The other parts will be published later. If you cannot wait, then check out my other channel (in Mandarin): th-cam.com/channels/rONDbyO94HIrJyfhS6qfjA.htmlvideos
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.
@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
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)
@@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...
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♾️🙂
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
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
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.
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
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.
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
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)
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.
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.
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
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)
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!
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.
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)
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.
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?
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?
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
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.
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
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).
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
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...
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
Good day I found one of your video explanations of sin3x in terms of sinx to help my daughter who is in matric. She has difficulty in another Trigonometry problem, can you maybe help us ? It was such an eye opener for my daughter, once again thank you, even if you cant help with the other problem. Regards Lourens Scholtz
He has several videos of it already, plus you can check the Wolfram Alpha page on the function and the Wikipedia page on it for an introduction to the topic. In essence, though, all it is, it is the inverse "function" of f(x) = xe^x.
Merci pour cette mis🎉e en application de la fonction de Lambert😀 Mais je ne suis pas d'accord avec votre postulat de départ, implicite, selon lequel: x^x^3 = 2 = x^(x^3) En effet, si on peut écrire: x^x^3 = (x^x)^3 👍, il est clair que: x^x^3 ≠ x^(x^3 ) ⛔ On peut aisément vérifier par exemple que: 2^3^4 = (2^3)^4 = 8^4 = 4096 ≠ 2^(3^4) = 2^81 = 2417851639229258349412352 Cela vient de ce que l'exponentiation n'est pas associative, et donc que: x^x^3 ≠ x^(x^3). Ainsi votre résultat: x = exp(W(3ln2)/3) ≈ 1.3367 ne vérifie pas l'équation x^x^3 = 2 Car: 1.3367^1.3367^3 ≈ 3.201907... En revanche: 1.3367^(1.3367^3) ≈ 1.999934925...
The other parts will be published later.
If you cannot wait, then check out my other channel (in Mandarin): th-cam.com/channels/rONDbyO94HIrJyfhS6qfjA.htmlvideos
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.
@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
i just know that you have other channel in mandarin. thanks a lot for making that channel!
Why don't you feature your two other channels?
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)
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.
College is not "scientific" enough...
Alright, I am already expecting Lambert function here
Ok
Lambert function is so cool!
You thought right
You already know what blackpenredpen will say....hello! lovely voice.
Lol thanks!
@@blackpenredpen you are so cute 😍😍
I like it how you say "Let's do some Math for fun."
Isn't all Math fun!
Only pure, pure math. The things they make seniors study in Greece are outrageous...
@@erikkonstas,
What's wrong with math education in Greece? Your country once pioneered mathematics.
Math by itself is fun but when you're forced to memorize stuff for exams, it loses its value
@@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...
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♾️🙂
I'm 11
Woah the new TH-cam UI on this video is so cool! I love seeing the timebar all sectioned and labelled ❤️
I smell Lambert W function incoming
I was right
Yes
You can substitute x^3 as other variable and then solve
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
Whoa!! I didn't know you could divide videos into sections like that!
Yea it’s a new feature. I think it’s so cool!
Same , i think its cool too
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
You really explain every problem very well.
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
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
@@Vivek-io3gj could you not swap around the exponents?
@@adamwright4634 you cant because it’s x^(x^3)^1/3 not x^((x^3)^1/3)
I wonder what’s going to be in part 2,maybe a infinite power tower?
*an
Yes. Will release that this coming Friday
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.
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
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.
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
Could you do a video showing that if a Sequence is "Cauchy" that it converges? 🙂 One of my favorite Theorems.
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)
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.
umm
didn't he do this in the video ?
I like how you manipulate ln, e, W to come up with great solutions...
Man I love you! You are probably the most friendly guy on youtube!
Your the best mathematician PLZZ can u make a video on question of Indian prmo questions they're very tough.. thankyou sir
In the Observation, where does the 3= ( x to 3rd power) come from?
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.
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:)）
That last demonstration was HOLY
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😊😊
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
Thank you, I was looking for this kind of solution.
3:25 We say kinda same in Kazakh
I like the little timestamp chapters. When did YT add this?
6:01 this reminds me infinitely nested Michael Jordans!
Hi sir great job.
I am from India
Love you sir♥️♥️♥️♥️♥️
btw, in 6:49 we can do it until the number gets bigger than e^(1/e), this is kinda cool
Yea if u only have finite amount of x
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)
You can use it in bith
Are you allowed to derivate both sides of the equation instead of using the W function?
I don't know why but I love this guy
It's the fish,
and the smile,
and lately, that little beard ...
... "isn't it?"
You rarely get any of that with math teachers.
Pre-celebration for your channel getting 500k subs!!!!
Where is your old board?
Hey BPRP, do you know any good resources for analysis?
Dr. Peyam and Prof. Penn's channels : )
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)
I love this channel
Excellent work
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)
Blackpenredpen you solved the question elegantly. Very much impressed!
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!
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.
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.
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)
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.
Wow thanks guys.This community is the best!
Nearly HALF A MILLION subs!!!!
so in general x^x^c=c --> x=c^(1/c) for all c>0 (as that is the domain of ln)
yes it is working even you add more powers x^x^x^..x^c=c
x=c^1/c
Yes.
Really interesting... Is there any other methods of solving this question?
Newton's method is quite fast
1.5
1.40
1.348
1.3371
1.33671
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?
