Did a bit of research and now I know that the function x^x^x^x^..... (which can also be viewed as the solution for y in the equation x^y=y or x=y^(1/y)) is called "tetration of infinite height" and this function converges for e^(-e) < x < e^(1/e), roughly the interval from 0.066 to 1.44. Also this function is related to the Lambert W-function. In fact y = WC-In x)/ (In x) where Wis the Lambert W-function, which is defined as the functional inverse of xe^x with respect to x.
@@manuelferrer6501 no, x * ln(x^x^ . . .) = ln((x^x^ . . .)^x) (1), which is not ln(x^x^ . . .) (2) (e. g. if we substitute √2 for x in (1) we'll get ln(2^√2) ≈ 0.980, and if we substitute that in (2) we'll get ln(2) ≈ 0.693.
The infinite power tower converges on: { e^(-e) < x < e^(1/e) }, approx. { 0.066 ~< x ~< 1.445 } Also fun fact, the square root of 2 raised to its own power infinitely many times is actually equal to 2. Try it in a calculator!
Maybe I made an error, but my efforts to evaluate sqrt(2)^sqrt(2)^sqrt(2)^... aren’t giving me a clear answer. Here’s what I did: Let x = sqrt(2)^(sqrt(2)^(sqrt(2)^...)) x = sqrt(2)^x x = 2^(x/2) Squaring both sides gives: x^2 = 2^x Which has two positive solutions: x = 2 or x = 4 I know that there is another point where the functions x^2 and 2^x intersect, where x < 0, but since sqrt(2)^sqrt(2)... > 0, I disregarded it. I assume there is a similar process of elimination that rules out x = 4 as a solution, but what is it?
@@Riolupai the third root of 3 gets you two solutions: The true solution, 2.478 The second solution that you have to rule out, 3 The fourth root of 4 is sqrt(2) 😂 The fifth root of 5, is 1.7649 (true solution) Or 5 (other solution) I'm refering by the "true solution", to the result of a^a^a^... But if you take for example 5=x^x^x^... Then you will find x=5^1/5 So then 5 would be either "the true solution" or "the other solution" to 5^1/5 (And we found out it is the "other solution")
Interesting! I tried on Georgebra to add more and more Xs in the denominator, but I observed that the value of f(0) was oscillating between 1 and 0 when I add more Xs, so it is not converging. But if you plug 0 in your incredible derivative, we see that it gives a division by 0, which correlates to the impossible value of f(0)! Awesome mate!
this is 2 when x=sqrt(2): let Y= x^x^x^x^... -> Y=x^Y, take ln -> ln(Y)=Y*ln(x), now let Y be 2, then ln(2)=2*ln(x) ln(2)/2=ln(x) take the exponential and swap places x=e^(ln(2)/2) x=(e^ln(2))^(1/2) x=(2)^(1/2)=sqrt(2)
because that what is what he ask us in the video... and as this is a infinite exponentiation there is a value of x to which this no longer a constant, I don't know exactly where this thing explode, but is obvious that it does... maybe for anything greater than sqrt(2)? but it sure does for x>=2
I believe it blows up for x > e^(1/e) The function y = x^y Can be rewritten as x = y^(1/y) Of which the maximum is at x = e^(1/e), y = e He did found this maximum on his video about e^pi vs. pi^e.
