they don’t teach these kinds of expoential equations in algebra

[@blackpenredpen]

The other parts will be published later. If you cannot wait, then check out my other channel (in Mandarin): kzbin.info/door/rONDbyO94HIrJyfhS6qfjAvideos

[@pauljackson3491]

Well that answers my question. What "dialect" of "Chinese" do you speak? They are in quotes because Chinese isn't just one language and things like Mandarin and Cantonese aren't dialects.

[@youkaihenge5892]

@Michael Bishop I speak and write japanese and I love how its simplified Kanji from the Chinese Dialect 😃 Bad part is Japanese read each Kanji with two or more readings called 音読み and 訓読み

[@alfiangunawan5946]

i just know that you have other channel in mandarin. thanks a lot for making that channel!

[@erikkonstas]

Why don't you feature your two other channels?

[@prateekmourya9567]

I have really good question for you if alpha and beta are roots of equation x^2-x-1(Fibonacci equation) we define a function f(n)=alpha^n+beta^n then prove that f(n+1)=f(n)+f(n-1)

[@douro20]

We never talked about W in algebra class when I was in college. I think it would had been a good subject for the class.

[@thorstenbrandt8739]

College is not "scientific" enough...

[@maxamedmuuse4882]

You already know what blackpenredpen will say....hello! lovely voice.

[@blackpenredpen]

Lol thanks!

[@vietphamhung3290]

@@blackpenredpen you are so cute 😍😍

[@pauljackson3491]

I like it how you say "Let's do some Math for fun." Isn't all Math fun!

[@erikkonstas]

Only pure, pure math. The things they make seniors study in Greece are outrageous...

[@erik-ic3tp]

@@erikkonstas, What's wrong with math education in Greece? Your country once pioneered mathematics.

[@pbj4184]

Math by itself is fun but when you're forced to memorize stuff for exams, it loses its value

[@erikkonstas]

@@erik-ic3tp Man, you don't want to know, it's horrible. And, yes, this is a great shame some of us think could land the country in total despair very soon. Thales's, Pythagoras's, Archimedes's, Euclid's and the other great mathematicians' bones are shaking...

[@jofx4051]

Alright, I am already expecting Lambert function here

[@youmemeyou]

Ok

[@mohammadfahrurrozy8082]

Lambert function is so cool!

[@Nylspider]

You thought right

[@jcers]

I smell Lambert W function incoming I was right

[@blackpenredpen]

Yes

[@gigachadkartik]

You can substitute x^3 as other variable and then solve

[@Meilo0110]

You're the reason why I'm good at math, I'm only 17 and I already mastered algebra and some parts of calculus thanks to you. Keep up the amazing work. You deserve 1 million subs or even more. Greatest math teacher ever♾️🙂

[@Careerhumresource]

I'm 11

[@Jupiterninja95]

Whoa!! I didn't know you could divide videos into sections like that!

[@blackpenredpen]

Yea it’s a new feature. I think it’s so cool!

[@nintendoswitchfan4953]

Same , i think its cool too

[@ryderpham5464]

Woah the new KZbin UI on this video is so cool! I love seeing the timebar all sectioned and labelled ❤️

[@advaitanand1864]

You really explain every problem very well.

[@Nylspider]

I love the abrupt starts! You always are so excited to get into the math! Are you going to do some videos on the AOIME problems? #YAY

[@integralboi2900]

I wonder what’s going to be in part 2,maybe a infinite power tower?

[@integralboi2900]

*an

[@blackpenredpen]

Yes. Will release that this coming Friday

[@leif1075]

Important question. I took ln of both sides. Then i divided both sides by x^3 so i have ln x equals ln 2/x^3..Then i took thebderviative with respect to x of both sides to get rid of the ln x and be able to solve for x. Taking the derivative of both sides gets me 1/x equals ln 2(-3x^-4)..therefore solving for x you get x^3 equals -3(ln 2)..so taking the cibe root yiu get x equals the cube root of -3(ln2). Didnt a lot of people do it this way? it's totally valid i dont see why not? Especially if you dont k ow Lambert's function. It's still the same x variable even though i took the derivative so it's the correct value.

