To my dear recent patrons, This video was uploaded 3 months ago (as an unlisted video) and I just published it today. So if you don't see your names on the outro card, I am sorry about it. I will be sure to include your names in my new videos. Thank you, blackpenredpen
@devanshsharma8638
3 жыл бұрын
Please make a video on integration xsquare-1/lnx dx from zero to one The answer was ln3 My professor did it by taking a function integration x power b-1/lnx from 0 to 1 and then differentiated with respect to b and then at last taking b=2 then answer came ln3 is there any other approach
@oximas
3 жыл бұрын
hey, what is the lambert W function? it feels like mathmaticians just define functions that solve certain equations and say thats it. like lets say I want to solve x^x =2 for x, I will define G(x^x) = x for any x taking the G function on both sides you get x = G(2) wow I solved for x genius wow. I hope you get the point, we can't just define a function as the inverse of another function to help us and say that this solve our equation, letting the computer do the rest of the work, this is stupid.
@benextinction__144
3 жыл бұрын
@@oximas bprp defined what the function does, but its strict formula would seem unnecessary as its so complex. Here it is if you want to try solving with it by hand :) images.app.goo.gl/q6txbn3FwC7X5Mrk8
@sakshitandel8572
3 жыл бұрын
@@md.faisalahamed5202 well all of these is the game of orientable and non orientable surface
@sakshitandel8572
3 жыл бұрын
@@oximas functions are defined not just when the need for them arises but they become all too common in mathematics like equations for conics etc.
@anthonyguerrera191
4 жыл бұрын
It’s not a black pen red pen video without the Lambert W function 😅
@GaussianEntity
4 жыл бұрын
Name a more iconic duo
@ffggddss
3 жыл бұрын
And at least one green or blue pen. Fred
@carultch
2 жыл бұрын
What does the W mean in the LambertW function?
@Podriman
2 жыл бұрын
@@carultch Wambert
@santiagonoya5702
4 жыл бұрын
Plz someone gifts this man a big whiteboard XD
@MathIguess
4 жыл бұрын
*cries in 3 wooden planks*
@tanmayshukla5330
3 жыл бұрын
Brpr is capable of writing the whole solution on a single sticky notes and you rae saying small whiteboard is not enough??
@plislegalineu3005
3 жыл бұрын
@@tanmayshukla5330 *bprp
@carultch
2 жыл бұрын
@@plislegalineu3005 What does PL stand for that is legal in the EU?
@plislegalineu3005
2 жыл бұрын
@@carultch 🇵🇱
@JSSTyger
4 жыл бұрын
The Lambert W(tf) function again...
@EsperantistoVolulo
3 жыл бұрын
*Cries in elementary functions*
@alexmason6839
4 жыл бұрын
You really must continue with the math for fun series 😍
@juancruzftulis3249
4 жыл бұрын
What kind of series? P-series? Ok... bad joke xd
@alexmason6839
4 жыл бұрын
@@juancruzftulis3249 amen bro
@peglor
4 жыл бұрын
1:30 approximating an 'approximately equal to' sign.
@garyhuntress6871
4 жыл бұрын
#2 is my fav because I'm 58 and had never heard of W(fish) function until I learned it here!
@SurfinScientist
4 жыл бұрын
I am 59, and I never heard about it either. And I am a mathematician-computer scientist. Never too old to learn. Cool function!
@skycocaster
3 жыл бұрын
This is simply a name for the inverse of x*exp(x), which can be proven to be a bijection, that's why W exists. Basically this is just naming some inverse function. The more traditional method would consist in proving that the function we are trying to invert is bijective and thus that the solution is unique.
@dexter2392
3 жыл бұрын
@@skycocaster xexp(x) is not always a bijection though, at 0>= x >= -1/e the inverse function forms a second branch
@skycocaster
3 жыл бұрын
@@dexter2392 Yes, but we have ln(x) in the equation, thus restricting the considered domain to R+*, on which the function is bijective :) If x*exp(x) was not a bijection on some considered domain, W wouldn't be a function.