You can find a formula for when it does converge. x^...^x = y_n has y_n = x^y_(n-1). In order for each of these to converge, it must be the case that y_(n+1) = y_n, meaning x^y_n = y_n. So, if it does converge, it'll converge to a solution of that. Now, y is actually is the Lambert-W function, the inverse function of xe^x, of -ln(x) over -x. This is because -W(-ln(x)) / ln(x) = y, and -y * ln(x) = W(-ln(x)), so x^(-y) = e^(W(-ln(x))), and thus x^(-y) * W(-ln(x)) = W(-ln(x)) * e^(W(-ln(x))) = -ln(x), so you get -W(-ln(x)) / ln(x) * x^(-y) = 1. or y = x^(y). You find that there are no solutions to the Lambert-W function outside of the interval [-1/e, e], See here: math.stackexchange.com/questions/1693561/for-what-values-does-this-method-converge-on-the-lambert-w-function. So, we must have -1 / e
Okay, I'll throw here my approach to the max. value where this function converges. First of all let's call y=x^x^x..... Therefore, y=x^y like he does in the video. After that, we take natural logarithms on both sides, so we get ln(y)=y*ln(x). Isolate ln(x) and we have ln(x)=ln(y)/y. Let's call f(y)=ln x = ln(y)/y and differentiate respect to y. f'(y)=(1-ln(y))/y^2 (after some simplification). The maximum of this function, is when f'(y)=0, (1-ln(y))/y^2=0 --> 1-ln(y)=0 --> y=e With the second derivative we can see it's a maximum and not a minimum. Coming back to ln(x)=ln(y)/y substitute y=e to obtain x=e^(1/e). Therefore x=e^(1/e) is the biggest value of x where the function y=x^x^x... converges, to y=e.
Link_Z - Hardcore Indie Game Achievements You have only shown that x^x^x^... does not converge when x>e^(1/e), but you have not shown when it does converge. It turns out, for example that it does not converge for 0,05. To be exact, x^x^x^... converges exactly if 1/e^e≤x≤e^(1/e) or if x=-1.
love your videos, theyre pretty much the main reason i love calculus. slight request if u havent already done so, integrate it (if its possible im not sure)
It is, he didn’t prove the necessary assumptions to do it. I think this one requires uniform convergence, and this function series doesn’t even converge pointwise for most inputs in R. Edit: After further thought, maybe pointwise convergence is enough for this approach, I’m not sure though since I haven’t tried it on paper.
It's not even math. Firstly, he had do define what is x^x^x.... , because it's not a function from R to R. Then he had to define what is d/dx, and only after that he is allowed to do math
You can also rewrite xy ln(x) at the end as x ln(x^y), which also equals x ln(y) because y = x^y. That would make the derivative in terms of x be [x^(x^(x^...))]^2 / [x - x ln(x^(x^(x^...)))].
Can’t it still be simplified to dy/dx = ( x^(2 * x^x^x....) )/( x - ln(x^x^x...) ) If you bring the 2 into the top tower and you bring the “xy” part into the ln and then use exponent laws to get the tower again
I think y = x^y is only a valid function for 0< x ≤ 1 and x = e^(1/e). For 1 < x < e^(1/e) there are two possible values of y for each x, and for x > e^(1/e) y doesn't exist. You can see this by considering the inverse function, which is simply y = x^(1/x).
Totally expecting this! Haha, I really love the humor of your channel and how enthusiastic are you in your videos, greetings from Colombia, keep it up :P
Only converges when X = 1. When X=1 dy/dx = 1 If X is 0 the bottom will be undefined due to the ln function. If X = -1 the bottom becomes imaginary (also due to ln function)
I would love some comments about which functionspace this function belongs to. It is not clear to me that it is differentiable. It looks to me like it is only defined between 0
I plotted the first 71 tetrations of x, really quite interesting. Also, the first time I tried I made the mistake of iterating (((x^x)^x)^x ..... Is there a term for this? The limit is equal to 1 for 0
Say you have an even number of infinite exponentials, if you plug in x = 0, then y = 1, however if there are an odd number of exponentials than if you plug in x = 0, then y = 0. It's like saying whats the answer to Grandi's series if you start with 0 as he first term.
Weird. In the denominator of the final answer, if you bring the factors of ln(x) inside as its exponents, won't you just end up with ln(y) ? I feel like there's an infinite amount of answers
The function y only exists on the interval (0,1]. And it appears to always be equal to 1. Therefore it's a constant and the derivative is zero. Note: pay attention to the domain of the function.
I cam up with y^2/(y^(1/y)*(1-ln(y)) I did it by drawing a line tangent to y=x^y (since that is equivalent to the derivative) and then making a right hand triangle out of it with dx and dy as the legs. From their I just imagined dx/dy(x=y^(1/y)). It turned out to be equivalent to 1/tan(z) where is the angle formed by the dx leg going through the tangent line. From their I just worked out tan(z) so that I get dy/dx. That was equivalent to just taking the reciprocal of dx/dy(x=y^(1/y)). And that gave me dy/dx(y=x^y)=1/(1/y^2*y^(1/y)*(1-ln(y))) I feel skeptical of my answer but it checked out with the 2 tests I did. Can anyone confirm that my trigonometry method works?
