I did the limit, got 1, watched the video. He says the answer is 0. I pause rught before he writes the factorial. Now im fucking panicked. I go back to my whiteboard, keep looking for errors. 5 minutes later, now 2 of my friends are involved. All of us frantically looking for where we went wrong. Math degree ego on the line. After an hour an 4 different methods all leading to the same thing, we give up, and look at the video. I resume it on my phone. And there. 1 second later. The greatest treachery I've faced since 12th December 2014. An hour of my life ill never get back.
@@nanamacapagal8342 Doyle's constant: e to the (e + 1/e) power, which is a paradigm for a photon and is the ratio of potential energy over kinetic energy at the most dense state of a Big Bounce event. Consider only the exponent as the vertical asymptote and vertical tangent. This connects the strong nuclear forces in a Big Bang paradigm to reduce complexity in calculations of synaptic functions in computer science.
At least include the blue pen first!! it should be bprpbpgp, not just bprpgp! Unless the b at the beginning stands for both black and blue at the same time
The most advanced mathematics I ever did was limitations and mechanics. Logs always confused me and I never learned the Lambert W function. So this video gave me an actual headache 😂
@9:45: I too think this is really, really cool. There can never be too much Lambert W function content on RUclips. Now that you've computed the derivative of W(x), can you compute its antiderivative? I'll give it a try, and leave another comment if I succeed.
It took me a couple of hours, but I finally got it. On my first attempt, I solved W'(x) = W(x)/(x*(W(x)+1)) for W(x) to get x*W'(x)/(1-x*W'(x)), and integrated x*W'(x)/(1 - x*W'(x))*dx through a series of substitutions, starting with u = x*W'(x) (which introduced an exponential in W(x) to remove an x), culminating in a polynomial in t, times e^t. Unfortunately, I must have made a sign error somewhere, because the result did not have a derivative equal to W(x) (instead getting (W(x)^2 + 2*W(x) - 1)/(W(x) + 1) - 1). But it was close enough that I was able to deduce that the true antiderivative of W(x) was likely a quadratic in W(x), times e^W(x), plus a constant, and starting from there, I was able to find h(x) = (W(x)^2 - W(x) + 1)*e^W(x) + C, which is a function such that h'(x) = W(x).
Of course, after a few more minutes of playing around with this, I realized I should have *started* with the substitution u = W(x), because that would give me dx = (u+1)*e^u*du, and integrating u*(u+1)*e^u*du is easy....
Yeah, reverse function integral makes life so much easier W(x) is just f^-1(xe^x) so what we can do is. Int(f^-1(x))= xf^-1(x) - F(f^-1(x)) + C where F is the integral of f(x). So, the integral of xe^x can be done by parts, and we'll just skip to the DI table to get xe^x-e^x. Substituting W(x) into every X, we get x-x/W(x). So, the end result is x(W(x)+(1/W(x))-1)+C. You can turn it into the quadratic you gave but this is simpler to get.
Hi!, Im in senior year of hs and I need major help for a school project. I need to calculate the arc length for polinomials of 2nd, 3rd and 4th power. Using symbolab and wolfram i was able to find the derivative of a general parabola, but with cubics it doesnt say anything. Let me explain The formula for the arc length is length=bounded_integral(sqrt(1+f'(x)²)) Where f(x) is the function you want to calculate the arc length of. In parabolas u first substitute u=f'(x), so du=f''(x)dx=number*dx So you can move it around. However in higher powers f"(x) is no longer just a number, it contains "x" so you are much more limited. Any alternatives to the original precess would be of immense help (u-substotution, then trig-substitution), you can see it when plugging f(x)=x²+x+1 in the formula. Any tips or other programa that might be able to calculate it would help too. I also tried desmos but im afraid it uses a numerical method to calculate nounded integrals, since it only allows for those. Thank you!!
While the limit is correct, these functions do not really become the same at large x. For large x, W(x)=ln(x)-ln(ln(x)+O(1). Hence as x->Infinity, W(x)/ln(x) ->1 because ln(x) grows faster than ln(ln(x)). However, as x-> infinity also ln(x)-W(x) -> ln(ln(x)) -> Infinity. Thus the difference between log and product log becomes infinite at large x. It's just that this difference grows slower than the functions themselves, so the result of dividing them tends to 1 at large x...
