Can we find the value of (1-i) ^ (3+2i) just using binomial identity expansions? (euler constant e is not allowed) . I will subscribe to your channel if you answer this question with explanation.
It is possible to use W to solve a = x^(x^(x+1)). Solution follows: 1) Take the log of both sides: ln a = x^(x+1) ln x = x^x ln x^x. 2) Apply W: W(ln a) = ln x^x = x ln x. 3) Apply W: W(W(ln a)) = ln x. 4) Finally: x = e^W(W(ln a)). Done! For example, this solves 4 = x^(x^(x+1)) as x ~ 1.559610469.
These were fun! This was not the video for it, but it came up twice, so I mention it here: When we encounter a new function, we never have original intuition about the actual values it takes. This is certainly true of things like the logarithm, but there was even a time when you did not know what f(x) = x looks like. Sometimes, we can just plug-and-chug some example inputs. But, usually, we get our best intuition by plotting the function. We can be confident in that intuition and get a more-detailed view of the function's properties by doing things like finding domain, range, intercepts, end-behavior, growth patterns, extrema, saddle points, points of inflection or undulation, etc. Similarly, approximations (Newton's method, Taylor expansion for analytic functions, etc.) can help a student get an actual feel for specific outputs (especially for non-special points). After all of that is completed, people tend to feel that they have a good grasp of the function and no longer question things in the manner like "Isn't W(2) cheating?" (not exactly a paraphrase - more like a stereotypical example). It is a good question to have, but we should be asking it about all functions - and the reason for our not doing so is because we have that improved intuition.
I really like this video, it helped me understand the ways to use the W(x) function better than any other sources I could find. I think your way of working through these problems helps teach and demonstrate the usefulness of the W() function tool without getting deep into the weeds like other sources (Wikipedia) do. To comment on something you said in the middle of the video, I think people consider it "cheating" because it feels like mathematicians just "made up" a function to be the opposite of x*exp(x) instead of "solving it for real." But in reality there are plenty of these types of functions that exist. The "error function" for the integral of exp(-x^2) or the Gamma function for non-real-integer factorials. If you think about it, even the "logarithm" itself is just made up to be the opposite of exponentation. I think we're more comfortable with logs because it feels intuitive, but really it's no different. Thanks for the great videos.
ln(W(ln(x)))=1...........1 ln(W(ln(x))) e =e¹.......2 W(ln(x)) =e................3 Finally the answer 1+e e X=e. 😂We really need good luck for writing but ....😂😂😂
damn, wish I knew about this stream yesterday. I had to derive wien's displacement law from max planck's law, and i arrived at an equation that I thought could not be solved analytically... turns out I needed to use the Lambert W function which I only JUST learned about by googling today
Hey @bprp There is such a nice content on your channel. I just love seeing your videos. Today, my mind got stuck at a question. Can you please try to figure out what could the possible strategy be to solve the following equation: ln(x+8).ln(2) = ln(x).ln(10) P.S: x=2 seems to be very obvious, but I can't get how can I reach to that (and other real solutions, if any).
I like these, but I feel like there’s a more interesting class of equations to be solved (you I believe have done 3 of these) Equations that do not involve W(x), but can only be solved with W(x)
You should have noticed that the expression (1- (x+1)(e^(x+1)^2 = 0 is impossible because (x+1)^2 is always >= 0. Hence, e^(x+1)^2 will always be >=1. Additionally, no matter whatever value of x might be , (x+1) (e^(x+1)^2 ) will never ever = 1 Checking: For x = -1, (x+1)e^(x+ 1)^2 = 0---> LHS =1 - 0 = RHS = 0 is nonsense or impossible For x < -1 , say x= -2----> (-2+1)e^(-2+1)^2 = -1(e) = -e. Plugging e in the eq will give: LHS =1- (-e) = RHS = 0 , which is again impossible For x > -1, say x = 1----> (1+1)(e^(1+1)^2 = 2(e^4). Plugging 2(e^4) in the eq will give: LHS = 1 - e^4;= RHS = 0, which is again impossible Therefore, the eq (1-(x+1)(e^(x(1)^2 = 0 is false. In other words, no more solution can be found from that equation. The only solution is x = -1 You guys will need to learn how to solve probs with logical reasoning to understand such notions of mathemstics, as imaginary or complex numbers, Lambert function, and suchlike, which can be misleading or false, instead of copying them so blindly nonsensically, like a photocopying machine.