I have a question. I haven't studied that kind of derivative yet so I came up with this: f'(x)=((f^(-1))')^-1 (x) The inverse of f(x)=x^x^x^x^x^... -> y=x^y -> x=y^x; f^-1(x)=y=e^(lnx /x) y=f^-1 ' (x)=e^(lnx /x) * (1-lnx)/x^2 f'=((f^-1)' )^-1-> x=e^(lny /y) * (1-ln y)/y^2 How horrendous is that answer? (It actually works if you want to find the tangend line on y=a)
Why not factor out the common factor of x in the denominator and then make all those red x’s the power of the x that is the argument of the ln function. Then the denominator would be x(1 - ln [infinite tetration of x]). You could also move that 2 in your numerator inside the parentheses and subtract 1 by factoring the x out of the denominator. The numerator would be x to [(twice the infinite tetration of x) - 1)]
Is it not the case you can simplify that expression a little? When we have xyln(x), if we move the y to the exponent of the ln(x), we get xln(x^y). But as we stated earlier, x^y = y, so I believe we can simplify xyln(x) to just xln(y).
Can you solve x^x^x^...=2 vs. x^x^x^...=3 ua-cam.com/video/WAjgupg3hDk/v-deo.html
Girl : So how many exes do you have?
Me : Well, I have an infinite amount of Xs
Meme patch update 2020
Me: Yes
I don't want to find my x, and don't ask me about y, either.
Hromosomes
Don't worry, I can differentiate them all.
You two must integrate well
Did a bit of research and now I know that the function x^x^x^x^..... (which can also be viewed as the solution for y in the equation x^y=y or x=y^(1/y)) is called "tetration of infinite height" and this function converges for e^(-e) < x < e^(1/e), roughly the interval from 0.066 to 1.44. Also this function is related to the Lambert W-function. In fact y = WC-In x)/ (In x) where Wis the Lambert W-function, which is defined as the functional inverse of xe^x with respect to x.
Knew the first part, but the relation with the W function blew my mind🤯❤️
Wis is the mentor to Beerus in Dragonball Super
Instructions unclear; impressed too many girls
Instructions were clear, mate
Instructions clear: mate
How stupid you are. Lol
lol
How many?
4:52
You was smiling and that all. But all of sudden you became serious and said "Blue."
Lmao haha
Was looking for this comment
":) I will do this in..."
":| Blue. Of course."
Bprpbp of course 😂😂
3:37 now that is what I cal huge integration
Alfonso J. Ramos hahhahhahaha!!!!!
LMAO
Integration?
@@clyde__cruz1 that arrow he draws is like a huge integration symbol
I had the same initial wtf thought xD
In the rendez-vous :
girl : "So, what can you do?"
me : "I can differentiate x^x^x^x^... "
*she falls in your arms*
I went and she slapped me and said not maths
ua-cam.com/video/E29xoux3i_Y/v-deo.html
I doubt it
This works, can confirm. It took 12 hours of saying x^x though
BPRP does know the right kind of girls^^
Hahahaha, thanks!!!!
And the rare quadruple factorial
You’re profile picture is Paul Dirac. You are now one of my favorite people on the internet.
@@blackpenredpen I have differentiated x[4]x with respect to x, where this a[b]c represents infix notation.
Freshman boys, if you want to have the girls rolling in, show them the full solution. I cannot stress this enough. 👏
lol!!!
You can technically reduce that even more, as ylnx is equal to lny, so in the denominator you'd simply have x-xln(x^x^x^...)
Same though, but x ln(x^x^...) could also just be = ln( x^x^x...) ?
@@manuelferrer6501 no, x * ln(x^x^ . . .) = ln((x^x^ . . .)^x) (1), which is not ln(x^x^ . . .) (2) (e. g. if we substitute √2 for x in (1) we'll get ln(2^√2) ≈ 0.980, and if we substitute that in (2) we'll get ln(2) ≈ 0.693.
