Actually, the domain for love(u) is surely infinite, but the domain for love(x) is indeterminate. Some postulate that love(x)'s domain is the empty set, others say it's an infinite set of finite measures all over the complex plane, and then there are others still that say it should get real. Similar functions with a set of real values for x include loved(x)=true, past(love(u))=x, and future(drunk(text(x)))=regret.
It's worth noticing that for complex values every z satisfying z=ln(z) (there are infinitely many) is a solution. Would be a great idea for your next video!
It's amazing how it doesn't converge as a real function for any x, but converges as a complex function almost everywhere (oh, except e ^^ n for n = -2, -1, 0, 1, 2, ...) to -W(-1, -1) or 0.3181... + 1.3...i
It would have been nice if you had generalised fn(x) first before moving to the case of infinity. It appears to be the x > "the (n-2)th tetration of e", where n>=2. Then show that the "infinite tetration of e" diverges.
ChaosPod indeed or with that same n, it could be written in terms of functional composition: n-ln's composed together have a domain of (n-1)-e^x's composed together and evaluated at 0. Where n>=2 (or 1... depending on your compositional definition). There should also be an expression that can include all cases including zero though.
Earthbjorn Nahkaimurrao hmmm, tetration when defined using Knuth up-arrow notation is defined recursively, so the value of a -1 tetration would depend on an exponentiation where the exponent is the value of a -2 tetration (which would turn into an exponentiation with -3 =\ ) so we never end up with a base case: a 1 tetration Perhaps another definition allows us to define negative tetrations more usefully :)
Windsor Mason You’re wrong. 1. The notation does not define the operator itself. 2. The recursive relation necessarily implies that the -1 tetration of any number for which tetration is defined is 0.
I thought of it a different way: As you increase the ln(x) chain, the output keep decreasing because ln(x) < x. Because the range of ln(x) goes down to -infinity, we know that the ln(x) chain will at least get as low as e (though it doesn't have to hit e for every x). ln(x)
Solution for problem at the end: 0. First understand that you can square both sides and move the 6 to the other side to verify the expression. 1. Simplify the expression as N = Sqrt(x+N) where N is an integer. 2. Square expression as N^2-N-x = 0. 3. Solve for N using Quadratic Formula: N = 1/2+-sqrt(1+4x)/2 = 1/2+-sqrt(1/4+x). This needs to be an integer (any integer), so sqrt(1/4+x) needs to be half-integer (any half integer), so we can introduce: i+1/2 = sqrt(1/4+x), where i is any integer. 4. Squaring gives i^2 + i +1/4 =1/4 +x, or x = i^2 + i. 5. We can insert any integer in i to generate an x for which N is integer. By solving i^2+i=1000 we get i = 31.13 or i = -32.13, so the maximum input for i is 31 and minimum -32. Including zero this gives us 64 inputs for i. HOWEVER because the expression for x(i) is parabolic, the negative values for i will give the same x(i) as the positives. The amount of unique x values thus is half the integer domain: 32. Example values: Take i = -4, then x = 12, and N = 4. Take i = 3, then x = 12, and N = 4.
I'm looking forward to seeing the proof that e^e^e^e^... diverges. As some other have mentioned, x^x^x^x^... actually converges for some values of x, such as x = 1 or x = root 2. I can see why it makes sense to say that e^e^e^... diverges but proofs are always nice to look at haha.
You can look up and find that the boudnary of convergence is e^(1/e). I'm not sure why, it looks like a fun exercise trying to prove it. I'd possibly try with induction. Maybe transfinite induction (same principle really)
The problem is that I am absolutely awful with divergent and convergent series and anything that involves those two words. I can look at something and understand it fairly well, but when it comes to doing it, my brain just stops working until I start doing something else. lol
To show that a recursively defined sequence converges what you often need to do is show that it is bounded and monotonic. That is sufficient. To show divergence it is sufficient to show it's monotonic and NOT bounded. You'd normally try induction and usually the proof is quite straightforward, in other cases you might want to try assuming the negation and prove by contradiction.
Avana: 2^2^2^2^... clearly diverges. Taking each element in order gives you 2^2=4, then 2^4 = 16, then 2^16 = 65536, then 2^65536, etc. - the number are increasing really fast. Since e > 2, it follows that e^e^e^e^... also diverges.
Lets see... F(x)=y F(x)=e^y y=e^y Ln(y)=y Therefore we should find some x which, when applied to ln, equals itself. This would be a zero point of the function ln(x)-x, which has two complex solutions involving the W function
I need an adblocker that blocks ads embedded at the beginning of otherwise good videos. Also, the question in the video seems like a good place to introduce tetration.
About the last question, i think i found the answer. Let y be the whole expression: y=sqrt(x+y) y²=x+y y=(1±sqrt(4x+1))/2 Since y and x are both integers it must be 4x+1=n² for some n. if so, 1±sqrt(4x+1) is for sure even, so we don't have to worry about that. Therefore 4x=(n+1)(n-1) x has to be ≤1000, so (n+1)(n-1)/4≤1000 or n
If you think about that, it can be written as the infinite tetration of e. Therefore, the numbers approach to some large number. Therefore, there’s no answer to this domain.