Hi blackpenredpen! I watched this video and your other video on the lambert w function. Thanks for taking the time to make these videos! They are very helpful! If it wouldn’t be too much trouble, could you please help me with this one? Maybe post a video? x^x + 2x + 3 = 50
I just made a program for calculating lambert w function in my scientific calculator, though numbers too large might not work :( and it takes about 3-5 seconds to evaluate lol
I got everything right until the last one, when I overlooked the possibility of two solutions. I got e^(-W(2)/2)-1, which is equivalent to your second solution.
For x^x^x=a xlnx^x=lna from lnx^a=Alnx x*xlnx=lna x^2lnx=lna 2*x^2lnx=2lna x^2lnx^2=2lna e^lnx^2*lnx^2=2lna lnx^2=W(2lna) Take exponential then take root, hard typing on a tablet and it is 3.40 am I hope it's correct LOL took me like 5 minutes to come up with it
hi friends. hi teacher We know that W(xlnx) = lnx And how about *W[(lnx)/x]* ? Is there some "formula" in this case? For example We know W(17ln17) = ln17 But to find/calculate W[(ln17)/17], can we find an exact value, a perfect value, without using approximations, without using things like Wolfram Alpha?
Sir , I am a big fan of you . I really like your videos and the way you explain everything . I am class 12th and I have to say that you are doing a great job ☺ . Sir can you reply me please 😊
In applied math, when a sum is involved, the units much match each terms, so we more often find equations like: x + a * exp ( b * x) = c than x + exp (x) = c (where a, b and c are constants, and in the case of electronic, a and c are positive, b negative, in the case of a P-N junction of a diode/transistor for example). Is there a "trick" to bring the first equation implying the W-Lambert's function? Or to solve analytically: K = x + exp( a* x ) which seems "close" to the Lambert W function. Side note: Since x has units, and that the exponent of the exponential should not have units (thus constant "b" has (unit^ -1) of x, and b*x becomes a pure number) and the result of the exponential is a pure number, so the constant "a" has units of x.
While we can get rif of the constant "a" with a simple substitution like a*u = x, I haven't found a way to then bring the exponent to the proper template.
The tough part of this problem is that you’ve got addition along with exponentiation separating the xs. The W function is able to “bridge the gap” between exponentiation and multiplication, but I’m not sure what function would do that for adding and exponentiation. Maybe try defining a new function F(x+e^x)=x and work from there? (I think I’ve got something here but it’s a lot of vibes and not much rigor lol)
@@redpepper74 Thanks a lot for your time. Really appreciated. I just recently noted that a relatively new section was added in Wikipedia in the Diode Modeling article, specific to a diode in series with a resistor, where the explicit resolution for the current in terms of the Lambert W function is obtained.
@@snnwstt I only have a high-school-level understanding of circuits and electricity so I don’t really get what’s going on there. But looking into that sum log function was fun :)
No 6 - x ln(x) = e^2 +1, W(x ln x = e^2 +1, where the 1st x gone from the previous line? Then follows the line W (ln x e ^(ln X) where W (ln x e ^(ln X) = ln x Who can help?
At 32:20 why did he need to square both sides. He could've just done it normally like (x+1)(e^2(x+1))=1 And then 2(x+1)(e^2(x+1))=2 2(x+1)= W(2) x+1= W(2)/2 x= (W(2)/2)-1 It produces a different answer tho 😭😭
Is it simple enough to say that the third solution to the eighth question must be ruled out because the value is not in the domain of W(x+1)? Since -sqrt(W(2)/2) < -1/e
Yes. You can find them in the Wolfram Alpha article for the Lambert W function, and in the Wikipedia article as well. There are also a number of integral formulas you can derive and differential equations you can derive for this function and with this function.
