The range being from e raised to its additive inverse, to e raised to its multiplicative inverse, is really aesthetic.

That blue pen is crying about not being in the channel name. It did so much work for no recognition. We salute you, blue pen.

I love how honest you were about not knowing the = sign part! That's great. You still had something to teach me and I appreciate it when people don't try to cover up parts they don't know. First video I've seen of yours and I just subscribed!

A question still remains, why did you not already factored pen in blackpenredpen ? sorry dont judge me ... love ya

I really wish that i could give you a hug seriously your the most amazing and kind teacher ive ever seen ....... im about to cry cause you really helped me a lot ....

You really are an amazing teacher! Thank you

You are my favorite youtuber I never miss your videos

2 of my fav channel 1st one mind your decision and 2 one black pen red pen 🔥🔥

Very good presentation!

Another mind boggling video!

I am not sure but I remember 3b1b asked a question about this as a homework in his lockdown math series, so this video answered the question,great!!

Can't wait to try this out on my calculator! Thanks for the Interesting video.

Steve sir(blackpenredpen) pls reply to this Thank u sir for such tricky questions I am in 8th standard U made my interest in calculus I learnt calculus because of u I am from india Keep challenging us with such questions My name is prittish

Nice chain chomp

Cool beard! I love your chnnel, youre amazing

By looking at thumbnail I was like..I know this😀..after watching full video I realised I knew only 40% of this..Thank You!🙏🏻👍🏻

4:22 The equality is missing because think of g(x) = f(x) - x. f(x) is the recurrence function. If g(x_0) = 0 and dg/dx > 0, in physics it would be an unstable equilibrium as if you got away from x_0, the repeated application of the recursion will take you further from x_0, hence it is not a convergent value. If dg/dx < 0, in physics it would be an stable equilibrium because if you got away from x_0, the repeated application of the recursion will take you towards from x_0, hence it is a convergent value. But if dg/dx is Zero, it is only converging from one side, but diverging from the other Edit : 3b1b's video explaining ne it much better, obviously kzfaq.info/get/bejne/ecyHa5dmsbSzkoE.html

There is a very interesting research paper called “exponential reiterated” that answers your questions among others.

Love the Mario Chomper Mic!

About what happens at the equalities: the upper limit does not generally belong to the domain (or at least it depends on your starting value), but the lower one does. Motivation: We can look at how the function y_n+1 = x ^ y_n behaves to justify that. We look at how a small perturbation eps around the solution behaves (with eps arbitrarily small): - First, for the upper limit where x = e^(1/e) and y = e. You may check that if y_n = e, then y_n+1 = (e^(1/e)) ^ e = e^1 = e as well, so it is indeed a fixed point, consistent with the video. Let's now look at a perturbed value y_n = e+eps. Then y_n+1 = (e^(1/e))^(e+eps) = (e^(1/e))^e * (e^(1/e))^eps = e * e^(eps/e) = e * (1 + eps/e + 1/2*(eps/e)^2 + ...) = e + eps + 1/2*eps^2/e + ... . So we see that a perturbed y_n by an amount eps is mapped onto a perturbed y_n+1 by an amount eps+1/2*eps^2/e (plus higher-order terms that are negligible for sufficiently small eps). That extra quadratic term in eps is strictly positive. Therefore, negative perturbations eps become less perturbed, but positive perturbations become more perturbed. Therefore the solution is stable from one side but unstable from the other. This is called semi-stable, but that *is not* generally speaking a stable fixed point. (Although it does converge for the case where it is specified that y_1 = x, like in the video, since that approaches from below, albeit slowly.) - Next, for the lower limit where x = e^(-e) and y = 1/e. You may check that if y_n = 1/e, then y_n+1 = (e^(-e)) ^ (1/e) = e^(-1) = 1/e as well, so it is again a fixed point, as derived in the video. For the perturbed value y_n = 1/e+eps however, y_n+1 = (e^(-e))^(1/e+eps) = (e^(-e))^(1/e) * (e^(-e))^eps = 1/e * e^(-eps*e) = 1/e * (1 - eps*e + 1/2*(eps*e)^2 + ...) = 1/e - eps + 1/2*eps^2*e + ... . So we see that a perturbed y_n by an amount eps is mapped onto a perturbed y_n+1 by an amount -eps+1/2*eps^2*e (plus higher-order terms). At first sight, this seems similar to the previous case. However, interestingly, the sign of the perturbation flips, from eps towards -eps. So also the perturbations flip back and forth between the stable and unstable side and become alternately smaller and larger in magnitude! To resolve the total effect of that, we look at an additional term of the power series to find that y_n = 1/e+eps is mapped onto y_n+1 = 1/e - eps + 1/2*eps^2*e - 1/6*eps^3/e^2 + ... , and repeating the same idea for another iteration this is mapped onto y_n+2 = (e^(-e))^(1/e - eps + 1/2*eps^2*e - 1/6*eps^3*e^2 + ...) = (e^(-e))^(1/e) * (e^(-e))^(-eps) * (e^(-e))^(1/2*eps^2*e) * (e^(-e))^(-1/6*eps^3*e^2) * ... . Keeping only up to third-order terms in eps, this amounts to 1/e * (e^(eps*e)) * (e^(-1/2*eps^2*e^2)) * (e^(1/6*eps^3*e^3)) = 1/e * (1+eps*e+1/2*eps^2*e^2+1/6*eps^3*e^3) * (1-1/2*eps^2*e^2) * (1+1/6*eps^3*e^3) = 1/e + eps - 1/6*eps^3*e^2. Note that the quadratic terms cancel, but when including third-order terms a positive perturbation is mapped onto a slightly smaller perturbation, and a negative one to a slightly bigger one (since the third-order term -1/6*eps^3*e^2 always has the opposite sign compared to eps). So on both sides the perturbation tends towards zero now. Therefore, this *is* a stable fixed point. Together, the lower limit of the domain converges, but the upper limit only "semi-converges" (which is therefore not generally convergent, and depends on the starting value for y_1). The domain includes the lower limit but not necessarily the upper. #QED Just to be sure, I checked this with some numeric computations, and it seems to work out as derived. Might be a nice followup video in itself? With apologies for the cumbersome plain text notation... ;-)

