This presentation is fantastic. Not a dull moment, and it is performed with humor. I have never seen such a beauty in demonstrating a mathematical truth. Thank you from my whole heart.
For all those asking why 4 isn't the solution, when it it a solution to √2^x = x, it can be shown that each term of the sequence {a_n} that defines the infinite super power is less than 2 with a simple induction argument. To see this, define g(x) = √2^x. Note that a_(n+1) = g(a_n). The induction proof begins with a_1 = √2 < 2. Now suppose a_n < 2. We need to show that a_(n+1) < 2. First notice that a_(n+1) = √2 ^ (a_n). Also let me remind you that g(x) = √2^x is an increasing exponential function. With a_n < 2, then g(a_n) < g(2). That is, a_(n+1) < √2^2 = 2. Note, one can also show that {a_n} is increasing as an induction proof, which I will omit here as it is a straight forward proof using the fact that g(x) is an increasing function. So {a_n} is a monotonic increasing function bounded by 2. Therefore it converges. Its limit must be ≤ 2. This would rule out 4 as a possible solution. I get why he didn't include all this, as he wanted to streamline the presentation over being very meticulous an losing a good portion of the audience.
Indeed but our host has shown that x=2 is "a" solution to the problem, not that there may be "other" solutions. We might call it the _principle_ solution if you want to be semantic. Just like the √ 4 =2 , is usually what my students write, but we all know that √ 4 ={2, -2}. BTW the first step in you inductive proof was a_1 = √2 < 2, could easily have started a_1 = √2 < 4?
@@tomctutor sqrt is a well-defined function that is the inverse of x -> x² for x > 0 so in fact sqrt(4)=2, not sqrt(4)={2,-2}. If you teach your students that sqrt(x) has multiple solutions for a real x, you're teaching them something wrong. It is correct though that {-2,2} is the set of solutions to the equation x²-4=0. This doesn't mean that sqrt isn't well-defined, though.
@@nigerianprinceajani I know you are talking about Reals, but lets extend this to complex numbers for a moment, whats the √1? =1^(1/2)=e^i(0+2nπ)/2=e^i(nπ)={-1,1} √4= 2e^i(2nπ)/2={-2,2} I know what you are saying though, if we interpret the radical √, meaning the positive root then you would be correct. But, unless you are wanting to abandon √4 ☰ 4^(1/2) and it's extension into complex field, then you can't ignore the -ve branch of the root. All that said there is nothing wrong with asserting √x ≥ 0 ( only the principle branch) as also there is nothing wrong with choosing a ≤√x ≤ b for some arbitrary choice of real interval [a,b], thus a restriction on your domain, which you need to do in some geometrical problems.
I just discovered your channel from your popular tetration video! Thank you for making all these videos, I'll definetly be using them in the future for learning math
After experimenting with tetration, the e root of e seems to be the limit of infinite tetration. Any number higher than this infinitely tetrated will grow till infinity, whereas the e root of e infinitely tetrated settles at the constant e. Very interesting stuff. Thank you Mr. Prime!
@mari-with-a-gun It was fairly straight forward to show that what you say (that is, that the tetration converges to n) is true for 1e) there are two solutions to the equation x^(1/x)=n^(1/n), and there exists 1
2/20/24 I have a 69 year old brain and I'm having some concerning brain issues. This topic intrigues me, I know nothing about tetration. I never saw the concept. Your video shows that you are a teacher. With that, I am certain. Thank you for this, I'll watch it over and over as well as watching related vidoes. I will understand this. I plan to beat down cognitive issues using leaning events. Thank you for pushing me toward this path. It's a good one
For simplicity let’s denote root(2)=r. Now, by r^r^r^r^… we mean the value that the following sequence converges to: a_1 = r a_2 = r^r a_3 = r^r^r … a_{n+1} = r^(a_n). First of all, this sequence is monotonically increasing and bounded from above by 2. a_1 = r < 2. a_n < 2, so a_{n+1}=r^(a_n)
WAY to complicated a solution for such a simple problem. Just do the following: √2 = 2^(1/2) √2^√2 = 2^(1/2 * 2)^(1/2) = 2^1^(1/2) But 1^x is always 1, therefore √2^√2 = 2^1 = 2 It doesn't matter, how many √2 you stack on top of each other. Anything beyond the second √2 can simply be ignored, as it is just an exponent for 1.
