@@That_One_Guy... you can copy and paste the sigma (Σ) from online sources or use a Greek keyboard; for the powers (¹²³⁴⁵⁶⁷⁸⁹⁰) you can get those from copying and pasting as well. For the indices (n=0 at the bottom of the Σ), you can get them to look like they're right below the sigma by using superscripts (aka exponents) on the line beneath, so that they look like they're actually under the sigma. See this example: Σ ⁿ
Not only that, but he will be able to continue calculating solutions for all of his lifetime, since there are infinitely many. Not sure what he will eat, but that problem takes care of itself eventually.
@@blackpenredpen what do you think about "one minus zero-point-nine*repeating"? ( 1-0.(9)=? ) I think matematicians should invent new numbers like 0.(0)1 with the rule that they can not be converted like this: A.b = ab/10
@@Icy-ll5ie Lol, but 0.(9) is equal to 1. It's not a some kind of magic, neverending irrational number, but it's 1. It's simple proof: 0.(9)*10 = 9.(9) 0.(9)*9=0.(9)*10-0.(9)=9.(9)-0.(9)=9 So, when we divide this by 9, we get 1.
@@Icy-ll5ie You can readily prove it via the nested intervals and lower bound, or alternatively, Dedekind cuts, if a simplistic algebraic proof doesn't exactly appeal to you.
A more careful and rigorous way of handling the equation z^2 = 2^z is by noticing that if z is not an integer, then 2^z is inevitably multivalued. Namely, 2^z := exp[ln(2)·z + 2nπi·z] for any integer n. This implies z^2 = exp([ln(2) + 2nπi]·z). Now, a few cases must be considered. The first case is that |Arg(z)| < π/2, and the second case is that π/2 < |Arg(z)| < π. It is notable that if |Arg(z)| < π/2, then |Arg(z^2)| < π, and if π/2 < |Arg(z)| < π, then also |Arg(z^2)| < π. In the first case, z^2 = exp([ln(2) + 2nπi]·z) implies z = exp([ln(2)/2 + nπi]·z), and z = exp([ln(2)/2 + nπi]·z) implies -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = -[ln(2)/2 + nπi]. Therefore, -[ln(2)/2 + nπi]·z = W(m, -[ln(2)/2 + nπi]), equivalent to z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]). Notice that if n = m = 0, then this simplifies to z = 2. In the second case, z^2 = exp([ln(2) + 2nπi]·z) implies -z = exp([ln(2)/2 + nπi]·z), equivalent to -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = ln(2)/2 + nπi. Therefore, -[ln(2)/2 + nπi]·z = W(m, ln(2)/2 + nπi), hence z = -W(m, ln(2)/2 + nπi)/[ln(2)/2 + nπi]. With this, the remaining cases are Arg(z) = π, Arg(z) = π/2, and Arg(z) = -π/2. In the first of these three, z = -r, with r = |r|, so -r = exp(-[ln(2)/2 + nπi]·r), thus [ln(2)/2 + nπi]·r·exp([ln(2)/2 + nπi]·r) = -[ln(2)/2 + nπi], and [ln(2)/2 + nπi]·r = W(m, -[ln(2)/2 + nπi]). Therefore, z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. If Arg(z) = -π/2, then z = -ri, with r = |r|. Hence -ri = exp(-[ln(2)/2 + nπi]·ri), and [ln(2)/2 + nπi]·ri·exp([ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi]. Therefore, [ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), and z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. Finally, if Arg(z) = π/2, then z = ri, with r = |r|, so ri = exp([ln(2)/2 + nπi]·ri), and -[ln(2)/2 + nπi]·ri·exp(-[ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi], hence -[ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), so z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. In summary, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi], in both cases for arbitrary integers n and m. This gives the complete family of complex solutions with no extraneous solutions. The solution families can be compactified. Notice that exp[W(t)] = t/W(t), so exp[-W(t)] = W(t)/t. Therefore, -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = +exp[-W(m, -[ln(2)/2 + nπi])], and -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = -exp[-W(m, +[ln(2)/2 + nπi])]. As such, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = +exp[-W(m, -[ln(2)/2 + nπi])], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -exp[-W(m, +[ln(2)/2 + nπi])]. Again, this solution is complete and with no extraneous solutions.
Okay, you expressed the result in terms of the Lambert function. But we wanted the missing root, 0.77, expressed in rationals and complex roots as well.
@@MindYourFunds you didn't pay attention to the video.... The third root is a negative real number -0.7666.... the fourth fifth and so on - Solutions are imaginary, and they aren't intuitive
I've been trying out this problem since I was in 10th grade of school. Now I'm in the third year of engineering. Thanks to you I got to know about the new function "Lambert W function"
One can also use the identity 2^x = e^((ln2)*x) and take square roots on both sides of the equation. I think it is a little more simple to work out then. I get the solutions x=-(2/ln(2))*W_0(-ln(2)/2) = 2, x= -(2/ln(2))*W_1(-ln(2)/2) = 4 and -(2/ln(2))*W_0(ln(2)/2) = -0.766664... The terms look a little different but yield the same values as in the video.
@@blackpenredpen No. It is not awesome. To make assumptions about what your student "already" knows is a poor teaching technique. Yes I can do the algebra in my head also. But you might soon learn that most people cannot. How fast you learn that depends upon your IQ.
@@24kGoldenRocket true but I think the line must be drawn where he stops explaining every detail and assumes the viewers know what he is talking about, because this is a (mostly) calc-based channel. The instance mentioned in which he states that "we know that" is quite basic, though. I doubt anyone watching this wouldn't understand how to turn a negative (x < 0) value into a positive value by multiplying by a negative value.
@@24kGoldenRocket this channel is not for noobs he wont spend his time explaining u what log and exponentials ate in every video he have to pre assume that u know basic maths if not go and learn some basics first
Great video. I completely forgot about Lambert, this was an excellent case, where I could have used it if I was creative enough to think of the way out.
