ALL solutions to x^2=2^x

This is my "equation of the year" in 2019.

[poprockssuck]

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.

[taflo]

"This is not a fish yet, but it's almost a fish." Sounds like evolution at work.🤣

[angelmendez-rivera]

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.

[daniel-fich]

12:57 Look at the way he looks at his math with passion in his eyes :-D

[dihydrogenmonoxid]

I should read a book for english but pen(black+red) math videos are sooo much nicer💯

[HasXXXInCrocs]

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!

[phyricquinn]

Thank you for producing all of this unique content! I really love these videos!

[PlutoTheSecond]

BPRP: It's probably irrational...
Wolfram: It's transcendental.

[JyothiSwaroopM]

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"

[nibblesdotbas]

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?

[MathswithMuneer]

Respect from one math teacher to another math teacher <3

[josephsmith]

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.

[Bayerwaldler]

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.

[nou]

My teacher explaing math: so x is equal to y

Blackpenredpen explaning math: *_F I S H_*

[lumi]

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.

[spelunkerd]

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.

[NektoYa-wn]

Can you make a video explaining LambertW indexes?

[atheoristspointofview]

8:45 "we know that-we know that"(pro jump cuts)

[byronvega]

From this point on I'll start writing functions that depend on fish rather than x

[InvaderMixo]

I love learning, especially math. I love your enthusiasm. It's so uplifting to be in a math moment with you!