37:57 Uh... we expected that you try. We never expected you to know, but we wanted to at least try. Oh well, maybe you didn't want the livestream to go too long. I'll think about it!
y = sin(x)^(1/2) dy = cos(x)/[2·sin(x)^(1/2)]·dx, so sin(x)^(1/2)/cos(x)·dx = sin(x)/cos(x)^2·cos(x)/sin(x)^(1/2)·dx = 2·sin(x)/cos(x)^2·cos(x)/[2·sin(x)^(1/2)]·dx = 2·sin(x)/cos(x)^2·dy = 2·sin(x)/[1 - sin(x)^2]·dy = 2·y^2/(1 - y^4)·dy 2·y^2/(1 - y^4) = [(1 + y^2) - (1 - y^2)]/[(1 + y^2)·(1 - y^2)] = 1/(1 - y^2) - 1/(1 + y^2) = 2/[2·(1 - y^2)] - 1/(1 + y^2) = [(1 - y) + (1 + y)]/[2·(1 + y)·(1 - y)] - 1/(1 + y^2) = 1/2·1/(y + 1) - 1/2·1/(y - 1) - 1/(y^2 + 1). Before proceeding to antidifferentiate, here is something that needs to be understood. sin(x) is necessary an element of [-1, +1] for every x, and as such, sin(x)^(1/2) is an element of [0, 1] for every x that is an element of the union over all integers m of the intervals [2·m·π, (2·m + 1)·π]. Furthermore, cos(x) = 0 for every x = m·π + π/2. As such, the antiderivatives of sin(x)^(1/2)/cos(x) only exist at the intervals (2·m·π, 2·m·π + π/2) and (2·m·π + π/2, 2·m·π + π), all of which are mutually disjoint and disconnected. Thus y is an element of the interval (0, 1), and in this interval, 1/(y + 1) has antiderivatives ln(1 + y) + C0, 1/(y - 1) has antiderivatives ln(1 - y) + C1, and 1/(y^2 + 1) has antiderivatives arctan(y) + C2. Therefore, 1/2·1/(y + 1) - 1/2·1/(y - 1) - 1/(y^2 + 1) has antiderivatives 1/2·ln(1 + y) - 1/2·ln(1 - y) - arctan(y) + 1/2·C0 - 1/2·C1 - C2 with respect to y. However, since y is a function with a domain that is a disjoint union of disconnected intervals, the constants of integration with respect to y are sums of step function with respect to x, so the antiderivatives of sin(x)^(1/2)/cos(x) with respect to x are given by 1/2·ln[1 + sin(x)^(1/2)] - 1/2·ln[1 - sin(x)^(1/2)] - arctan[sin(x)^(1/2)] + C(x), where C(x) = A(m) if x is an element of the interval (2·m·π, 2·m·π + π/2), C(x) = B(m) if x is an element of the interval (2·m·π + π/2, 2·m·π + π), where A and B are families of real numbers indexed by the integers.
Please i have an equation to resolve ! 🙏 exp ( -x²) - 2√π *( 10^-4)* x = 0 Please someone help me i need the solution ! Can you make a short video for it ????
You can rule out the solution with the minus sign by plugging it into your earlier equation (x+1)e^[(x+1)^2]=1 if you plug in x=-1-sqrt((W(2))/2), you will get (-sqrt((W(2))/2))e^[(W(2))/2] = 1 Assuming we are looking at W(2) in the real world: -sqrt((W(2))/2) is a negative number. e^[(W(2))/2] is a positive number. when you multiply them together, you get a negative number. How can a negative number equal 1? It can't, so that isn't a valid solution.
Unfortunately I couldn't solve x^x^x = a, but I have for you a surprise: a demonstration that x=x. Yee! So, for the sake of clarity, let's make explicit the parenthesis: x^(x^x) = a Let x = e^t (e^t)^((e^t)^(e^t)) = a e^(t((e^t)^(e^t))) = a e^(t(e^(te^t))) = a t(e^(te^t)) = ln(a) e^(te^t) = ln(a)/t te^t = ln(ln(a)/t) t = W(ln(ln(a)/t)) Let's return to x, remember that x = e^t and t = lnx x = e^W(ln(ln(a)/ln(x))) x = e^W(ln(log x (a))) x = e^W(ln(x^x)) x = e^W(xlnx) x = e^W(lnx e^lnx) x = e^lnx x = x
0:00 Getting ready!