Another way to show this is, the process eventually maps every number below some number M into the void. Numbers less than 0 for starters. For any M and number less than e^M will be mapped under M. So there is no M big enough to have numbers that aren't mapped into the void.
The Brilliant one has 31 solutions, where let k be in the integers {1, ..., 31} and x = k * (k+1), and the end result is (1+sqrt(1+4x))/2, which equals k+1
And even if there weren't the problem with the domain, there's another reason you could never have an x for which f_n(x) converges as n goes to infinity. If you did, by the continuity of the ln function, taking the limit as n goes to infinity in f_{n+1}(x) = ln f_n(x), you'd get f_{infinity}(x) = ln f_{infinity}(x), which is not possible since ln(x) < x - 1 < x for all x.
If that function were to converge for a values of x, it would converge such that the corresponding y value satisfies y = ln(y). But there is no solution to y = ln(y). So there cannot be any x values for which the function converges. So the function has no domain and is thus just the null function.
ln is a real function though, so there arent really any complex solutions. If you want to extend ln to compelx numbers you get into a whole lot of other troubles (namely that there does not exist a continuous function "ln" with e^(ln(z)) = z for all z, which is defined on all C. You can only define that on parts of C)
NAMEhzj the lack of continuity is not a problem here. Also "ln" can be both the real logarithm, or the complex logarithm, it's hard to tell without giving context. The complex logarithm can be either defined as a multi-valued function. Or if that's not proper enough, it can be defined as a function where the value is a set of complex numbers. So the range is not C, but P(C). But the question is about the domain of such function, and complex logarithm defined either way, the domain contains at least two complex numbers. About ln(x) = x makes less sense, when it's defined as having a set as a result. Then it could be understood as x being an element of ln(x). Or if we go by definition, ln(x) is just an inverse of e^x, similarly to arcsin of sin (where arcsin is not a proper inverse). So we could say that ln(x) = x is true when x = e^x is true. (This is bit less accurate though). There is one unaddressed thing: what is ln(ln(x)) if the value is a set? Well, usually it means apply the function to all element in the set, and the result is a set of sets (or sometimes just an union), so now asking for the domain makes sense.
Oh nice.. yesterday i thought about this a little and was like "well |ln(z)| = ln(|z|) anyways, so any complex fuxed point z would give a real fixed point |z|" but thats just wrong. So it could actually be very interesting to think of a complex solution...
So if the approach in this video makes any sense, you could just do the same for complex numbers and conclude that ln is defined on C\{0,1,e,e^e, ...} but its values are only these two fixed points (which, as i read, dont really have a closed form but one is the conjugate of the other, right?). Cool stuff...
let f(x) = root(x + root(x + ...)) f(x) = root(x + f(x)) f(x)^2 = x + f(x) x = f(x)^2 - f(x) x = f(x) * ( f(x) - 1) From this, we can see that x must be of form (a - 1) * a, where a is an integer. x is restricted from 1 to 1000, so by checking, a is restricted from 1 to 31. (31 * 32 = 992, so it is the last possible number.) Since there are 31 - 1 + 1 = 31 such a and no answer overlaps, there are 31 values of x.
But what is even more interesting is that if you use complex numbers that function converges to two values, the Lambert W function of -1 and its complex conjugate (not counting x=0)
What I thought of was, if you plug in a positive number greater than zero, the ln(that number) will make it smaller and if you do that infinitely many times it will become zero and ln(0) DNE
What's cool about this is you could use tetration to denote pattern for an n amount of nested natural logarithms, where ln(ln(...(x)...)) for n an amount of times, x must be e^^(n-2), where n≥2
No matter how many times you square root (even infinitely!) the index of the radical will be positive. That is, it will always look like x^(1/2n). The domain of an even degree root is always x is greater than or equal to zero.
You can think about the natural log problem visually... The domain breaks at 0, and the range includes 0 at 1. So next time you plug it in, that 1 becomes the new domain problem point and the 0 in the range slides further to the right (where it will become the next problem point... And so on) Sqrt doesn't have this kind of behavior. The domain is x>=0, and the range is as well. It's a 1-1 mapping.
Here is different way: the expression can be defined as the limit of the relation a_(n+1)=ln(a_(n)) with arbitrary a_1, first let's check what the limit will be if it exists: assuming the limit exists lim a_n=lim a_(n+1) hence lim a_n-a_(n+1)=0 hence we need to solve A=lnA, there is no answer to this in the reals, assuming the limit exists we got to contradiction hence there is no limit to any a_1 and hence the domain is nothing
Interesting proof. I essentially did it in the opposite direction. I proved that x > ln(x) using calculus. If you plot y = ln(x) and take an arbitrary starting x value, the y value obtained is always smaller. Use this smaller value as a new x coordinate and you obtain yet another smaller y value. Repeat this infinitely and you will eventually get ln(negative). Therefore ln(ln(ln...(x)...) has no domain. *Edit* See replies. It has been pointed out to me that you must also prove that ln(x) < x - c for c > 0.