[@OriginalSuschi]

Leif ok after taking the derivative we have 1/x=-3ln(2)/x^4. After multiplying by x^4 we get x^3=-3ln(2) so: x=cbrt(-3ln(2)) The only things possible here are: 1. We did a mistake 2. There are more than one answer for this. 3. Both solutions equal to each other. When this is right, you've proven that e^(W(3ln(2))/3)=cbrt(-3ln(2))... But it's sadly not :( It's really weird. I can't spot the mistake here... bprp pls help

[@angelmendez-rivera351]

Leif No, it isn't valid. The fact that two functions are equal to each other at a point does not imply their derivatives are equal to each other at that point. In other words, f(x) = g(x) at x = c does not imply f'(x) = g'(x) at x = c. I have very simple counterexample to prove this. sin(x) = cos(x) if x = π/4. Taking the derivative yields cos(x) = -sin(x), and substituting x = π/4 implies 1/sqrt(2) = -1/sqrt(2), which is false. This just proves that taking the derivative is not a valid operation here.

[@ffggddss]

Ah, I think I can see how this will go. First take ln of both sides, then a little sleight of hand to make it fit the Lambert W function. Same approach for both problems; call the constant on the RHS, "a" : x^(x³) = a x³ lnx = ln a e^(3 lnx) lnx = ln a . . . . . . [from this line on, is a correction of what I had initially.] e^(3 lnx) 3 lnx = 3 ln a 3 lnx = W(3 ln a) x = e^(⅓W(3 ln a)) With a = 2, 3, the results are: 3 ln 2 = 2.07944154...; W(3 ln 2) = 0.8706335...; x = e^(⅓W(3 ln 2)) = 1.336709735... 3 ln 3 = 3.29583687...; W(3 ln 3) = ln 3 = 1.0986...; x = e^(⅓ ln 3) = 1.442249570... Post-watch: AHA! I should have gone . . . x = e^(⅓ ln 3) = ∛3 = 1.442249570... Fred

[@adamwright4634]

I just raised both sides to 1/3, then took the natural log, and then the product log, and then just solved for x to get, x=e^w(ln(2^1/3)) for the first one, same method for the other one as well

[@Vivek-io3gj]

Year late reply but that doesn’t work because then it would be 1/3 x^3 in the exponent and would not cancel the cubed

[@adamwright4634]

@@Vivek-io3gj could you not swap around the exponents?

[@Vivek-io3gj]

@@adamwright4634 you cant because it’s x^(x^3)^1/3 not x^((x^3)^1/3)

[@youkaihenge5892]

Could you do a video showing that if a Sequence is "Cauchy" that it converges? 🙂 One of my favorite Theorems.

[@arctan-k]

3:25 We say kinda same in Kazakh

[@rob876]

Thought you might be interested in this :- written in a language that won't die too soon: -- cannot declare variables in postgresql - we're using 'from ( select 5.0*exp(5.0) as z ) as declarations' instead with recursive lambert_w as ( select z, 1 as n, case when z 1.0e-41 ) select n, w, w*exp(w) from lambert_w And, in a language that is about to die (VBA): ---------------------------------------------------------------------- ' The Lambert W function is the function W(x) such that W(x)*exp(W(x)) = x ' or W(x*exp(x)) = x, since W(W*exp(W)) = W if we take W of both sides of the above equation. Public Function LAMBERTW(x as double) As Double Dim W, eW, WeW, WeW_x, dW As Double ' First guess for Lambert W If x

[@sobhansyed4482]

you can also make e^[(1/3)w(3ln3)] into cube root 3 by making the first 3 in w(3ln3) e^ln3 cuz it becomes w(ln(3)e^ln3) they cancel so it becomes ln3 then e^(1/3 ln3) can become e^(ln(3^(1/3)) e and ln cancel so you end up with 3^(1/3)

[@angelmendez-rivera351]

Mmm... yes and no. The Lambert W map is multivalued. So while one of the two branches -1 and 0 will evaluate to this, its objectively more useful to leave it in terms of the Lambert W expression because of the branches.