1:11 Some main properties of W(x)
5:00 Q1, W(W(x))=1
7:51 Q2, W(x)=ln(2x)
12:00 Q3, W(e^x)=(W(e))^x
15:20 Q4, 1+W((e^2-1)x)=e^2
17:48 Q5, W(9x-7)e^W(9x-7)=-1/e
20:29 Q6, W(e^(e^2+1+x^x))=x^x
26:01 Q7, W(e^(x+1))=x+1
28:41 Q8, W(x+1)=(x+1)^2
37:57 Solve x^x^x=a?
Can we find the value of (1-i) ^ (3+2i) just using binomial identity expansions? (euler constant e is not allowed) . I will subscribe to your channel if you answer this question with explanation.
where are the 100 questions?
It is possible to use W to solve a = x^(x^(x+1)). Solution follows:
1) Take the log of both sides: ln a = x^(x+1) ln x = x^x ln x^x.
2) Apply W: W(ln a) = ln x^x = x ln x.
3) Apply W: W(W(ln a)) = ln x.
4) Finally: x = e^W(W(ln a)). Done!
For example, this solves 4 = x^(x^(x+1)) as x ~ 1.559610469.
That’s a beautiful one!
@@blackpenredpen Thanks! :)
Aku HATE SHARKBRAINZ!
These were fun!
This was not the video for it, but it came up twice, so I mention it here: When we encounter a new function, we never have original intuition about the actual values it takes. This is certainly true of things like the logarithm, but there was even a time when you did not know what f(x) = x looks like. Sometimes, we can just plug-and-chug some example inputs. But, usually, we get our best intuition by plotting the function. We can be confident in that intuition and get a more-detailed view of the function's properties by doing things like finding domain, range, intercepts, end-behavior, growth patterns, extrema, saddle points, points of inflection or undulation, etc. Similarly, approximations (Newton's method, Taylor expansion for analytic functions, etc.) can help a student get an actual feel for specific outputs (especially for non-special points). After all of that is completed, people tend to feel that they have a good grasp of the function and no longer question things in the manner like "Isn't W(2) cheating?" (not exactly a paraphrase - more like a stereotypical example). It is a good question to have, but we should be asking it about all functions - and the reason for our not doing so is because we have that improved intuition.
I have never though of lambert W function to be so interesting , thank you for teaching me that ! , i have developed a new thinking ability !
I love it! I could solve the first 7 questions on my own thanks to your fish 🐠 analogy.
Excellent!
@@blackpenredpen Excellet videos!
H t
B
B m s d. 2
I really like this video, it helped me understand the ways to use the W(x) function better than any other sources I could find. I think your way of working through these problems helps teach and demonstrate the usefulness of the W() function tool without getting deep into the weeds like other sources (Wikipedia) do.
To comment on something you said in the middle of the video, I think people consider it "cheating" because it feels like mathematicians just "made up" a function to be the opposite of x*exp(x) instead of "solving it for real." But in reality there are plenty of these types of functions that exist. The "error function" for the integral of exp(-x^2) or the Gamma function for non-real-integer factorials. If you think about it, even the "logarithm" itself is just made up to be the opposite of exponentation. I think we're more comfortable with logs because it feels intuitive, but really it's no different. Thanks for the great videos.
vous etes le plus admirable de par la methode, rigueur et maitrise des maths! sincerement vous honnorez le youtube! think you
HW: Solve ln(W(ln(x)))=1
Good luck on typing up the answer : )
x=e^(e*e^e)
Haha love the answer! e isn't just the most common letter in english, it seem's like in math as well xD
ln(W(ln(x))) = 1
W(ln(x)) = e
ln(x) = e*e^e
ln(x) = e^(e+1)
X = e^(e^(e+1))
I hope i have been careful enough in my typing :)
@@yohangross5518 You're right, but e^(e*e^e) looks funnier to me xD
@@yohangross5518 Can confirm! Good job!
ln(W(ln(x)))=1...........1
ln(W(ln(x)))
e =e¹.......2
W(ln(x)) =e................3
Finally the answer
1+e
e
X=e.