It won’t suffice to show, that applying x -> ln(x) will result in the value getting smaller and smaller. You actually need to prove that it gets substantially smaller, showing something like for example, there is some constant c such that ln(x) < x - c holds for all x. Then you know that starting with some value a, after a/c (rounded up to the next integer) steps you will hit the negatives. Otherwise, a sequence like 1+1/2 ; 1+1/3 ; 1+1/4 ; ... ; 1+1/n ; ... gets smaller and smaller but never hits negatives, not even 0, not even something ≤1. Fortunately you can get such a constant c for ln(x). Take the difference x-ln(x). That’s a differentiable function (with the same domain as ln(x)), so you can find a lower bound (our constant c) for x-ln(x) [such that x-ln(x) ≥ c holds, so ln(x) ≤ x - c, and ln(x) < x - (c-ε) for any ε>0] by looking for potential minima. Taking d/dx (x-ln(x)) = 1 - 1/x [on the domain of numbers >0], we find out, 0 = 1 - 1/x => 1=1/x => x=1 is the only candidate. 1-ln(1) = 1 is a candidate for a minimum. Since for 1/e 1 and for e>1, e-ln(e)=e-1>2.71-1 = 1.71 > 1, is is a minimum indeed. So c=1 is such a constant that ln(x) ≤ x-1, or for example ln(x) < x-0.99 if you want a proper less than relation.
I've assumed that this expression approaches y and then realized that e^y has to be equal to y which is false for any real y. My solution is valid right? ^^' (We can prove e^y cant equal y by realising that e^y>y for y=0, for any positive y the derivative of e^y>the derivative of 1 and for any negative y the opposite is true)
My solution: If f(x) = ln(ln(...(x)...), then f(x) = ln(f(x)), right? It is well known (and easy to show) that the equation x = ln(x) has no solution, so the same goes for f(x) = ln(f(x)). Therefore, there’s no value for f(x) (and, equivalently, x) that satisfy the given equation.
I’m guessing a proof looks like: assuming the limit exists and is L, then L = e^L, which is false for any L. The limit doesn’t exist anywhere, so the domain has to be empty.
We assumed our domain had to be a subset of the reals (although this was never outright stated, it is the normal case). No real number gives a defined value for ln(ln(...(x)...)), so our “domain” is the empty set.
misotanni Dunno... seems like mathematical abstractions that aren’t “real” manage to show up in physics shockingly often for things that don’t exist. If we’re restricting ourselves to “normal” real numbers anyway, though, I concede the point.
Since we are watching math videos, I imagine many of us have seen some infinite tetration thing, and those of us who have looked a bit more into it may know they do converge between e^(-e) and e^(1/e). Since the "answer" to this requires infinite tetration, would that mean a variation with a different base between e^(-e) and e^(1/e) would have a valid answer?
Hey sir,why not answered my question- value of sine^-1(sine 12),where 12 is in radian?plz answer my question with graphical n traditional methods by making a video lecture.
Not possible, lnx has a domain of 0, lnlnx has domain of 1, lnlnlnx has e, each time we have to raise the ln to the e to cancel out so we end up raising e to the prev. domain. so you are left with an infinite stack of e’s Edit after watching: #YAY
This problem now leads me to another similar one. Here, we looked at real values of x. What if we extended the domain of ln(x) to the complex numbers? Probably limiting ourselves to the principal branch for simplicity. Since the domain of the natural log in the complex world is the whole plane (except 0), would we end up with a non-empty domain this time? I’m not sure, I’m looking for insight.
Adam Kangoroo Its x = 0 as arcsin0=0, so it will just repeat that. Every iteration of arcsin, arcsin(x)>x (x>0). So after infinetly many iterations, x will go outside arcsines domain which is [-1,1]. The same thing will happen for negative but itll cross -1 instead.
There would be countably infinite outputs of the first arcsin, uncountably infinite outputs of the second arcsin, so you can see where this going aka out of number theory
Gold161803 but that for convenience and sanity. Pure math will not be biased like that towards inputs from a small subset of real numbers when they give the same output
Let θ: C -> C , f |--> f'' be a function, that returns the second derivative of a function. C are all REAL functions, that are arbitrarily often differentiable. Proof: 1) All real numbers x greater or equal zero are Eigenvalues of θ 2) -1 is a Eigenvalue of θ
Hello... I followed ur chanel and i LOVED IT!! but i have a small question which i haven't found an answers for it yet... How can i solve (x^x=c) where c is a constant...I'll will be happy if u aswered it an thank u...
If you draw the graph of cos(x^-x^2) you will see that from x>2 the output ~1, so it would be like integrating x from 2 to infinity, which explodes to infinity. This means cos(x^-x^2) dx from 0 to infinity will not converge either. as x->inf, x^-x^2 approaches 0, so cos(x^-x^2) approaches 1. However, if you use sin(x^-x^2), this will converge. And it converges fast. int 0 -> inf sin(x^-x^2) ~ 1.32561639224 A slightly more interesting question could be the area between x^-x^2 and sin(x^-x^2), since the small angle approximation actually comes into play for large values of x and they form the same curve.