@@fullfungo Where exactly did I say that? Please check the math presented before you say something that simply isn't true. I replaced ALL √2 with 2^(1/2), so there is no 1/2*√2 anywhere in my calculations. There is only 1/2 * 2 = 1, which you should agree is correct.
@@m.h.6470 as you said √2 = 2^1/2 So √2^√2 = 2^(1/2•√2) which is *NOT* 2^1. For some weird reason you concluded that 1/2•√2 = (1/2•2)^1/2, but this is not correct. 1/2•√2 = 1/2•(2^1/2) which is completely different. You would know this if you weren’t so smug and just used a calculator √2^√2 = 1.6325…
I love ur videos!! As a high school student, I would appreciate it sm if u could do a video on Calculus introduction :) I wish all my maths lessons were like urs haha
Wow! I just stumbled across your videos yesterday and i don't know why but they reminded me immediately on "Bob Ross and the Joy of Painting". I adore the way you present the topics with passion and joy. Just fascinating! You should go to high schools and teach young people maths: They'll start loving it :-). Greetings from Germany. Alex.
is the idea that the powers are raised from the right to left? if so this makes sense. Imagine this: the rightmost sqrt2 is raised to the sqrt2, then sqrt2^(sqrt2^sqrt2), keep in mind that every time, you end up raising a sqrt2 to a power less than 2, I say this because sqrt2^sqrt2 < 2 because sqrt2^n where n is less than 2 is less than 2, just logic, and we repeat this process: sqrt2^(sqrt2^(sqrt2^sqrt2)) keep in mind that the inner bracket are less than 2, and as such the outer brackets are less than 2, thus this pattern continues, and the power approaches 2 as the reiteration approaches infinity. This is beautiful.
That is so amazing! I audibly gasped at so many moments during this video and your attitude towards math is so beautiful! 9:40 and a beautiful checkmate indeed.
Would have been worthwhile to check that there is a limit and that it is finite. Producing one upper bound would have been valuable, as the process produces an increasing sequence. An upper bound of 16 should be producible by induction.
Earlier I made a comment because I made the classic order-of-operations-mistake. Rather than deleting I edited it to say what I am saying now in case anyone read the comment earlier.
Why is this the case? I see what you mean by having an upper bound and using induction, but if I solve the same way for the infinite power tower of (3)^(1/3), I get 3 as the answer. However if I plug my answer into a calculator, I don't get 3. I know that working with infinity, weird things can happen, but where in the steps does this break down and not work?
An exceptional articulation on the stuff. The way it is being presented it's really awsome.Though I am a retired person (Petroleum Exploration & Production Business) with Chemistry background I have become regular viewer of this channel ever since I discovered it in UA-cam.By viewing this ,I recall all those good memories of doing these sums with great efforts while in our college days.Thanks to Prime Newtons.... and we do hope that it will keep on educating us
I give a like to all your videos even before I see everything, cuz I know how good it's gonna be. In this one, I understood the math, but I still opened an excel sheet to test it out and check :D
In right hand side of the equation the number is 2. So according to the question we have to show the value of x which satisfy the equation. We are not here to find the possible value of that equation. I think that's what he tried to prove here.
When I make videos. I try to make it as easy to understand as possible. I know the tetration converges to 2. I just needed to show the only valid answer. It is like solving for the sides of a triangle. If you get +2 and -3, you only keep the valid answer, in this case +2. Hope this clarifies it.
I also found out that sqrt(2) ^ 4=4 holds and I was wondering where it a step which is technically illegal. Turned out that this is the second from the end step. If W-function of something equals to W-function of something else, these somethings aren't necessarily equal. (W(x) = W(y) does not mean x=y)
As I said above, W delivers 2 values in that range! W_0(y) = - ln 2 and W_1(y) = - ln 4. This implies the correct results 2 and 4. You may call 2 a principal branch result, just as for W() function, but I'd say, it's just not clever to use the W() function for real values < 0. The only value < 0, that doesn't have zero or two results, is W(- 1/e) = -1.