Want daily doses like michael penn

You are my favorite youtuber

the application of the fixed point theorem is clear for the case z>1; in this case, for zz>e^(-e), and for the bipolar oscillation for z

You are an amazing inspiration. How about doing some multivariate analysis?

I think that in this video kzfaq.info/get/bejne/m9KBib19sc_TlGg.html 3b1b does a really nice job explaining a nice (visual) solution to the puzzle (especially around 19:55). If you watch it, you immidiately understand why the equalities hold, and not only the inequalities. His approach is based on cobweb diagrams: the infinite power tower converges whenever the cobweb diagram between the functions y=a^x and y=x, starting from the input value x, converges to an intersection point of the two functions. The upper bound of this interval of convergence is found when and where y=a^x is tangent to y=x, i.e. there's only one point of intersection between the two functions in which both have slope 1.

I really do appreciate the fact that you are willing to say that you don't know how to get the the equalities So many people just pretend they know, and in my opinion that's just bad teaching

Amazing

My attempt to show that the infinite exponentiation tower converges for x = exp(1/e). It is based on showing that the sequence is monotonically increasing and bounded, hence it converges. This argument does not work for x = exp(-e), because the sequence is not monotonic. I tend to be verbose (maybe tedious!) so this comment is quite long! Let define recursively the tetration sequence as: y(1) = x, y(n) = x^[y(n-1)] for n > 1. Note that it's a parametric family of sequences, and we limit to the case 1 ≤ x ≤ e^(1/e). 1a) Let's first show that the sequence y(n) is monotonically increasing, that is y(n+1) ≥ y(n). For the base case (n = 1) we need to prove that y(2) ≥ y(1), which reduces to x^x ≥ x, or x^x - x ≥ 0. This is a calculus exercise, and one can relatively easily show that a) x^x - x = 0 for x = 1 (by direct substitution) and that b) the derivative of x^x - x is given by x^x·[ln(x)+1] - 1 ≥ ln(x) + 1 - 1 ≥ 0 for x ≥ 1 (since x^x ≥ 1 and ln(x) ≥ 0 for x ≥ 1). This implies that x^x - x is non-negative for x ≥ 1, that is x^x ≥ x, which proves the induction base case. One can also double-check graphically using any plotting tool. 1b) Moving on to the general induction step, let's assume that y(n) ≥ y(n-1), and prove that y(n+1) ≥ y(n). We have y(n+1) = x^y(n) ≥ x^y(n-1) = y(n), which is what we want. I have used the induction hypothesis and the fact that the function x^t is monotonically increasing (as a function of t) for a fixed x ≥ 1. 2) Let's now prove that y(n) is bounded; remember that 1 ≤ x ≤ e^(1/e) ≈ 1.445. First, y(n) is obviously always positive, that is bounded from below. We now show, again by induction, that y(n) ≤ e. The base case is trivial, as y(1) = x ≤ e^(1/e) < e. For the general induction step, let's assume that y(n) ≤ e, and prove that y(n+1) ≤ e. We have y(n+1) = x^y(n) ≤ x^e ≤ [e^(1/e)]^e = e^(e/e) = e. This last inequality sheds some light on why the upper bound of the x convergence range is e^(1/e): it is the largest value of x such that x^e ≤ e. Points 1) and 2) show that y(n) is a bounded, monotonically increasing sequence. Since ℝ is a complete metric space, y(n) converges to a limit L, which concludes the proof. Furthermore, if L = lim y(n) for n → +∞ , by taking the limits for n → +∞ of both sides of the recursive definition of y(n), we have L = lim[y(n)] = lim[x^y(n-1)] = x^lim[y(n-1)] = x^L, which shows that L is the solution to the equation L = x^L. Consistently, this equation has solutions only if x ≤ e^(1/e). PS: step 1a) would become simpler by defining y(0) = 1 (carrying exponentiation "0 times" yields 1) and y(n) = x^[y(n)] for n > 0. This still recovers y(1) = x as it should be. But the base induction step becomes y(1) ≥ y(0), that is x ≥ 1, which is trivially true. The general induction step does not change because it still holds y(n) = x^[y(n-1)] for n ≥ 1. In point 2), the base induction step is similarly trivial: y(0) = 1 < e^(1/e) < e.