I did expect it to have infinitely many solutions after I remember W(x) was a multi-valued function, but it still surprised me when I found out it was actually true.
Thanks a lot! I Just finished school, but never heard of the LamberW equasion. Everything else I understood perfectly. I liked it so much, I found solutions for the x>0 myself.
Alternative solution: x = 2^(x/2), so rearranging gives -ln(sqrt(2)) x exp(- ln(sqrt(2)) x) = -ln(sqrt(2)). Taking W of both sides gives (by rearranging) x = W(-ln(sqrt(2))) / -ln(sqrt(2)). No assumptions on signs needed here.
Lambert W comes up in calculations of current-voltage relationships in solar cells and diodes. Blew my mind the first time I saw it and fun that it comes up in many places.
Lambert W! Amazing that 1) The function is useful in all kinds of modelling of actual real physical behaviours, e.g. how long will it take for a bucket of water with a hole in it to drain to a certain depth. 2) The function is not implemented in floating point libraries or FPUs.
x^n = n^x -> x^n = e[log(n)·x] -> xω(n)^m = e^[log(n)·x/n], where ω(n) = e^[(2π/n)i], and m = 0, 1, ..., n - 1, meaning there are n cases to consider, one for each value of m. Regardless, xω(n)^m = e^[log(n)·x/n] -> xe^[-log(n)·x/n] = ω(n)^m, since 1/ω(n)^m = ω(n)^(n - 1 - m), which we can reindex to be ω(n)^m, since m is a variable. Then [-log(n)·x/n]e^[-log(n)·x/n] = -log(n)·ω(n)^m/n -> -log(n)·x/n = W[-log(n)·ω(n)^m/n] -> x = -n·W[-log(n)·ω(n)^m]/log(n).
OMG man, your videos are great. Please keep doing these kind of videos, they are so interesting. Calculus is much more interesting when You explain it. Love you
idk why but the piece wise function for absolute value of x still seriously fucks up my mind. like on a good day im able to make sense out of it, and on other days its just a brainfuck.
I think it is related to Euler's formula: e^(z) = cosx + isinx Where z is a complex number, if that's the case then there are infinite imaginary solutions.
I'm still waiting for the proof of the cubic equation I think you can do it in 15 minutes or less. I did it in 20 minutes (because I'm not good at explaining Lol)
Aaryan Bhatia What I did was give other values to the variables x. if you want to know the extended formula you just have to change those final values at the roots of x
Aaryan Bhatia The answer is you *cannot.* Not in general, anyway. Only for certain values of the coefficients is the result simplifiable. Whenever it is not, this is known as casus irreducibilis. If you have ever wondered why an angle of 1° is not constructible, the answer is because 3° is constructible, and third angles of nontrivial constructible angles are not constructible usually, because the third-angle formula in general is unsolvable, as it requires solving a cubic equation that results in casus irreducibilis. This is why the cubic formula is not taught as opposed to the quadratic formula being taught. It is usually not useful, because casus irreducibilis is more common than not.
Aaryan Bhatia Well, no, it is not like the quadratic formula. The quadratic formula is useful and practical. The quadratic formula suffers from no casus irreducibilis. The quadratic discriminant also does not suffer from case deficiency. The cubic formula is, in every sense, analogous to the quadratic formula, but strictly inferior.
I followed you up until 7:05. In math it’s dangerous to say “of course...” Also, you introduced this lambert function without providing an explanation as to how to work with it
Equations like x^2=2^x always fascinated me, but solving them felt like hitting a wall at times. What really helped me was using tools that walk you through the problem-solving process. SolutionInn’s AI study tool is fantastic for breaking down equations like this, making complex solutions easier to understand. Highly recommend it if you’re tackling challenging problems like this one.
Ah, man... This is very unsatisfying: How do we know what the Lambert W function is - without having to use Wolfram Alpha? And what are these "indexes", what do they mean - can you please do a video about them? Would we need to use Wolfram Alpha to calculate them? Because changing them obviously gets you different answers to the question/problem.
@@Gulyus Thank you. However, even after viewing that video, I still have questions. For example, how do we know that when using f(x) to calculate the inverse: "f inverse" or "f -1(x)" that f(x) should equal "xe^x" based on the equation given in the video - which is "x^x = 2"? In addition, the video did not explain anything about "indexes". So I would not have even known that they exist otherwise, without them being mentioned in THIS video.
This is basically the method I used to investigate the question at the end of your “Are you tired of this kind of question” video. We can compare a^b and b^a by considering the Lambert W function. Unfortunately it does not seem that this answer helps because I do not believe we can evaluate the Lambert W function without a calculator. So for example, this method does not help us decide if (e-1)^π is greater or less than π^(e-1).
You can't evaluate many other functions without a calculator (e.g. exponentials for non-integer powers and/or irrational base, logs or trigonometric functions). They are just more familiar than W, so more people have an "approximate" idea of their behaviour.
@@dlevi67 exponentials, natural logs, trig, inverse trig and hyperbolic trig as well as many others can be estimated to arbitrary accuracy by hand with Taylor series (I point out those ones because I know them off the top of my head). Non-integer and negative powers can also be estimated to arbitrary accuracy by hand using binomial theorem, which I have encountered some smug bastards who have done that to calculate things like cube roots or things like that in their head with reasonable accuracy. For these things you just have to be a bit more patient than using a calculator.
@@almachizit3207 W can be calculated by hand too. Lambert and Euler tabled it in the 1750s - no calculators then. But it's a slog and using a calculator makes it significantly easier. What I'm pointing out is that many people commenting here seem to think it "weird" and "difficult" and "different". It's not; it's just not something that is typically taught in high school (or even initial degree-level math courses).
tough topic .. good explanation. I have not met the Lambert W function .. clearly it has some fascinating properties .. to be continued!! Thank you again.