The infinite power tower converges on: { e^(-e) < x < e^(1/e) }, approx. { 0.066 ~< x ~< 1.445 }
Also fun fact, the square root of 2 raised to its own power infinitely many times is actually equal to 2. Try it in a calculator!
Maybe I made an error, but my efforts to evaluate sqrt(2)^sqrt(2)^sqrt(2)^... aren’t giving me a clear answer. Here’s what I did:
Let x = sqrt(2)^(sqrt(2)^(sqrt(2)^...))
x = sqrt(2)^x
x = 2^(x/2)
Squaring both sides gives:
x^2 = 2^x
Which has two positive solutions:
x = 2 or x = 4
I know that there is another point where the functions x^2 and 2^x intersect, where x < 0, but since sqrt(2)^sqrt(2)... > 0, I disregarded it. I assume there is a similar process of elimination that rules out x = 4 as a solution, but what is it?
@@paulchapman8023 when you squared, you created an extra solution
@@GhostHawk272he didnt, 4 satisfies second equation (before squaring) as well
What about 3rd root of 3, 4th root of 4, etc (or alternatively sqrt(3), sqrt(4), etc)
@@Riolupai the third root of 3 gets you two solutions:
The true solution, 2.478
The second solution that you have to rule out, 3
The fourth root of 4 is sqrt(2) 😂
The fifth root of 5, is
1.7649 (true solution)
Or
5 (other solution)
I'm refering by the "true solution", to the result of a^a^a^...
But if you take for example 5=x^x^x^...
Then you will find x=5^1/5
So then 5 would be either "the true solution" or "the other solution" to 5^1/5
(And we found out it is the "other solution")
Interesting! I tried on Georgebra to add more and more Xs in the denominator, but I observed that the value of f(0) was oscillating between 1 and 0 when I add more Xs, so it is not converging. But if you plug 0 in your incredible derivative, we see that it gives a division by 0, which correlates to the impossible value of f(0)! Awesome mate!
It oscillates because 0^0 = 1 and 0^1 = 1. So at the top-most x^x, it's 0^0 = 1, then 0^1 = 0, and it repeats until you get all the way down.
this is 2 when x=sqrt(2):
let Y= x^x^x^x^... -> Y=x^Y, take ln -> ln(Y)=Y*ln(x), now let Y be 2, then
ln(2)=2*ln(x)
ln(2)/2=ln(x) take the exponential and swap places
x=e^(ln(2)/2)
x=(e^ln(2))^(1/2)
x=(2)^(1/2)=sqrt(2)
That's cool, why specifically 2 though? This in general is x = (y)^(1/y) , y > 0, isn't it?
because that what is what he ask us in the video...
and as this is a infinite exponentiation there is a value of x to which this no longer a constant, I don't know exactly where this thing explode, but is obvious that it does... maybe for anything greater than sqrt(2)? but it sure does for x>=2
I believe it blows up for
x > e^(1/e)
The function
y = x^y
Can be rewritten as
x = y^(1/y)
Of which the maximum is at
x = e^(1/e), y = e
He did found this maximum on his video about e^pi vs. pi^e.
🤔 that sound very interesting, yeah, that must be it...
other commentator said the same...
PD: you mean x>e^(1/e) right?
David Franco Whoops, yeah I did mean that haha. Thanks for pointing it out.
You can find a formula for when it does converge. x^...^x = y_n has y_n = x^y_(n-1). In order for each of these to converge, it must be the case that y_(n+1) = y_n, meaning x^y_n = y_n. So, if it does converge, it'll converge to a solution of that.
Now, y is actually is the Lambert-W function, the inverse function of xe^x, of -ln(x) over -x. This is because -W(-ln(x)) / ln(x) = y, and -y * ln(x) = W(-ln(x)), so x^(-y) = e^(W(-ln(x))), and thus x^(-y) * W(-ln(x)) = W(-ln(x)) * e^(W(-ln(x))) = -ln(x), so you get -W(-ln(x)) / ln(x) * x^(-y) = 1. or y = x^(y).
You find that there are no solutions to the Lambert-W function outside of the interval [-1/e, e], See here: math.stackexchange.com/questions/1693561/for-what-values-does-this-method-converge-on-the-lambert-w-function. So, we must have -1 / e
"So, we must have -1 / e
Yep, you're right!
