Given is the function
I want to calculate the maximums auf this function using the FCN>EXTR function of the 50g. This fails with the message 'Bad Guess(es)', because the derivate of f is complex. Why is f' complex, whereas f only returns real values?
Adding a >NUM to the end of the YVar program slows down the plot and sends the calculator into an infinite loop when trying to find the maximum of f.
The TI83 fails at this, too. It finds the maximum at the guessed one.
Has anyone an idea how to find the maximum of f (using the 50g)?
Finding extremums of function with complex derivate

10152011, 06:18 AM
10152011, 08:44 AM
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.
10152011, 10:55 AM
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
10152011, 10:56 AM
My URL didn't work either; the parentheses get lost. Oh, well.
10152011, 11:20 AM
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 :(
10152011, 07:24 PM
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.
10152011, 07:35 PM
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)?
10172011, 06:06 AM
It seems that: Re(z)!=z^2 My two cents... Regards 
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