I really wish you had plotted the solutions for us.
@XJWill1
Ай бұрын
See my other comment for the equations you can plot to visualize the infinite number of solutions.
@XJWill1
Ай бұрын
To visualize in the complex plane the infinite number of complex z values that solve the equation, exp( Log(z) / z ) = i plot the following sets of curves and observe the intersection points 0 = x * ln(x^2 + y^2) + 2*y * atan2(y , x) 1 = tan( (1/2 *x*atan2(y , x) - 1/4 *y*ln(x^2 + y^2) ) / (x^2 + y^2) ) If using Desmos, note that atan2(y , x) is entered as arctan(y , x)
@XJWill1
Ай бұрын
The curves become difficult to resolve as you zoom in, like fractals. Desmos sometimes creates artifacts in the graphs, but if you keep zooming in you can get better resolution. One thing that is interesting to see from the graph, that I think is difficult to realize if the solution is given in terms of the product-log function, is that there is only 1 solution with negative real part (real part is approximately - 1.86). Any of the other infinite number of solutions have positive real part.
@Skyler827
Ай бұрын
I tried it and got a very different wrong answer.... 😅 but still, great problem and great solution!
@aplusbi
Ай бұрын
Thanks!
@DonEnsley
Ай бұрын
problem z^(1/z) = i Recall polar definition of i: i = e^((iπ/2)(1+4k)) Take both sides to z power. z = i^z Replace i in polar form: z = [e^((iπ/2)(1+4k))]^z z e^[(-iπ/2)(1+4k)z] = 1 (-iπ/2)(1+4k)z e^[(-iπ/2)(1+4k)z] = (-iπ/2)(1+4k) (-iπ/2)(1+4k)z = W[ (-iπ/2)(1+4k) ] -(1+4k)z = -2 i /π W[ (-iπ/2)(1+4k) ] z = (2 i /π) W[ (-iπ/2)(1+4k) ] /(4k+1) The Lambert's W function solution: z = (2 i/[π(4k+1)]) W(-i (π/2)(4k+1)), k an integer >= 0 A few solutions for different values of k: k = 0 z = (2 i / π) W(-i (π/2)) (Approx 0.43828 + 0.36059 i ) k=1 z = 2i W(-5 i π/2) /(5π) (Approx. 0.12539 + 0.18893 i ) k=2 z = [2 i/(9π)] W[-i (9π/2)] (Approx. 0.07485 + 0.13298 i) k = 3 z = [2 i/[13 π)] W[-i (13π/2) ] (Approx. 0.053720 + 0.104764 i) . . There are infinitely many solutions. THESE CURVES ARE AWESOME LOOKING! They look like pkasma ion distributions (real = proton density, imaginary = electron density COOL!! 😎 😅😅❤❤ 👍🏼👍🏼👍🏼🙏🏻🙏🏻🙏🏻🙏🏻😃
@aplusbi
Ай бұрын
Where did you plot them?
@hafizusamabhutta
Ай бұрын
Which app do you use?
@aplusbi
Ай бұрын
Notability
@scottleung9587
Ай бұрын
Cool!
@mcwulf25
Ай бұрын
Not as straightforward as it first appears. I rearranged to get u = 1/z u^u = -i Which didn't simplify!
@aplusbi
Ай бұрын
A nice way to approach it, though
@XJWill1
Ай бұрын
@mcwulf25 It appears that you assumed (1/u)^u = 1 / u^u I wonder if you proved that to be the case. Because in general (1 / w)^p is not equal to 1 / w^p For example, w = - 1 and p = 1/2 demonstrates the inequality of the more general case For (1/u)^u and 1 / u^u , try u = - 1/2
@rishy773
Ай бұрын
It does actually simplify!! There is this super cool function called the "super square root," which is defined such that ssrt(x) is the inverse of x^x (it can also be calculated through ssrt(x) = e^(W(lnx)).) Therefore, you can say: u^u = -i u = ssrt( -i ) 1/z = ssrt( -i ) z = 1/(ssrt( -i )) = 1/ (e^(W(ln(-i)))) = e^-(W(-i*pi/2)) (which is the same as the the answer he gave! - it is just the 'principal value' or where n = 0) ≈ 0.438 + 0.361i
@Nobodyman181
Ай бұрын
U are fun❤😄
@aplusbi
Ай бұрын
u is funner than me 😜
@XJWill1
Ай бұрын
Your mistake is equating ln( z ^ (-1) ) = -1 * ln(z) That is not true in general. It is only an identity for real z > 0 For other values of z, equality does not always hold
@xleph2525
Ай бұрын
Can you point out a particular value of z which does not hold for ln( z ^ (-1) ) = -ln(z)?
@XJWill1
Ай бұрын
@@xleph2525 The most obvious failure is - 1
@82rah
Ай бұрын
How is the problem solved without making this mistake?
@xleph2525
Ай бұрын
@@XJWill1 I see your point, but nothing like -1 was being operated on in the string of reasoning. In fact, it is very apparent that no real z < 0 can be a solution. I suppose there could have been a disclaimer in the video with that observation, but I don't believe that would have been necessary. On a side note, I think -ln(z) = ln( z^-1 ) does hold true for z = a + bi. I encourage you to try a few complex numbers and see what happens.
@XJWill1
Ай бұрын
@@xleph2525 You obviously do not understand identities and real and complex numbers.
Пікірлер: 28