@MathNerd1729
4 жыл бұрын
Since blackpenredpen didn't go through too much details into proving that u has to equal 1, I will do it for those reading this! Looking back at the original equation, there is no way u=ln(x) can be negative otherwise the left hand side would be positive and the right hand side would be negative. This justifies taking the log of both sides in the first step and taking ln(u). Now, once we reached u² - u = ln(u), we can move everything to the left hand side to get u² - u - ln(u) = 0. Let the left hand side represent a function: f(u) = u² - u - ln(u) We want to prove that the only root of this is u=1. If we can prove that u=1 is an absolute minimum then that will be a much stronger claim than what was wanted to know. That's what I'll prove. Proof by Second Derivative Test The first derivative of f is: f'(u) = 2u - 1 - 1/u f'(u) = (2u² - u - 1)/u f'(u) = (2u + 1)(u - 1)/u Since u>0, the ONLY critical point is at u=1. Now, looking at the second derivative: f(u)= 2 + 1/u² This is clearly 3 at u=1, which is greater than 0, so we've proven that u=1 is a minimum. Further more, since u=1 was the ONLY critical point according to the first derivative, then we've shown it's an ABSOLUTE minimum as well. Conclusion: Yes, u=1 is the only way we can get u² - u to equal ln(u), but it also follows from our proof that if u≠1, u² - u is greater than ln(u). Filling in the last steps of that part will be left as an exercise to either the reader or blackpenredpen's calculus students
@quantumsoul3495
4 жыл бұрын
Thanks
@JasonOvalles
4 жыл бұрын
Nice algebraic proof!
@seroujghazarian6343
4 жыл бұрын
The left hand side CAN be negative for abs(u)
@MathNerd1729
4 жыл бұрын
@@seroujghazarian6343 Ah sorry, my mistake! Thanks for catching it! :)
@Nithesh2002
4 жыл бұрын
Hmm what about solutions on the complex plane?
@JasonOvalles
4 жыл бұрын
To show that u²-u and ln(u) only intersects once, I think it's enough to know that u²-u is convex and ln(u) is concave. That, combined with showing that they are tangent at u=1, is enough to show that this tangent point is their only intersection.
@pedrosso0
2 жыл бұрын
Thank you for some actual proof
@WilliametcCook
3 жыл бұрын
Teacher: talking about Lambert W function Me, not paying attention: 2 + 2 = fish
@carultch
2 жыл бұрын
Have you been watching Flippin' Physics? They call alpha the fishy thing.
Q2 - I got the same anawer but I did it in another way: x + lnx = 2 e^(x + lnx) = e^2 e^x * e^lnx = e^2 e^x * x = e^2 x * e^x = e^2 W(x * e^x) = W(e^2) x = W(e^2)
@skylerpretto1221
4 жыл бұрын
@@damianbla4469 That's how I did it, too. Both methods are great though.
@nicolascalandruccio
4 жыл бұрын
Ahah spoiled! For the 2nd equation, you could have started with the fact that exponential transforms sums into products. Btw, same answer. x+lnx=2 e^(x+lnx)=e^2 e^x.e^lnx=e^2 xe^x=e^2 x=W(e^2) Same as @damian bla
@AlgyCuber
4 жыл бұрын
q2 also : xlnx = 2 take e to both sides --> x^x = e^2 take super sqrt --> x = ssrt(e^2)
@spelunkerd
2 жыл бұрын
This video reminds me of how much I love maths. I wish I had done more of these when in school, practicing helps to cement fundamentals.