#YAY Well why cant we say that the domain of x is infinity itself.... since e^e^e..... is approaching infinity but its really infinte...???? Also there is another question which I would like to ask you... Can you tell me where can I send you that question?
Infinity isn't a number, so you can't have it as a domain. e^e^e^e^e is infinite if you have infinitely many e's, if you look at the lim x->e for x^x^x^x^x, it won't converge. That function converges for x
I m in the common core but i understood limits , integrals , complex number , modular form , arithmetic .....etc.........and lot of hight school lecons cz of ure videos #yaaaaaaaaay yeeh!
Isn't this an example of tetration application? Just use Principle of Finite Induction to prove Domain_n equals to " ] e tetrated by n-2 ; +∞ ) " for n >= 2
![[Sub Thai] APT. - ROSÉ & Bruno Mars](http://i.ytimg.com/vi/bYjhzP8x_uc/mqdefault.jpg)
But what is the domain of love(x)? I postulate that it is *U* because love knows no bounds.
Skippy the Magnificent : )
Actually, the domain for love(u) is surely infinite, but the domain for love(x) is indeterminate.
Some postulate that love(x)'s domain is the empty set, others say it's an infinite set of finite measures all over the complex plane, and then there are others still that say it should get real.
Similar functions with a set of real values for x include loved(x)=true, past(love(u))=x, and future(drunk(text(x)))=regret.
It's worth noticing that for complex values every z satisfying z=ln(z) (there are infinitely many) is a solution. Would be a great idea for your next video!
It's a good idea to move in complex world, very good!
It's not obvious without a proof that x^x^x^x^... Diverges to infinity for x=e . For example for x=sqrt(2) it converges to 2.
śledź śledźowski
True. Stay tuned : )
(Just a preview for u, anything bigger than e^(1/e) diverges)
woow that sounds interesting can't wait for the proof
F
blackpenredpen that s awesome XD
Oh? Neat! Does that hold in general? Like x^x^x^... converges to N if x = sqrt(N) for anything less than e^(1/e)?
It's amazing how it doesn't converge as a real function for any x, but converges as a complex function almost everywhere (oh, except e ^^ n for n = -2, -1, 0, 1, 2, ...) to -W(-1, -1) or 0.3181... + 1.3...i
my guess is the empty set. let's watch the video
called it
AndDiracisHisProphet yup, of course!
Im from turkey. i can understand integral now. Thanks 👍
Ulavatur Ardıç great!
If you want an easy way to write e^e^e^e^..., you could write it in tetration, as the infinity-th tetration of e.
It would have been nice if you had generalised fn(x) first before moving to the case of infinity. It appears to be the x > "the (n-2)th tetration of e", where n>=2. Then show that the "infinite tetration of e" diverges.
ChaosPod indeed or with that same n, it could be written in terms of functional composition: n-ln's composed together have a domain of (n-1)-e^x's composed together and evaluated at 0. Where n>=2 (or 1... depending on your compositional definition). There should also be an expression that can include all cases including zero though.
is the -1 tetration of e equal to zero?
Earthbjorn Nahkaimurrao hmmm, tetration when defined using Knuth up-arrow notation is defined recursively, so the value of a -1 tetration would depend on an exponentiation where the exponent is the value of a -2 tetration (which would turn into an exponentiation with -3 =\ ) so we never end up with a base case: a 1 tetration
Perhaps another definition allows us to define negative tetrations more usefully :)
Windsor Mason You’re wrong. 1. The notation does not define the operator itself. 2. The recursive relation necessarily implies that the -1 tetration of any number for which tetration is defined is 0.
I thought of it a different way:
As you increase the ln(x) chain, the output keep decreasing because ln(x) < x. Because the range of ln(x) goes down to -infinity, we know that the ln(x) chain will at least get as low as e (though it doesn't have to hit e for every x). ln(x)
Can you do this for complex values next?
f(x) = ln(f(x))
e^(f(x)) = f(x)
-1 = -f(x)*e^(-f(x))
f(x) = -W(-1)
Roberto Hernández, does this imply that f(x) is a constant function or am I missing something?
Faris Yazdi i think so, but W(n) is multivalued so i dont know how to take that...
Isn't the ln function defined on all the complex plane? (Except 0+0i)
Roberto Hernández it's a fixed point at +/- W(-1), depending on initial conditions and assuming that only the principle branch of ln is selected.
Wow Brilliant has gotten pretty popular recently! Congrats on the sponsorship!
Eduardo Rivera thanks!
We did start to collab since last year already. : )
In the math world, limits = magic
Very few people understand them
: )
Sure if by the math world you mean the world of people who haven't taken an introductory class on analysis or topology.
