the most fun derivative of x^x^x^...
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- Опубліковано 6 жов 2017
- Derivative of infinite power tower x^x^x^...
Can you solve x^x^x^...=2 vs. x^x^x^...=3 • Video
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#calculus #blackpenredpen #math
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
Instructions unclear; impressed too many girls
Instructions were clear, mate
Instructions clear: mate
How stupid you are. Lol
lol
How many?
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
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__cruz that arrow he draws is like a huge integration symbol
I had the same initial wtf thought xD
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.
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
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.
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.
thank you so much, you make integration/differentiation fun
I love maths now
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")
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.
that is just awesome
no wait a minute its better than that
can't wait for next video
This was very entertaining, thanks!
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
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.
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
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
Ya boi finna go impress some girls
this is gold you're great!
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
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
Cant wait to impress the guys with this ;)
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 🙂
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.
Very cool. Thanks!
I got this as one of my practise questions when I was preparing for my uni entrance exams in a course...
This made me sub. Thank you sir
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)
blackpenredpenbluepen
I made that joke before...
yet, reading it from you makes me want a to see a BlackPenRedPen & 3Blue1Brown collaboration
@@Theraot BluePenBluePenBluePenPencil
Wonderful explonation ,Sir. Thanks
thank you!
you are broking my mind) it's good
I have a question. How does it works the dx/dy as a constant or variable? I've had half year of calculus but never seen it before
can you do a video on statistics or factorials or matrices?
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.
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^...)))].
You need to simplify more
dy/dx=y²/x (1-lny)
😇😇😇great job
Some say the x’s still go on to this day
"I have no idea what this is anymore"
Your videos are awesome!!
趣的視頻關於導數﹗Again a wonderful and interesting video about derivatives 加油﹐加油
And which is the convergence domain of this function? Isnt it identically equal to 1?
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) }
Could we use implicit differentiation using Newton's notation? Because it seems like Leibniz's notation is way more useful.
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 "_"?
This guy is great , my new holiday hobby.
May you do a video about the nth term of the sequence {-1/3,1/3,0,-1/3,1/3,0….}
I would be grateful for a similar explanation for the antiderivative of x^x^x...
Lots of fun and completely AWESOME! I'd love to show Gottfried Leibniz this. He could taunt Newton with it!
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
Lol I love watching videos about stuff I have learned yet.
Goddamnit, I love this.
fantastic sir
3:38 that sign look like a big integral 😂
the range of this function is from 0 to e^(1/e) or "e-th" root of e
That is insane
I believe it converges for any value between one over e and the e root of e
Well, it does converge for 0
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
Including derivate of this infinite tetration is also calculate the main function -W(-ln(x))/(ln(x))
🤣🤣🤣 I love that the motivation for this derivative is to impress girls. I'm definitely stealing that!
at the end: y*ln(x)=ln(y) tho, it looks nicer that way
especially when you do it as
(x^x^x...)^2 / (x * (1 - ln(x^x^x...))
Convergent for all x such that 0
It converges for values of x between 0 and about 1.445
Hey man, could you please do - arcsin(ι)? I tried but i could do it with my high school maths
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.
i started studying mechanical engineering but since we also have math classes I kinda regret not simply going for pure MATH :D
enjoying night here in india
but i think its day there yeaah
this creates an awkward situation cause you uploads during your day but i gets them when i am up during my day
No idea what all that means but you sounded smart.
But in all seriousness, I am trying to self-teach myself calculus in hopes of getting a degree in physics. Although, it is very confusing hopefully, one day I will be able to understand all of your videos. Until then keep up the awesome videos ^-^
Thank you! And I wish you the best of luck!
Calculus isn't hard (espesially differentiation). I think the best way to self-teach calculus is watch whole calculus course somewhere like MIT OpenCourseWare or coursera.org
Good luck!
Im teaching myself with the AP calculus BC course on khanacademy.org
The calc 1 topics are really cool and not difficult, Im gonna start infinite series soon
If you want a more college-like course watch lectures and use the college course on khan
Good luck!
Watch '3 Blue 1 Brown' to get a good understanding of Calculus concepts, and 'Dr. Physics A' for the gist of many physics topics.
Do you understand it now?
You can also write the final answer as y / x (1 - ln(y)), since y ln(x) = ln(xʸ) = ln(y).
Can you do a series expansion of e^x^x^x?
So seeing as the derivative of x^(x^x) includes the derivative of x^x in its solution, would it be right to assume one just follows on from the previous, and then would it be possible to form some sort of general equation for the derivative of [x to the x to the x to the x....] with 'n' number of x's in the power tower?
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)
Sir, any idea how can we integrate the same?
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 wonder what this function would represent in real life;
probably some kind of recursion, where a function is taken to the x:th power, and then mapped back to itself.
Or something like that.
Instead of implicit differentiation I just move all to one side and do -partial x/ partial y = dy/dx
Much Love
I got my class to study the function y = the x-th root of x, ie y = x^{1/x}. that satisfies x = y^x.
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)
You're very brillant for math
You won my like
Thanks!
blackpenredpen You're welcome
Pretty sure the range -1 to 1 converges but idk if anything else does.
my math teacher had me do this, i had no idea how
Lovely, a simply beautiful derivative. Another cool and "easy" derivative would be derivative of f(x)=x^ln(x)^x ... Try it, please :3
You forgot to check if x-xyln(x)=0 because if it is, your have to exclude values of x and y.
Please do the limit of this same function (y=x^x^x^x...) as x approches i (the imaginary unit)
Perfect ❤
Wow ... that`s great
And i think you should differentatied all infinite expression you done ;)
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).
4:35 hahah impressing girls be like, hey honey, look at this huge exponential function
the max value on which converge is e^(1/e) and it converges to e
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?
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.
Bro this guy is straight up hacking math
Since the exponent is infinitely repeating, I think this can be simplified by factoring x in the denominator, which devides into (x^x^x^x^... )^2 which is x^2x^x^x^x to form x^(2x^x^x^...^(x-1)) , the y can further be shifted into the ln to form ln(y) since (x^y=y), leaving x^2x^x^..^(x-1)/(1-ln(y))=x^2y^(x-1)/(1-ln(y)) or am I too stoned for infinities at the moment?
That's my boy,👍👍👍