[@hamiltonianpathondodecahed5236]

umm didn't he do this in the video ?

[@michaelroditis1952]

Man I love you! You are probably the most friendly guy on youtube!

[@DanBurgaud]

I like how you manipulate ln, e, W to come up with great solutions...

[@einsteingonzalez4336]

3:23 So I analyzed the Chinese that you spoke in your channel, and found out that it's 二分的[what number] or 三分的[what number], and so on. That means a half, a third, or a specifc number of times that the number in the numerator is being divided.

[@JordHaj]

6:01 this reminds me infinitely nested Michael Jordans!

[@Kdd160]

很棒的視頻！我先解決了問題，然後看了您的視頻。您解決的方法是完全相同的:)）

[@CC--qn4gf]

That last demonstration was HOLY

[@robsbackyardastrophotograp8885]

I like the little timestamp chapters. When did YT add this?

[@christianmanuba539]

Hey! I know this is a bit irrelevant to the topic of this video but can you try and solve inverse laplace using contour method? Thanks😊😊

[@prashantshukla6018]

Your the best mathematician PLZZ can u make a video on question of Indian prmo questions they're very tough.. thankyou sir

[@aditya_01_jha]

Hi sir great job. I am from India Love you sir♥️♥️♥️♥️♥️

[@arctan-k]

btw, in 6:49 we can do it until the number gets bigger than e^(1/e), this is kinda cool

[@blackpenredpen]

Yea if u only have finite amount of x

[@hilmi-litv1179]

I don't know why but I love this guy

[@ejb7969]

It's the fish, and the smile, and lately, that little beard ... ... "isn't it?" You rarely get any of that with math teachers.

[@alanclarke4646]

In the Observation, where does the 3= ( x to 3rd power) come from?

[@eriktruong9856]

Are you allowed to derivate both sides of the equation instead of using the W function?

[@2ndtik]

Pre-celebration for your channel getting 500k subs!!!!

[@ralphocava4130]

Best outro music for math... 2020 is extracting the good juice from people...

[@cosmolbfu67]

x^(x^3)=2 (x^3)^(x^3) = 2^3 (x^3)•ln(x^3) = 3ln2 In(x^3)•e^ln(x^3) = 3ln2 ln(x^3) = w(3ln2) x = e^(w(3ln2)/3)

[@OmnipresentPotato]

Since e^(⅓W(3ln3)) is the same as ³√3, which is basically 3^⅓, does that mean that e^W(3ln3) is 3?

[@nasserdawood2171]

For eq(1) x^x^3 = 2 We can also use ssrt method. (X^3)^(x^3) = 2^3 x^3 = ssrt(8) x = (ssrt(8))^(1/3)

[@lgooch]

You can use it in bith

[@cavansirhasanzada1755]

Excellent work

[@Chill----]

Blackpenredpen you solved the question elegantly. Very much impressed!

[@musicisthefoodofthesoul]

I’m new to this channel. The π at 0:29 looks like the π in 3blue1brown 😁

[@blackpenredpen]

Because it is! : ) I got it from his merch store!

[@tungstengaming3440]

I love this channel

[@gigachadkartik]

Say x^x^3 = k Put x^3 = t this means x=t(1/3) (t(1/3))^t =k t^t/3 =k t^t=k^3 If k=3 by observation t=3 aka x=3^(1/3) If k=2 solve t^t =8, t is about 2.9 aka x=1.33

[@R2242V]

Thank you, I was looking for this kind of solution.

[@equalcell1555]

Where is your old board?

[@danikhan5185]

Helooo blackpenredpen.I love ur videos and they make math a lot more interesting. can you plz help with this trigonometric question: 2cos^2x - sin^2x - 2sinx - 1 = 0. Thanks!