What is "_"?
Okay, I'll throw here my approach to the max. value where this function converges. First of all let's call y=x^x^x..... Therefore, y=x^y like he does in the video.
After that, we take natural logarithms on both sides, so we get ln(y)=y*ln(x). Isolate ln(x) and we have ln(x)=ln(y)/y.
Let's call f(y)=ln x = ln(y)/y and differentiate respect to y.
f'(y)=(1-ln(y))/y^2 (after some simplification).
The maximum of this function, is when f'(y)=0, (1-ln(y))/y^2=0 --> 1-ln(y)=0 --> y=e
With the second derivative we can see it's a maximum and not a minimum.
Coming back to ln(x)=ln(y)/y substitute y=e to obtain x=e^(1/e).
Therefore x=e^(1/e) is the biggest value of x where the function y=x^x^x... converges, to y=e.
If I've made any mistake please tell me!
that sound about right, and as sqrt(2)
The singular of maxima and minima are maximum and minimum.
Doug R. Thanks! I didn't know
Link_Z - Hardcore Indie Game Achievements You have only shown that x^x^x^... does not converge when x>e^(1/e), but you have not shown when it does converge.
It turns out, for example that it does not converge for 0,05. To be exact, x^x^x^... converges exactly if 1/e^e≤x≤e^(1/e) or if x=-1.
Ya boi finna go impress some girls
it converges if x=1.
Its 1.
Then it would just be a constant function...
converges for -1
But what's the sign of y (or dy/dx, for that matter) for x
Converges for 0
love your videos, theyre pretty much the main reason i love calculus. slight request if u havent already done so, integrate it (if its possible im not sure)
Louis Moore you can't even integrate X^X and keep it in elementary functions. I doubt it can be done.
semi awesomatic ah yes i remember now. thanks :) same goes for x^y then i assume
Louis Moore unless it's a multivariable integration? Honestly don't know. But yeah in single variable it's impossible.
If you integrate the derivative you will just obtain y again.
@@tracyh5751 + C
this feels like cheating...
Why?
@Pete Berg Yeah. Basically, all derivatives are cheating.
All math looks like cheating. And it is the beauty of the math.
It is, he didn’t prove the necessary assumptions to do it.
I think this one requires uniform convergence, and this function series doesn’t even converge pointwise for most inputs in R.
Edit:
After further thought, maybe pointwise convergence is enough for this approach, I’m not sure though since I haven’t tried it on paper.
It's not even math. Firstly, he had do define what is x^x^x.... , because it's not a function from R to R. Then he had to define what is d/dx, and only after that he is allowed to do math
You can also rewrite xy ln(x) at the end as x ln(x^y), which also equals x ln(y) because y = x^y. That would make the derivative in terms of x be [x^(x^(x^...))]^2 / [x - x ln(x^(x^(x^...)))].
blackpenredpenbluepen
I made that joke before...
yet, reading it from you makes me want a to see a BlackPenRedPen & 3Blue1Brown collaboration
@@Theraot BluePenBluePenBluePenPencil
you could set x^x^x^x^... to be 2.
then you could plug all the x^x^x^x which are not the base to be two.
then you get x^2=2, or x=sqrt(2)
Cant wait to impress the guys with this ;)
3:38 that sign look like a big integral 😂
Some say the x’s still go on to this day
thank you so much, you make integration/differentiation fun
I love maths now
Really cool video, I wondered about this when I learned calculus last year. I wasn't to find as nice of a solution as you
Your videos are fascinating. Brings back my Calculus memories.
Can’t it still be simplified to
dy/dx = ( x^(2 * x^x^x....) )/( x - ln(x^x^x...) )
If you bring the 2 into the top tower and you bring the “xy” part into the ln and then use exponent laws to get the tower again
I think y = x^y is only a valid function for 0< x ≤ 1 and x = e^(1/e). For 1 < x < e^(1/e) there are two possible values of y for each x, and for x > e^(1/e) y doesn't exist. You can see this by considering the inverse function, which is simply y = x^(1/x).