@pedromello5854
4 жыл бұрын
Q1: e^(e^(W(ln2))) Q2: W(e^2)
@joshmcdouglas1720
3 жыл бұрын
That’s what I got too
@Pankaw
3 жыл бұрын
yup
@KRO_VLOGS
7 ай бұрын
After applying the W function add 2 and exponentiate both sides so answer will be (e^(2-W(e^2)
@jibiteshsaha4392
4 жыл бұрын
Before even starting the video I knew he Will use Lambert W and I stopped the video😂😂
@ffggddss
3 жыл бұрын
Pre-watch: (1) The first one appears to be the only one of the three that's soluble in "standard" functions. x^(ln x) = 2 - - - take ln() of both sides ... (ln x)² = ln 2 ln x = √(ln 2) x = e^√(ln 2) = 2.299184767... Curiously, if this value were to satisfy eq. 2, it would automatically also satisfy eq. 3 (just equate the LHS's of eqs. 1 & 2, and that is eq. 3). Alas, it does not. My guess is that eqs. 2 & 3 can be solved using the Lambert W function. (2) x ln x = 2; let y = ln x; x = eʸ then yeʸ = 2 - - - take W() of both sides ... y = W(2) x = e^(W(2)) (3) x^(ln x) = x ln x This one doesn't seem as cooperative ... let's see how you do it. Post-watch: Very nice! I missed the 2nd solution on eq. 1; it works, too. And for eq. 3, your graphical solution seems to be the only way to solve it. Graphical solutions usually don't cut it, but your argument (along with concavity) can make it rigorous. There can be no other (real) solutions. Fred
@blackpenredpen
3 жыл бұрын
Thanks for the answers, Fred! Hope you have been well : )
@ffggddss
3 жыл бұрын
@@blackpenredpen Thanks! I can't complain. Except about the weather and the 'lock-down,' of course. And thanks for continuing to do these, and I hope you (and your piano-playing gf) are well, too! (You seem to be...) Fred
@blackpenredpen
3 жыл бұрын
@@ffggddss Thanks, we are : )
@Austin101123
3 жыл бұрын
Q2: turn it side into exponents of e, and we get e^(x+lnx)=e^2 This becomes x*e^x=2, and then x=W(2) Q1: Let a=lnx, we have a^a=2 .Take the ln of both sides and we have alna=ln2 -->e^lna *lna=ln2, take lambert W of both sides and get lna=W(ln2), so a=e^W(ln2), replace a with lnx and have lnx=e^W(ln2), so x=e^(e^W(ln2)) lna=W(log2)-->lna*e^lna=lna*a=ln2
@d4slaimless
2 жыл бұрын
You have wrong solution for Q2. You have e(x+lnx)=e^2 -> xe^x=e^2 , so if you take W-function from both sides you will have x=W(e^2). I got the same answer with logarithms: x+lnx=lne^x+lnx=lne^2 ->ln(x*e^x)=lne^2 -> x*e^x=e^2 -> x=W(e^2) = 1,557 . Now 1,557+ ln1,557 = 2
@Austin101123
2 жыл бұрын
@@d4slaimless yes looks like I messed up. I put x+lnx in the exponent on the left side but they should be multiplied not added
@rodrigomarinho1807
4 жыл бұрын
Let u because I love u was THE line of this video. Great job
@naushaadvanderveldt5342
4 жыл бұрын
Q1: Ln(x)^ln(x)=2 Taking the natural log on both sides: Ln(ln(x)^ln(x))=ln(2) which simplifies to ln(x)*ln(ln(x))=ln(2) Let u=ln(x) u*ln(u)=ln(2) You can write u as e^ln(u) so, ln(u)*e^ln(u)=ln(2) Taking the Lambert W function on both sides: W(ln(u)*e^ln(u))=W(ln(2)) ln(u)=W(ln(2)) u=e^W(ln(2)) Since we defined u to be ln(x): ln(x)=e^W(ln(2)) x=e^(e^W(ln(2))) Q2: x+ln(x)=2 You can write x as e^ln(x) so, e^ln(x)+ln(x)=2 e^ln(x)=2-ln(x) Multiply by e^-ln(x): 1=(2-ln(x))*e^-ln(x) Multiply by e^2 so we can use W Lambert function: e^2=(2-ln(x))*e^(2-ln(x)) Taking W Lambert function on both sides: W(e^2)=W((2-ln(x))*e^(2-ln(x))) W(e^2)=2-ln(x) ln(x)=2-W(e^2) x=e^(2-W(e^2)) This was pretty hard to type so please let me know if I have made any mistakes.