Hi, I worked out y=x^x (x to the power of x) and found for x=0, y approaches 1. Also for x=1, y will be 1. For x=1/e=~0.3678, y will have its minimum which will be 0.6922. beyond x=1, y increases continuously until +infinity. Here's my problem and I need your help, perhaps a video: for x= -infinity, y=0. For all even negative integers, y>0. For all odd negative integers, y
You probably already got an answer but for negative x the real and the imaginary values oscilate around the x axis converging to 0 as x goes to -infinity. The imaginary part is 0 when x is an integer and negative for x in the interval (-(2n+1),-2n) and positive for x in the interval (-2n,-(2n+1)) for every natural number n. The real part will be zero at x = e^W(log(i b) + 2 i π k) i think (probably not all the zeros and there are some conditions for wich this isnt zero, for full solution check wolframalpha i guess) the function will never have real part and imaginary part both equal to 0 at the same time. For the question why you cant write (-2.3)^(-2.3) as (-2.3)^(-46/20) is because this is like writing sqrt(-1)=4th root of (-1)^2 = 4th root of 1. You create extra solutions wich aren't solutions to the previous equations (4th root of 1 has 4 solutions while sqrt(-1) only has 2)...
I think you could compare the growth of e^x and x^e. When x tends to +inf e^x will tends to +inf way faster than x^e.In fact for any a,b > 0 (e^ax)/(x^b) will always tends to +inf if x tends to +inf (so in our case b = e and a = 1)
Since e^x is an exponential and x^e is a polynomial it will probably diverge but you can work like this: Since your question is a infinity over infinity situation, you can use l'Hopital's rule and differentiate the top and bottom Since the top is e^x it will stay like this and to differentiate x^e you just minus 1 the expontent and get: e^x/x^(e-1) This will still be infinity over infinity so you do it again which will lead to: e^x/x^(e-2) So you do it again... e^x/x^(e-3) But this time the exponent of x^(e-3) is negative so you can put it on numerator and get: e^x× x^|e-3| which diverges
If you replace the absolute value with the magnitude of a complex number z, this approach still works and the indices of the product log gives all of the (infinitely many!) complex solutions. :)
A more careful and rigorous way of handling the equation z^2 = 2^z is by noticing that if z is not an integer, then 2^z is inevitably multivalued. Namely, 2^z := exp[ln(2)·z + 2nπi·z] for any integer n. This implies z^2 = exp([ln(2) + 2nπi]·z). Now, a few cases must be considered. The first case is that |Arg(z)| < π/2, and the second case is that π/2 < |Arg(z)| < π. It is notable that if |Arg(z)| < π/2, then |Arg(z^2)| < π, and if π/2 < |Arg(z)| < π, then also |Arg(z^2)| < π. In the first case, z^2 = exp([ln(2) + 2nπi]·z) implies z = exp([ln(2)/2 + nπi]·z), and z = exp([ln(2)/2 + nπi]·z) implies -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = -[ln(2)/2 + nπi]. Therefore, -[ln(2)/2 + nπi]·z = W(m, -[ln(2)/2 + nπi]), equivalent to z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]). Notice that if n = m = 0, then this simplifies to z = 2. In the second case, z^2 = exp([ln(2) + 2nπi]·z) implies -z = exp([ln(2)/2 + nπi]·z), equivalent to -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = ln(2)/2 + nπi. Therefore, -[ln(2)/2 + nπi]·z = W(m, ln(2)/2 + nπi), hence z = -W(m, ln(2)/2 + nπi)/[ln(2)/2 + nπi]. With this, the remaining cases are Arg(z) = π, Arg(z) = π/2, and Arg(z) = -π/2. In the first of these three, z = -r, with r = |r|, so -r = exp(-[ln(2)/2 + nπi]·r), thus [ln(2)/2 + nπi]·r·exp([ln(2)/2 + nπi]·r) = -[ln(2)/2 + nπi], and [ln(2)/2 + nπi]·r = W(m, -[ln(2)/2 + nπi]). Therefore, z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. If Arg(z) = -π/2, then z = -ri, with r = |r|. Hence -ri = exp(-[ln(2)/2 + nπi]·ri), and [ln(2)/2 + nπi]·ri·exp([ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi]. Therefore, [ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), and z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. Finally, if Arg(z) = π/2, then z = ri, with r = |r|, so ri = exp([ln(2)/2 + nπi]·ri), and -[ln(2)/2 + nπi]·ri·exp(-[ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi], hence -[ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), so z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]. In summary, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi], in both cases for arbitrary integers n and m. This gives the complete family of complex solutions with no extraneous solutions. The solution families can be compactified. Notice that exp[W(t)] = t/W(t), so exp[-W(t)] = W(t)/t. Therefore, -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = +exp[-W(m, -[ln(2)/2 + nπi])], and -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = -exp[-W(m, +[ln(2)/2 + nπi])]. As such, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = +exp[-W(m, -[ln(2)/2 + nπi])], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -exp[-W(m, +[ln(2)/2 + nπi])]. Again, this solution is complete and with no extraneous solutions.
Solve this by the super square root ruclips.net/video/F_XC9_XSw7k/видео.html
what happend if x=0, becouse I can't see 0 in any formula (I mean xe(-∞;0)u(0,∞)
@@krucyferariusz1813 Shut up
@@Juan-yj2nn LOL
0:09 MaLeFiC lAuGh
and really? a complex power? brah
according to geogebra, the 3rd root is +- = -0.766664744346434 (346434 repeats infinitely or many times)
Best variables for math:
X
Y
Fish
JZ Animates .....what about star, square, lightning bolt? ........I have seen that before.