Well, it does converge for 0
4:35 hahah impressing girls be like, hey honey, look at this huge exponential function
Convergent for all x such that 0
that is just awesome
no wait a minute its better than that
can't wait for next video
Thank you
jeromesnail You are actually wrong. For x=0,05 it will not converge for example.
It's actually only for { e^(-e) < x < e^(1/e) }
Including derivate of this infinite tetration is also calculate the main function -W(-ln(x))/(ln(x))
This was very entertaining, thanks!
🤣🤣🤣 I love that the motivation for this derivative is to impress girls. I'm definitely stealing that!
Totally expecting this! Haha, I really love the humor of your channel and how enthusiastic are you in your videos, greetings from Colombia, keep it up :P
Only converges when X = 1. When X=1 dy/dx = 1
If X is 0 the bottom will be undefined due to the ln function.
If X = -1 the bottom becomes imaginary (also due to ln function)
"if you want to make this slightly more interesting so you can impress girls" honestly one of the funniest things I've ever heard.
4:54 you said 'blue' so seriously xD
Fun fact: x^x^x^... can be expressed by solving y=x^y for y, using Lambert's W function!
Result:
x^x^...=e^(-W(-ln(x)))
for x in (0, e^(1/e)]
I would love some comments about which functionspace this function belongs to. It is not clear to me that it is differentiable. It looks to me like it is only defined between 0
Okey, now integrate it
x^x^x^x... converges when x < e^(1/e) because if you set x^x^x^x^x...= b then x = b^(1/b) and the maximum value of b^(1/b) is b=e.
You can also write the final answer as y / x (1 - ln(y)), since y ln(x) = ln(xʸ) = ln(y).
y^2=x^(2y)
x*y=x^(y+1)
Bro this guy is straight up hacking math
I plotted the first 71 tetrations of x, really quite interesting.
Also, the first time I tried I made the mistake of iterating (((x^x)^x)^x ..... Is there a term for this? The limit is equal to 1 for 0
That equals x ^ (x ^infinity)
This made me sub. Thank you sir
this is gold you're great!
Man, you're just amazing and your content is both enjoyable and useful at the same time
Not so useful for me but enjoyable yes 🙂
Zero dislikes. You've somehow hacked UA-cam.
Bc of Oreo!
Who doesn't like bunnies? :p
71
Akshay Gavini shouldn’t have said it
just added one dislike for fun
And which is the convergence domain of this function? Isnt it identically equal to 1?
x converges between 1/(e^e) and the eth root of e.
It converges for values of x between 0 and about 1.445
Me: I can take the Derivative of infinite power tower x^x^x^...
Girl: That's cool. Let's integrate together, baby!
Say you have an even number of infinite exponentials, if you plug in x = 0, then y = 1, however if there are an odd number of exponentials than if you plug in x = 0, then y = 0. It's like saying whats the answer to Grandi's series
if you start with 0 as he first term.
Weird. In the denominator of the final answer, if you bring the factors of ln(x) inside as its exponents, won't you just end up with ln(y) ? I feel like there's an infinite amount of answers
My hunch is that it converges for
0 < x < e^(1/e)
0 < y < e
Cameron Pearce it’s actually the inverse relation of y=x^(1/x)
Your domain is spot on, but you will actually get 2 values for 1
TheGinginator14 I understand that. I was trying to paraphase because I am incredibly tired and I didn't think many would care.
You get Lim(x→0+)(x^x^x^x...)=0 from that.
Yes.
The domain of x is e^-1 < x < e^e^-1
The function y only exists on the interval (0,1]. And it appears to always be equal to 1. Therefore it's a constant and the derivative is zero. Note: pay attention to the domain of the function.
On closer inspection, it only seems to be defined for x=1. Since the function only exists at one point, it has no derivative.
calculate the limit of this derivative as x approaches 0
Infinite Exponentials converge in, [0.06,1.4466], Barrows ~1936. His notation is, k = a ^ k, for , a, in, [0.06,1.4466], as i remember it…
Aka derivative of x tetrated to infinity
d/dx( ᪲x)
Also, ᪲x is equal to the inverse of ˣ√x
The more formal form of result is W(-ln(x))^2/(x ln^2(x) (W(-ln(x)) + 1)). W(x) is product log function, also called Lambert function.
