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S Jun 24, 2015 at 16:20 history suggested CommunityBot CC BY-SA 3.0
Corrected two small typo's.
Jun 24, 2015 at 16:04 review Suggested edits
S Jun 24, 2015 at 16:20
Aug 4, 2011 at 1:10 comment added Michael Hardy Oh. I was missing the forest for the trees.
Aug 3, 2011 at 22:28 comment added Noam D. Elkies @M.Hardy: it's the logarithmic-derivative form of the derivative of a product of $n$ functions. I already evaluated the factor $P(X)$ at $X=1$.
Aug 3, 2011 at 1:04 comment added Michael Hardy Where does the second equality after $-S$ come from?
Aug 1, 2011 at 0:29 comment added Noam D. Elkies :-) But this isn't "engineering_overflow".
Aug 1, 2011 at 0:22 comment added Michael Hardy BTW, if you had left the index as $i$ and let $j$ be the square root of $-1$, that would have been just the notation in which I (as the son of an electrical engineer) first learned about complex numbers.
Jul 31, 2011 at 23:53 comment added Noam D. Elkies Thanks --- and yes, you're right, those $\cos \alpha_j$'s should have been in the numerator, and the ones coming from $1 + i \tan \alpha_j$ in the denominator!, I corrected this now.
Jul 31, 2011 at 23:51 history edited Noam D. Elkies CC BY-SA 3.0
Put $\cos \alpha_j$ factors in numerator and denominator where they belong, rather than vice versa, correcting error noted by M.Hardy
Jul 31, 2011 at 22:53 comment added Michael Hardy Nice. (Looks to me as if the product of cosines should be in the numerator, so that −S=∏jcosαjReP′(1). I'll finish working through some further details shortly.....)
Jul 31, 2011 at 6:18 history edited Noam D. Elkies CC BY-SA 3.0
inserted \cdot in $i \tan \alpha_j \cdot X$ (twice)
Jul 31, 2011 at 3:40 history edited Noam D. Elkies CC BY-SA 3.0
Corrected a sign error in the initial formula for $S$
Jul 31, 2011 at 3:01 history edited Noam D. Elkies CC BY-SA 3.0
A bit more copy editing, mostly to highlight and explain the key point $\sum_j \alpha_j = \pi$
Jul 31, 2011 at 2:26 history answered Noam D. Elkies CC BY-SA 3.0