@mahmoudalbahar1641
4 жыл бұрын
very cool.....your answer is 100% true
@naushaadvanderveldt5342
4 жыл бұрын
@@mahmoudalbahar1641 Thanks!
@imirostas4920
2 жыл бұрын
I did Q2 like this: x+ln(x)=2 e^(x+ln(x))=e^2 Notice: a^(b+c)=(a^b)*(a^c) (e^x)*(e^ln(x))=e^2 Notice: e^ln(x)=x --> the exponential and logarithm cancel each other x*e^x=e^2 x=W(e^2) so x is approximately: 1.55714559899761... plugging this in for x in x+ln(x) yields 2
@naushaadvanderveldt5342
2 жыл бұрын
@@imirostas4920 very smart! On closer inspection of my final answer (e^(2-W(e^2))), it is indeed equal to yours (W(e^2)) by the properties of the Lambert W function. Following is a proof of this for those who are curious: Rewrite e^(2-W(e^2)) as (e^2)/(e^W(e^2)) Remember the definition of the LambertW function as the inverse of xe^x, so W(x)e^W(x)=x which can be rewritten as: e^W(x)=x/W(x). In our case: e^/e^W(e^2) = e^/(e^2/W(e^2)) in which the term e^2 cancels out and so does the double fraction so what you’re left with is just W(e^2). Edit: upon rereading this, i realized there was a much easier way to do this and I actually feel kind of dumb for not seeing this earlier: Remember the definition of the LambertW function as the inverse of xe^x, so W(x)e^W(x)=x Multiply by e^-W(x) to get xe^-W(x)=W(x) In our case: e^(2-W(e^2))=e^2*e^-W(e^2) which precisely fits our above proven property and thus equals W(e^2).
@maliciousmarka
3 жыл бұрын
Him: Let u, because I like you guys, be = ln(x) Us: It’s only *Natural* he says that about us
@plislegalineu3005
3 жыл бұрын
US United States? XD
@danny1504-g2d
10 ай бұрын
Wow, your math problems and the way you solve it really gives me inspiration to learn more about this subject, Math, that i used to hate a lot.
@itaycohen7619
4 жыл бұрын
I did 3 like that : X^lnx = xlnx e^ln^2(x) = (e^lnx)*lnx Flipping the e to the powers to the other sides we get : e^-lnx = (lnx)e^-ln^2(x) Multiply both sides by -lnx we get : -lnx(e^-lnx)=-ln^2(x)*(e^-ln^2(x)) And now, all that is left is to take the lambert w fucntion on both sides to get : -lnx = -ln^2(x) a simple equation with the solution x=e. Also gives x=1 but we check to see it dosent work for 1.
@pedrosso0
2 жыл бұрын
Same
@louislecam7773
4 жыл бұрын
Here is a very interesting video, thank you !!!
@masonhyde9411
4 жыл бұрын
@blackpenredpen I had a math problem that I would love for you to do. I have had people tell me "it's impossible" and it "can't be simplified" but I know there must be a way. x^(x+1)=ln 3
@learnwithfun-swayamgupta5836
4 жыл бұрын
I had got two answers, can you tell correct answer for this question please 1st answer-0 and; 2nd answer-3.
@Mutlauch
4 жыл бұрын
Dear bprp, First let me tell you that what you are doing is amazing and an incredible inspiration for me and most certainly for many others as well :) Furthermore, I want to point out that for all real numbers r the equation x(x-1)=r*ln(x) has a unique solution at x=1 : 1(1-1)=r*ln(1) => 1*0=r*0 and this is true for all real numbers r. But it could be very difficult, if at least one of the coefficients of the polynomial on the left side is not equal to 1 (and obviously if those coefficients are not equal to each other) or the parabola is shifted horizontally. Greetings from Germany 😊
@ilyashick3178
Жыл бұрын
Great ! some interesting staff: x intercepts as u^2-u=0 and u=0 and u= 1 then only ln_eX= 1 and then E=X.
@richardfredlund3802
4 жыл бұрын
that was awesome. i like how all three had quite different solutions.
