So, I took 11 semesters of advanced math in college and grad school, and I am constantly amazed at how much I don't know when I watch your excellent videos.
@@blackpenredpen and the lam ert finctiom anyway? Does anyone actually understand what it means? And it soubds like youncintradict yiurself in the video later when you say cube root of 3 isna solution..Hope you can respond.
For anyone lost watching comments from 3 months ago: This video was unlisted and its link was put in the description of another video. So, just a few people watched this part 4. Yesterday, Blackpenredpen changed the settings and now it is public. Have a great day!
My mind blew at 5:12 So the issue was transforming the equation to another equation does not work because their domain:ranges are different. What kind of equations can this happen with? Clearly when infinity comes into play, also whenever imaginary roots are possible. But how do you find functions with limited domains? It seems like an easy trap if you don't know this could happen.
Very suspicious results, lol. If I wasn't into maths (such as being interested, obviously not super high level), I would probably assume that these things are made up. But honestly, it's super satisfactory even when you consider easy questions and discuss them through as if you were tackling a major problem. So satisfactory, as usual, bprp!
i love how thanks to you i can improve my math skills and im still in 10 grade i started learning complex numbers with you and you inspired me to think out of the box and more abstractly i hope one day i may know at least a fourth of what you know
One can figure that 3 is a value that cannot be converged to because the function f = x |-> x^(1/x) attains a global maximum at x = e in the positive real numbers. Hence if x = 3, then 3^(1/3) < e^(1/e).
Blackpenredpen you are a true inspiration for me, taking a Calc 1 class for my BA bachelor ( Idk why they include it) I have motivation from your videos that I did not have before. I am certain that I will do everything to pass in my class, I am stuggling with some things but your story and your enthusiasm brother blackpenredpen , God will help me on the way to succes hopefully! Thank you for sharing your sweet knowledge brother !
10:14 so what you're saying, is that the two graphs (and therefore the two equations) are only equivalent on certain intervals, and 3 is outside of this interval, but this bags the question, how do you find the interval in which the two equations are equivalent?
For the infinite power tower, I am curious why it doesn't converge at values between 0 and e^-e. I tested 0.05 up to 30 copies of 0.05 and it jumps back and forth between ~0.1 and ~0.7 getting closer towards some value between those two numbers so it seems like it would converge to some specific value. I wonder if values between 0 and e^-e exhibit some kind of bifurcation chaos where they stay finite but bounce between two values and thus diverge in that way.
Here is a question for you: Given: A=(1/2)sqrt((1/2)(3-sqrt(5))) B=(1/4)(sqrt(5)-1) It can be shown that A=B Question is, given A, how do you turn it into B?
So I suppose the different values for W come out when you use different cube roots of 3. All these multivalued functions are countable, so it should be able to be bijected.
f(z) = ze^z is not monotonically increasing and thus the inverse f^-1(z) = W(z) is multivalued. Quite easy. Same with sin, cos, tan except those are periodic and easy to understand.
Hello! Can you make the video where you get the link between the Riemann zeta function and Bernoulli numbers kind of B(t) = -t*Zeta(1-t). How is it gotten? I triied to get this equotion but I couldn't - only my brain was broken! Thanks, I hope for this video will appear soon...
I came to same equation 3^(⅓k) = k, but then I took ln() of both sides ⅓k ln(3) = ln(k) Divide both sides by k ln(3)/3 = ln(k)/k k = 3 Why is it wrong?
I am really trying to find the derivation of the fixed point convergence criteria I tried everywhere Didn't find anything Not even in lot of papers Plz someone help me
@ 2:20 - when you have (3^(1/3))^k = k, couldn't you take the k-th root on both sides, so the k exponent disappears on the left (k/k=1), and the equation becomes 3^(1/3) = k^(1/k), therefore k=3? where is my logic flawed?
Edit: just watched the rest of the video, so 3 can be a solution. This is technically a way to solve the problem, but if you solve this way without using W function, the other solutions are not intuitive.
