Talk:Cosine Exponential Formulation

The current formulation is a bit strange and uncommon. I immediately thought a typo was made, but it had been applied consistently. I suggest that:


 * $\displaystyle \frac 1 2 i \left({e^{-i x} - e^{i x}}\right) \quad\rightsquigarrow\quad \frac 1 {2i} \left({e^{i x} - e^{-i x}}\right)$

effectively using that $\dfrac 1 i = -i$. --Lord_Farin (talk) 11:30, 3 December 2012 (UTC)


 * I don’t mind that, but you’ll have to fix Sine and Cosine of Sum/Proof using Exponential Formulation accordingly then. — Timwi (talk) 11:39, 3 December 2012 (UTC)


 * Can somebody explain "Integrate into the flow of Binmore"? I've looked throughout but I can't find it in there. --prime mover (talk) 19:56, 3 December 2012 (UTC)


 * I put that up after following the links prev and next, and noticing that they didn't link back again. --Lord_Farin (talk) 20:08, 3 December 2012 (UTC)


 * You're all right, I know what happened. The page itself was cut-n-pasted from somewhere completely different and then wasn't cleaned up afterwards. Sorted now. --prime mover (talk) 21:28, 3 December 2012 (UTC)

Thinking about this, I believe the presentation of this result looks amateurish and grade-school: my view is that it looks better like $\dfrac {e^{i x} - e^{-i x}} {2i}$ - considerably more compact and neat. What does anyone else think? Is there merit in extracting the half into a separate fraction? --prime mover (talk) 22:59, 13 December 2012 (UTC)

“Left-right errors”
What are “left-right errors”? — Timwi (talk) 15:23, 13 December 2012 (UTC)


 * Illustration by example. Consider the following excerpt of the page (first eqn template, r column):

\frac 1 2 \left( \sum_{n \mathop = 0}^\infty \frac {(-ix)^n}{n!}                          + \sum_{n \mathop = 0}^\infty \frac {(ix)^n}{n!}                      \right)


 * As you can read in the house style guide, ( ... ) and \left( ... \right) should actually be \left({ ... }\right) (i.e., a left-right pair and braces). This is what I meant. With this convention, the above becomes:

\frac 1 2 \left({ \sum_{n \mathop = 0}^\infty \frac { \left({-i x}\right)^n } {n!}                          + \sum_{n \mathop = 0}^\infty \frac { \left({i x}\right)^n } {n!}                      }\right)


 * where I have also applied some code spacing niceties. I hope it's clear. --Lord_Farin (talk) 15:28, 13 December 2012 (UTC)


 * Ah, OK. Thanks. — Timwi (talk) 15:30, 13 December 2012 (UTC)
 * Done. — Timwi (talk) 15:33, 13 December 2012 (UTC)

Issue with Euler's Formula
It is only valid for real $x$. Thus the statements and proofs for real and complex $x$ are best separated. --Lord_Farin (talk) 22:01, 13 December 2012 (UTC)