∞
∑🙂ⁿ = 1+🙂+🙂²+🙂³+🙂⁴+...
ⁿ⁼⁰
I present emoji variable in infinite series.
@@Pete-Prolly i see only squares lol
@@Pete-Prolly how did you type the sigma symbol along with the index and number ?
@@That_One_Guy... you can copy and paste the sigma (Σ) from online sources or use a Greek keyboard; for the powers (¹²³⁴⁵⁶⁷⁸⁹⁰) you can get those from copying and pasting as well. For the indices (n=0 at the bottom of the Σ), you can get them to look like they're right below the sigma by using superscripts (aka exponents) on the line beneath, so that they look like they're actually under the sigma. See this example:
Σ
ⁿ
Give a man a fish, and he'll eat for a day. Teach a man WITH fish, and he'll be able to solve for the solutions of x^2=2^x.
Aaron L Hahahahahaha!!!
Give PIERRE DE FERMAT a fish, and he'll FIND A BEAUTIFUL SOLUTION THIS FISH IS NOT ABLE TO CONTAIN
Not only that, but he will be able to continue calculating solutions for all of his lifetime, since there are infinitely many. Not sure what he will eat, but that problem takes care of itself eventually.
lol
I am confusion.
"This is not a fish yet, but it's almost a fish." Sounds like evolution at work.🤣
Hahahahahaha definitely!!
Evolution of equations
@@blackpenredpen what do you think about "one minus zero-point-nine*repeating"?
( 1-0.(9)=? )
I think matematicians should invent new numbers like 0.(0)1 with the rule that they can not be converted like this:
A.b = ab/10
@@Icy-ll5ie Lol, but 0.(9) is equal to 1. It's not a some kind of magic, neverending irrational number, but it's 1. It's simple proof: 0.(9)*10 = 9.(9) 0.(9)*9=0.(9)*10-0.(9)=9.(9)-0.(9)=9 So, when we divide this by 9, we get 1.
@@Icy-ll5ie You can readily prove it via the nested intervals and lower bound, or alternatively, Dedekind cuts, if a simplistic algebraic proof doesn't exactly appeal to you.
A more careful and rigorous way of handling the equation z^2 = 2^z is by noticing that if z is not an integer, then 2^z is inevitably multivalued. Namely, 2^z := exp[ln(2)·z + 2nπi·z] for any integer n. This implies z^2 = exp([ln(2) + 2nπi]·z). Now, a few cases must be considered. The first case is that |Arg(z)| < π/2, and the second case is that π/2 < |Arg(z)| < π. It is notable that if |Arg(z)| < π/2, then |Arg(z^2)| < π, and if π/2 < |Arg(z)| < π, then also |Arg(z^2)| < π.
In the first case, z^2 = exp([ln(2) + 2nπi]·z) implies z = exp([ln(2)/2 + nπi]·z), and z = exp([ln(2)/2 + nπi]·z) implies -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = -[ln(2)/2 + nπi]. Therefore, -[ln(2)/2 + nπi]·z = W(m, -[ln(2)/2 + nπi]), equivalent to z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]). Notice that if n = m = 0, then this simplifies to z = 2.
In the second case, z^2 = exp([ln(2) + 2nπi]·z) implies -z = exp([ln(2)/2 + nπi]·z), equivalent to -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = ln(2)/2 + nπi. Therefore, -[ln(2)/2 + nπi]·z = W(m, ln(2)/2 + nπi), hence z = -W(m, ln(2)/2 + nπi)/[ln(2)/2 + nπi].
With this, the remaining cases are Arg(z) = π, Arg(z) = π/2, and Arg(z) = -π/2.
In the first of these three, z = -r, with r = |r|, so -r = exp(-[ln(2)/2 + nπi]·r), thus [ln(2)/2 + nπi]·r·exp([ln(2)/2 + nπi]·r) = -[ln(2)/2 + nπi], and [ln(2)/2 + nπi]·r = W(m, -[ln(2)/2 + nπi]). Therefore, z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi].
If Arg(z) = -π/2, then z = -ri, with r = |r|. Hence -ri = exp(-[ln(2)/2 + nπi]·ri), and [ln(2)/2 + nπi]·ri·exp([ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi]. Therefore, [ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), and z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi].
Finally, if Arg(z) = π/2, then z = ri, with r = |r|, so ri = exp([ln(2)/2 + nπi]·ri), and -[ln(2)/2 + nπi]·ri·exp(-[ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi], hence -[ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), so z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi].
In summary, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi], in both cases for arbitrary integers n and m.
This gives the complete family of complex solutions with no extraneous solutions.
The solution families can be compactified. Notice that exp[W(t)] = t/W(t), so exp[-W(t)] = W(t)/t. Therefore, -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = +exp[-W(m, -[ln(2)/2 + nπi])], and -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = -exp[-W(m, +[ln(2)/2 + nπi])]. As such, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = +exp[-W(m, -[ln(2)/2 + nπi])], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -exp[-W(m, +[ln(2)/2 + nπi])]. Again, this solution is complete and with no extraneous solutions.
Wow man
You simply answered all my questions about this solution and much more. Thank you very much.
Can't wait for the day where I can understand half the crap that you wrote- if I remember, I might come back here in a few years.
@@M1m1sNah bro you coming back today
bro took it personally
Lambert W function intro: ruclips.net/video/sWgNCra93D8/видео.html
I can't wait to watch this and learn in a class setting. I recently took my differentiation exam (calculus 1) and I got 96.6% woo
How do we show the following is a good approximation:
xe^x~(1-7/9)^(-x)-(1+7/9)^x
That’s very well done!! I am happy to hear that!
Thanks!
Okay, you expressed the result in terms of the Lambert function. But we wanted the missing root, 0.77, expressed in rationals and complex roots as well.
7:48 How does one know in advance what indices (in this case 0 and -1) to use for the Lambert W function in order to get real answers?