I think that assumption is how you get the false solution. tower = x^tower = 3, and plug in 3 for tower. Well, the tower built from 3 actually diverges toward infinity.
So interesting pardon my naive, Never thought about X^(1/X) when X =3 that (3)^(1/3) equals to when X =~2.478 that (~2.478)^(1/~2.478), indeed wonder how many numbers between that range has two real solutions as pairs🤔, very interesting!
Here is a related problem where cube root of 3 does seem to be the answer: suppose start with a reasonably large number, say 100. Then you start taking log base some fixed number x greater than 1. Then you notice that the more you take log base x the closer and closer it gets to 3. What was the x?
@@JSSTyger Well this video and discussion are about mathematics - not the peculiarities of spreadsheet applications. (There's probably a historical reason those programs do it that way - to not break compatibility with legacy software.) If you're talking about mathematics, "x^x^x" doesn't mean "input this string into Excel/programming language", but the abstract mathematical expression of two exponentiations stacked without parenthesis. Can't really write that as is in these comments. Furthermore, actually (a^b)^c = a^(bc) that is "b times c" in the exponent.
If you do it like this, then you get 3^k = k^3 which means k = log3 of k^3 which equals 3* log3 of k and if you repeat this, you get k = 3 * log3 of 3 * log3 of 3 * log3 of... which evaluates to 1? There doesn’t seem to be another solution to k that makes the RHS true :/
Evan Lewis You wouldn't use logarithm base 3, you would simply use the natural logarithm. k^3 = 3^k implies 3·log(k) = k·log(3), which implies log(k)/k = log(3)/3, which is solved by using the Lambert W map.
@Ajay singh Video was unlisted for ~3 months, but with link provided in the description of a previous video in the series. That's how some people could find it before it was published a few days ago.
@@SoWe1 no i^2 = (-1) , sqrt(i) is different z = sqrt(i) means z^2 = i so there are 2 solutions : (z = sqrt(2)/2 + i*sqrt(2)/2 ) or (z = -sqrt(2)/2 - i*sqrt(2)/2 )
@@SoWe1 i'm not sure but i would say : - a square root of a complex number (not real) is always is a complex number (not real) - therefore sqrt(x) = -2 only have solutions in R - but sqrt(x) >= 0 in R so there are no solutions at all
Hey man Cube root of 3 to power cube root of 3 Infinite times Let it be x Cubing both sides 3power cube root of 3 power cube root if 3 Infinite times=x³ Dividing both sides by 3 1power cube root of 3 Infinite times=x³/3 1=x³/3 X³=3 X=cube root of three Solved.
And this is why I cringe slightly every time you (or anyone else) goes from the square root of a value to that value to a one half power without addressing that they're slightly different.
Pretty sure that the x^(1/2) is not how we represent the multivalued root of x.... i mean that x^(1/2) = sqrt(x)... same domain, same range, same function.
The Lambert W function is a pathway to many abilities some consider to be unnatural.
But will it save padme?
¿що ти думал? From a certain death?
But can it express x as a function of y that's elementary ? 😏
@@dihydrogenmonoxid1337 you will fall in love with the lambert w function instead of padme so that question is irrelevant
So, I took 11 semesters of advanced math in college and grad school, and I am constantly amazed at how much I don't know when I watch your excellent videos.
Thank you : )
@@blackpenredpen Why didnt you tskenthe natural log of both sides at 1:26 first?
@@blackpenredpen and the lam ert finctiom anyway? Does anyone actually understand what it means? And it soubds like youncintradict yiurself in the video later when you say cube root of 3 isna solution..Hope you can respond.
@Michael Bishop just look at the graph of x=y×e^y... that's the graph of y=W(x)....
@@leif1075 Lambert w function is inverse of any function f(x) which in this video is xe^x
For anyone lost watching comments from 3 months ago:
This video was unlisted and its link was put in the description of another video. So, just a few people watched this part 4. Yesterday, Blackpenredpen changed the settings and now it is public. Have a great day!
