Could you just use the sigmoid function as a mapping from all real numbers to the interval (0,1)
@drpeyam7 жыл бұрын
Xander Gouws That would work too!
@TheRealSamSpedding7 жыл бұрын
Very good, would like to see more analysis videos like these in the future.
@TheRedfire217 жыл бұрын
Great! More topology and set theory please this are great
@jeromesnail6 жыл бұрын
Great video as usual! The notation (0,1) is really confusing IMHO, it can be mixed with the ordered pair (0,1) if not enough context is given. In France we use the notation ]0,1[ for open intervals and [0,1] for closed ones. I wonder where else those notations are used in the rest of the world :)
@drpeyam6 жыл бұрын
I agree, I went to a French school as well :P
@jeromesnail6 жыл бұрын
-Also couldn't we have simply studied the function (2/π)*tan(x) ?- Ignore my question it was a brain fart...
@jeromesnail6 жыл бұрын
Dr. Peyam's Show ahah I saw the video where you talk in like 10 different languages, you're really awesome!!
@theoleblanc97616 жыл бұрын
Personally, (I am french) and every time I see (1,0) I don't known if it is [0;1] or ]0;1[
@eleonard0o4 жыл бұрын
In Peru we use for open intervals and [0,1] for closed intervals. Some of us use the same notation you mentioned :')
@adumont4 жыл бұрын
I love all your videos Dr PiM. It reminds me when I studied Math Spe M' at Ginette, Versailles, back in 1993-1995.
@drpeyam4 жыл бұрын
Spé Maths 🥰 Good old times
@adumont4 жыл бұрын
@@drpeyam thanks I studied that I can enjoy your videos! They are like little pills of Spe Math. Refreshers. I am usually nostalgic of what I remember I managed to learn and actually understand at that time, and now so long after I feel like an idiot, not remembering nor using any of it... So this pills 💊 of Math you give us are very welcome! Thanks. And I like your style.
@ShubhayanKabir4 жыл бұрын
So glad you quoted "Eat, Pray, Love". That was really romantic. 😍
@supertren5 жыл бұрын
Please Dr: a brief introduction of Topology will be great!.
@drjoaoventura4 жыл бұрын
The sound of math, blackboard, and chalk is so relaxing... #ASMR
@RealEverythingComputers2 ай бұрын
Awesome tutorial! helped me out for a math proofs course. Wondering if there is an alternative method to this proof - for starters, is there another function aside from tan(x)?
@MrRyanroberson17 жыл бұрын
by 8:30 it's just like "now that we proved 1=2, we can prove 1=infinity", since 1=2=1+1, and 1=2, 1=4, 1=8, 1=2^(set of all integers), and also 1=2^infinity=infinity
@9wyn7 жыл бұрын
Good camera angle. 🙏
@drpeyam7 жыл бұрын
I’ll let the cameraman know :)
@Shadow47077 жыл бұрын
I agree.
@BuddyVQ3 жыл бұрын
Keep up the great work! I learned a lot!
@thesakinator76399 ай бұрын
can you not adjust the tan function so that its bounds can be 0,1 and have it approach -inf at 0 and inf at 1? such as tan(pi(x+0.5))
@Charm98thanks7 жыл бұрын
Amazing video as always! I just have a couple comments to point out some minor technicalities in this proof. First, when you proved that f was onto, you should also evaluate f(x) after finding x to prove that f(x)=y. You essentially still want to show that it actually equals y (the way it is written, you have proven that if an x exists such that f(x)=y, then it must take the given form; however you must also show that the given x actually satisfies the equation). Moreover, the derivative of tan(x) is equal to 0 at x=0, so the increasing argument (correct me if I'm wrong) does not hold. Perhaps a couple more theorems might have made this result more rigorous. I think this could have been shown with the definition fairly easily as done for f. Regardless, I love your videos and your enthusiasm. This was a great video, and I can't wait to see more of your brilliant, enthusiastic work!
@drpeyam7 жыл бұрын
PcRules1 Thanks so much, I really appreciate it! :) And actually when I solved for x, all the steps I did were reversible, so there’s no need to show f(x) = y, it automatically holds in this case! And sec^2 is always positive, so tan is strictly increasing!
@Charm98thanks6 жыл бұрын
Right! Thanks for the clarifications. I also for some reason thought sec^2(0) was 0 haha. My mistake! Keep up the excellent work :)
@TheMauror227 жыл бұрын
But what's so special about (0,1)? Because, in that sense its possible to show that an interval (a,b) with a and b contained on the real numbers and different from each other, is isomorphic with R, doesn't it?
@drpeyam7 жыл бұрын
You’re absolutely right, there’s nothing special about (0,1), any interval (a,b) would do!
@TheMauror227 жыл бұрын
Thank you Dr. Peyam! I love your videos, I always see them all!!
@wherestheshroomsyo6 жыл бұрын
Can we simplify this by first proving that, (0,1) is isomorphic to (-r,r) and then obtain any real number by allowing r to increase to any size? The limit as r -> infinity of (-r,r) approaches the set of real numbers by definition I think.