I have a math question that I haven't really been able to find an answer for. When integrating why does the dx 'disappear' for a lack of a better word? Like why is dx or whatever differential gone when you do the integral? Hope I'm making sense with that
At the end you show the ln(x) and the W(x) functions plotted on the same graph. If the limit of their ratio for large numbers goes to one, why the two functions do not seem to sit one on another? The convergence is so slow?
Does this work in general with inverses of functions like this? If f(x) goes to infinity as x goes to infinity and g(x)=xf(x), will their inverses always have this limit?
@@bjornfeuerbacher5514thanks I tried to compute it on wolfram alpha but it fails. I had observed that ln(x)-W(x) grows extremely slowly, although it diverges
I did it with a variable change x = te^t -> Lím(Ln(x) / W(x), x -> inf) = Lím(Ln(te^t) / W(te^t), t -> inf) = Lím((Ln(t) + Ln(e^t)) / t, t -> inf) = Lím(Ln(t)/t, t -> inf) + Lím(t/t, t->inf) = 0 + 1 = 1
A function f has an inverse if for every x there is a unique y so that f(y) = x. For that to happen, it has to be bijective - one-to-one and onto. The function x e^x can be defined for all real x, but you’ll find that there are values of x less than -1 and values greater than -1 that give the same value of the function, meaning you can’t pick a unique inverse across that domain. By restricting the domain of the function to [-1,infinity), you force it so that there’s only one value in the domain that corresponds to each value in its range.
No that is not true. The +1 with the infinity makes it a limit question again. Those sums do eventually diverge and if you use very large numbers to look at them like a Graham's number then the natural log wins with the greater growth.
@@blackpenredpen well we are looking not only at the limit originally, but a limit of limits. while infinity and and infinity+1 are both infinity they are not equal. You had another infinity over infinity and you needed to perform L'H again. We cannot draw a conclusion when it is infinity over infinity. That +1 will matter as if you look that the delta between the changes of change the ln while exceedingly the product log is even slower. As you like to say you have to do more work.
I am not a mathematician but a casual math viewer, The lambert W is only loved because it requires a lot of creativity to use(which is where all the fun lies) and its something new Ofcourse, for you, this all must be basic, so it's understandable why you would feel that it's just a boring function 😅
I don't use the Lambert W function for much in my personal work, but I like using it in Calculus classes. Apparently, it has uses in some natural sciences, but the interesting thing, to me, is that we can't find a "nice formula" for it in terms of elementary functions, but we can still do calculus with it. We can use implicit differentiation to calculate its derivative. We can use things like Newton's Method to calculate values of W(x) to arbitrary precision. So the fact that we can do so much with a function that we don't have an "nice formula" for shows the power of calculus theory.
Whenever i see the w function i automatically i lose interest. I am not sure what you fixation with it is. Its not something that's thought here and we are lucky to be spare of it.
Hello, I know this might be an absurd idea. But i am a small minecraft youtuber, If you would be interested. I think it would be cool to explain equations Utilizing minecraft. Let me know.
I substituted x with ue^u so I will get (u+ln(u)) /u which its limit goes to 1
That’s an extremely smart way to do it!!!
that's what I did too !
It genuinely threw me off that he didn't do that. We have a W(x), we probably will make life easier by getting rid of it.
Same method I got! (I suspect a ton of people did it this way as well...)
Everyone here who did not do this obvious but BRILLIANT approach should be ashamed! Edit: _I'm_ ashamed!
I did the limit, got 1, watched the video. He says the answer is 0. I pause rught before he writes the factorial. Now im fucking panicked. I go back to my whiteboard, keep looking for errors. 5 minutes later, now 2 of my friends are involved. All of us frantically looking for where we went wrong. Math degree ego on the line. After an hour an 4 different methods all leading to the same thing, we give up, and look at the video. I resume it on my phone. And there. 1 second later. The greatest treachery I've faced since 12th December 2014. An hour of my life ill never get back.
Get trolled bro
AHAHAHAH THAT'S FUCKING INSANE DUDE
Where is thr fish?