😂We really need good luck for writing but ....😂😂😂
damn, wish I knew about this stream yesterday. I had to derive wien's displacement law from max planck's law, and i arrived at an equation that I thought could not be solved analytically... turns out I needed to use the Lambert W function which I only JUST learned about by googling today
Here's a cute identity. Easy to prove but still nice.
W[ ln(a^a) ] = ln(a)
Bro congrats this is like the first video of yours which has absolutely easy questions
Numbers 2,4 5,7...I only solved this 4 questions completely...6 and 8 halfly ....thanks for this amezing vedio ...✨
My first time that I participated, this was fun!
lmao 'its e^(1+e) not e^(e+1)' you dumb?
For Q8, I got ln(x+1) = W(-(e)^2), then solved for x to get
x = e^(W(-(e)^2)) - 1
Thank you so much master! From the philippines here
Hey @bprp
There is such a nice content on your channel. I just love seeing your videos.
Today, my mind got stuck at a question. Can you please try to figure out what could the possible strategy be to solve the following equation:
ln(x+8).ln(2) = ln(x).ln(10)
P.S: x=2 seems to be very obvious, but I can't get how can I reach to that (and other real solutions, if any).
Very interesting! But why should expression (x + 1) e ^2(x + 1) in an exercise 8 be squared? You can solve:
1 = (x + 1) e ^2(x + 1)
2 = 2 (x + 1) e^ 2(x + 1)
W(2) = W [2(x + 1) e^ 2 (x + 1)]
2 (x + 1) = W(2)
x = W(2)/2 -1
Nice, first you exit Lumbert and then you enter Lumbert from a different door
うおお 夢のW関数ライブ40分！！
必見すぎる
ちょっとずつ何日かに分けて味わって見ます！！！
うひょひょ
6番とか、包み紙をたくさん剥がした中にいいもの入ってるみたいで、やらしい（笑）
8番とか要注意なんですね～
I like these, but I feel like there’s a more interesting class of equations to be solved (you I believe have done 3 of these)
Equations that do not involve W(x), but can only be solved with W(x)
You should have noticed that the expression (1- (x+1)(e^(x+1)^2 = 0
is impossible because (x+1)^2 is always >= 0. Hence, e^(x+1)^2 will always be >=1.
Additionally, no matter whatever value of x might be , (x+1) (e^(x+1)^2 ) will never ever = 1
Checking:
For x = -1, (x+1)e^(x+ 1)^2 = 0---> LHS =1 - 0 = RHS = 0 is nonsense or impossible
For x < -1 , say x= -2----> (-2+1)e^(-2+1)^2 = -1(e) = -e. Plugging e in the eq will give:
LHS =1- (-e) = RHS = 0 , which is again impossible
For x > -1, say x = 1----> (1+1)(e^(1+1)^2 = 2(e^4). Plugging 2(e^4) in the eq will give:
LHS = 1 - e^4;= RHS = 0, which is again impossible
Therefore, the eq (1-(x+1)(e^(x(1)^2 = 0 is false. In other words, no more solution can be found from that equation.
The only solution is x = -1
You guys will need to learn how to solve probs with logical reasoning to understand such notions of mathemstics, as imaginary or complex numbers, Lambert function, and suchlike, which can be misleading or false, instead of copying them so blindly nonsensically, like a photocopying machine.
Hi blackpenredpen! I watched this video and your other video on the lambert w function. Thanks for taking the time to make these videos! They are very helpful! If it wouldn’t be too much trouble, could you please help me with this one? Maybe post a video?
x^x + 2x + 3 = 50
I just made a program for calculating lambert w function in my scientific calculator, though numbers too large might not work :( and it takes about 3-5 seconds to evaluate lol
bprp, i thought about the final question you gave with the superroot and the actual answer might be: x = W(lna)
Awesome video bprp, I learned a lot
I got everything right until the last one, when I overlooked the possibility of two solutions. I got e^(-W(2)/2)-1, which is equivalent to your second solution.