@stapler942
3 жыл бұрын
Here's how I did the u^2 - u = ln(u) step without the graph: -Take e to the power of both sides. e^(u^2 - u) = u -Differentiate both sides to get e^(u^2 - u) * (2u - 1) = 1 -Divide both sides by (2u - 1). -Now you have the identity e^(u^2 - u)= u = 1/(2u - 1) -Using the quadratic formula you can get: u = 1 or -1/2 -Since ln(u) is not defined for negative numbers we discard -1/2 and say that u = 1.
@SKMathPosnMathsk-fg3fu
2 жыл бұрын
Nice.
@d4slaimless
2 жыл бұрын
I am sure you can't use differentiation this way. For example if you have equation x^2=x the obvious answer is 1. But if you differentiate both sides you will have 2x=1 and it gives you x=1/2. Derivatives for functions are not necessary equal at the same point where functions equal. They are equal where tangents are the same.
@yaxeenrahman
4 жыл бұрын
BPRP IS MY FAVOURITE
@Ethiomath16
4 жыл бұрын
Really fun and excellent way of solving
@jjpswfc
4 жыл бұрын
Thank you for all your videos that you do, they're always really helpful for my maths even though I'm from the UK :)
@SKMathPosnMathsk-fg3fu
2 жыл бұрын
These videos are really helpful 😊. I love logarithm now 😀.
@ikntc514
Жыл бұрын
I feel so happy that I solved the second equation at the end🔥🔥 i basically did: x+lnx=2 ln(e^x)+lnx=2 sum of logs = log of the product =ln(xe^x)=2 xe^x=e^2 so the answer is the Lambert W Function of e^2
@garavelustagaravelusta9717
4 жыл бұрын
For the second question, can't you just solve it like this: ln(xlnx) = ln(2) ln(x) + ln(lnx) = ln(2) Let ln(x) = u; u = ln(2) - ln(u) u = 0.693 - ln(u) Finally, solve out for the common points of two sides of the equation (like in part 3). Great Channel btw.
@navman2014
4 жыл бұрын
How would you ever find the common point? It's not nice like 1 its a irrational number.
@prathampatel1740
3 жыл бұрын
mans gotta make a video on the lambert W function, like an explanation
@aubertducharmont
2 жыл бұрын
On problem 2 you said that for calculating lambert w function you need wolfram alpha or mathematica. But thats not true. You need just a calculator and use Newton-Raphson or Hailey iteration formula.
@AllInOne-nz5kg
3 жыл бұрын
Please make a full course on logarithms if you already made then please give me it's link
@gbc1827
3 жыл бұрын
Wow this lecture reminds me of my calculus class when i was in high school XD
@d4rk_1egend
2 жыл бұрын
An easier way to solve the 1st log problem would be to rewrite each side so they have the same base so then we would rewrite the x in x^ln(x) as e^ln(x), so then we would have e^(ln(x))^2, then we would rewrite the 2 as e^ln(2), thus we have e^(ln(x)^2=e^ln(2), we can see they have the same base so then we have the equation (ln(x))^2=ln(2), we then take the square root (positive and negative) on both sides to get ln(x)=sqrt(ln(2)), then we exponentiate it to find that the solutions are x=e^sqrt(ln(2)), and x=e^-sqrt(ln(2)).
@dushyanthabandarapalipana5492
2 жыл бұрын
I am so greatful to you in this trouble so me time in sri Lanka!
