Hey, what do you think? 5x^6 + 2x^4 +x^2 - (4/ (e^x)) = ln(i)
There are complex solutions in terms of transcendental and complex numbers and integers?

Question

Hey, what do you think?  5x^6 + 2x^4 +x^2 - (4/ (e^x)) = ln(i)
There are complex solutions in terms of transcendental and complex numbers and integers?

kainui · Answer

I have no idea, I don't know, I could try some crazy stuff on this to try to get an approximate answer, plus there's also wolfram alpha we could check with. Beats me, any ideas?

daniel.ohearn1 · Answer

I worked it out algebraically to change it some using properties of log and i.  But how to solve for x when it's complex..  I now have -x^2(x^2 -2)(x^2+2)-e^2i=pi^2 -199

daniel.ohearn1 · Answer

Isn't an exact result possible?

kainui · Answer

Wolfram alpha is spitting out numerical solutions, that doesn't mean it's not exactly solvable but it's definitely not a good sign. http://www.wolframalpha.com/input/?i=5x%5E6+%2B+2x%5E4+%2Bx%5E2+-+(4%2F+(e%5Ex))+%3D+ln(i)&t=crmtb01 I'm not sure, why do you suspect it's exactly solvable?

kainui · Answer

Also, since the exponential is periodic, for any integer n:

$$i = e^{i(\pi/2+2 \pi n)}$$

This means there can be infinitely many logarithms, dunno how important that is to you in finding solutions or not:

$$\ln i = i(\pi/2+2 \pi n)$$

daniel.ohearn1 · Answer

Wolfram Alpha doesn't give step by step solutions with complex solutions in pro for students, I guess, does someone know about Pro?

kainui · Answer

Well there might not be a step-by-step solution here since the answers it gives are decimal approximation. You can approximate these with perturbation methods to arbitrary precision I believe though.

daniel.ohearn1 · Answer

perturbation?