Thanks ,,, i was shocked
Those graphs are really cool!
Love your vids :)
How is ur comment 3 years ago wtf ?
how's your comment 3 months ago?
😱😱😱are you Gost
Wait yo what
My mind blew at 5:12 So the issue was transforming the equation to another equation does not work because their domain:ranges are different. What kind of equations can this happen with? Clearly when infinity comes into play, also whenever imaginary roots are possible. But how do you find functions with limited domains? It seems like an easy trap if you don't know this could happen.
That was a brilliant, beautiful explanation. Thank you!
Thanks!
@@AstroB7 what is going on....?
Very suspicious results, lol. If I wasn't into maths (such as being interested, obviously not super high level), I would probably assume that these things are made up. But honestly, it's super satisfactory even when you consider easy questions and discuss them through as if you were tackling a major problem. So satisfactory, as usual, bprp!
i love how thanks to you i can improve my math skills and im still in 10 grade i started learning complex numbers with you and you inspired me to think out of the box and more abstractly i hope one day i may know at least a fourth of what you know
4:43 easy, k = w
🤦♂️
XD
🤦♂️🤦♂️🤦♂️
@@quirtt bhai tu idhar?
@@quirtt har jagah se padhte ho kya xd
One can figure that 3 is a value that cannot be converged to because the function f = x |-> x^(1/x) attains a global maximum at x = e in the positive real numbers. Hence if x = 3, then 3^(1/3) < e^(1/e).
Blackpenredpen you are a true inspiration for me, taking a Calc 1 class for my BA bachelor ( Idk why they include it) I have motivation from your videos that I did not have before. I am certain that I will do everything to pass in my class, I am stuggling with some things but your story and your enthusiasm brother blackpenredpen , God will help me on the way to succes hopefully! Thank you for sharing your sweet knowledge brother !
When you said "power tower," it made me really happy.
Unbelievable! Extraordinary! Marvelous!
10:14 so what you're saying, is that the two graphs (and therefore the two equations) are only equivalent on certain intervals, and 3 is outside of this interval, but this bags the question, how do you find the interval in which the two equations are equivalent?
I'm in class 12 cbse 2020-21 batch
I don't know what's happening but yes it looks cool to solve your questions ❤️❤️🥰
For the infinite power tower, I am curious why it doesn't converge at values between 0 and e^-e. I tested 0.05 up to 30 copies of 0.05 and it jumps back and forth between ~0.1 and ~0.7 getting closer towards some value between those two numbers so it seems like it would converge to some specific value. I wonder if values between 0 and e^-e exhibit some kind of bifurcation chaos where they stay finite but bounce between two values and thus diverge in that way.
Here is a question for you:
Given:
A=(1/2)sqrt((1/2)(3-sqrt(5)))
B=(1/4)(sqrt(5)-1)
It can be shown that A=B
Question is, given A, how do you turn it into B?
djttv
It’s similar to this one ruclips.net/video/afz3t8R34r4/видео.html
@@blackpenredpen I learned something new! Thank you
HAPPY TEACHERS' DAY ❤️ thank you so much for your videos, I wan't you to know that we really appreciate you and what you are doing ☺️
So I suppose the different values for W come out when you use different cube roots of 3. All these multivalued functions are countable, so it should be able to be bijected.
f(z) = ze^z is not monotonically increasing and thus the inverse f^-1(z) = W(z) is multivalued. Quite easy. Same with sin, cos, tan except those are periodic and easy to understand.
Is three (3) the only misbehaving number in these problems???
Excellent!
@blackpinredpin Looking fresh as always! Great video.
i love how you explain it
1:00
Well, that escalated quickly
From India watching you nice concept loved it
Hello! Can you make the video where you get the link between the Riemann zeta function and Bernoulli numbers kind of B(t) = -t*Zeta(1-t). How is it gotten? I triied to get this equotion but I couldn't - only my brain was broken! Thanks, I hope for this video will appear soon...