[@angelmendez-rivera351]

cos(x)^2 = 1 - sin(x)^2, hence 2·cos(x)^2 = 2 - 2·sin(x)^2. Therefore, 2·cos(x)^2 - sin(x)^2 - 2·sin(x) - 1 = 0 is equivalent to 2 - 2·sin(x)^2 - sin(x)^2 - 2·sin(x) - 1 = 1 - 2·sin(x) - 3·sin(x)^2 = 0, which is equivalent to 3·sin(x)^2 + 2·sin(x) - 1 = 0. Hopefully that helps you finish it.

[@integralboi2900]

That = cos(2x) + cos^2(x) -2sin(x) -1 = cos(2x) - sin^2(x) - 2sin(x) = 1-2sin^2(x)-sin^2(x) - 2sin(x) = 1-3sin^2(x)-2sin(x)=0 I’m pretty sure you know how to use the quadratic equation, so I’ll leave it here.

[@jofx4051]

Plugged identity cos^2(x)=1-sin^2(x); you would solve for sin x = 1 or sin x = -1/3 The answer for it would be (in degree; it has periodic answer): x = 90 + 360*k or x = arcsin (-1/3) + 360*k (I cannot reverse sin 1/3 except using calculator, except your teacher allow it, I would leave it in this form)

[@yassinezaoui4555]

I'm not bprb obviously but I suggest that u can turn the expression into à quadratic equation in terms of sin(x) then use the quadratic formula to seek for solutions. Meaning: 2cos²(x)-sin²(x)-2sin(x)-1=0 2(1-sin²(x))-sin²(x)-2sin(x)-1=0 -3sin²(x)-2sin(x)-1=0 We may notice that a-b+c=0 so sin(x)=-1 or sin(x)=-c/a=-1/3 So x=-pi/2 + 2kpi or x=arcsin(-1/3) + 2kpi where k is an integer.

[@danikhan5185]

Wow thanks guys.This community is the best!

[@KiNG_282_]

I don't know why I can't sleep without watching your videos 😍😍

[@2false637]

Hey BPRP, do you know any good resources for analysis?

[@blackpenredpen]

Dr. Peyam and Prof. Penn's channels : )

[@DanDart]

So it looks like a property of W is that W(xlnx) = lnx as well.

[@copperfield42]

so in general x^x^c=c --> x=c^(1/c) for all c>0 (as that is the domain of ln)

[@behzat8489]

yes it is working even you add more powers x^x^x^..x^c=c x=c^1/c

[@angelmendez-rivera351]

Yes.

[@ttchip8464]

The thing that kinda irks me tho is if you need to use the Lambert W function, you are likely using software like wolfram alpha anyway to solve it, which means why not just plug the original in wolfram alpha anyway since u need to use it either case

[@tobyzxcd]

Nearly HALF A MILLION subs!!!!

[@buler5441]

W E C A N D O THE Follow yng 5:16

[@prabhakarmishra9042]

Really interesting... Is there any other methods of solving this question?

[@frankk2231]

Newton's method is quite fast 1.5 1.40 1.348 1.3371 1.33671

[@andreimiga8101]

Hey bprp! How do you integrate W(x)?

[@angelmendez-rivera351]

To integrate W(x), you use a u-substitution. Let u = W(x). Hence x = ue^u, hence dx = (u + 1)e^u·du. Therefore, when you convert everything "to the u-world" and simplify the integrand, this becomes the antiderivative of u(u + 1)e^u with respect to u. Integrate this by parts by differentiating u(u + 1) to 2u + 1 and antidifferentiating e^u to e^u. This results in u(u + 1)e^u minus the antiderivative of (2u + 1)e^u. (2u + 1)e^u can itself be antidifferentiated by parts by differentiating 2u + 1 to 2 and antidifferentiating e^u to e^u, resulting in (2u + 1)e^u minus the antiderivative of 2e^u, which is just 2e^u. Therefore, the antiderivative of u(u + 1)e^u with respect to u is given by u(u + 1)e^u - (2u + 1)e^u + 2e^u. Notice that u = W(x) and x = ue^u. Therefore, u(u + 1)e^u - (2u + 1)e^u + 2e^u = [W(x) + 1]x - 2x - e^W(x) + 2e^W(x) = x[W(x) + 1] - 2x + e^W(x) = x[W(x) + 1] - 2x + x/W(x). With all the simplifications completed, don't forget to add your +C. Therefore, the antiderivative of W(x) is x[W(x) + 1] - 2x + x/W(x) + C.