@drpeyam6 жыл бұрын
But then you’d have to prove that for any set if An goes to A (which you have to define rigorously), then the size of An goes to the size of A, which is a nontrivial Statement (and might not even be true)
@wherestheshroomsyo6 жыл бұрын
I think I see what you're saying. So, the size of the interval (-r,r) having the same size as the reals is nontrivial even if (-r,r) does approach the reals. It would need to be proven, but not using the theorem from your video or else I'd be trying to prove a theorem using itself. My idea seemed so solid though. proving (0,1) is isomorphic to (-r,r) is straightforward, it uses the same argument as the one from your video with different numbers. The parameterization I used was f(x)=-r+2rx The derivative is 2r, and r is positive so f is always increasing. f is continuous on the closed interval [0,1] so by the intermediate value theorem f attains all values from -r to r. So if r is an arbitrary positive real (0,1) could be mapped to any finite interval centered on 0. Is this formulation not strong enough to prove that (0,1) is isomorphic to the set of real numbers? Thanks Dr. Peyam!
@drpeyam6 жыл бұрын
That’s just half of the proof, basically the same thing as me showing that (0,1) has the same size as (-pi/2,pi/2). The nontrivial part is the second part, showing that the latter has the same size as R.
@drpeyam6 жыл бұрын
That’s the tricky thing about analysis btw, just because it’s true for every r doesn’t mean it’s true for R. For example, any continuous function has a maximum on [-r,r] for finite r, but this property is not true for R.
@wherestheshroomsyo6 жыл бұрын
I see, rigor is annoying sometimes. I thought that f being bijective and lim of f(0) -> -infinity and f(1) -> infinity as r -> infinity was good enough, but it needs something more. Can't wait for |R|=|R^n| video!
@renzalightning60086 жыл бұрын
Can you do a video showing examples or even proofs of when composite functions are bijective given the initial functions are injective and/or surjective? i.e. does an injective function composed with a surjective function end up as an overall bijective function?
@drpeyam6 жыл бұрын
It doesn’t, unfortunately, take f(x) = e^x, which is injective, and g(x) = x, which is surjective; then g o f is e^x, which is not bijective!
@renzalightning60085 жыл бұрын
@@angelmendez-rivera351 Why not both? XD That's an interesting one though, does the order of functions matter? I'd imagine it does.
@yugerten_a4 жыл бұрын
f(x)=ln(-ln(x)) is a bijection from (0,1) to R
@drpeyam4 жыл бұрын
Correct but tan is easier
@МаксимЮрченков-ы5ь7 жыл бұрын
(2x-1)/(1-abs(2x-1)) is easier, isn't it?
@drpeyam7 жыл бұрын
Both ways are equally nice :) Yours might be a bit harder to prove that it’s 1-1, but it’s doable!
@kamehamehaDdragon6 жыл бұрын
Epsilon is indeed like casting a spell!
@random199110045 жыл бұрын
A first I thought of just using logistic function (or the inverse of logistic). Was wondering why you went through changing it to (-pi/2 to pi/2). Logistic regression maps (-inf, inf) to a probability that is in (0,1). I guess they are both good.
@random199110045 жыл бұрын
I have a statistics background, not a pure math background.
@Sam_on_YouTube6 жыл бұрын
Haven't watched the video yet, but I'm noticing it is 20 minutes long. Can't you do this in 2 minutes by drawing a semicircle of length 1 and projecting rays from the center through it into the real number line?
@drpeyam6 жыл бұрын
Yep, that works too! The video would still be 20 mins long, though, if you want to prove everything rigorously!
@Sam_on_YouTube6 жыл бұрын
Dr. Peyam's Show If you want to do it rigourously, 2 minutes is too fast. But I don't think it would take that long. You don't need 2 separate functions that way, for example. You don't even need to know the function. Indeed, there are infinitely many of them depending on how far away you place the real number line. You just need to show that as the ray approaches 0 degrees and pi degrees, it approaches positive and negative infinity. Also, it is continuous and increasing. And since the length of the line is irrelevant, you can do the general case off the bat with no need to first do a specific length like 0 to 1.
@drpeyam6 жыл бұрын
Yeah, but actually if you think about it, it’s basically the same function, tan maps the semicircle to the whole real line :)
@Sam_on_YouTube6 жыл бұрын
Dr. Peyam's Show Yes. The advantage of your method is that you do get a known function out of it. The advantage of the other method is you don't need to. You can just show that a function exists.
@emanuellandeholm56572 жыл бұрын
Just a nitpick, but don't you also have to prove that the relation R(x, y) "x is isomorphic to y" is transitive, ie. that the composition of two bijections is also a bijection.
@emanuellandeholm56572 жыл бұрын
I googled "bijection of bijections" and I found a nice proof that, alas, I cannot fit into this youtube comment. It was like a two page pdf. I really appreciate that once you took a flight to enjoy a torus tho.