Beautiful... Thanks🌹
Can you pls tell me which company markers do you use.they seem pretty easy to hold 2 in a hand and switch
Hi, love your videos, can you take the integral from -2 to 2 sqrt(1-x^2) dx ?
You know you're dealing with advanced math when blackpenredpen doesn't summon the fish.
Nice song at the end
I came to same equation 3^(⅓k) = k, but then I took ln() of both sides
⅓k ln(3) = ln(k)
Divide both sides by k
ln(3)/3 = ln(k)/k
k = 3
Why is it wrong?
Please someone explain me how does the fixed point convergence criteria come
How does that interval come
Wow that function is cool
I am really trying to find the derivation of the fixed point convergence criteria
I tried everywhere
Didn't find anything
Not even in lot of papers
Plz someone help me
Is there a number such that when applied to an infinite power tower it converges on pi?
very nice!
@ 2:20 - when you have (3^(1/3))^k = k, couldn't you take the k-th root on both sides, so the k exponent disappears on the left (k/k=1), and the equation becomes 3^(1/3) = k^(1/k), therefore k=3? where is my logic flawed?
Edit: just watched the rest of the video, so 3 can be a solution. This is technically a way to solve the problem, but if you solve this way without using W function, the other solutions are not intuitive.
dont you also assume the the power tower of x can be written as 3 when replacing it, i.e. assuming it converges?
I think that assumption is how you get the false solution. tower = x^tower = 3, and plug in 3 for tower. Well, the tower built from 3 actually diverges toward infinity.
Exactly. That's exactly the problem.
So interesting pardon my naive,
Never thought about X^(1/X) when X =3 that (3)^(1/3) equals to when X =~2.478 that (~2.478)^(1/~2.478), indeed wonder how many numbers between that range has two real solutions as pairs🤔, very interesting!
Downloaded a curve equation generator, between 1 and e
How many numbers? Infinite....
The chomper microphone!
This guys is really addicted to the W function.
Here is a related problem where cube root of 3 does seem to be the answer: suppose start with a reasonably large number, say 100. Then you start taking log base some fixed number x greater than 1. Then you notice that the more you take log base x the closer and closer it gets to 3. What was the x?
So.. ln(100)/ln(x) = 3? Is that what we are trying to solve? Because that's trivial....
Is that analytic continuation?
Holy shit I was just thinking of this earlier did you read my mind?
Why does the power tower oscillate if you chose x smaller than e^-e?
Well, well, well... excellent!
Pendyala education and eentertainment 👍
Amazing!
Sir please provide the PDFs of 100 integrals, 100 derivatives, 100 series...
Humble request👍👍
I have a question. Does x^x^x equal x^(x^x) or (x^x)^x?
x^x^x = x^(x^x) always. You need parenthesis on (x^x)^x.
@@Silvar55x MicroSoft Excel says 3^3^3 = 19683
@@JSSTyger Well this video and discussion are about mathematics - not the peculiarities of spreadsheet applications. (There's probably a historical reason those programs do it that way - to not break compatibility with legacy software.)
If you're talking about mathematics, "x^x^x" doesn't mean "input this string into Excel/programming language", but the abstract mathematical expression of two exponentiations stacked without parenthesis. Can't really write that as is in these comments.
Furthermore, actually
(a^b)^c
= a^(bc)
that is "b times c" in the exponent.
ooh
Need your help.
What is the answer here?
x^x^x^x^... = 10
My head is going to expode.
I asked this in the previous video
I want a proof for the fixed point convergence criteria
2020 raise to the power 2019 - 2020 divided by 2020 square + 2021=N
then find the sum of digits of n
bro plz solve this?? trying from last 5 weeks
Regular videos
Woooooow
I just thought for a moment,why his beard was shorter than usual. But later I realized that this video was shot months before.