@@antonvakhitov9477 Actually, it's the other way around. W(x) = W(y) does mean x = y , but u*(e^u) = v*(e^v) does not automatically mean u = v . As a consequence, -ln(2)*e^[-ln(2)] = -ln(4)*e^[-ln(4)] does not automatically mean -ln(2) = -ln(4) Note: the lefthandside is -ln(2) * e^[-ln(2)] = = -ln(2) / e^[ln(2)] = -ln(2) / 2 while the righthandside is -ln(4) * e^[-ln(4)] = = -ln(4) / e^[ln(4)] = -ln(4) / 4 = -ln(2^2) / 4 = -2 * ln(2) / 4 = -ln(2) / 2 which matches the lefthandside.
J'aime beaucoup votre style (surtout la casquette), votre dynamisme. Et cerise sur le gâteau, votre diction est très claire et facile à suivre pour un non anglophone.
Hm, but why can't x=4? It is definitely a solution to the equation "rad2^x=x" and you lost that solution, when using the W-"function". (You seem to treat it as the inverse to the function f(x)=x*e^x which isnt bijective..) If you look at the graphs y=x and y=rad2^x they clearly intersect in exactly 2 points: x=2 and x=4.To determine which solution for x is correct you'd first have to define what you mean by infinite tetration. For example, you could define it as the limit of rad2^rad2^...^rad2 (n times) as n goes to infinity. In that case 2 seems to be the correct solution (you'd have to prove that though), but there might be other ways to define it and get 4 as an answer.
he also treated lnx as the inverse of e^x, negating all solutions which require ln(x) = ln(x+2πn) but thats not what he was proving, he was proving that √2 ^^ inf = 2, not 4
@@Bradley2016_ Well I dont think we need complex numbers here, and over the real numbers ln(x) IS the inverse to e^x, so I think thats fine. But his prove that it equals 2 just ignores 4 as a possible solution, he doesnt show, that it can't be 4. What he would have to do is show that rad2^rad2^...^rad2 (n times) approaches 2 as n goes to infinity (which seems to be the case). All his approach can show is that the answer is either 2 or 4, or indeed infinity, i forgot that possibility in my original comment: If x=rad2^^inf=inf then the equation "rad2^x=x" also holds.
@@Marcel-yu2fw no, im not disagreeing with you, that IS a solution, 3b1b has demonstrated it himself, but my point is that there are infinitely many solutions, hes trying to prove ONE of which, not two, not 786.5, ONE
For the purpose of calculating the infinite tetration for any positive real number, let a > 0, and define the sequence S(n) by S(1) = a, and S(n) = a^S(n-1) for n > 1. It turns out (to be proved below) that for 0 < a < (1/e)^(e), the sequence diverges and for a > e^(1/e), the sequence diverges to infinity. For (1/e)^(e) 0, that is, f is strictly increasing (2) f(0) = a and f(1) = a^a (3) f has one point of inflection, at xo = [ln(ln(1/a))]/ln(1/a)
Good video, thanks! The only thing, which somehow escaped from the discussion, is the fact that equation 2^(x/2)=x has two solutions: 2 and 4... which would be probably valid to address, at least, and also it somehow compromises the idea that sqrt(2) power infiniti has a fixed value, as it probably can't take 2 values the same time
Wow! I was actually using the lambert function here, but i didnt see it that way. I just had an equation with a lambert function that you had to conpute. That's really smart!
I love thissss. Is there any chance you can explain why the infinites are acctually the equal despite the minute difference? I just feel like since this proof depends on this fact, itd be great to understanding it. Is it similar logic as why 0.9999999......... =1?
I understand the use of W function. But for the youngsters who possess basic knowledge of log functions, there's an easier way. We have x ln (root 2)= ln(x) => x ln(2)^0.5=ln(x) =>x/2*ln(2)=ln(x) =>x ln(2)=2ln(x) This is a one on one function unless x itself is a function. So by observation x=2 Or rearrange it like X/ln(x) =2/ln(2) And we get the same result. Discovered this channel 5 minutes ago, but am hooked with your energy man. Great work 👌🏻
@@Notthatkindofdr Yup, you are right. I wrongly assumed it as a one on one function. My bad. X will still be equal to 2, since if the result is 4 then that's equal to 2^2. I have no idea how to explain this mathematically but the function will not be greater than or equal to 2^2.