Are there integral and derivative problems, and differential equations, with the Lambert W function? That would be fun!
ua-cam.com/video/Rg3dBosfZ3Y/v-deo.html
Try finding the derivative of the W function in terms of the W function :)
They were easy if u know W(xe^x) =x, but the 8th one was a little tough.
For x^x^x=a
xlnx^x=lna
from lnx^a=Alnx
x*xlnx=lna
x^2lnx=lna
2*x^2lnx=2lna
x^2lnx^2=2lna
e^lnx^2*lnx^2=2lna
lnx^2=W(2lna)
Take exponential then take root, hard typing on a tablet and it is 3.40 am
I hope it's correct LOL took me like 5 minutes to come up with it
second line is already wrong.
ln(x^x^x) = (x^x)lnx, not xln(x^x)
Thanks! Best wishes from Brazil!
hi friends. hi teacher
We know that W(xlnx) = lnx
And how about *W[(lnx)/x]* ?
Is there some "formula" in this case?
For example
We know W(17ln17) = ln17
But to find/calculate W[(ln17)/17], can we find an exact value, a perfect value, without using approximations, without using things like Wolfram Alpha?
Very fun video!
I came in to this video thinking I wouldnt understand it, but it actually turned out to be pretty easy and interesting.
Thanks for the knowledge
Glad to hear it!
Sir , I am a big fan of you . I really like your videos and the way you explain everything . I am class 12th and I have to say that you are doing a great job ☺ . Sir can you reply me please 😊
Fabulous. Thanks.
In applied math, when a sum is involved, the units much match each terms, so we more often find equations like:
x + a * exp ( b * x) = c
than
x + exp (x) = c
(where a, b and c are constants, and in the case of electronic, a and c are positive, b negative, in the case of a P-N junction of a diode/transistor for example).
Is there a "trick" to bring the first equation implying the W-Lambert's function? Or to solve analytically: K = x + exp( a* x ) which seems "close" to the Lambert W function.
Side note: Since x has units, and that the exponent of the exponential should not have units (thus constant "b" has (unit^ -1) of x, and b*x becomes a pure number) and the result of the exponential is a pure number, so the constant "a" has units of x.
While we can get rif of the constant "a" with a simple substitution like a*u = x, I haven't found a way to then bring the exponent to the proper template.
The tough part of this problem is that you’ve got addition along with exponentiation separating the xs. The W function is able to “bridge the gap” between exponentiation and multiplication, but I’m not sure what function would do that for adding and exponentiation. Maybe try defining a new function F(x+e^x)=x and work from there? (I think I’ve got something here but it’s a lot of vibes and not much rigor lol)
Ok after messing with it for a bit I figured out that if
x + ae^(bx) = c, then
x = [F(bc + ln(ab)) - ln(ab)]/b.
@@redpepper74
Thanks a lot for your time. Really appreciated.
I just recently noted that a relatively new section was added in Wikipedia in the Diode Modeling article, specific to a diode in series with a resistor, where the explicit resolution for the current in terms of the Lambert W function is obtained.
@@snnwstt I only have a high-school-level understanding of circuits and electricity so I don’t really get what’s going on there. But looking into that sum log function was fun :)
Ooh. Should I get popcorn?
And soda too! : )
How do you evaluate to find some value for x in question 6? x = e^(W(e^2 + 1))
symphatique de plus!!!!!
awesome. thank you
The man is great!
No 6 - x ln(x) = e^2 +1, W(x ln x = e^2 +1, where the 1st x gone from the previous line? Then follows the line W (ln x e ^(ln X) where W (ln x e ^(ln X) = ln x Who can help?