I had in mind the exact same question...
According to Wikipedia, seems like the indices 0 and -1 are standard when working with real answers with the Lambert W
8:47 We know that we know that
Just a composite function f(f(x)) such that f(x)=we know that
Eminem in disguise
J
Make it by 2x
I didn't see that
12:57 Look at the way he looks at his math with passion in his eyes :-D
Thank you for producing all of this unique content! I really love these videos!
BPRP: It's probably irrational...
Wolfram: It's transcendental.
Hahahah yea I know!! I saw that when I was editing the video and I was like hmmmm should I edit that part out. Haha
I mean...you're not wrong.
@@blackpenredpen All real transcendental numbers are irrational. So you are still right.
@@jamboree1953 wrong, e*i its still transcendental but not irrational
@@Phantlos if it's not irrational, wouldn't it be representable with division of two intervals and this not transcendental?
Who else doesn’t understand anything but still watches
Me, me
Volcanic i dont speak english but i see it aniway
Bruh I'm still learning about line gradients
Meee
Me
Respect from one math teacher to another math teacher
Thank you!!!
The third root from the graph is obviously a negative x...how come it became imaginary... Perhaps the two math teacher will explain it to me
@@MindYourFunds you didn't pay attention to the video.... The third root is a negative real number -0.7666.... the fourth fifth and so on - Solutions are imaginary, and they aren't intuitive
I've been trying out this problem since I was in 10th grade of school. Now I'm in the third year of engineering.
Thanks to you I got to know about the new function "Lambert W function"
Glad to hear!! Thank you for the comment too
the fish is a paid actor
It is!
אח יקר
Is he in the Stream Actors Guild?
I guess I'll have to look it up on the Ichthyological Movie Data Base.
אח יקר
Prove the Lambert W function! I've never heard of it (haven't taken analysis yet, not sure if thats covered there) and i think its awesome!
Matt Heitmann I have an introductory video in the description already.
My teacher explaing math: so x is equal to y
Blackpenredpen explaning math: *_F I S H_*
I should read a book for english but pen(black+red) math videos are sooo much nicer💯
That's a nice way to write his name...
Can you make a video explaining LambertW indexes?
Yeah, plsss
Yes
Yes no Also asks for more
Yeah I missed this last step.
True. He never describes the LambertW index. I had to use a numerical algorithm to solve it. It converges quite slowly.
One can also use the identity 2^x = e^((ln2)*x) and take square roots on both sides of the equation. I think it is a little more simple to work out then. I get the solutions x=-(2/ln(2))*W_0(-ln(2)/2) = 2, x= -(2/ln(2))*W_1(-ln(2)/2) = 4 and -(2/ln(2))*W_0(ln(2)/2) = -0.766664... The terms look a little different but yield the same values as in the video.
8:45 "we know that-we know that"(pro jump cuts)
Definitely awesome!
@@blackpenredpen No. It is not awesome. To make assumptions about what your student "already" knows is a poor teaching technique. Yes I can do the algebra in my head also. But you might soon learn that most people cannot. How fast you learn that depends upon your IQ.
@@24kGoldenRocket true but I think the line must be drawn where he stops explaining every detail and assumes the viewers know what he is talking about, because this is a (mostly) calc-based channel. The instance mentioned in which he states that "we know that" is quite basic, though. I doubt anyone watching this wouldn't understand how to turn a negative (x < 0) value into a positive value by multiplying by a negative value.
@@24kGoldenRocket this channel is not for noobs he wont spend his time explaining u what log and exponentials ate in every video he have to pre assume that u know basic maths if not go and learn some basics first
o@@anand.suralkar LOL. I was a University Math Instructor.
Great video. I completely forgot about Lambert, this was an excellent case, where I could have used it if I was creative enough to think of the way out.
Thank you!!
I did expect it to have infinitely many solutions after I remember W(x) was a multi-valued function, but it still surprised me when I found out it was actually true.
From this point on I'll start writing functions that depend on fish rather than x
Leave endangered species alone, sir. You have done enough harm to our oceans with plastic bags.
Applies Lambert to fish ( e)^ (fish)
Le fish:
Allow me to introduce myself
why is this comment pure gold
No, I won't allow
lol
I love learning, especially math. I love your enthusiasm. It's so uplifting to be in a math moment with you!
I have to say I thoroughly enjoy YT suggesting older BPRP vids so I can watch the epic beard/goatee evolution :)
Thanks a lot! I Just finished school, but never heard of the LamberW equasion. Everything else I understood perfectly.
I liked it so much, I found solutions for the x>0 myself.
"The fish" lmao, cracks me up every time.
Alternative solution: x = 2^(x/2), so rearranging gives -ln(sqrt(2)) x exp(- ln(sqrt(2)) x) = -ln(sqrt(2)).
Taking W of both sides gives (by rearranging) x = W(-ln(sqrt(2))) / -ln(sqrt(2)). No assumptions on signs needed here.
You already made an implicit assumption on signs.
x² = y
x = ±sqrt(y)
So your first 2 lines should be:
x² = 2ˣ
x = ±2^(x/2)
Cool, very cool. I never knew that Herr Lambert was working on photometry, colours etc. and that his W function has so many other uses! Thank you!
The fish is So Cool !!! Great illustration!
This question was in my 11th grade math book. They seriously expect 11th grade students to solve this omg.
Probably you are mistaken. Probably they asked just for positive values.
Lambert W comes up in calculations of current-voltage relationships in solar cells and diodes. Blew my mind the first time I saw it and fun that it comes up in many places.
Problem: contain exponents
blackpenredpen: it's Lambert-W time
Thank you blackpenredpen for answering my question
You’re welcome. In fact many ppl have asked this question in the past. It’s a very popular question
Bro You are Godfather of Mathematics Very Impressed by your knowledge..KEEP POSTING SUCH GREAT 😊
Thank you!!
