I Solved A Quick And EZ Exponential Equation

[@Qermaq]

I'm curious if there are complex solutions.

[@justyceleague698]

Probably because as the function approaches negative values it becomes undefined, like if you put it in desmos and don't zoom in, there's points but LOTS of open circles, so when you do zoom in they disappear. Also the graph doesn't touch the x-axis so that's another indicator.

[@n8cantor]

Even if you allow for all the infinite solutions to the multi-valued complex exponentials, there are still no complex solutions outside of x = 1 and -1 [edit: there was an error in my solution and this isn't correct, see below]

[@Qermaq]

@@n8cantor Cool. I don't know a lot about z^z. I know that the parity of the imaginary part is irrelevant, you get the same value from (a+bi)^(a+bi) and (a-bi)^(a-bi). I know that i^i is about 0.2 with no imaginary part. But are there any other complex numbers which when put as z^z will have an imaginary part of 0? We could also consider what values of z^z have a modulus of 1.

[@n8cantor]

​@@Qermaq I take that back, there was a wrong assumption in my solution. There are complex solutions to the equation! The modulus and argument of the solutions must solve the equations: (r cosθ -1) lnr - θ r sinθ = 0 and r lnr sinθ + (r cos θ - 1) θ = 2 pi n (for some integer n). One solution is at approximately 2.86293 + 3.2233 i

[@DeadJDona]

yes, and it's weird )) let x = i*a (a is some variable) then we have i*a^(i*a) = i*a, let rename a back to x for simplicity and simplify by dividing both parts by i x^(i*x) = x wolframalpha gave next solution to this: x ≈ -2.606627112782256075145609011×10^-17 - 1.000000000000000085910417883 i...

[@stvp68]

Is the graph of y^(x-1) a parabola or a different kind of curve?

[@Qermaq]

Can't really graph y^(x-1) as it's not an equation. You could set it equal to z and graph it in 3d. You'll get some sort of 2d curve.

[@JWQweqOPDH]

Not sure. Note y=(x-1)^2 is a parabola.

[@stvp68]

@@Qermaq so what graph did he show in this video? I was wondering if the graph he showed is parabolic or some other kind of curve…

[@ShortsOfSyber]

the graph shown is y = x^x and it's not a parabola

[@Abid_Ibn_Ashraf]

2:53 You is the exponent here

[@honestadministrator]

Trivial answer is x = UNITY for a small number h ( 1 - h) ^( 1 - h) > (1 - h) and ( 1 + h) ^( 1 + h) > (1 + h) Here by x^x attains a minima at x=1

[@byrnman]

Why does the graph of x^x not graph -x values, but we said x=-1 is a solution?

[@nitayderei]

Since it's not defined there properly (over the reals), only for specific values x. For example (-1/2)^(-1/2)=i*sqrt(2).

[@Qermaq]

There are some values that work but most don't. So the negative side of the graph will just be disassociated points that simply won't show up. On the positive side the function is continuous so we see a curve.

[@ShortsOfSyber]

Well said!

[@kianmath71]

X = 1, -1

[@Fire_Axus]

x^(x-1)=1 x-1=0 x=1