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?
I don't know why I can't sleep without watching your videos 😍😍
I’m new to this channel. The π at 0:29 looks like the π in 3blue1brown 😁
Because it is! : )
I got it from his merch store!
Please upload factorisation of cyclic expressions using factor theorem and How it works and What is the real life apllication of cyclic polynomial ?
Since e^(⅓W(3ln3)) is the same as ³√3, which is basically 3^⅓, does that mean that e^W(3ln3) is 3?
Love the chain chomp
Love the vids !!
Best outro music for math... 2020 is extracting the good juice from people...
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
How do you know there are no negative solution? Also, Since Lambert W function is multivalued, can’t there be more positive real answers?
Almost 500k!
Hey bprp! How do you integrate W(x)?
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.
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
Hello @blackpenredpen, whats about x^x^x=2 ?
Beautiful
What is the brand of your pen please ? 😅
I am great fan you sir! You are my cool! 😎 Maths teacher 💯
Very nice video !!!
Does it mean that exp(1/2*W(2ln2)) = sqrt(2)??
Yes
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).
Simply comparing x^3=1
Taking cube roots of unity
W, w^2,1 and further simplification
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
So it looks like a property of W is that W(xlnx) = lnx as well.
there are two more solutions as well for the right side: (-1)^(2/3) and (-1)^(4/3) times the cuberoot of 3
But you need the solutions to be in real world.
@@Shreyas_Jaiswal they are in real world, they just happen to not fit in the set of real numbers.
@@MrRyanroberson1 real world means on earth. But complex numbers are on moon. 😂😂
Please make a video on ith derivative of x^i
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...
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?
Man this was a great video
Can you do a video for Kids?
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
didnt you also have a video for x^x^x^2017...... x=2017^1/2017
Hey is your Chinese channel 黑比红比 abandoned? There's no video.
I don't use that anymore. But here
th-cam.com/channels/rONDbyO94HIrJyfhS6qfjA.html
blackpenredpen gr8! just followed😁
Thanks! How did u know that channel btw? It’s been so long lol.
blackpenredpen Google'd "blackpenredpen Chinese" and found this m.th-cam.com/video/8j611oiDCoI/w-d-xo.html
Can you solve for x:
x^(x/(1-x))=1
Sir,, can you solve this problem:
integral of [e^(tan3x)]*(secx)^2 dx
Thank"s
bprp has an exponential that you don’t know how to solve? It must be Lambert W!
Can I have the link for your Chinese channel as I am from Hong Kong in which Chinese will be more familiar with me.
An Easy peasy question for you-
Find the sum of series: sin x+3sin 3x+5sin5x +...... +(2k-1) sin(2k-1) x.
(Using calculus)
Good day
I found one of your video explanations of sin3x in terms of sinx to help my daughter who is in matric. She has difficulty in another Trigonometry problem, can you maybe help us ?
It was such an eye opener for my daughter, once again thank you, even if you cant help with the other problem.
Regards
Lourens Scholtz
W E C A N D O THE Follow yng 5:16
What's it???
does that kind of equation always have 1 solution?
What course is this taught in? I haven't covered this
Ideally, in a precalculus course.
i love your concepts i am from india
How can I get te W funtion , in other words where come de W funtionfu, how is it derivided??
Please
I have that in the description.
He has several videos of it already, plus you can check the Wolfram Alpha page on the function and the Wikipedia page on it for an introduction to the topic. In essence, though, all it is, it is the inverse "function" of f(x) = xe^x.
If you look at the duration of the video, it says Hello next to it.
you are from hong kong or mainland?
I tried to derivate it all the way but then x^x^3 = 1 not 3. Can it be solved without that "W" thing?
Merci pour cette mis🎉e en application de la fonction de Lambert😀
Mais je ne suis pas d'accord avec votre postulat de départ, implicite, selon lequel:
x^x^3 = 2 = x^(x^3)
En effet, si on peut écrire:
x^x^3 = (x^x)^3 👍,
il est clair que: x^x^3 ≠ x^(x^3 ) ⛔
On peut aisément vérifier par exemple que:
2^3^4 = (2^3)^4 = 8^4 = 4096 ≠ 2^(3^4) = 2^81 = 2417851639229258349412352
Cela vient de ce que l'exponentiation n'est pas associative, et donc que:
x^x^3 ≠ x^(x^3).
Ainsi votre résultat:
x = exp(W(3ln2)/3) ≈ 1.3367 ne vérifie pas l'équation x^x^3 = 2
Car:
1.3367^1.3367^3 ≈ 3.201907...
En revanche:
1.3367^(1.3367^3) ≈ 1.999934925...
He said "one-third" correctly and then corrected himself..
Eh, when?
@@blackpenredpen 3:13 and 4:20
@@vari1535 He wrote the denominator first. I find it adorable that he corrected himself for that!
@@michaelroditis1952 Ohh, I thought it was what he _said_ lol
Just use the super root, after letting u=x^3. (tetration)
e^{3ln(x)} = x^3 for positive x
5:38 時 為什麼 3 = X^3 ? 這段是怎麼得來的 ?
反推
你從那先看的話
你會發現你可以一直把x^3放上去
What about super quadratic equations?
It works more or less the same.
三分之一 3 parts take 1 haha! This feels like watching a translated version lol
What'the link for the chinese channel?