How does he manage to always teach me something new every single vid?
In 100th episode, please show your left biceps. I can only imagine how shredded it is after this hard holding-ball workout
I love your channel, man! thanks for your videos. Cheers from Chile
0:20 i have subtitles on and i cant see .-.
There is litterelly a tower of subtitles 💀☠
Unexpected, that was. 😅
Lambert W! Amazing that 1) The function is useful in all kinds of modelling of actual real physical behaviours, e.g. how long will it take for a bucket of water with a hole in it to drain to a certain depth. 2) The function is not implemented in floating point libraries or FPUs.
Well now you have to find all the values of x for a given n such that x^n=n^x
Hahahahaha. Replace all the 2 by n then we are done!
x^n = n^x -> x^n = e[log(n)·x] -> xω(n)^m = e^[log(n)·x/n], where ω(n) = e^[(2π/n)i], and m = 0, 1, ..., n - 1, meaning there are n cases to consider, one for each value of m. Regardless, xω(n)^m = e^[log(n)·x/n] -> xe^[-log(n)·x/n] = ω(n)^m, since 1/ω(n)^m = ω(n)^(n - 1 - m), which we can reindex to be ω(n)^m, since m is a variable. Then [-log(n)·x/n]e^[-log(n)·x/n] = -log(n)·ω(n)^m/n -> -log(n)·x/n = W[-log(n)·ω(n)^m/n] -> x = -n·W[-log(n)·ω(n)^m]/log(n).
@@angelmendez-rivera351 That is intuitively obvious. :{ ) (Respect!)
ejbejbphone Thank you
I was working on this problem but I didn't knew this vid existed and after few minutes it ramlndomly popped up and seems like my FBI did a good job 😊
This is anxiety-inducing as now I know some of the things I'll have to learn when I'm older.
UNDEADWARLORD 17 for me it is anxiety-inducing because it shows me things I have forgotten now that I am older
It's not so bad. When it's your turn to learn them, you'll be more prepared than you were when exposed to this video for the first time.
Sir... You are truly VALUABLE to our society.
A more elegant way to write the solution would be -1/ssqrt(√2), where ssqrt() is the super square root from tetration.
I just recorded a video on this today! Thank you so much for the idea!
blackpenredpen Wow! Looking forward to it!
It is a more *concise* way of writing it, yes, but the ssqrt operator is strictly defined in terms of W(x) in the first place.
"The Lambert W function will give you the fish back" is something I never thought I'd hear
Everyone gangsta until lambert w function
OMG man, your videos are great. Please keep doing these kind of videos, they are so interesting. Calculus is much more interesting when You explain it. Love you
dear teacher, could you give a more detailed lesson on the Lambert function?
idk why but the piece wise function for absolute value of x still seriously fucks up my mind. like on a good day im able to make sense out of it, and on other days its just a brainfuck.
Can you do a video on what the Lambert W function actually is and how it was derived?
I have a video in description
I really enjoyed it! Your enthusiasm is contagious and spurred me on too!
Could you explain this index of W? Just for curiosity
It s a complex number
@@anas8183 but how it works
@@5000jaap i d ont know my current level in math is not very good i am just 16 yrs old
I think it is related to Euler's formula:
e^(z) = cosx + isinx
Where z is a complex number, if that's the case then there are infinite imaginary solutions.
13:00 and lnsqrt2=O
x=-e^-W(O)
End -1= H
x=He^HW(O)
all the possible variables you could have chosen and you chose FISH!?!?!
respected.
Hahahah thanks!
omg teddy what u doing here??!!!
InfinityGaming omg infinity what r u doing here !?!?!?!!?!?
@@TheSavageTeddy idk!!!
InfinityGaming I’m watching youtube!!!!!
I feel so much better. I spent so much time thinking about this trying to solve it. Who knew it would be something so crazy?!!
I'm still waiting for the proof of the cubic equation
I think you can do it in 15 minutes or less. I did it in 20 minutes (because I'm not good at explaining Lol)
@@aaryanbhatia4939 you aint
Aaryan Bhatia What I did was give other values to the variables x.
if you want to know the extended formula you just have to change those final values at the roots of x
Aaryan Bhatia The answer is you *cannot.* Not in general, anyway. Only for certain values of the coefficients is the result simplifiable. Whenever it is not, this is known as casus irreducibilis. If you have ever wondered why an angle of 1° is not constructible, the answer is because 3° is constructible, and third angles of nontrivial constructible angles are not constructible usually, because the third-angle formula in general is unsolvable, as it requires solving a cubic equation that results in casus irreducibilis. This is why the cubic formula is not taught as opposed to the quadratic formula being taught. It is usually not useful, because casus irreducibilis is more common than not.
Aaryan Bhatia Well, no, it is not like the quadratic formula. The quadratic formula is useful and practical. The quadratic formula suffers from no casus irreducibilis. The quadratic discriminant also does not suffer from case deficiency. The cubic formula is, in every sense, analogous to the quadratic formula, but strictly inferior.
@Aaryan Bhatia I'm going to guess the answer is 'try and fail'
at case 1 we get 1/x ln(x)= 1/2 ln 2 so by compiring x=2
I followed you up until 7:05. In math it’s dangerous to say “of course...” Also, you introduced this lambert function without providing an explanation as to how to work with it
There's plenty other stuff he didn't define thoroughly first. A video can't be 10 years long to replace your entire school.
Chris Koperniak He defined the function for what it is. I'm not sure what else you think needs to be explained.
So is that the definition of the Lambert function?, I.e. as the solutions to that equation?
I have a masters in engineering, and I think this is the first time I've heard of the Lambert w function.
It comes up in diode characteristic curve equations, when you know the current and want to solve for the voltage.