Solution for problem at the end:
0. First understand that you can square both sides and move the 6 to the other side to verify the expression.
1. Simplify the expression as N = Sqrt(x+N) where N is an integer.
2. Square expression as N^2-N-x = 0.
3. Solve for N using Quadratic Formula:
N = 1/2+-sqrt(1+4x)/2
= 1/2+-sqrt(1/4+x).
This needs to be an integer (any integer), so sqrt(1/4+x) needs to be half-integer (any half integer), so we can introduce: i+1/2 = sqrt(1/4+x), where i is any integer.
4. Squaring gives i^2 + i +1/4 =1/4 +x, or x = i^2 + i.
5. We can insert any integer in i to generate an x for which N is integer. By solving i^2+i=1000 we get i = 31.13 or i = -32.13, so the maximum input for i is 31 and minimum -32. Including zero this gives us 64 inputs for i.
HOWEVER because the expression for x(i) is parabolic, the negative values for i will give the same x(i) as the positives. The amount of unique x values thus is half the integer domain: 32.
Example values:
Take i = -4, then x = 12, and N = 4.
Take i = 3, then x = 12, and N = 4.
I forgot the exercise said to have x at least 1, so x = 0 doesnt count, reducing the total number of x to 31.
My whole explanation can be done in a much easier way!
Just use N^2 - N = x to generate x directly...
I'm looking forward to seeing the proof that e^e^e^e^... diverges. As some other have mentioned, x^x^x^x^... actually converges for some values of x, such as x = 1 or x = root 2. I can see why it makes sense to say that e^e^e^... diverges but proofs are always nice to look at haha.
You can look up and find that the boudnary of convergence is e^(1/e). I'm not sure why, it looks like a fun exercise trying to prove it. I'd possibly try with induction. Maybe transfinite induction (same principle really)
The problem is that I am absolutely awful with divergent and convergent series and anything that involves those two words. I can look at something and understand it fairly well, but when it comes to doing it, my brain just stops working until I start doing something else. lol
x=e^e^e^e^...
x=e^x
ln(x)=x
no solution
To show that a recursively defined sequence converges what you often need to do is show that it is bounded and monotonic. That is sufficient.
To show divergence it is sufficient to show it's monotonic and NOT bounded.
You'd normally try induction and usually the proof is quite straightforward, in other cases you might want to try assuming the negation and prove by contradiction.
Avana: 2^2^2^2^... clearly diverges. Taking each element in order gives you 2^2=4, then 2^4 = 16, then 2^16 = 65536, then 2^65536, etc. - the number are increasing really fast. Since e > 2, it follows that e^e^e^e^... also diverges.
#yay your videos should converge to infinity
Varad Mahashabde You mean diverge?
Lets see...
F(x)=y
F(x)=e^y
y=e^y
Ln(y)=y
Therefore we should find some x which, when applied to ln, equals itself. This would be a zero point of the function ln(x)-x, which has two complex solutions involving the W function
I need an adblocker that blocks ads embedded at the beginning of otherwise good videos. Also, the question in the video seems like a good place to introduce tetration.
About the last question, i think i found the answer.
Let y be the whole expression: y=sqrt(x+y)
y²=x+y
y=(1±sqrt(4x+1))/2
Since y and x are both integers it must be 4x+1=n² for some n.
if so, 1±sqrt(4x+1) is for sure even, so we don't have to worry about that.
Therefore 4x=(n+1)(n-1)
x has to be ≤1000, so (n+1)(n-1)/4≤1000 or n
Another GREAT educational video!!! Keep making M-O-R-E !!!
Thank you so much , that is really helpful 🙏😄
Thoughts on doing some difficult Brilliant(dot)org questions for the channel? Could be interesting!
If you think about that, it can be written as the infinite tetration of e. Therefore, the numbers approach to some large number. Therefore, there’s no answer to this domain.
Another way to show this is, the process eventually maps every number below some number M into the void. Numbers less than 0 for starters. For any M and number less than e^M will be mapped under M. So there is no M big enough to have numbers that aren't mapped into the void.
The Brilliant one has 31 solutions, where let k be in the integers {1, ..., 31} and x = k * (k+1), and the end result is (1+sqrt(1+4x))/2, which equals k+1
And even if there weren't the problem with the domain, there's another reason you could never have an x for which f_n(x) converges as n goes to infinity. If you did, by the continuity of the ln function, taking the limit as n goes to infinity in f_{n+1}(x) = ln f_n(x), you'd get f_{infinity}(x) = ln f_{infinity}(x), which is not possible since ln(x) < x - 1 < x for all x.
If that function were to converge for a values of x, it would converge such that the corresponding y value satisfies y = ln(y). But there is no solution to y = ln(y). So there cannot be any x values for which the function converges. So the function has no domain and is thus just the null function.