[@aspectator3680]

Hello @blackpenredpen, whats about x^x^x=2 ?

[@purpleontop2133]

Love the chain chomp

[@tasninnewaz6790]

Please upload factorisation of cyclic expressions using factor theorem and How it works and What is the real life apllication of cyclic polynomial ?

[@prateekmourya9567]

I have really good question for you if alpha and beta are roots of equation x^2-x-1(Fibonacci equation) we define a function f(n)=alpha^n+beta^n then prove that f(n+1)=f(n)+f(n-1)

[@volodymyrgandzhuk361]

What about super quadratic equations?

[@angelmendez-rivera351]

It works more or less the same.

[@Exachad]

Almost 500k!

[@Mephisto707]

How do you know there are no negative solution? Also, Since Lambert W function is multivalued, can’t there be more positive real answers?

[@daurenyermenov4766]

Does it mean that exp(1/2*W(2ln2)) = sqrt(2)??

[@blackpenredpen]

Yes

[@angelmendez-rivera351]

Yes. You probably already see the pattern: exp(W[n·ln(n)]/n) = n^(1/n) for any positive integer n. This also implies that W(n·ln(n)) = ln(n) for any n. In fact, more generally, W[x·ln(x)] = ln(x) for any x satisfying the domain restrictions (and if you allow complex numbers, then for all nonzero complex x).

[@manishkumartangri2521]

Please make a video on ith derivative of x^i

[@martinzone8153]

What'the link for the chinese channel?

[@mathematicalworld4063]

didnt you also have a video for x^x^x^2017...... x=2017^1/2017

[@denielalain5701]

Hello! I jumped to it the following way, and my solution got very peculiar compared to yours, and i could only wonder wether i made a mistake, or i could bring a product out from the Lambert W function. I got it like this: (x^x)^3=2 x^x=2^(1/3) x*lnx = ln(2^(1/3)) e^(lnx) *lnx = (1/3)*ln2 W(e^(lnx) *lnx) = W((1/3)*ln2) lnx = W((1/3)*ln2) x = e^W( (1/3)*ln2 ) instead of your x = e^( (1/3)*W(3 * ln2 ) ) Is it a coincidence, or have i made a mistake, or what is going on?

[@zhendai]

Love the vids !!

[@theultimateg.1109]

If you look at the duration of the video, it says Hello next to it.

[@moregirl4585]

Is it allowed to still say x=e^(W(3ln3)/3)=e^(W(ln3 e^ln3)/3)=e^(ln3/3) in an exam where W is not taught?

[@palmolive2005]

what about x^x^y=y?

[@JoseVieira-hs9qo]

HI @blackpenredpen can you tell me how I get, using Lambert as you did, the negative value of the result in this case?: x^x^2=16, the result should be +/-2 or am I missunderstanding this?

[@AllanPoeLover]

5:38 時 為什麼 3 = X^3 ? 這段是怎麼得來的 ?

[@blackpenredpen]

反推 你從那先看的話 你會發現你可以一直把x^3放上去

[@mathadventuress]

What course is this taught in? I haven't covered this

[@angelmendez-rivera351]

Ideally, in a precalculus course.

[@RogerGirardon]

x^(x-1)=243 find x ?

[@ericherde1]

bprp has an exponential that you don’t know how to solve? It must be Lambert W!

[@Yazoon_S]

This reminded me of the good ol’ school days EDIT: good thing that i still have the school notebook , so i can relook at the ln laws if (somehow) needed

[@alfiangunawan5946]

does that kind of equation always have 1 solution?

[@thorstenbrandt8739]

What's the shark for...?