At 32:20 why did he need to square both sides. He could've just done it normally like (x+1)(e^2(x+1))=1
And then
2(x+1)(e^2(x+1))=2
2(x+1)= W(2)
x+1= W(2)/2
x= (W(2)/2)-1
It produces a different answer tho 😭😭
Hello, is there a way to post all 100 questions regarding what u said at the first of your stream, plz
Sir plz make video on integrals which including w function 😘😘😘
Can you make a video about the Barnes G-function please?
wow, 24:00 was crazy
10:42 am
Is it simple enough to say that the third solution to the eighth question must be ruled out because the value is not in the domain of W(x+1)? Since -sqrt(W(2)/2) < -1/e
X = -1 +sqrt(W(2)/2) is also a baseless statement . You must prove it's a correct answer
je mexcuse admirable et respectueux prof: comment on calcule linverse de la fonction f(x)=x^n.y^m, n et m quelconques, il sagit dun cas general
Thanks from France !
hello.
i have a question.
how to find real solutions of the equation x^x=a, if a
Hoy do we get the shirt from this video?
Would Q5 be w(-1/e) ? Then solve for x?
why the product of slope of x- axis and y - axis is not -1 as the product of slope of two perpendicular lines is -1
lambert w function of a number less than -1/e does exist but in the complex world
6:33 look at the clock
10:42 am I the only person who thinks w(x)=1/2 is a better answer?
10:42 am
What is the deal with sqrt(x) = - 1?
Is there a list of properties involving the Lambert W function (like the ones for the natural log)?
Yes. You can find them in the Wolfram Alpha article for the Lambert W function, and in the Wikipedia article as well. There are also a number of integral formulas you can derive and differential equations you can derive for this function and with this function.
Hi guys. Can you help express imaginary part of z^((1-z)/(1+z)) as a fuction?
Please..Can you find Im (i/z)?
Can you please make a video on the inverse of the Lambert W function? That means, if W(x . e^x) = x, what is W(x) ?
Can we find the value of (1-i)^(2+i) just using binomial identity extensions ? (Euler constant e is not allowed)
I need solution for complimentary error function of negative x
Actually the question is erfc(-x) =1+erfc(x)
How i can contact you...?because i have a problem in second order differential equation(non linear),it is very important to solve it...
Just post it here
I wanna see some integrals with W(x)
37:57
Uh... we expected that you try.
We never expected you to know, but we wanted to at least try.
Oh well, maybe you didn't want the livestream to go too long.
I'll think about it!
I have tried it before: ua-cam.com/video/ef-TSTg-2sI/v-deo.html
Where did you bought the pokeball involving the mic
Hlo plz help im stuck
: integrate( (sinx)^1/2)/cos x dx
y = sin(x)^(1/2) dy = cos(x)/[2·sin(x)^(1/2)]·dx, so sin(x)^(1/2)/cos(x)·dx = sin(x)/cos(x)^2·cos(x)/sin(x)^(1/2)·dx = 2·sin(x)/cos(x)^2·cos(x)/[2·sin(x)^(1/2)]·dx = 2·sin(x)/cos(x)^2·dy = 2·sin(x)/[1 - sin(x)^2]·dy = 2·y^2/(1 - y^4)·dy
2·y^2/(1 - y^4) = [(1 + y^2) - (1 - y^2)]/[(1 + y^2)·(1 - y^2)] = 1/(1 - y^2) - 1/(1 + y^2) = 2/[2·(1 - y^2)] - 1/(1 + y^2) = [(1 - y) + (1 + y)]/[2·(1 + y)·(1 - y)] - 1/(1 + y^2) = 1/2·1/(y + 1) - 1/2·1/(y - 1) - 1/(y^2 + 1).
