@DanielLichtblau Wolfram got back to me and confirmed it's an issue with Series[]. Btw, I missed your joke on the first read. LOL

@DanielLichtblau I see. I sent a note to Support though. The fact that Limit gets it but Series doesn't is definitely a software fail even if it's not a full on bug.

@DanielLichtblau Ah. So, is this a bug I should report to Wolfram?

@DanielLichtblau Limit[D[mdel, theta], theta -> 0] gives me 0, the correct answer. You can also plug in small values and see that for small theta, mdel is proportional to theta^2. For example, mdel /. {q -> .1, theta -> .01} yeilds -.000249 while mdel /. {q -> .1, theta -> .001} yield -.00000249, both of which are very good matches to my 2nd order expansion 0 + 0 -(cot(q)/4)*theta^2

@DanielLichtblau Thanks for the reply. I get exactly what you get for mdel. But I'm expanding the Taylor series in theta not q. The 0th and 1st derivs of mdel in theta are both zero. The 2nd deriv of mdel wrt theta at theta==0 is -cot(q)/2. So the series in theta should simply be 0 + 0 + (-Cot[q]/4)*theta^2 + O(theta^3)

I just rewrote the question to clarify what I was asking. Hopefully that helped.

The Limits I calculated above are the 0th, 1st, & 2nd derivates at theta = 0, which are the coefficients of the first three terms of a Taylor Series. So the first three terms of the Taylor Series of my function is 0 + 0 + (cot[q]/4)*theta^2. Series[] is not correctly calculating that. I'm trying to figure out how to make Series work correctly, not just find a different way to calculated the individuals limits I already calculated. Do you understand that Series[f,{x,0,n}] is supposed to give you an nth order Taylor Series?

Right, but I'm still trying to understand why the Series[] function isn't working properly they way I used it.

Okay, I see you used different code than I did for the Series and got the right result. But could you give some idea why mine didn't work? Seems like it should have. Is what you've written here just a workaround to something MMa fails at? Or did I use it wrong? Thanks for the reply.