[@andreemery4964]

For the second, you can note that W(3 ln3) = W(e^ln 3 * ln 3) = ln 3 and e^(1/3 * ln 3) = 3^(1/3) edit: i got too far ahead of myself, and it turns out he had this in the end of the video

[@funnyyoushouldsaytha]

Man this was a great video

[@egillandersson1780]

Very nice video !!!

[@MrRyanroberson1]

there are two more solutions as well for the right side: (-1)^(2/3) and (-1)^(4/3) times the cuberoot of 3

[@Shreyas_Jaiswal]

But you need the solutions to be in real world.

[@MrRyanroberson1]

@@Shreyas_Jaiswal they are in real world, they just happen to not fit in the set of real numbers.

[@Shreyas_Jaiswal]

@@MrRyanroberson1 real world means on earth. But complex numbers are on moon. 😂😂

[@supersaiyan2]

Can you solve for x: x^(x/(1-x))=1

[@mamadou3076]

Beautiful What is the brand of your pen please ? 😅

[@ayushgangrade2443]

Simply comparing x^3=1 Taking cube roots of unity W, w^2,1 and further simplification

[@nuklearboysymbiote]

三分之一 3 parts take 1 haha! This feels like watching a translated version lol

[@curiousaboutphysics8605]

An Easy peasy question for you- Find the sum of series: sin x+3sin 3x+5sin5x +...... +(2k-1) sin(2k-1) x. (Using calculus)

[@vari1535]

He said "one-third" correctly and then corrected himself..

[@blackpenredpen]

Eh, when?

[@vari1535]

@@blackpenredpen 3:13 and 4:20

[@michaelroditis1952]

@@vari1535 He wrote the denominator first. I find it adorable that he corrected himself for that!

[@vari1535]

@@michaelroditis1952 Ohh, I thought it was what he _said_ lol

[@R2242V]

I tried to derivate it all the way but then x^x^3 = 1 not 3. Can it be solved without that "W" thing?

[@OriginalSuschi]

We can prove something with these solutions: e^(W(3ln(3))/3)=3^(1/3) W(3ln(3))/3=ln(3)/3 W(3ln(3))=ln(3) Like already said in the video we can plug in any number threre, so we have something really interesting: W(x(ln(x))=ln(x) Yeah but this makes totally sense since we can do x=e*ln(x), so we get W(ln(x)*e^ln(x)=ln(x)... Didn't watch your video till the end hahaha

[@isaacgai3301]

you are from hong kong or mainland?

[@apfelsamenn]

Hey is your Chinese channel 黑比红比 abandoned? There's no video.

[@blackpenredpen]

I don't use that anymore. But here kzbin.info/door/rONDbyO94HIrJyfhS6qfjA

[@apfelsamenn]

blackpenredpen gr8! just followed😁

[@blackpenredpen]

Thanks! How did u know that channel btw? It’s been so long lol.

[@apfelsamenn]

blackpenredpen Google'd "blackpenredpen Chinese" and found this m.kzbin.info/www/bejne/bpuZYmSlnql2pas

[@navaneethnani1481]

I am great fan you sir! You are my cool! 😎 Maths teacher 💯

[@sthinvc]

Can I have the link for your Chinese channel as I am from Hong Kong in which Chinese will be more familiar with me.

[@matron9936]

Just use the super root, after letting u=x^3. (tetration)

[@coronaman365]

What's it???

[@lilichavez4770]

Can you do a video for Kids?

[@Oskar-zt9dc]

but what are the complex solutions

[@angelmendez-rivera351]

The complex solutions are given by the complex branches of the logaritm and the W map respectively.

[@educaticher]

In the first case I got a different solution. Is it right? x^x^3 = 2 x^x = 2^(1/3) xLn(x) = Ln(2^(1/3)) Ln(x)e^(Ln(x)) = Ln(2^(1/3)) Ln(x) = W(Ln(2^(1/3))) x = e^(W(Ln(2^(1/3)))) x = 1.210334...

[@satyamverma101]

Okay but do you now what 3*430!! is?

[@angelmendez-rivera351]

3·430!! = 3·(2·215)!! = 3·2^215·215!

[@moshadj]

e^{3ln(x)} = x^3 for positive x