I'm still early in my freshman year in a university, so I could be very wrong, but I wondered: could you simply after xln(sqrt(2)) = ln(x) divide both sides by x, so that you get ln(sqrt(2)) = (1/x)*ln(x), then write it ln(sqrt(2)) = ln(x^(1/x)) and by removing the natural log from both sides get sqrt(2) = x^(1/x), which is 2^(1/2) = x^(1/x)-where you get x = 2 trivially?
@@cwldoc4958 Oh yes, but since the solution can be showed to converge towards 2 and not even close to 4, x = 2 is the only valid result. The same explanation was omitted from the main video, too.
Pay attention. W(-ln(sqrt2)) gives two results, not just one. Since -ln(sqrt2))=1/2 ln(1/2), it follows that W(-ln(sqrt2))=ln(1/2). On the other hand, since -ln(sqrt2))=1/2 ln(1/2))=1/4 ln(1/4), it also follows that W(-ln(sqrt2))=ln(1/4). Hence, actually the equation (sqrt2)^x=x has two solutions which are x=2 and x=4. Nevertheless, the infinite tetration of sqrt2 cannot be equal to 4 because the sequence a1=sqrt2, an=(sqrt2)^an-1 is increasing and an
I’ve only found this channel an hour ago but he’s making me love math more and more every minutes
same, yet i always hated math and found it uninteresting. hes the only person that can explain math that is interesting to me.
@kevinclark3086 oh math and music are both great :D
Same here 😅
This presentation is fantastic. Not a dull moment, and it is performed with humor. I have never seen such a beauty in demonstrating a mathematical truth. Thank you from my whole heart.
For all those asking why 4 isn't the solution, when it it a solution to √2^x = x, it can be shown that each term of the sequence {a_n} that defines the infinite super power is less than 2 with a simple induction argument.
To see this, define g(x) = √2^x. Note that a_(n+1) = g(a_n).
The induction proof begins with a_1 = √2 < 2. Now suppose a_n < 2. We need to show that a_(n+1) < 2. First notice that a_(n+1) = √2 ^ (a_n). Also let me remind you that g(x) = √2^x is an increasing exponential function. With a_n < 2, then g(a_n) < g(2). That is, a_(n+1) < √2^2 = 2.
Note, one can also show that {a_n} is increasing as an induction proof, which I will omit here as it is a straight forward proof using the fact that g(x) is an increasing function.
So {a_n} is a monotonic increasing function bounded by 2. Therefore it converges. Its limit must be ≤ 2. This would rule out 4 as a possible solution.
I get why he didn't include all this, as he wanted to streamline the presentation over being very meticulous an losing a good portion of the audience.
Indeed but our host has shown that x=2 is "a" solution to the problem, not that there may be "other" solutions. We might call it the _principle_ solution if you want to be semantic.
Just like the √ 4 =2 , is usually what my students write, but we all know that √ 4 ={2, -2}.
BTW the first step in you inductive proof was a_1 = √2 < 2, could easily have started a_1 = √2 < 4?
@@tomctutor sqrt is a well-defined function that is the inverse of x -> x² for x > 0 so in fact sqrt(4)=2, not sqrt(4)={2,-2}. If you teach your students that sqrt(x) has multiple solutions for a real x, you're teaching them something wrong.
It is correct though that {-2,2} is the set of solutions to the equation x²-4=0. This doesn't mean that sqrt isn't well-defined, though.
@@nigerianprinceajani sqrt(4) =/= -2 then?
@@tomctutor Indeed.
@@nigerianprinceajani I know you are talking about Reals, but lets extend this to complex numbers for a moment, whats the √1?
=1^(1/2)=e^i(0+2nπ)/2=e^i(nπ)={-1,1}
√4= 2e^i(2nπ)/2={-2,2}
I know what you are saying though, if we interpret the radical √, meaning the positive root then you would be correct.
But, unless you are wanting to abandon √4 ☰ 4^(1/2) and it's extension into complex field, then you can't ignore the -ve branch of the root.