@donaldmcronald2331
7 ай бұрын
Q1 was easy: You can also use a substitution. Before that you take the ln on both sides and get u*ln(u)=ln(2). Just change that to ln(u)*e^(ln(2))=ln(2). Finally, you use W(x) and do the resubstitution and you get x=e^e^((W(ln(2)))) Q2: do e on both sides: e^(x+ln(x))=e^2. Then use properties of exponents and pull them apart. e^(ln(x)) simpplifies to x and you can use W(x) again, so x*e^x=e^2 x=W(e^2)
@humanoidtyphoon991
3 ай бұрын
For Q2 you can also rewrite x as ln(e^x) and then use the properties of ln to have ln(xe^x)=2, you take e on both sides and you get xe^x=e^2 and then use W(x) so x=W(e^2)
@rom2089
3 жыл бұрын
btw for the 3rd problem: using properties of logarithms (ln(X))^2 - ln(X) = (ln (X/X))^2= ln(1)^2=0 so you have 0=ln(ln(X)) 1=ln(X) and finally X=e
@peterharmouche8662
3 жыл бұрын
this is incorrect, ln(x)^2 - ln(x) is not equal to ln(x/x)^2 because the property of logarithms you're referring to (ln(a) - ln(b) = ln(a/b)) can't be used if one of the ln is squared Here's the full answer: x*ln(x) = x^ln(x) (divide by x) (x can't be equal to 0) ln(x) = (x^ln(x))/x = x^(ln(x)-1) (replace x by e^ln(x) on the RHS) ln(x) = e^((ln(x)*(ln(x)-1)) = e^(ln(x)^2)*e^(-ln(x)) (multiply by -ln(x)*e^(-ln(x)^2)) (ln(x) can't be equal to 0) (-ln(x)^2)*e^(-ln(x)^2) = (-ln(x))*e^(-ln(x)) (take the lambert w function on both sides) -ln(x)^2 = -ln(x) (re-arrange) ln(x)^2 - ln(x) = 0 (take ln(x) common factor) ln(x)*(ln(x)-1) = 0 (2 possibilities) ln(x)-1 = 0 or ln(x) = 0 (but we already know from earlier that ln(x) can't be equal to 0) ln(x) = 1 x = e
@yuvalnachum4190
4 жыл бұрын
Hey blackpenredpen i really love your math for fun videos. Try this math for fun question : In rectangle ABCD the area is equal to the perimeter and to the diagonal . find AB and BC.
@anuraagrapaka2385
4 жыл бұрын
0:16 reminded me of WII (mariokart) and those joyful moments
@mehmeteminconkar2590
Жыл бұрын
Q1 sol equals e^e^w(ln2)
@manuelruizsanchez2053
4 жыл бұрын
Could you teach some tricks for optimization and real problems with them in the current life? Kisses from Spain and thanks for the video, as always you're excellent ❤️🤙
@liab-qc5sk
4 жыл бұрын
Q1 the sol: exp(exp(W(ln(2)))) Q2 the sol: exp(2-W(exp(2)))
@erikestrella7240
4 жыл бұрын
When sacas la gráfica jeje, la vieja confiable
@Kdd160
4 жыл бұрын
This is cool
@ilyas6601
2 жыл бұрын
I did the first one in my head before watching and i found x = √(2e) as a good approximation (~2.33). As some teacher said someday: "I dont know, but you used the wrong formula and got the correct answer."
@jayktomaszewski8738
8 ай бұрын
3 is my favourite
@a.osethkin55
3 жыл бұрын
Thanks
@orenfivel6247
3 жыл бұрын
this is 2 cool sol.: BTW can u solve (*) y' =-ay⋅(1-y)^2, i.e., WTF y(x) s.t solves (*) for a>0? I Think LambertW function is involved. maybe W(0,x) or W(-1,x).
@renielescopete2558
2 жыл бұрын
Reniel escopete R 11-humms B 2nd pandemic
@estera742
3 жыл бұрын
Q1:x=e^e^w(ln2)
@qqqquito
Жыл бұрын
The solution to the 3rd equation, X^(lnX)=XlnX, is an interesting one.
@pedrosso0
2 жыл бұрын
Here's how to find the value without looking at a graph: x^lnx = xlnx e^ln^2(x)=lnx e^lnx e^(-lnx)=lnx e^(-ln^2(x)) -lnx e^(-lnx)=-ln^2(x) e^(-ln^2(x)) -lnx=-ln^2(x) 1=lnx x=e All solutons would have something to do with the lambert W function
@astraestus8828
4 жыл бұрын
I love this channel.
@egillandersson1780
4 жыл бұрын
I got the first and second but failed with the third :-( Thank you !
@That_One_Guy...