But how would you calculate W lambert values manually?
Same as the natural log, you basically can't
Or with trial and error
@@savitsios i mean there are many ways to approximate....
Nice
Integration of 1 /1+secx .....
PLEASE SUGGEST SIR......😔😔😔😔😔😔😔😔☹️
@3head individual done sir... THANKS a lot ..
@3head individual where are you from sir. Japan??
cool.
Brilliant XD
Can’t you just raise k to the 3rd power to get: 3^k = k^3 and solve as if it was x^y = y^x?
If you do it like this, then you get 3^k = k^3 which means k = log3 of k^3 which equals 3* log3 of k and if you repeat this, you get k = 3 * log3 of 3 * log3 of 3 * log3 of... which evaluates to 1? There doesn’t seem to be another solution to k that makes the RHS true :/
Evan Lewis You wouldn't use logarithm base 3, you would simply use the natural logarithm. k^3 = 3^k implies 3·log(k) = k·log(3), which implies log(k)/k = log(3)/3, which is solved by using the Lambert W map.
Thank u maths teacher
Your pens surely hold infinite power!
@Ajay singh Video was unlisted for ~3 months, but with link provided in the description of a previous video in the series. That's how some people could find it before it was published a few days ago.
ok I'm stupid why does -2 = sqrt(x) not have any solutions even in the complex plane? i*4?
sqrt(4*i) = 2 * sqrt(i) = 2 * ( sqrt(2)/2 + i*sqrt(2)/2 ) = sqrt(2) + i*sqrt(2)
@@aeropoulpe7818 sqrt(4*i) = 2 * sqrt(i) = 2*(-1) = -2
@@SoWe1 no i^2 = (-1) , sqrt(i) is different
z = sqrt(i) means z^2 = i so there are 2 solutions : (z = sqrt(2)/2 + i*sqrt(2)/2 ) or (z = -sqrt(2)/2 - i*sqrt(2)/2 )
@@aeropoulpe7818 rightrightright, yes
but still, why does that equation not have any solutions?
@@SoWe1 i'm not sure but i would say :
- a square root of a complex number (not real) is always is a complex number (not real)
- therefore sqrt(x) = -2 only have solutions in R
- but sqrt(x) >= 0 in R so there are no solutions at all
How come some comments be from 1 month ago ??
This video was available by link
It was unlisted in my playlist
@@blackpenredpen Ok thank you , great video btw
Now I am in future
whatislambertwfunctionpleasemakeavideoonthatthanks
Hey man
Cube root of 3 to power cube root of 3 Infinite times
Let it be x
Cubing both sides
3power cube root of 3 power cube root if 3 Infinite times=x³
Dividing both sides by 3
1power cube root of 3 Infinite times=x³/3
1=x³/3
X³=3
X=cube root of three
Solved.
Wait what? You must be joking.....
Hello bprp!!!!
e = 3
U're the best
0:27 not according to physicists
how did I get here
🔒
Hi.. I have a challange question for you.. 😂😂
hi
6:22 how could complex numbers infinite power look?
3
How do you learned English?
henlo
And this is why I cringe slightly every time you (or anyone else) goes from the square root of a value to that value to a one half power without addressing that they're slightly different.
I dont get what you're saying, how are they different?
@@yarakharam5343 x^2 and x^(1/2) are inverses of each other. The square root of x is a function that takes the principal root of x^(1/2)
Pretty sure that the x^(1/2) is not how we represent the multivalued root of x.... i mean that x^(1/2) = sqrt(x)... same domain, same range, same function.
@@shashankambone6920 not what I was taught. The radix is a function, a non-integer exponent is not
@@joshuahillerup4290 maybe we have different conventions...
うーん むずかしかった
3週間ぶりにようやく理解しました😀
こりゃ学部1年終えるのに8年かかっちゃうね😭😭
Who came here from bprp fast
Technically sqrtx=+-sqrtx
no
First!
no
Nice