All that said there is nothing wrong with asserting √x ≥ 0 ( only the principle branch) as also there is nothing wrong with choosing a ≤√x ≤ b for some arbitrary choice of real interval [a,b], thus a restriction on your domain, which you need to do in some geometrical problems.
The flat cap, the mesmerizing accent, the theatrics, the MATH, this guy is perfect
your channel is very underrated Mr. Prime. I respect you more than I do my actual math teacher
Your math teacher deserves all the respect. I appreciate the compliment, though.
I just discovered your channel from your popular tetration video! Thank you for making all these videos, I'll definetly be using them in the future for learning math
After experimenting with tetration, the e root of e seems to be the limit of infinite tetration. Any number higher than this infinitely tetrated will grow till infinity, whereas the e root of e infinitely tetrated settles at the constant e. Very interesting stuff. Thank you Mr. Prime!
The other bound is 1/(e^e) before things get unstable :)
yes, this is due to the derivative of the graph x^x or x^-x I can't remember which
I don’t have a formal proof but I think the infinite tetration of the nth root of n always converges to n (and the max of n root n is e root of e)
@mari-with-a-gun It was fairly straight forward to show that what you say (that is, that the tetration converges to n) is true for 1e) there are two solutions to the equation x^(1/x)=n^(1/n), and there exists 1
@@mari-with-a-gun I believe your statement holds for 1
2/20/24 I have a 69 year old brain and I'm having some concerning brain issues. This topic intrigues me, I know nothing about tetration. I never saw the concept. Your video shows that you are a teacher. With that, I am certain. Thank you for this, I'll watch it over and over as well as watching related vidoes. I will understand this. I plan to beat down cognitive issues using leaning events. Thank you for pushing me toward this path. It's a good one
For simplicity let’s denote root(2)=r.
Now, by r^r^r^r^… we mean the value that the following sequence converges to:
a_1 = r
a_2 = r^r
a_3 = r^r^r
…
a_{n+1} = r^(a_n).
First of all, this sequence is monotonically increasing and bounded from above by 2.
a_1 = r < 2.
a_n < 2, so a_{n+1}=r^(a_n)
WAY to complicated a solution for such a simple problem.
Just do the following:
√2 = 2^(1/2)
√2^√2 = 2^(1/2 * 2)^(1/2) = 2^1^(1/2)
But 1^x is always 1, therefore
√2^√2 = 2^1 = 2
It doesn't matter, how many √2 you stack on top of each other.
Anything beyond the second √2 can simply be ignored, as it is just an exponent for 1.
That's the correct solution
@@m.h.6470 did you seriously just say that 1/2•√2 = 1???
Please check your math before you call my method “WAY to [sic] complicated”.
@@fullfungo Where exactly did I say that? Please check the math presented before you say something that simply isn't true.
I replaced ALL √2 with 2^(1/2), so there is no 1/2*√2 anywhere in my calculations. There is only 1/2 * 2 = 1, which you should agree is correct.
@@m.h.6470 as you said √2 = 2^1/2
So √2^√2 = 2^(1/2•√2) which is *NOT* 2^1.
For some weird reason you concluded that 1/2•√2 = (1/2•2)^1/2, but this is not correct.
1/2•√2 = 1/2•(2^1/2) which is completely different.
You would know this if you weren’t so smug and just used a calculator √2^√2 = 1.6325…
The bob ross of maths. Calling Phi a flower is just poetic.
I love ur videos!! As a high school student, I would appreciate it sm if u could do a video on Calculus introduction :) I wish all my maths lessons were like urs haha
Wow! I just stumbled across your videos yesterday and i don't know why but they reminded me immediately on "Bob Ross and the Joy of Painting". I adore the way you present the topics with passion and joy. Just fascinating! You should go to high schools and teach young people maths: They'll start loving it :-). Greetings from Germany. Alex.
is the idea that the powers are raised from the right to left? if so this makes sense. Imagine this: the rightmost sqrt2 is raised to the sqrt2, then sqrt2^(sqrt2^sqrt2), keep in mind that every time, you end up raising a sqrt2 to a power less than 2, I say this because sqrt2^sqrt2 < 2 because sqrt2^n where n is less than 2 is less than 2, just logic, and we repeat this process: sqrt2^(sqrt2^(sqrt2^sqrt2)) keep in mind that the inner bracket are less than 2, and as such the outer brackets are less than 2, thus this pattern continues, and the power approaches 2 as the reiteration approaches infinity. This is beautiful.
