Recent content by Brendy

Ah, thanks guys. I confused myself with the exponential for a bit.

Just thought a bit more on that solution... Isn't it undefined everywhere since e2in\pi=1 and therefore the denominator is 0?

That's exactly what he did. It's a nice way to eliminate that annoying exponent that I didn't think of until I saw him do it. The next part is just rearranging what he had to get it in the form z= It isn't an easy subject. Unfortunately, the prerequisite subject for complex analysis at my uni...

Thanks, jackmell. I didn't think of doing that. Once I did, the rest of the problem proved to be simple.

It would, wouldn't it! That equation should have an =0 at the end. My apologies.

Alright, I've got (r23ei\theta-ei\pi)23-r23e23i\theta I don't recall any way to combine the two exponents. Expanding those brackets are going to be a mess, I think, and probably unnecessary since I'm not trying to find the solutions exactly.

Homework Statement Consider the equation (z-1)^23 = z^23 Show that all solutions lie on the line Re(z)=1/2 How many solutions are there Homework Equations The Attempt at a Solution Really have no idea. I figured polar form might be helpful somehow so I converted it and got...

Oh, you mean light emitting diodes? I can't speak for OP but in my lab we had curtains that closed off each bench from the rest of the lab so there was very little light other than that from the sodium lamp. There may have been a little red light on the power switch on the power supply but...

I'm not sure I know what a red led indictor lamp is...

I think I'm doing the same experiment as OP and the lab manual states that the sodium D-line is the yellow line which is actually a doublet. I'm not sure what the red line is either as it was too faint to get an accurate measurement. As for Vanadium 50's hint, it seems that he's implying that...

After rewriting it so that x_new... line is now x_new= atan (pi + x_old) it converged within 6 iterations. So is that just a fluke then? It is giving me the correct number to find the roots though. I don't follow post #51 at all I'm sorry.

I didn't write the code. I modified an existing program that was solving x_n+1 = e^-x_n Which solution are you referring to? Even changing the initial guess to something further away from the root such as 0.5 sees it converging within 6 iterations.

Thanks a lot, you've been a massive help. One last thing, what are f_old and f_new doing in my code? After I edit the line x_new... to the expression for theta, it will converge on a root but I'm not sure what those two lines after that mean. Are they needed? Also, is it supposed to converge so...

Do you mean at the end of running the algorithm? Also, when I solve for x, the x represents the roots, right? Not just another initial guess? Or do I use what I find x to be as the next initial guess?

\theta\ = arctan (n\pi+\theta) So that's it? Do I try and get everything in terms of x now or is this what the algorithm will use?