The fish: 🐟
my reply got deleted lol, google free speech platform.
google deleted both my fish lol, frxe-spexch = orangeman? lol.
It swam away
I ate it
Nothing better than a fresh limit on a Saturday morning
Yes!!!
he really just dropped a green pen out of nowhere like it isn't a huge deal
Thanks
Thank you for the super thanks!
We got green pen on bprp before GTA6 release
Imagine how fun GTA6 will be after SweatBabbyInk "fixes" it! lolfml.
ohhhh google canceled facts again lol.
Nah his old vids had green pen
if x+y=8, find the max of x^y (Lambert W function)
ruclips.net/video/zdAJXil-NvA/видео.html
I've always loved how organized your equations are
GREEN PENCIL???
I was shocked also lol
Yup, legends says that when BPRP uses the FOURTH mighty color will be a signal of The Advent
Now I just need to see THE PURPLE PEN!
Wait until he pulls out the orange pen
@@nanamacapagal8342 Doyle's constant: e to the (e + 1/e) power, which is a paradigm for a photon and is the ratio of potential energy over kinetic energy at the most dense state of a Big Bounce event. Consider only the exponent as the vertical asymptote and vertical tangent. This connects the strong nuclear forces in a Big Bang paradigm to reduce complexity in calculations of synaptic functions in computer science.
@@nanamacapagal8342 Maclaurin sectrix.
Yooo we need more "FISH" vids. (the w function)
I love the LambertW. It holds a special place with me, since highschool, leading me on a wonderful goose chase.
Ahh yes! Welcome to another very cool video of *"BlackpenRedpenBluepenGreenpen"* litterelly
This is the first time I got an idea of a real world property of the W fuction. Thanks!
bprp should change his name to bprpgp 😂
At least include the blue pen first!!
it should be bprpbpgp, not just bprpgp! Unless the b at the beginning stands for both black and blue at the same time
RGB Pens.
@@SilviuBurceaDev rgbp
@@nanamacapagal8342Don't forget purple, bprpbpgppp
The most advanced mathematics I ever did was limitations and mechanics. Logs always confused me and I never learned the Lambert W function. So this video gave me an actual headache 😂
@9:45: I too think this is really, really cool. There can never be too much Lambert W function content on RUclips. Now that you've computed the derivative of W(x), can you compute its antiderivative? I'll give it a try, and leave another comment if I succeed.
It took me a couple of hours, but I finally got it. On my first attempt, I solved W'(x) = W(x)/(x*(W(x)+1)) for W(x) to get x*W'(x)/(1-x*W'(x)), and integrated x*W'(x)/(1 - x*W'(x))*dx through a series of substitutions, starting with u = x*W'(x) (which introduced an exponential in W(x) to remove an x), culminating in a polynomial in t, times e^t. Unfortunately, I must have made a sign error somewhere, because the result did not have a derivative equal to W(x) (instead getting (W(x)^2 + 2*W(x) - 1)/(W(x) + 1) - 1). But it was close enough that I was able to deduce that the true antiderivative of W(x) was likely a quadratic in W(x), times e^W(x), plus a constant, and starting from there, I was able to find h(x) = (W(x)^2 - W(x) + 1)*e^W(x) + C, which is a function such that h'(x) = W(x).
Of course, after a few more minutes of playing around with this, I realized I should have *started* with the substitution u = W(x), because that would give me dx = (u+1)*e^u*du, and integrating u*(u+1)*e^u*du is easy....
@@jimschneider799hey just so you know, e^W(x) can just be written as x/W(x) instead
Use inverse integration formula
Yeah, reverse function integral makes life so much easier W(x) is just f^-1(xe^x) so what we can do is. Int(f^-1(x))= xf^-1(x) - F(f^-1(x)) + C where F is the integral of f(x). So, the integral of xe^x can be done by parts, and we'll just skip to the DI table to get xe^x-e^x. Substituting W(x) into every X, we get x-x/W(x). So, the end result is x(W(x)+(1/W(x))-1)+C.
You can turn it into the quadratic you gave but this is simpler to get.