No real solution, that is. There are two complex solutions.
ln is a real function though, so there arent really any complex solutions. If you want to extend ln to compelx numbers you get into a whole lot of other troubles (namely that there does not exist a continuous function "ln" with e^(ln(z)) = z for all z, which is defined on all C. You can only define that on parts of C)
NAMEhzj the lack of continuity is not a problem here. Also "ln" can be both the real logarithm, or the complex logarithm, it's hard to tell without giving context. The complex logarithm can be either defined as a multi-valued function. Or if that's not proper enough, it can be defined as a function where the value is a set of complex numbers. So the range is not C, but P(C). But the question is about the domain of such function, and complex logarithm defined either way, the domain contains at least two complex numbers. About ln(x) = x makes less sense, when it's defined as having a set as a result. Then it could be understood as x being an element of ln(x). Or if we go by definition, ln(x) is just an inverse of e^x, similarly to arcsin of sin (where arcsin is not a proper inverse). So we could say that ln(x) = x is true when x = e^x is true. (This is bit less accurate though). There is one unaddressed thing: what is ln(ln(x)) if the value is a set? Well, usually it means apply the function to all element in the set, and the result is a set of sets (or sometimes just an union), so now asking for the domain makes sense.
Oh nice.. yesterday i thought about this a little and was like "well |ln(z)| = ln(|z|) anyways, so any complex fuxed point z would give a real fixed point |z|" but thats just wrong. So it could actually be very interesting to think of a complex solution...
So if the approach in this video makes any sense, you could just do the same for complex numbers and conclude that ln is defined on C\{0,1,e,e^e, ...} but its values are only these two fixed points (which, as i read, dont really have a closed form but one is the conjugate of the other, right?). Cool stuff...
If the left side converges to a number, say A, then exping both sides yields A = ln(A) = exp(A) but ln and exp do not intersect.
I don't know who Brenda Worg is, but I'm glad she sponsored your video.
The question is does the
lim_{x -> inf} ln^x(x) exists ?
Where f^n(x) denote functional composition
Is that argument valid:
V= lim_{x => inf} ln^x(x)
= ln( lim_{x => inf} ln^x(x) )
So V=ln(V)
e^V = V No solution
So limit doesn't exist
If the base of the logs were different, I'm pretty sure it would coverage. Haven't tried it out though
let f(x) = root(x + root(x + ...))
f(x) = root(x + f(x))
f(x)^2 = x + f(x)
x = f(x)^2 - f(x)
x = f(x) * ( f(x) - 1)
From this, we can see that x must be of form (a - 1) * a, where a is an integer.
x is restricted from 1 to 1000, so by checking,
a is restricted from 1 to 31. (31 * 32 = 992, so it is the last possible number.)
Since there are 31 - 1 + 1 = 31 such a and no answer overlaps,
there are 31 values of x.
You can not put inside e^e^e... because it says x>e^e^e... not x≥e^e^e...
infinitely nested logarithm doesnt exist, it cant hurt you
But what is even more interesting is that if you use complex numbers that function converges to two values, the Lambert W function of -1 and its complex conjugate (not counting x=0)
The point is that if y=x^x^x^x...... then y=x^(x^x^x......) but this is also equal to x^y and then x=y^(1/y)
Thanks for the video!
Greetings from Germany
FadingCode 😊
What I thought of was, if you plug in a positive number greater than zero, the ln(that number) will make it smaller and if you do that infinitely many times it will become zero and ln(0) DNE
What's cool about this is you could use tetration to denote pattern for an n amount of nested natural logarithms, where ln(ln(...(x)...)) for n an amount of times, x must be e^^(n-2), where n≥2
Lol i literally had this question in my homework xD
Wfreestyle wow! Cool.
Homework for what class?
@@stanrocks123 for e^e^e^...^e^eth grade.
@@Kitulous lol
have you already done domain of sqrt(sqrt(...(x)...))
Soufian 27 x is greater than or equal to 0
No matter how many times you square root (even infinitely!) the index of the radical will be positive. That is, it will always look like x^(1/2n). The domain of an even degree root is always x is greater than or equal to zero.
You can think about the natural log problem visually... The domain breaks at 0, and the range includes 0 at 1. So next time you plug it in, that 1 becomes the new domain problem point and the 0 in the range slides further to the right (where it will become the next problem point... And so on)
Sqrt doesn't have this kind of behavior. The domain is x>=0, and the range is as well. It's a 1-1 mapping.
Soufian 27
I have a similar one th-cam.com/video/aaWGhxTXKg8/w-d-xo.html
it's [0;inf[ and the image is 1.
I love your thumbnails😂
Here is different way:
the expression can be defined as the limit of the relation a_(n+1)=ln(a_(n)) with arbitrary a_1, first let's check what the limit will be if it exists: assuming the limit exists lim a_n=lim a_(n+1) hence lim a_n-a_(n+1)=0 hence we need to solve A=lnA, there is no answer to this in the reals, assuming the limit exists we got to contradiction hence there is no limit to any a_1 and hence the domain is nothing
It converges to complex values
Try this with a logarithmic base greater than 1/e and less than e^(1/e) (which Is been calling "t" for tetration)
Elliot Dotson 1/(e^e) *
Wow pretty crazy
In general, fₙ(x) has domain x > ⁿ⁻²e.