That is so amazing! I audibly gasped at so many moments during this video and your attitude towards math is so beautiful! 9:40 and a beautiful checkmate indeed.
Would have been worthwhile to check that there is a limit and that it is finite. Producing one upper bound would have been valuable, as the process produces an increasing sequence. An upper bound of 16 should be producible by induction.
Earlier I made a comment because I made the classic order-of-operations-mistake. Rather than deleting I edited it to say what I am saying now in case anyone read the comment earlier.
Why is this the case? I see what you mean by having an upper bound and using induction, but if I solve the same way for the infinite power tower of (3)^(1/3), I get 3 as the answer. However if I plug my answer into a calculator, I don't get 3.
I know that working with infinity, weird things can happen, but where in the steps does this break down and not work?
An exceptional articulation on the stuff. The way it is being presented it's really awsome.Though I am a retired person (Petroleum Exploration & Production Business) with Chemistry background I have become regular viewer of this channel ever since I discovered it in UA-cam.By viewing this ,I recall all those good memories of doing these sums with great efforts while in our college days.Thanks to Prime Newtons.... and we do hope that it will keep on educating us
I give a like to all your videos even before I see everything, cuz I know how good it's gonna be.
In this one, I understood the math, but I still opened an excel sheet to test it out and check :D
You brought me back to 40 years to my university ages. A big smile on my face. Thank you. You are very good at teaching btw. 👏👏👏
It's so nice an explanation! Really good job, my friend!
Love your enthusiasm! Great video
Good manipulation.
Good articulation.
Thank you for reasonable moving bit by bit (w/t the charcoal, but not ink).
Thanks for your style & pace.
The equation (sqrt(2))^x = x has two solutions which are 2 and 4. Then why does the infinite tetration of the square root of two does not equal 4?
In right hand side of the equation the number is 2. So according to the question we have to show the value of x which satisfy the equation. We are not here to find the possible value of that equation. I think that's what he tried to prove here.
When I make videos. I try to make it as easy to understand as possible. I know the tetration converges to 2. I just needed to show the only valid answer. It is like solving for the sides of a triangle. If you get +2 and -3, you only keep the valid answer, in this case +2. Hope this clarifies it.
I also found out that sqrt(2) ^ 4=4 holds and I was wondering where it a step which is technically illegal.
Turned out that this is the second from the end step. If W-function of something equals to W-function of something else, these somethings aren't necessarily equal. (W(x) = W(y) does not mean x=y)
As I said above, W delivers 2 values in that range! W_0(y) = - ln 2 and W_1(y) = - ln 4. This implies the correct results 2 and 4. You may call 2 a principal branch result, just as for W() function, but I'd say, it's just not clever to use the W() function for real values < 0. The only value < 0, that doesn't have zero or two results, is W(- 1/e) = -1.
@@antonvakhitov9477 Actually, it's the other way around. W(x) = W(y) does mean x = y , but u*(e^u) = v*(e^v) does not automatically mean u = v .
As a consequence,
-ln(2)*e^[-ln(2)] = -ln(4)*e^[-ln(4)]
does not automatically mean
-ln(2) = -ln(4)
Note: the lefthandside is
-ln(2) * e^[-ln(2)] =
= -ln(2) / e^[ln(2)]
= -ln(2) / 2
while the righthandside is
-ln(4) * e^[-ln(4)] =
= -ln(4) / e^[ln(4)]
= -ln(4) / 4
= -ln(2^2) / 4
= -2 * ln(2) / 4
= -ln(2) / 2
which matches the lefthandside.
That handwriting is so gorgeous
The presentation is way more interesting than the problem itself
J'aime beaucoup votre style (surtout la casquette), votre dynamisme. Et cerise sur le gâteau, votre diction est très claire et facile à suivre pour un non anglophone.
Knowledge is there for everyone to grab. But it takes a great teacher to guide you where to find it.