3 жыл бұрын
Don't worry, I don't think there's mechanical method to get value of u
@jaymsf4895
3 жыл бұрын
I got for Q1: x = e^[e^(W(2))] And for Q2: x = W(e^2)
@shinkolan1985
2 жыл бұрын
Can you find the integral int((1+x^2)^(-1/3))dx
@serenityteachings
3 жыл бұрын
You sir are a legend
@ChessGrandPasta
3 жыл бұрын
5:45 oh you like uranium?
@montsaintleondr7491
4 жыл бұрын
Cool pockemon ball microphone and also some nice explanations!
@felipediaz232
4 жыл бұрын
Sigue con esos videos resolviendo ecuaciones cortas y extrañas 👍🏼👍🏼
@danielmendes2923
4 жыл бұрын
"I feel the pressure" 😂😂😂
@Dhanush-zj7mf
4 жыл бұрын
I tried in a different way And got diiferent result😥😥 2 problem solution-- Xln(X) = 2 ln(x) = 2/x x = e^(2/x) 1 = 1/x (e^2/x) Multiply with 2 on both sides 2 = 2/x (e^2/x) Now take Lambert W function W(2) = 2/x X = 2/W(2) I dont know why i got a different answer anyone please tell me where I am wrong..💓💓
@rob876
4 жыл бұрын
You are correct. W(2)e^W(2) = 2, so 2/W(2) = e^W(2)
@Dhanush-zj7mf
4 жыл бұрын
@@rob876 Thanks it helped : )
@Lamiranta
4 жыл бұрын
3:33 Yeah, I got my fish back!
@Kevsunsix
Жыл бұрын
There is a very good algebraic way that I found to solve the last question without using the concept of graphs. x^ln(x) = xln(x) Call y to be ln(x). (e^y)^y = y*e^y e^(y^2) = y*e^y Divide e^(y^2) and e^y on both sides: 1/(e^y) = y/(e^(y^2)) Since 1/n can always be described as n^-1. We can say that y/(e^(y^2)) = y(1/e^(y^2)) = y * e^-(y^2). The same applies with 1/(e^y) which is just equal to e^-y. As such, we have the following equality: e^-y = y * e^-(y^2) Finally, multiply by -y on both sides. -y * e^-y = -y^2 * e^(-y^2) Take the Lambert W function on both sides. This way, the equation simplifies to be: -y = -(y^2) We can now ultimately solve for y. y = y^2 -> y^2 - y = 0 -> y(y-1) = 0 y cannot be 0 as ln(x) is not defined at x=0. So we only have one solution. ln(x) = 1 => x = e This, in my opinion, is the best way to solve this question as it doesn’t even require you to know about the Lambert W function nor any graphical concepts. It comes down to just raw algebra; albeit, quite tricky.
@frapassante
3 жыл бұрын
For the third equation, when you get to u^2 - u = ln(u), you can just differentiate both of them, and you get a 2nd degree equation which has two solutions: u = 1 and u = -1/2. Since you have ln(u) in the equation, u must be > 0, so the only acceptable solution would be u = 1 :)
@vbprogrammer95
2 жыл бұрын
Except deriving both sides is not something you can do to solve an equation. Take x²=4. Its solutions are 2 and -2, but if you derive it you get 2x=0, so x=0, which is absurd. Also if you have something like x=lnx the derivative changes (and by a lot) if you move one side to the denominator of the other.
@frapassante
2 жыл бұрын
Leggere il commento che ho scritto ora, dopo aver seguito analisi 1, mi fa sentire davvero un cretino. Vorrei cancellarlo, ma magari qualcuno che avrebbe fatto il mio stesso errore potrà correggersi con la tua risposta, quindi grazie!