7:58 Hah, jokes on you. I have a pink, light blue orange and purple pen
Lets go, comeback of the lambert W function
Approaching equality with ln(x), that's a real W for large x, right there.
Calculus is so neat, I love it
You’re the goat BPRP
This is really really cool.
I though that it would a bigger number
I guess not(but it actually makes sense)
So good to know this, because the Lambert W() function has been mysterious to me.
9:18 missaying: he want to say 1/W(x) goes to zero as z goes to inf.
oh yeah baby show me the limit
Se armó la grande en RUclips.
So good!
Hi!, Im in senior year of hs and I need major help for a school project. I need to calculate the arc length for polinomials of 2nd, 3rd and 4th power. Using symbolab and wolfram i was able to find the derivative of a general parabola, but with cubics it doesnt say anything. Let me explain
The formula for the arc length is length=bounded_integral(sqrt(1+f'(x)²))
Where f(x) is the function you want to calculate the arc length of.
In parabolas u first substitute u=f'(x), so du=f''(x)dx=number*dx
So you can move it around. However in higher powers f"(x) is no longer just a number, it contains "x" so you are much more limited.
Any alternatives to the original precess would be of immense help (u-substotution, then trig-substitution), you can see it when plugging f(x)=x²+x+1 in the formula. Any tips or other programa that might be able to calculate it would help too. I also tried desmos but im afraid it uses a numerical method to calculate nounded integrals, since it only allows for those.
Thank you!!
Excellent video.
Why is this so good?
AWESOME VIDEO! Really interesting. When will you make a quartic equation formula derivation?
Where's your *"PurplePen"* from the old videos? XD
that is cool math(s)
BONUS: the surprise green marker
My thought before substituting is to just let x -> xe^x. Then we have lim x->inf (lnx + x)/x = 1
09:33 - This plot with (x, y) confused me, then I made similar plot with (exp(x), y), and now it's obvious.
the natural next question: limit of (ln(x) - W(x)) / ln(ln(x)) as x -> infinity
WAIT WHAT A GREEN PEN :0
thats a great surprise
since W(x)->inf, W(x)+1->inf. applying L'Hospitals rule, the top and bottom become the same, so the limit is 1
While the limit is correct, these functions do not really become the same at large x. For large x, W(x)=ln(x)-ln(ln(x)+O(1). Hence as x->Infinity, W(x)/ln(x) ->1 because ln(x) grows faster than ln(ln(x)). However, as x-> infinity also ln(x)-W(x) -> ln(ln(x)) -> Infinity. Thus the difference between log and product log becomes infinite at large x. It's just that this difference grows slower than the functions themselves, so the result of dividing them tends to 1 at large x...
Fun challenge: what's the minimum of W(x)/ln(x)? Yes, it has a "nice" answer.
@@ingobojak5666 e/(e+1)?
Yeah, I think it's good to point out that the ratio going to 1 does not mean the difference is going to zero.
It depends on what you mean by behave the same. If we're talking Big O then they are both O(ln x).
Yes, that really threw me off when I learned Thermodynamics! XD
I have a math question that I haven't really been able to find an answer for. When integrating why does the dx 'disappear' for a lack of a better word? Like why is dx or whatever differential gone when you do the integral? Hope I'm making sense with that
Limits can never be cool!
But they do get as close as you could want.
At the end you show the ln(x) and the W(x) functions plotted on the same graph. If the limit of their ratio for large numbers goes to one, why the two functions do not seem to sit one on another? The convergence is so slow?
I was curious and checked inverses of x^n*exp(x) and apparently all of them also behave like ln(x)
Well, yes, it's because e^x grows much faster than any polynomial so it dominates.
7:58 surprise, he have a green pen
It's 4am why am I watching this lol. Notification gang?
Are you at US?
@Naman_shukla410 probably central US, maybe Mexico or Central America. Most likely US though.
It was 10am here when he posted the video.
Does this work in general with inverses of functions like this?
If f(x) goes to infinity as x goes to infinity and g(x)=xf(x), will their inverses always have this limit?
Black holes would grow infinitely if not checked by other factors.
What about a limit or an integral with logarithms in variable base? For example logx(some function in x)
You can simplify log_x(f(x)) to ln(f(x)) / ln(x).
