Finding extremums of function with complex derivate  Printable Version + HP Forums (https://archived.hpcalc.org/museumforum) + Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum1.html) + Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum2.html) + Thread: Finding extremums of function with complex derivate (/thread201353.html) 
Finding extremums of function with complex derivate  Lukas K.  10152011 Given is the function Re: Finding extremums of function with complex derivate  Allen  10152011 For most values of x the function would approach 4. You can see in this function , can you use the estimation?
Edited: 15 Oct 2011, 8:56 a.m.
Re: Finding extremums of function with complex derivate  Crawl  10152011 I think the URL you pasted didn't work right,how about this? Yeah, the exponential will be of the form (cos(t)+isin(t)). So that plus 1 has a maximum absolute value when it equals 2. (when t=0, 2pi, 4pi, etc) So you need to solve for (x+3.1e6)/x = an integer. Or x = 3.1e6/(n1), with n an integer.
I'm not sure why it seems like the derivative is complex; it shouldn't be. I'd have to check what the HP50g gives for that later. Edited: 15 Oct 2011, 10:58 a.m. after one or more responses were posted
Re: Finding extremums of function with complex derivate  Crawl  10152011 My URL didn't work either; the parentheses get lost. Oh, well.
Re: Finding extremums of function with complex derivate  Lukas K.  10152011 Well, plotting and finding it's maximums did the trick. Why didn't I think of this earlier? :(
f' seems to be complex, because the derivate of a+bi is signum(a+bi), which returns a complex value :(
Re: Finding extremums of function with complex derivate  Crawl  10152011 I tried playing around with the HP50g with that expression, and it can give wildly varying forms depending on what you do with it. Some of which do not contain complex numbers at all. At least, make sure rigorous is on, so it doesn't simply return abs(x) as x.
This is probably a situation for which the TI89 gives a more straight forward result. Edited: 15 Oct 2011, 7:25 p.m.
Re: Finding extremums of function with complex derivate  Han  10152011 Quote: Are you sure that Re(f) = f ? If this is true, then you can graph Re(f) and get your answer that way.
How about using the fact that z^2 = z * conj(z) and e^(i*t) = cos(t)+i*sin(t)?
Re: Finding extremums of function with complex derivate  Bunuel66  10172011 It seems that: Re(z)!=z^2 My two cents...
Regards
