I am in grade 10 and not "groomed" in competition math.
You state that the following are equivalent: "f^{-1}(x) = 1/(f(x)) f(f(x)) = 1/x".
While I see why "==>" holds, maybe someone could help me why "
Let f(x) = y; then x = f^{-1}(y), so f(f(x)) = 1/x => f(y) = 1/f^{-1}(y) => f^{-1}(y) = 1/f(y) => f^{-1}(x) = 1/f(x).
Oh, just every now and then.
What is the units digit of 2^2021. Uganda math contest
(1) 2^5 = 32 = 2 (mod 10); (2) 2021 = 3x5^4 + 5^3 + 4x5 + 1; therefore (3) 2^2021 = (2^{5^4})^3 x 2^{5^3} x (2^5)^4 x 2^1 = 2^3 x 2 x 2^4 x 2 (mod 10) = 2^9 (mod 10) = 512 (mod 10) = 2 (mod 10).
캬
규
Nope, it reflects from the blackboard and makes the text less contrast.
f(x)= x^i works
I'm baffled. U and V both appear to be countable sets but (0, infinity)/~ is uncountable... have I missed something?
While the examples were written as an infinite list, I think the point there was to show that they could be any arbitrary values. They should both be uncountable.
I had a go at this before watching the more-complicated-than-I-expected explanation. I managed to confuse myself at one point, and I'm clearly wrong, but could someone please explain to me why this doesn't work? (Hoping I'm not being too dumb... :)
If f(f(x)) = 1/x, then f(-1)( f(f(x)) ) = f(-1)(1/x). The LHS = f(x) and the RHS = 1/f(1/x) by the original condition. So f(x) = 1/f(1/x), or f(1/x) f(x) = 1. That's obviously not right, because (for example) f(x) = x^2 satisfies that condition but not f(f(x)) = 1/x.
It's a one-way implication: f(f(x)) = 1/x does imply f(1/x) f(x) = 1, but the reverse is not true.
z^(i) and z^(-i) should be the complex solutions
if f(f(x)) = 1/x and f is differentiable, then (f(f(x)))' = f'(f(x))*f'(x) = ln(x). I wonder if it would get us any far.
(1/x)' = -1/x^2, not ln(x).
Wait what... how did the ln come in?
@@ricardocavalcanti3343 oh, right... :)
Title has a typo
Perfect, fixed now
Am I the only one that felt very dissatisfied with the function being piece wise? Such a disappointing result after the glorious work to get there.
Lighting must be good. I can see every chalk mark.
Nice climbing
I wonder if there is a function satisfying f(f(x)) = f(x)
Two examples: (1) f(x) = constant; (2) f(x) = x.
One answer to the second question is obviously x². That's not satisfying, but there it is.
@@ricardocavalcanti3343 shit, I misread
f(x)=i*x is an example
Nope. For that, F^(-1)(x) = -i*x and 1/f(x) = -i/x. Clearly not equal.
@@justacutepotato2945 you're right. Thank you 👍🏻
x^(i)
That's wasn't easy :)
I'm going to say x^i and x^sqrt(2).
Why x^sqrt(2)? If f(x) = x^sqrt(2), then f^(-1)(x) = x^(1/sqrt(2)) and 1/f(x) = x^(-sqrt(2)). Clearly they are not equal functions.
@@justacutepotato2945 Sorry, that part was for f(f(x)) = x².
@@tomkerruish2982 oh i see. My bad. 👍
@@justacutepotato2945 Don't worry. I just make my posts way too terse.
I should've added x^-i and x^-sqrt(2) as additional functional square roots of 1/x and x².
I'm not sure whether you actually use the axiom of choice for your choice function. The axiom of choice only yields existence, but this should be clear, since you can choose e.g. C([x]) := max [x] = max {x, 1/x}. Maybe one needs the AOC to prove that the max-function exists on this subdomain of finite subsets of P(R).
However, as I understood your reasoning, You only state that any such choice function yields a desired function f. But this means that you show "C choice function ==> f has property", meaning at this point you don't even need existence.
I smell projective geometry :)
For me, a good place to stop was 1 min into the video.
wtf did I just watch - way above my head unfortunately
The problem is if y=f(x) And suppose that f is of class (at least) c¹ and apply f to both sides and differentiate u get the weird differential equation : 1= -y'²/y² which is impossible for a real valued function
This is a bit beyond my knowledge of math.
very sexy lighting