@RyanLucroy6 жыл бұрын
Well, first I wanted to ask, why you didn't use f(x) = - cot(π*x) till I found out that it's the same as tan(πx-π/2) which is the function you defined at the end :)
@rizkyagungshahputra2156 жыл бұрын
aleph 1 and |ℝ|, which one is bigger?
@drpeyam6 жыл бұрын
They’re the same :)
@yuvalpaz37527 жыл бұрын
I proved it using limits(i also showed that this is true for all open sets in the real numbers, excluding (a,a)) Let's construct a sequence a_n=(-n,n) (1)lim n-> infinity a_n=the real numbers (2)We can construct the function f(x)=kx+p, thus |(0,1)|=|(p,k+p)|, k=/=0. because k is arbitrary we have |(a,b)|=|(c,d)| for all a,b,c,d in the real numbers and a,b are not the same and c,d are not the same. (3)By (2) i know that |a_i|=|a_j| for all i,j (4)By (1) and (3) we have |the real numbers|=|lim n-> infinity a_n |=|a_i|. Using (4) and (2) again we have |the real numbers|=|a_i|=|(b,c)| which implies that |the real numbers|=|(b,c)| for all b,c in the real numbers and not the same
@drpeyam7 жыл бұрын
That’s a good idea BUT the issue is: how do you know that cardinality is preserved under limits? In other words, how do you know that |lim An| = lim |An| ? Stuff could break down at that point, unfortunately!
@yuvalpaz37527 жыл бұрын
the set {|a_n| ; for all n in the real numbers} is closed set, the sequence |a_n| is by definition inside of the set for all n in the real numbers therefore the limit is also inside of the set(i think this is enough to prove that)
@drpeyam7 жыл бұрын
Yuval Paz I’m not 100% sure, but consider for example f(A) = 1 if A is finite, and 0 if A is infinite, then f(An) = 1 but f(A) = 0, where An = (-n,n), and A = R, so you can’t really pass to limits like that even if the set of f(An) is closed
@kamoroso945 жыл бұрын
Hey Dr. Peyam, would f(x)=1/x-1/(1-x) work too?
@MrRyanroberson17 жыл бұрын
Begin: let the number 1 be in the set of natural numbers. Let all sums N1+N2 be in the set of natural numbers, for N1,N2 in natural numbers. Let all differences N1-N2 be in the set of integers. Let all quotients Z1/Z2 be in the set of rational numbers, for Z1,Z2 in integers and Z2 not = 0 Let all point values which divide the range of rational numbers into two sets of numbers respectively all greater or all not greater than R be in the set of reals. Let all sums a+bi be in the set of complex numbers, for a,b in the reals, and where i is the "principal" solution to x^2 +1=0, noting that -i should remain distinct from i.
@eipiwau6 жыл бұрын
Please use the terms injective for 1-1 and surjective for onto and bijective for both of them.
@RJYounglingTricking6 жыл бұрын
Loved it! Ty brother
@ericlizalde53627 жыл бұрын
Love also this type if videos
@theoleblanc97616 жыл бұрын
Easier: Use inverse tanh (For all x in ]0;1[, argtanh(x)=0.5ln((1+x)/(1-x))
@drpeyam6 жыл бұрын
Ummmm, not sure if it’s easier...
@gregorio88276 жыл бұрын
tan(pi × X) or something like that isnt it?
@gregorio88276 жыл бұрын
For the interval [a,b] it could be tan(pi ×((X-a)×b)) ?
@fadhlurrahmanal-akbar79462 жыл бұрын
thx for the explanation Dr Peyam (^.^)
@Hepad_7 жыл бұрын
Can we say that (0,1) = {floor(x)/x, x€R}?
@drpeyam7 жыл бұрын
I think that should be [0,1] because if x = 1/2, then 0 is in your set and if x = 1, then 1 is in your set.
@Hepad_7 жыл бұрын
Then what does (0,1) represent? I thought (0,1) and [0,1] meant the same thing
@willnewman97837 жыл бұрын
Hepad (0,1) does not include 0 and 1, while [0,1] does
@Hepad_7 жыл бұрын
Oh ok, I've always seen ]0,1[ for 0 and 1 excluded. Different countries, different ways.
@ekadria-bo49627 жыл бұрын
Proof that Cardinality of all dimension is same as (0,1)
@AvinashtheIyerHaHaLOL7 жыл бұрын
Enrian Sholeh I want that as well
@drpeyam7 жыл бұрын
By that do you mean that the cardinality of Rn is the same as R?
@AvinashtheIyerHaHaLOL7 жыл бұрын
Dr. Peyam's Show Yes
@ekadria-bo49626 жыл бұрын
yes, i think will realy cool. i have read some pdf on cantor's proof but i don't got it well...
@Jonathan_Jamps6 жыл бұрын
Dr.Payam's...you're awesome and looks like Rajesh of The Big Ben Theory hahaha...