Before proceeding to antidifferentiate, here is something that needs to be understood. sin(x) is necessary an element of [-1, +1] for every x, and as such, sin(x)^(1/2) is an element of [0, 1] for every x that is an element of the union over all integers m of the intervals [2·m·π, (2·m + 1)·π]. Furthermore, cos(x) = 0 for every x = m·π + π/2. As such, the antiderivatives of sin(x)^(1/2)/cos(x) only exist at the intervals (2·m·π, 2·m·π + π/2) and (2·m·π + π/2, 2·m·π + π), all of which are mutually disjoint and disconnected. Thus y is an element of the interval (0, 1), and in this interval, 1/(y + 1) has antiderivatives ln(1 + y) + C0, 1/(y - 1) has antiderivatives ln(1 - y) + C1, and 1/(y^2 + 1) has antiderivatives arctan(y) + C2. Therefore, 1/2·1/(y + 1) - 1/2·1/(y - 1) - 1/(y^2 + 1) has antiderivatives 1/2·ln(1 + y) - 1/2·ln(1 - y) - arctan(y) + 1/2·C0 - 1/2·C1 - C2 with respect to y. However, since y is a function with a domain that is a disjoint union of disconnected intervals, the constants of integration with respect to y are sums of step function with respect to x, so the antiderivatives of sin(x)^(1/2)/cos(x) with respect to x are given by 1/2·ln[1 + sin(x)^(1/2)] - 1/2·ln[1 - sin(x)^(1/2)] - arctan[sin(x)^(1/2)] + C(x), where C(x) = A(m) if x is an element of the interval (2·m·π, 2·m·π + π/2), C(x) = B(m) if x is an element of the interval (2·m·π + π/2, 2·m·π + π), where A and B are families of real numbers indexed by the integers.
Please i have an equation to resolve ! 🙏
exp ( -x²) - 2√π *( 10^-4)* x = 0
Please someone help me i need the solution ! Can you make a short video for it ????
After watching bprp #shorts then i come here to see what are you doing
Anyone else think of the Wronskian Fundamental Set when he writes W?
31:30 lol what???
It's around 3:00 am at night😓😓
@Cyril Scetbon opps 😅
Sorry Sir! Sin m^n=??? ( where m&n are real numbers, No expansion, it must be FINITE SERIES.)
when will you do 100 questions?????????????
You can rule out the solution with the minus sign by plugging it into your earlier equation (x+1)e^[(x+1)^2]=1
if you plug in x=-1-sqrt((W(2))/2), you will get (-sqrt((W(2))/2))e^[(W(2))/2] = 1
Assuming we are looking at W(2) in the real world:
-sqrt((W(2))/2) is a negative number.
e^[(W(2))/2] is a positive number.
when you multiply them together, you get a negative number.
How can a negative number equal 1? It can't, so that isn't a valid solution.
q3 was the easiest imo
Entiendo ,pero no sé cómo se halla:w(ln 3)
Why does W(e)=1?
W(x) = a aeª = x
Then:
W(e) = 1 1e¹ = e, which is True.
27:23 lmao
Please prove that the corresponding angles of parallel lines are equal.
If they were not, one would be less than the other, but then two different parallel lines would pass through the same point
why w(x)ew(x) =x ?
Think in W(x)
W(x)*e^W(x) = T
(Take W() on both sides)
W( W(x)*e^W(x) ) = W(t)
W(x) = W(t)
X = t
Are you a student or a teacher?
solved q8 with simultaneous equation 🤣
🌹🌹🌹🌹
I really want to know your name
1000th like
You need more preparation.
Ur store is broken
Unfortunately I couldn't solve x^x^x = a, but I have for you a surprise: a demonstration that x=x. Yee!
So, for the sake of clarity, let's make explicit the parenthesis: x^(x^x) = a
Let x = e^t
(e^t)^((e^t)^(e^t)) = a
e^(t((e^t)^(e^t))) = a
e^(t(e^(te^t))) = a
t(e^(te^t)) = ln(a)
e^(te^t) = ln(a)/t
te^t = ln(ln(a)/t)
t = W(ln(ln(a)/t))
Let's return to x, remember that x = e^t and t = lnx
x = e^W(ln(ln(a)/ln(x)))
x = e^W(ln(log x (a)))
x = e^W(ln(x^x))
x = e^W(xlnx)
x = e^W(lnx e^lnx)
x = e^lnx
x = x
Numbers 2,4 5,7...I only solved this 4 questions completely...6 and 8 halfly ....thanks for this amezing vedio ...✨