Thank you for your outstanding style of teaching.
guy, you are so charismatic !!
sir, the thing you did there was just unbelievable at all meanings. WOW. that's mindblowing thank you so much
You know what, I didn't listen your class. but I like your expression from your face so got a subscribe!
What a beautiful demonstration!
Hm, but why can't x=4? It is definitely a solution to the equation "rad2^x=x" and you lost that solution, when using the W-"function". (You seem to treat it as the inverse to the function f(x)=x*e^x which isnt bijective..) If you look at the graphs y=x and y=rad2^x they clearly intersect in exactly 2 points: x=2 and x=4.To determine which solution for x is correct you'd first have to define what you mean by infinite tetration. For example, you could define it as the limit of rad2^rad2^...^rad2 (n times) as n goes to infinity. In that case 2 seems to be the correct solution (you'd have to prove that though), but there might be other ways to define it and get 4 as an answer.
he also treated lnx as the inverse of e^x, negating all solutions which require ln(x) = ln(x+2πn)
but thats not what he was proving, he was proving that √2 ^^ inf = 2, not 4
@@Bradley2016_ Well I dont think we need complex numbers here, and over the real numbers ln(x) IS the inverse to e^x, so I think thats fine.
But his prove that it equals 2 just ignores 4 as a possible solution, he doesnt show, that it can't be 4. What he would have to do is show that rad2^rad2^...^rad2 (n times) approaches 2 as n goes to infinity (which seems to be the case).
All his approach can show is that the answer is either 2 or 4, or indeed infinity, i forgot that possibility in my original comment: If x=rad2^^inf=inf then the equation "rad2^x=x" also holds.
@@Marcel-yu2fw no, im not disagreeing with you, that IS a solution, 3b1b has demonstrated it himself, but my point is that there are infinitely many solutions, hes trying to prove ONE of which, not two, not 786.5, ONE
Why does Arsene Lupin have a UA-cam channel now
@@Bradley2016_i was thinking the same, you’re able to make a divergent series appear to be a lot of different solutions with various manipulations
i love your vids so much friend
I love specially the use of a tradicional blackboard❤ +1 subscriptor!!!
Glad you like it!
Wow man you just revived my love for maths. Love you man.
I love this guy's energy
ok tetrations are useful for taking the limits of tetrations got it. also convenient I got recommended this after ur lamber function video
What a wonderful lesson!
For the purpose of calculating the infinite tetration for any positive real number, let a > 0, and define the sequence S(n) by S(1) = a, and S(n) = a^S(n-1) for n > 1.
It turns out (to be proved below) that for 0 < a < (1/e)^(e), the sequence diverges and for a > e^(1/e), the sequence diverges to infinity. For (1/e)^(e) 0, that is, f is strictly increasing
(2) f(0) = a and f(1) = a^a
(3) f has one point of inflection, at xo = [ln(ln(1/a))]/ln(1/a)
this was great im surprised that i understood every line. if this is how you teach classes your students are lucky to have you.
What a joy. He is to mathematics as Bob Marley was to music.
Channel is worth subscribing 😊
This is a really surprising fact and an interesting proof; thank you for sharing!
Beautifully done
Good video, thanks! The only thing, which somehow escaped from the discussion, is the fact that equation 2^(x/2)=x has two solutions: 2 and 4... which would be probably valid to address, at least, and also it somehow compromises the idea that sqrt(2) power infiniti has a fixed value, as it probably can't take 2 values the same time
this is the most beautiful video ive ever had the privilege of seeing
"If you're still with me, give me a thumbs up"
Me, who has not understood one thing since the very beginning: *thumbs up*
What kind of mathematical magnificence did I just witness?
I love your channel !
Thank you so much!
Damn when i realised where its going this blew my mind
Great stuff! Thanks for sharing.
You've got to be a great father!
This is a nice content. You earned a subscriber 😊.
Welcome aboard!
Great. You're the TEACHER. A huge hug, to u & thanks a lot.....
Wow! I was actually using the lambert function here, but i didnt see it that way. I just had an equation with a lambert function that you had to conpute. That's really smart!
You have beautiful handwriting
Thank you so much 😀
I had an absolute joy watching this video!
I love thissss. Is there any chance you can explain why the infinites are acctually the equal despite the minute difference? I just feel like since this proof depends on this fact, itd be great to understanding it. Is it similar logic as why 0.9999999......... =1?