Interesting proof. I essentially did it in the opposite direction. I proved that x > ln(x) using calculus. If you plot y = ln(x) and take an arbitrary starting x value, the y value obtained is always smaller. Use this smaller value as a new x coordinate and you obtain yet another smaller y value. Repeat this infinitely and you will eventually get ln(negative). Therefore ln(ln(ln...(x)...) has no domain.
*Edit* See replies. It has been pointed out to me that you must also prove that ln(x) < x - c for c > 0.
It won’t suffice to show, that applying x -> ln(x) will result in the value getting smaller and smaller. You actually need to prove that it gets substantially smaller, showing something like for example, there is some constant c such that ln(x) < x - c holds for all x. Then you know that starting with some value a, after a/c (rounded up to the next integer) steps you will hit the negatives. Otherwise, a sequence like 1+1/2 ; 1+1/3 ; 1+1/4 ; ... ; 1+1/n ; ... gets smaller and smaller but never hits negatives, not even 0, not even something ≤1. Fortunately you can get such a constant c for ln(x). Take the difference x-ln(x). That’s a differentiable function (with the same domain as ln(x)), so you can find a lower bound (our constant c) for x-ln(x) [such that x-ln(x) ≥ c holds, so ln(x) ≤ x - c, and ln(x) < x - (c-ε) for any ε>0] by looking for potential minima. Taking d/dx (x-ln(x)) = 1 - 1/x [on the domain of numbers >0], we find out, 0 = 1 - 1/x => 1=1/x => x=1 is the only candidate. 1-ln(1) = 1 is a candidate for a minimum. Since for 1/e 1 and for e>1, e-ln(e)=e-1>2.71-1 = 1.71 > 1, is is a minimum indeed. So c=1 is such a constant that ln(x) ≤ x-1, or for example ln(x) < x-0.99 if you want a proper less than relation.
awsome bro...
Thanks for the correction. I've never come across a situation like that but at least I've learned something.
Once, I fooled around with nested ln(x), and I discovered this pattern.
Best part of the video: "Pause this video and give it a try".
Papai Pal why?
We get to work our brains.... That's why.
You can think that ln(x)
the domain may not be in the set of real numbers, but what about the hyperreals? It seems like it's in the set of hyperreals.
Domain
Nomain
Donomain
Donotmain
Don'tmain
Don'twhomst'vemain
Dtn'sey'twhomst'vemain
D'n
* insert expanding brain meme *
theiyr're
Dm
Domain't
Well, at least for every set X there's a function from {} (the empty set) to X. Initial objects are nice like that.
I've assumed that this expression approaches y and then realized that e^y has to be equal to y which is false for any real y. My solution is valid right? ^^'
(We can prove e^y cant equal y by realising that e^y>y for y=0, for any positive y the derivative of e^y>the derivative of 1 and for any negative y the opposite is true)
My solution:
If f(x) = ln(ln(...(x)...), then f(x) = ln(f(x)), right? It is well known (and easy to show) that the equation x = ln(x) has no solution, so the same goes for f(x) = ln(f(x)). Therefore, there’s no value for f(x) (and, equivalently, x) that satisfy the given equation.
And finally, e^e^e^this and that.
I’m guessing a proof looks like: assuming the limit exists and is L, then L = e^L, which is false for any L. The limit doesn’t exist anywhere, so the domain has to be empty.
nice video, very interesting problem, I get the correct solution #YAY!
What is your watch? I'm curious
Super table tennis it's by MVMT
Black classic
Which infinity is e^e^e^..., though? Aleph0? Aleph1? Couldn’t you constrain x to be “greater than that particular infinity”?
We assumed our domain had to be a subset of the reals (although this was never outright stated, it is the normal case). No real number gives a defined value for ln(ln(...(x)...)), so our “domain” is the empty set.
Sean Fraser Ah, ok. Thanks
misotanni Dunno... seems like mathematical abstractions that aren’t “real” manage to show up in physics shockingly often for things that don’t exist. If we’re restricting ourselves to “normal” real numbers anyway, though, I concede the point.
misotanni The real numbers are not cardinals. They are simply not transfinite, because they are finite in magnitude.
Sean Fraser The domain itself is less the problem, and more so the codomain.
Very helpful
You're making me like calculus, how is that possible?
Since we are watching math videos, I imagine many of us have seen some infinite tetration thing, and those of us who have looked a bit more into it may know they do converge between e^(-e) and e^(1/e). Since the "answer" to this requires infinite tetration, would that mean a variation with a different base between e^(-e) and e^(1/e) would have a valid answer?
Amazing
If this form approaches x^x^x^x^... where x=e for the log base e, then we can assume it is the same for x=n log base n. Would this mean that any n
Hey sir,why not answered my question- value of sine^-1(sine 12),where 12 is in radian?plz answer my question with graphical n traditional methods by making a video lecture.