Hummm for a guy who scored 6 out of 120 in mathematics... good job for recommending RUclips
i never thought there were people on the internet who enjoy math the way i do! man.... i wish you were my friend so we can talk about math forever!
The mic is back
Why is this enjoyable to watch...? if I had to sit in a class for this I wouldnt pay attention.
8:47 was absolutely unexpected
It's weird that, during the first 2 years of a maths degree, I have not encountered the lambert w function, but on RUclips, it's everywhere
LAMBERT OF 🐠 EAT 🐠 IS 🐠, SO COOL!
Wow. Never thought about it like that.
From Bangladesh....I love your videos🤗🤗
What is the negative real value?
-0.76666
@@blackpenredpen wow that was the fastest answer I ever got on RUclips😂 thank you
Lol. You are welcome
Or approximately -23/30. The actual value is irrational but -23/30 is rational.
Thank you for making me learn a new useful function I didn't know about, the Lambert function
Can I approximate the negative solution using maclaurin expansion of 2^x ?
Equations like x^2=2^x always fascinated me, but solving them felt like hitting a wall at times. What really helped me was using tools that walk you through the problem-solving process. SolutionInn’s AI study tool is fantastic for breaking down equations like this, making complex solutions easier to understand. Highly recommend it if you’re tackling challenging problems like this one.
Ah, man... This is very unsatisfying: How do we know what the Lambert W function is - without having to use Wolfram Alpha? And what are these "indexes", what do they mean - can you please do a video about them? Would we need to use Wolfram Alpha to calculate them? Because changing them obviously gets you different answers to the question/problem.
The equation is apparently covered in a video in the description. The indexes are different solutions to it in either real or conplex space.
@@Gulyus Thank you. However, even after viewing that video, I still have questions. For example, how do we know that when using f(x) to calculate the inverse: "f inverse" or "f -1(x)" that f(x) should equal "xe^x" based on the equation given in the video - which is "x^x = 2"?
In addition, the video did not explain anything about "indexes". So I would not have even known that they exist otherwise, without them being mentioned in THIS video.
This is basically the method I used to investigate the question at the end of your “Are you tired of this kind of question” video. We can compare a^b and b^a by considering the Lambert W function. Unfortunately it does not seem that this answer helps because I do not believe we can evaluate the Lambert W function without a calculator. So for example, this method does not help us decide if (e-1)^π is greater or less than π^(e-1).
Simple, e=3 & π=3 therefore e^π=π^e.
You can't evaluate many other functions without a calculator (e.g. exponentials for non-integer powers and/or irrational base, logs or trigonometric functions). They are just more familiar than W, so more people have an "approximate" idea of their behaviour.
@@almachizit3207 Fundamental theorem of engineering. To which one can add log(2) = 3/10.
@@dlevi67 exponentials, natural logs, trig, inverse trig and hyperbolic trig as well as many others can be estimated to arbitrary accuracy by hand with Taylor series (I point out those ones because I know them off the top of my head). Non-integer and negative powers can also be estimated to arbitrary accuracy by hand using binomial theorem, which I have encountered some smug bastards who have done that to calculate things like cube roots or things like that in their head with reasonable accuracy. For these things you just have to be a bit more patient than using a calculator.
@@almachizit3207 W can be calculated by hand too. Lambert and Euler tabled it in the 1750s - no calculators then. But it's a slog and using a calculator makes it significantly easier. What I'm pointing out is that many people commenting here seem to think it "weird" and "difficult" and "different". It's not; it's just not something that is typically taught in high school (or even initial degree-level math courses).
Lim Cot^2 (x) - 1/x^2
x->0
:)
tough topic .. good explanation. I have not met the Lambert W function .. clearly it has some fascinating properties .. to be continued!! Thank you again.
i tried x^-1 ln(lxl)=1/2 ln(2 on the desmos graphing calculator and the three answers i got were 2, 4, and -0.76666666666666...
Is -0.76(6) correct? because i also got that from my calculator and i wanna know if that's ok
Hi,
I worked out y=x^x (x to the power of x) and found for x=0, y approaches 1. Also for x=1, y will be 1. For x=1/e=~0.3678, y will have its minimum which will be 0.6922. beyond x=1, y increases continuously until +infinity. Here's my problem and I need your help, perhaps a video:
for x= -infinity, y=0. For all even negative integers, y>0. For all odd negative integers, y
You probably already got an answer but for negative x the real and the imaginary values oscilate around the x axis converging to 0 as x goes to -infinity. The imaginary part is 0 when x is an integer and negative for x in the interval (-(2n+1),-2n) and positive for x in the interval (-2n,-(2n+1)) for every natural number n. The real part will be zero at x = e^W(log(i b) + 2 i π k) i think (probably not all the zeros and there are some conditions for wich this isnt zero, for full solution check wolframalpha i guess) the function will never have real part and imaginary part both equal to 0 at the same time. For the question why you cant write (-2.3)^(-2.3) as (-2.3)^(-46/20) is because this is like writing sqrt(-1)=4th root of (-1)^2 = 4th root of 1. You create extra solutions wich aren't solutions to the previous equations (4th root of 1 has 4 solutions while sqrt(-1) only has 2)...
I want to see the w function
Buen video, ya dos meses viendo todos tua videos. Buen canal.
8:46 Yeah we really knew that
I love those videos. Best maths lessons I've watched in a while.