I cam up with y^2/(y^(1/y)*(1-ln(y))
I did it by drawing a line tangent to y=x^y (since that is equivalent to the derivative) and then making a right hand triangle out of it with dx and dy as the legs.
From their I just imagined dx/dy(x=y^(1/y)). It turned out to be equivalent to 1/tan(z) where is the angle formed by the dx leg going through the tangent line.
From their I just worked out tan(z) so that I get dy/dx. That was equivalent to just taking the reciprocal of dx/dy(x=y^(1/y)).
And that gave me dy/dx(y=x^y)=1/(1/y^2*y^(1/y)*(1-ln(y)))
I feel skeptical of my answer but it checked out with the 2 tests I did.
Can anyone confirm that my trigonometry method works?
4:34 geek to the power of infinity
"I have no idea what this is anymore"
Your videos are awesome!!
You missed the opportunity to define y of x using Lambert W function.
0:01 The oreo that kept by mathematician:
this is too easy for me (eat vegetable)
I have a question.
I haven't studied that kind of derivative yet so I came up with this:
f'(x)=((f^(-1))')^-1 (x)
The inverse of f(x)=x^x^x^x^x^... ->
y=x^y -> x=y^x; f^-1(x)=y=e^(lnx /x)
y=f^-1 ' (x)=e^(lnx /x) * (1-lnx)/x^2
f'=((f^-1)' )^-1-> x=e^(lny /y) * (1-ln y)/y^2
How horrendous is that answer?
(It actually works if you want to find the tangend line on y=a)
I believe it converges for any value between one over e and the e root of e
BlackpenRedpen can you integrate this ??!
fantastic sir
the derivative of y(x)=(x^x^x^x^...)*(lnx) seems much nicer and handier
: )
Hi, how would you plug in and solve for a value of x? I guess go into the definitions of infinities? I would love to see videos on that! :D
You're very brillant for math
You won my like
Thanks!
blackpenredpen You're welcome
Me thinking that the arrow means integration 😂😂😂
Pretty sure the range -1 to 1 converges but idk if anything else does.
This is even better than Abramowitz and Stegun.
Very cool. Thanks!
You need to simplify more
dy/dx=y²/x (1-lny)
😇😇😇great job
That x^x^x^.... looks crazy
Rabbits in the wild generally live about one year; domesticated rabbits commonly live up to ten years.
Why not factor out the common factor of x in the denominator and then make all those red x’s the power of the x that is the argument of the ln function. Then the denominator would be x(1 - ln [infinite tetration of x]).
You could also move that 2 in your numerator inside the parentheses and subtract 1 by factoring the x out of the denominator. The numerator would be x to [(twice the infinite tetration of x) - 1)]
Girls don't feel restricted by what he said, feel free to impress me with this (seriously I love math so much)
Could we use implicit differentiation using Newton's notation? Because it seems like Leibniz's notation is way more useful.
I would be grateful for a similar explanation for the antiderivative of x^x^x...
Instead of implicit differentiation I just move all to one side and do -partial x/ partial y = dy/dx
you are broking my mind) it's good
y=x^x^x^... is the same as y=x^y
x=y^(1/y)
When y=2, x=sqrt(2)
Try different values of x :) Has very interesting properties.
Wonderful explonation ,Sir. Thanks
May you do a video about the nth term of the sequence {-1/3,1/3,0,-1/3,1/3,0….}
You forgot to check if x-xyln(x)=0 because if it is, your have to exclude values of x and y.
How legitmate is to the subsitution of y =x^x^x^x^... @1:10 ?
For X^X^X^X... to converge, we have to choose a value of X between 1/(e^e) and e^(1/e)
If you want the proof, juste ask me
the range of this function is from 0 to e^(1/e) or "e-th" root of e
Is it not the case you can simplify that expression a little?
When we have xyln(x), if we move the y to the exponent of the ln(x), we get xln(x^y). But as we stated earlier, x^y = y, so I believe we can simplify xyln(x) to just xln(y).
i started studying mechanical engineering but since we also have math classes I kinda regret not simply going for pure MATH :D