Very intersting arxiv paper , simple enough for first year university students 👍

Maybe it's better to think backwards if f (x) is defined over [a,b] it's derivative it's defined over (a,b) it loses ending points

thanks man

Hello blackpenredpen, I looked at math on the internet (cuz it's fun) and I found a video that showed an amazing way to prove Euler's formula, plz put it in a video Let f(X)=e^(-ix)*(cos(X)+i*sin(X)) We'll take the derivative and after simplification (I won't write it because it will be so annoying to write in a comment) we will get that the derivative is 0 meaning that our original function is a constant c and after plugging into the function X=0 we will get c=1 meaning that: e^(-ix)*(cos(X)+i*sin(X))=1 Meaning that: e^ix = cos(X)+i*sin(X) And that's it, I think that this is so amazing and I hope you put this in a video (plz)

Why there are comments 2 months ago even the video is 5 hours ago

wooo very good

looks like i finally have a new favorite maths equation

You can have a limit of lim a→∞ (x↑↑a) maybe?

We should get you a bigger white board. Great stuff.

So basically this problem involves solving x^y = y for the domain and range where this equation is defined. If you are given y, you can the value for x, that being x = y^(1/y). What if you are given x? Can you find the value for y?

You are amazing 💙 sir😍😍😍👍👍❤️❤️

I couldn't understand anything but I still clicked the like button, because it's black pen red pen hence it must be good

Good thank you

that was cooool

I am from India. Your videos are excellent

Given other requirement we need as viewers, it's quite bizarre that you explained fixpoints to us. interesting video tho

You are just amazing 💖❤️, boss of Math

Mind blown

blackredpen Did you know you don't need the chain rule to calculate d/dx(log(9x)? You can just rewrite it as a sum as d/dx(log(9)+ log(x) because of the log property log(ab) = log(a)+ log(b). Log(9) becomes 0 because its a constant but log(x) becomes 1/ln(10)x. Other than that case and a couple of other cases, I recommend using the chain rule but that is also an optimized strategy.

Good

Only 420 people found this secret video?

Cool

Wow, you are wonderful. One question from myside plz. One person forgot his 4 Digit ATM pin but he remembers 4 things about his ATM pin... 1) First Digit is half of Third digit 2) sum of second and third digit is 8 3) Fourth digit is multiple of first & second digit 4) All four digits sum is 12. So, Generate the formula to build the ATM PIN

I really missing the marathon series, Lets do it again plzz

Is it possible to use analytic continuation or other techniques to extend the infinite power tower function?

Can I suggest the idea for your some future video? This idea was came to my mind at old school years. It is to find the function f(x,t) satisfied following conditions: f(1,t) = cos(t); f(2,t) = cos(cos(t)); f(3,t) = cos(cos(cos(t))); and so on.. an x of course is the real number in common case, not an integer! So, what about this ffffunction?:)

Blackpenredpen this is a challenge for you Find the value of x when x⁹⁹=99to the power x Note that x is not equal to 99

IN ORDER TO get the inequalites, for the domain, just plug a number bigger than the bigger number.

超可愛的麥克風

Is this function analytic in this range? If so, can we use analytic continuation to extend it further?

I like watching your videos because I can’t keep up. Does that make sense?