"most likely it's irrational" a proof on the rationality of this number would be pretty cool tbh
Brazil loves you man
=)
how to solve the limit as x tends to +infinty e^x/x^e
I think you could compare the growth of e^x and x^e. When x tends to +inf e^x will tends to +inf way faster than x^e.In fact for any a,b > 0 (e^ax)/(x^b) will always tends to +inf if x tends to +inf (so in our case b = e and a = 1)
I did it like that : x^e=e^(ln(x^e))=e^(e*ln(x))
Then (e^x)/(x^e) = e^(x-e*ln(x)) and so the limit as x tends to infinity is +∞
Since e^x is an exponential and x^e is a polynomial it will probably diverge but you can work like this:
Since your question is a infinity over infinity situation, you can use l'Hopital's rule and differentiate the top and bottom
Since the top is e^x it will stay like this and to differentiate x^e you just minus 1 the expontent and get:
e^x/x^(e-1)
This will still be infinity over infinity so you do it again which will lead to:
e^x/x^(e-2)
So you do it again...
e^x/x^(e-3)
But this time the exponent of x^(e-3) is negative so you can put it on numerator and get:
e^x× x^|e-3| which diverges
But that's kinda brutal in my opinion XD
@@CatchyCauchy This is why Taylor polynomials are so useful xD
I have never seen a fish with an antenna before. Must be a mathematical phenomenon.
My favourite part:
We can't see them because they are imaginary....12:28
😃
...like my friends.
@@roderickwhitehead I can be a real friend of yours....:)
@@IshaaqNewton - A real friend? Sounds like 1+2+3+4... = -1/12 witchcraft to me.
@@roderickwhitehead And for me it's
like
1^3+2^3+3^3+......
=(1+2+3+.........)^2
=(-1/12)^2
=1/144
😂😂😂
@@IshaaqNewton lol 😆
Someone: how do we solve this problem?
blackpenredpen: *have you heard of the Lambert W function?*
0:57 minecraft villager "hm"
Thanks for the video. I was just thinking about this problem for a few days and then this video popped up. That always seems to happen.
I need more about the LambertW function
Just go and check it out on Wikipedia...... म द फ क
Nunca imaginé que habriá más soluciones. Exploté❗
This is just the problem x^y = y^x you already solved right
Different style. I used a parametric approach in that video and lambert W function in this one.
Fish is my favorite variable now. Great video.
when i grow up i want to be that fish
8:47, Yes we know that, “We know that “we know that “if x is less than 0 so take out the absolute value...that's all.”””
I got a different answer: I got x=-2/ln(2)*W(-ln(2)/2)
If you replace the absolute value with the magnitude of a complex number z, this approach still works and the indices of the product log gives all of the (infinitely many!) complex solutions. :)
A more careful and rigorous way of handling the equation z^2 = 2^z is by noticing that if z is not an integer, then 2^z is inevitably multivalued. Namely, 2^z := exp[ln(2)·z + 2nπi·z] for any integer n. This implies z^2 = exp([ln(2) + 2nπi]·z). Now, a few cases must be considered. The first case is that |Arg(z)| < π/2, and the second case is that π/2 < |Arg(z)| < π. It is notable that if |Arg(z)| < π/2, then |Arg(z^2)| < π, and if π/2 < |Arg(z)| < π, then also |Arg(z^2)| < π.
In the first case, z^2 = exp([ln(2) + 2nπi]·z) implies z = exp([ln(2)/2 + nπi]·z), and z = exp([ln(2)/2 + nπi]·z) implies -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = -[ln(2)/2 + nπi]. Therefore, -[ln(2)/2 + nπi]·z = W(m, -[ln(2)/2 + nπi]), equivalent to z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi]). Notice that if n = m = 0, then this simplifies to z = 2.
In the second case, z^2 = exp([ln(2) + 2nπi]·z) implies -z = exp([ln(2)/2 + nπi]·z), equivalent to -[ln(2)/2 + nπi]·z·exp(-[ln(2)/2 + nπi]·z) = ln(2)/2 + nπi. Therefore, -[ln(2)/2 + nπi]·z = W(m, ln(2)/2 + nπi), hence z = -W(m, ln(2)/2 + nπi)/[ln(2)/2 + nπi].
With this, the remaining cases are Arg(z) = π, Arg(z) = π/2, and Arg(z) = -π/2.
In the first of these three, z = -r, with r = |r|, so -r = exp(-[ln(2)/2 + nπi]·r), thus [ln(2)/2 + nπi]·r·exp([ln(2)/2 + nπi]·r) = -[ln(2)/2 + nπi], and [ln(2)/2 + nπi]·r = W(m, -[ln(2)/2 + nπi]). Therefore, z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi].
If Arg(z) = -π/2, then z = -ri, with r = |r|. Hence -ri = exp(-[ln(2)/2 + nπi]·ri), and [ln(2)/2 + nπi]·ri·exp([ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi]. Therefore, [ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), and z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi].
Finally, if Arg(z) = π/2, then z = ri, with r = |r|, so ri = exp([ln(2)/2 + nπi]·ri), and -[ln(2)/2 + nπi]·ri·exp(-[ln(2)/2 + nπi]·ri) = -[ln(2)/2 + nπi], hence -[ln(2)/2 + nπi]·ri = W(m, -[ln(2)/2 + nπi]), so z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi].
In summary, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi], in both cases for arbitrary integers n and m.
This gives the complete family of complex solutions with no extraneous solutions.
The solution families can be compactified. Notice that exp[W(t)] = t/W(t), so exp[-W(t)] = W(t)/t. Therefore, -W(m, -[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = +exp[-W(m, -[ln(2)/2 + nπi])], and -W(m, +[ln(2)/2 + nπi])/[ln(2)/2 + nπi] = -exp[-W(m, +[ln(2)/2 + nπi])]. As such, if Arg(z) = π, |Arg(z)| = π/2 or |Arg(z)| < π/2, then z^2 = 2^z implies z = +exp[-W(m, -[ln(2)/2 + nπi])], and if π/2 < |Arg(z)| < π, then z^2 = 2^z implies z = -exp[-W(m, +[ln(2)/2 + nπi])]. Again, this solution is complete and with no extraneous solutions.