Not possible, lnx has a domain of 0, lnlnx has domain of 1, lnlnlnx has e, each time we have to raise the ln to the e to cancel out so we end up raising e to the prev. domain. so you are left with an infinite stack of e’s
Edit after watching: #YAY
This problem now leads me to another similar one. Here, we looked at real values of x. What if we extended the domain of ln(x) to the complex numbers? Probably limiting ourselves to the principal branch for simplicity. Since the domain of the natural log in the complex world is the whole plane (except 0), would we end up with a non-empty domain this time? I’m not sure, I’m looking for insight.
f(x) = -W(-1)
My answer is 63 integers numbers.
How about complex numbers? For example, will f(i) diverge (assuming taking the principal value for ln)?
I think that your latest video got deleted. I saw it this morning and now it is gone.
AYEEEEEEE YAY!
Calculus detention for you. When will you learn... #YAY~!
BPRP: Domain or Nomain?
Me: DonMain (NodMain). 😁👍
#Yay #Domain #NoMain
Ready For Music : )
I would have thought at least one of the numbers on the infinite tetration line would have been in the domain.
Guess I was wrong
Like e^e^this and that.
Did we just get a stealth intro to tetration?
Now what's the domain of arcsin(arcsin(arcsin(...(x)))) ? Haha.
Adam Kangoroo Its x = 0 as arcsin0=0, so it will just repeat that. Every iteration of arcsin, arcsin(x)>x (x>0). So after infinetly many iterations, x will go outside arcsines domain which is [-1,1]. The same thing will happen for negative but itll cross -1 instead.
filip S this is only if output of arcsin is limited between -π and π
There would be countably infinite outputs of the first arcsin, uncountably infinite outputs of the second arcsin, so you can see where this going aka out of number theory
Varad Mahashabde That's how arcsin works when written. Always take the principal value
Gold161803 but that for convenience and sanity.
Pure math will not be biased like that towards inputs from a small subset of real numbers when they give the same output
I know it's undefined regardless, but I'm curious if x has to be bigger than infinity here, as in infinity would be too small
I wonder if you could make this make sense with different infinite Cardinals
Let θ: C -> C , f |--> f'' be a function, that returns the second derivative of a function.
C are all REAL functions, that are arbitrarily often differentiable.
Proof:
1) All real numbers x greater or equal zero are Eigenvalues of θ
2) -1 is a Eigenvalue of θ
Hello... I followed ur chanel and i LOVED IT!! but i have a small question which i haven't found an answers for it yet... How can i solve (x^x=c) where c is a constant...I'll will be happy if u aswered it an thank u...
Ahmad Kalaoun
x^x = c
xlnx = lnc
lnx*e^lnx = lnc
lnx = W(lnc)
x = e^W(lnc)
Does this mean this function only exists in the complex plane?
#yay
Can you take the integral of cos(x^-x^2) from 0 to infinity? It seems like an interesting integral since it's a non elementary one :)
If you draw the graph of cos(x^-x^2) you will see that from x>2 the output ~1, so it would be like integrating x from 2 to infinity, which explodes to infinity. This means cos(x^-x^2) dx from 0 to infinity will not converge either.
as x->inf, x^-x^2 approaches 0, so cos(x^-x^2) approaches 1.
However, if you use sin(x^-x^2), this will converge. And it converges fast.
int 0 -> inf sin(x^-x^2) ~ 1.32561639224
A slightly more interesting question could be the area between x^-x^2 and sin(x^-x^2), since the small angle approximation actually comes into play for large values of x and they form the same curve.
#YAY Well why cant we say that the domain of x is infinity itself.... since e^e^e..... is approaching infinity but its really infinte...???? Also there is another question which I would like to ask you... Can you tell me where can I send you that question?
Infinity isn't a number, so you can't have it as a domain. e^e^e^e^e is infinite if you have infinitely many e's, if you look at the lim x->e for x^x^x^x^x, it won't converge. That function converges for x
Domain or Nomain, so strange! #yay
What if the output didn't have to be a real number?
Ie ln(-1) = pi*sqrt(-1)
5:13 missed one bracket
Andi Tafel
)
I close it for you😂
yea really annoying
you missed that he forgot one in the question itself
Christopher
)
I m in the common core but i understood limits , integrals , complex number , modular form , arithmetic .....etc.........and lot of hight school lecons cz of ure videos #yaaaaaaaaay yeeh!
You might want to go back and learn how to spell.
=[
Can I apply the ln function a fractional number n of times to x? The plot n vs ln_n(x) would be interesting :0
did u mean to say that e tetrated with e does not converge?
Why can't we just say y = ln(y), no number satisfies this equation?
In surreal numbers, x wouldn't be omega? Or I'm talking nonsense? I don't fully understand surreal numbers lol
#YAY you got a sponsor!
Would this function have a domain if we expand to the complex plane? I feel like it would, but I am unsure how to attack it
Zac Chaney yes
f(x) = -W(-1)
Why is this guy carrying grenade in his hands
Could ln^[inf](x) perhaps have a domain in the complex plane?
Isn't this an example of tetration application? Just use Principle of Finite Induction to prove Domain_n equals to " ] e tetrated by n-2 ; +∞ ) " for n >= 2
mah boi
What about complex numbers?
André Reis
f(x) = -W(-1)
8:04 is the answer 31 ?