I understand the use of W function. But for the youngsters who possess basic knowledge of log functions, there's an easier way.
We have x ln (root 2)= ln(x)
=> x ln(2)^0.5=ln(x)
=>x/2*ln(2)=ln(x)
=>x ln(2)=2ln(x)
This is a one on one function unless x itself is a function. So by observation
x=2
Or rearrange it like
X/ln(x) =2/ln(2)
And we get the same result.
Discovered this channel 5 minutes ago, but am hooked with your energy man. Great work 👌🏻
But your final equation also has the solution x=4. That is, 4/ln(4) = 2/ln(2). So does x=4 or x=2?
@@Notthatkindofdr Yup, you are right. I wrongly assumed it as a one on one function. My bad. X will still be equal to 2, since if the result is 4 then that's equal to 2^2. I have no idea how to explain this mathematically but the function will not be greater than or equal to 2^2.
Very good lecture Sir. Thanks 🙏🙏🙏🙏🙏🙏
How is he able to make me so interested in mathematics again. It’s almost 00:00 am and i was just watching this for fun
Though it's irrelevant, I kind of want to hear him say "YEAH BOI"
so glad YT recommended this to me!
never stop learning...you bring life back
Beautiful!
Just fabulous!
Marvelous presentation
I'm still early in my freshman year in a university, so I could be very wrong, but I wondered: could you simply after xln(sqrt(2)) = ln(x) divide both sides by x, so that you get ln(sqrt(2)) = (1/x)*ln(x), then write it ln(sqrt(2)) = ln(x^(1/x)) and by removing the natural log from both sides get sqrt(2) = x^(1/x), which is 2^(1/2) = x^(1/x)-where you get x = 2 trivially?
Yes!
@@PrimeNewtons Ok, thank you!
2^(1/2)=x^(1/x) does not imply x=2, because x=4 is also a solution.
@@cwldoc4958 Oh yes, but since the solution can be showed to converge towards 2 and not even close to 4, x = 2 is the only valid result. The same explanation was omitted from the main video, too.
absolute beautiful
Please will you make playlist of operations research begginer to advance all topics , it will be a huge help for students
まじ分かりやすい。
Pay attention.
W(-ln(sqrt2)) gives two results, not just one.
Since -ln(sqrt2))=1/2 ln(1/2), it follows that W(-ln(sqrt2))=ln(1/2).
On the other hand, since -ln(sqrt2))=1/2 ln(1/2))=1/4 ln(1/4), it also follows that
W(-ln(sqrt2))=ln(1/4).
Hence, actually the equation (sqrt2)^x=x has two solutions which are x=2 and x=4.
Nevertheless, the infinite tetration of sqrt2 cannot be equal to 4 because the sequence a1=sqrt2, an=(sqrt2)^an-1 is increasing and an
You're fantastic man!
Thanks, you too!
You are a great teacher !!!
Great teaching Sir🙏🙏🙏
Your blackboard writing is very good
He taught good lesson in new ways.
Wow I love your tutorials
As a math teacher, the misspeak at 6:30 is so relatable. 😂
Incredible !!
Thank you ❤ keep giving!
You are very gifted in teaching.
Beautiful solve.
"what does it cost to give your neighbor a dollar"
none cause i can have it back after hes done using it
Found you minutes ago and you're..................... *cannot express *respect
Muito bom!❤
He maths like a pro wrestler.
The last line!!✨
If only I had found your channel earlier
Bonjour professeur. Je suis fan des mathématiques, c est mon sport préféré. J aime beaucoup votre façon de nous enseigner cet art.
Beautiful
I believe we first need to discuss whether the tetration converges rather than diverges before letting x = tetration, and that's not so trivial.
Great video!👍
Nice demonstration of the W-function, though @4:06 it is easy to see that x=2 since sqrt(2)^2=2.
Nice Lecture
you sound like some mad scientist endlessly scribbling incoherent thoughts on a chalkboard and then getting a revolutionary idea
I love it!
Prime Newtons: flower
BPRP: fish
Where was this channel man, i found it late
Just here to comment before this video became a freaking famous
I love this guy