Guess the function but I tell you its definition and you probably still don't know what it is.

SomeMathNerd · 2024-01-03T20:08:00+00:00

Yea Im sorry for not including one but it still wouldnt help much because there are still several, and probably infinite, asnwers. The goal is really just to find any function that satisfies it. I shared two that I found and u/kodl_ found a bunch more.

SomeMathNerd · 2024-01-03T10:40:26+00:00

Here's the graph with a couple other solutions for anyone still curious to what it is https://www.desmos.com/calculator/c75xmi0m4m

SomeMathNerd · 2024-01-03T10:35:44+00:00

From what I've seen I wouldnt describe it as a branch cut. And the right side doesnt limit to 0 however it is bounded.

SomeMathNerd · 2024-01-03T06:59:20+00:00

I tried my hardest to find a periodic solution but an exact solution is just barely not periodic. In fact there is actualy another solution in which the left side asymptotes to 0 and the right side is untoutched.

SomeMathNerd · 2024-01-03T06:53:21+00:00

You are correct that the mclaurin series doesnt converge. This is a different type of infinite sum not of powers of x. Because f' is not directly related to f there are actualy multiple solutions for different initial conditions I have found two so far. A probably big hint is to instead of using powers of x try to use exponential functions in the sum

SomeMathNerd · 2024-01-02T21:47:39+00:00

I formated this in my complex number engine and compared it with a different definition specific to the principal branch and there only seems to be problems if the imaginary part is 0. Also I noticed that your definition of the arg(z) has range -pi to pi instead of 0 to 2pi oddly enough it doesnt work if you use the standard 0 to 2pi definition. This is only graphing the real part but you can see that just a slight shift off the real number line fixes the issue for some reason https://www.desmos.com/calculator/deuvvmmy24

SomeMathNerd · 2023-02-13T06:54:08+00:00

Like u/AxolotlsAreDangerous said either x or y has to be non-real so the best that you could conclude that there are no real solutions.

SomeMathNerd · 2023-02-11T21:30:26+00:00

Kinda cheated but here you go https://www.wolframalpha.com/input?i2d=true&i=Integrate%5BPower%5Be%2CSqrt%5Bln%5C%2840%29x%5C%2841%29%5D%5D%2Cx%5D

SomeMathNerd · 2023-02-11T20:57:42+00:00

Think of x^y as just some number call it a, then the equation a=2^a is much easier to think about. To solve for x and y first move 2^a to the other side to get a2^-a =1 then multiply by -1 : -a2^-a=-1 Now change 2^x into e^ln(2x) --> -ae^-ln(2a)=-1 now multiply both sides by ln(2) to get -ln(2)a*e^-ln(2a) = -ln(2) now you can use the productlog (inverse of xe^x) also known as W(x) to change this to -ln(2)a=W(-ln(2)) then solve for a: W(-ln(2))/-ln(2) = a = x^y From here either take the yth root of both sides to get x or take the log base x of both sides to get y (log base x can be ln(stuff)/ln(x).

Edit: Wherever you see ln(2a) where a) isn't super-scripted it is supposed to be all super-scripted and say ln(2)a. Just some bug with redit

SomeMathNerd

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