@frapassante
2 жыл бұрын
Also, I guess it would only make sense to derive both sides if the relation is an identity
@tarfarian
4 жыл бұрын
5:43 I'm definitely not using that on someone ;)
@Flinagus
3 жыл бұрын
shouldnt #3 be e if you have x=e on both sides it checks out e^(lne) = elne e^1 = e(1) e = e x = e? desmos graph agrees
@sandeepsharma-jw2rz
4 жыл бұрын
INCREDIBLE!!!!!!! Edit: adding my name at the last is impossible and you know why.😁
@marius3023
4 жыл бұрын
1/e^x=ln(x) ??? I cannot get this :^((
@starfir9745
3 жыл бұрын
Is the 2 possible without the W fonction ?
@jacobthecool3000
4 жыл бұрын
Cool vid but never in my life have I heard of the W function.
@barryomahony4983
4 жыл бұрын
Must be the first time you've watched this channel. 🤣
@jacobthecool3000
4 жыл бұрын
@@barryomahony4983 Actually no. I must have somehow avoided it in his previous videos.
@johndavecusi6990
3 жыл бұрын
(1-x^2)y'=1-xy-3x^2+2x^4 would you solve this differential equation i think it's non exact
@1510xdjan
3 жыл бұрын
Regarding the first equation. Didn't u forget about the other solution, namely x=e^{-sqrt[ln(2)]}. ?
@James2210
2 жыл бұрын
first one is x=e^(±sqrt(ln(2)))
@tokajileo5928
Жыл бұрын
what about complex solutions?
@vaibhavm2916
4 жыл бұрын
This is the fourth time you uploaded a previously shot video.but how did doctor peyam find your lost video
@mehmeteminconkar2590
Жыл бұрын
Q1 answer is e to the power of w of ln2
@rajns8643
4 жыл бұрын
In 2nd question, wouldn't taking Ln on both sides be more easier?
@humanoidtyphoon991
3 ай бұрын
The answer for Q2 is w(e^2)
@cadlag2010
3 жыл бұрын
for 2, you can directly apply W(xln(x))=ln(x)
@邱伊華-t6g
3 жыл бұрын
I found the other solution of the third equation.Differentiating u at the two sides of the equation,it will get another answer is e^-1/2.But actually,this is wrong.Can you tell me what is the mistake in my solution?
@MK-tb6rb
Жыл бұрын
Q1 X=e^e^W(ln2) Q2 X=W(e^2)
@IngeniusFool
4 жыл бұрын
0:06 pause here 🤔 and ponder 💭
@diyaael-shorbagy7056
3 жыл бұрын
Can u do a video on how to draw the gamma fun.
@kadenrobinson1993
2 жыл бұрын
e^e^w(ln2)
@sebastianbarbu8691
3 жыл бұрын
When and where have you learned those things?
@forgexgames
3 жыл бұрын
Only got the third one right before watching, as a 14 year old. Btw I love your videos, keep it up!
@mfol2374
3 жыл бұрын
Interesting question, is there the chance to solve with Lambert W function the equiation x^x = 2x to find second root?
@Bananananamann
3 жыл бұрын
For (2), why do I get a wrong value if I do lnx=2/x | d/dx; 1/x=-2/x^2; x=-2 ?
@Iamnotyou29
3 жыл бұрын
Q.1 & Q.2 both are just crushing🧐🧐
@klementhajrullaj1222
Жыл бұрын
(lnx)^x=x^(lnx)
@generaldarian1263
3 жыл бұрын
First time using W on my own but here's my answers for the last two. (ln x)^(ln x) = 2 1. ln both sides (ln x) * ln( ln(x) ) = ln 2 2. Let k = ln x, so k*ln k = ln 2 ---> ln k * e^(ln k) = ln 2 3. Apply W function and solve W(ln k * e^(ln k)) = W(ln 2) ---> ln k = W(ln 2) k = e^W(ln 2) ln x = e^W(ln 2) x = e^e^W(ln 2) x + ln x = 2 1. Power both sides by e e^(x + ln x) = e^2 (e^x) * (e^ln x) = e^2 x * e^x = e^2 2. Apply W function W(x * e^x) = W(e^2) ---> x = W(e^2)
@estera742
3 жыл бұрын
Q2:x=w(e^2)
@Jordan-zk2wd
4 жыл бұрын
Another good one: x^ln(y)=y^ln(x). Looks really complicated, until you realize...
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