Can you show please how to compare W(W(1)) and (W(1))^2 without calculator?
nice
How gosh he got a green one ! 😮
Before viewing, I guessed e^(1/e), which is actually not that far off! :)
very coooool
Now can you compute this: lim ( ln(x)-W(x) )
x→∞
ingobojak5666 already answered that in his comment.
@@bjornfeuerbacher5514thanks
I tried to compute it on wolfram alpha but it fails. I had observed that ln(x)-W(x) grows extremely slowly, although it diverges
i thought he was making a rap video for a moment when he kept saying "to the e to the y"
CERN collisions.
Neat!
I made the substitution let x = lna(e^lna) and came to yhe same solution: 1
I did it with a variable change
x = te^t
-> Lím(Ln(x) / W(x), x -> inf) = Lím(Ln(te^t) / W(te^t), t -> inf) = Lím((Ln(t) + Ln(e^t)) / t, t -> inf) = Lím(Ln(t)/t, t -> inf) + Lím(t/t, t->inf) = 0 + 1 = 1
64000 Rutherford Curve
Why is the domain [-1,inf)? xe^x accepts any number as input. Maybe i just dont kniw what "to have inverse" means exactly
A function f has an inverse if for every x there is a unique y so that f(y) = x. For that to happen, it has to be bijective - one-to-one and onto.
The function x e^x can be defined for all real x, but you’ll find that there are values of x less than -1 and values greater than -1 that give the same value of the function, meaning you can’t pick a unique inverse across that domain. By restricting the domain of the function to [-1,infinity), you force it so that there’s only one value in the domain that corresponds to each value in its range.
I tried and done in 2nd try ❤
9:16 Vai me dar zero? Não é infinito?
pls help why is the domain [-1;inf)????
Because f(x) = xe^x has the range (0, 1/e) in the domain (-inf, 0) and f(x) = f(y) doesn't imply x = y. It's not injective in that domain.
Black pen red pen blue pen green pen YAY
No that is not true. The +1 with the infinity makes it a limit question again. Those sums do eventually diverge and if you use very large numbers to look at them like a Graham's number then the natural log wins with the greater growth.
?
@@blackpenredpen well we are looking not only at the limit originally, but a limit of limits. while infinity and and infinity+1 are both infinity they are not equal. You had another infinity over infinity and you needed to perform L'H again. We cannot draw a conclusion when it is infinity over infinity. That +1 will matter as if you look that the delta between the changes of change the ln while exceedingly the product log is even slower. As you like to say you have to do more work.
@@kennethgee2004 x/x=1 provided that x≠0...
I think I'm the only mathematician that doesn't get the love of the Lambert W function. What's its purpose, other than being the inverse of x e^x?
I am not a mathematician but a casual math viewer, The lambert W is only loved because it requires a lot of creativity to use(which is where all the fun lies) and its something new
Ofcourse, for you, this all must be basic, so it's understandable why you would feel that it's just a boring function 😅
I don't use the Lambert W function for much in my personal work, but I like using it in Calculus classes. Apparently, it has uses in some natural sciences, but the interesting thing, to me, is that we can't find a "nice formula" for it in terms of elementary functions, but we can still do calculus with it. We can use implicit differentiation to calculate its derivative. We can use things like Newton's Method to calculate values of W(x) to arbitrary precision. So the fact that we can do so much with a function that we don't have an "nice formula" for shows the power of calculus theory.
❤❤
Ln for Latural Nog
zero…… factorial 😂
Hi
It is not a "natural log 🪵"! It is a "natural logarythm".
Woag
Painis
The 🐟 Function is here!
Whenever i see the w function i automatically i lose interest. I am not sure what you fixation with it is. Its not something that's thought here and we are lucky to be spare of it.
Highly effective click bait 👌
.
Oh man you spoiled the result :(
Hello, I know this might be an absurd idea. But i am a small minecraft youtuber, If you would be interested. I think it would be cool to explain equations Utilizing minecraft. Let me know.
x>inf (de hospital)(1/x)/W(x)/x(W(x)+1)=(W(x)+1)/W(x)..>1
.
.