I have one question, why don't you use a simple microphone which you don't need to carry in your hand? Doesn't this become a bit uncomfortable? Btw, awesome video as always!

If you are oriented to a rigorous theoretical approach to the convergence behaviour, that uses no graphic/computational hints (e.g. coboweb...), a complete work is "Reiterated exponentials" by Knoebel, 1981 (free on web); divergence, convergence and alternate oscillation are rigorously explained using classic real analysis theoretical tools. An accurate, deep study of the complete proof presented in this article provides all the necessary background if you are interested in a rigorous approach to further generalization of this problem, (e.g. infinite power tower with variable exponents) as is treated by more recent literature(*) (* see e.g.: F.J. Toledo "On the convergence of infinite towers of powers and logarithms for general initial data: applications to Lambert W function sequences")

but can you integrate this or find length of the curve

@blackpenredpen Can you explain how the convergence of fixed point rule was derived?

Cool..

Blackpenredpen 😎😎😎 cool

5:09 I thought this formula was only applicable if the base were a constant. Can someone explain how he used it for (x^y), even though x is dependent on y?

Least popular video of BPRP ever after so much time... Great Video, no crowds at all😄😄😃😁😁😂

nice pokemon

It's been a while since I've taken calculus, can someone remind me why we know the function converges if the absolute value of the function's derivative is less than one?

Did you bought the wireless mic???😂😂😂

Your mic is the best

*You are really my true teacher,I am doing Mathmatics at in India,first year*

3b1b has a Lockdown math video about this infinite power tower

Nice Mic!!!!!

I have a doubt related to integral of √tanx. For the integration of dx/x²-a², we can also use 1/2a ln|x-a/x+a| instead of hyperbolic tangent right. Edit: please clear my doubt. I hv an exam in my skl tomorrow

Wasn't this video posted earlier?

AZE..Salam aleykum...Halal olsun əmoğlu😁😊

Hi sir how can we suggest you or ask you intresting question's? Please tell

Given an acute non-isosceles triangle ABC with circumcircle Γ. M is the midpoint of segment BC and N is the midpoint of arc BC of Γ (the one that doesn't contains A). X and Y are points on Γ such that BX || CY || AM. Assume there exists point Z on segment BC such that circumcircle of triangle XYZ is tangent to BC. Let ω be the circumcircle of triangle ZMN. Line AM meets ω for the second time at P. Let K be a point on w such that KN || AM, ωb be a circle that passes through B, X and tangents to BC and be a circle that passes through B, X and tangents to BC and ωc be a circle that passes through C, Y and tangents to BC. Prove that circle with center K and radius KP is tangent to 3 circles ωb, ωc & Γ. Can you solve this question plz...

Lagrange resolvent...Please explain that

but y=x/2 has a slope smaller than 1 and it still doesnt converge. shouldnt the slope be zero at infinity for it to converge in infinity? but y is not infinity so at the border of our domain our range and our slope will be infinity or rather undefined

Hi sir i'm your student (sri lanka)

What is the marker's brand?

Don’t disturb me during PSG BAYERN

Can I ask a question? I wanted to solve an ODE: y'' + xy' + ky = 0, where k is constant.

Hey! Can you please make a video about the Zeta function at irrational number Such as 'e' Or pi'' Please try it Hey blackpenredpen please reply👍👍 or any body can

Can you name a best book of calculus (differentiation,calculus, area related to calclus ,and more .) Which can make me to zero to best? Please any one @ *blackpenredpen*

Sorry I'm not understand that how you tell about the fixed point, for x^y =y , you are considering the input as y ,the output is y too, then for the coordinates you just got (x,5) let say y=5 ,but x isn't fixed isn't it ? How come the point fixed ? For one more ,the x should be the input the y would be the output isn't it ? Is the y in the x^y is the input ,while the y on the right hand is output , the whats x suppose to be ?

Bprp what is the factorial of π

Misss you bprp

wait didnt you already post this video a few months ago?

Can anyone explain how did the inequality at 4:15 come from Any proof?? I really tried a lot to find it Please someone help How can we prove the criteria for convergence of fixed point?? 🙏🙏🙏 It's bugging me for days

Why is it unlisted??

This guy is really genius in maths

But why do you have a chain chomp?

0:52 that's what she said🥺😞😭😭😭

Doesn't d/dy(x^y) equals d/dy(y) which is 1, so the point is not convergent?

I have a challenge for you. Find the deriative of sin^x(x) (the sin(x) raised to the xth power).

5:25 i thought this formula works only if b is constant

Since when did he have a blue pen