¿Tengo la expresión $$\frac{1}{2}i\log\left(1-e^{\frac{2i\pi{x}}{b}}\right)-\frac{1}{2}i\log\left(1-e^{-\frac{2i\pi{x}}{b}}\right)$$ and I want to simplify it without getting any division-by-zero errors. Like when I convert this to $$\frac{1}{2}i\log\left(\frac{1-e^{\frac{2i\pi{x}}{b}}}{1-e^{-\frac{2i\pi{x}}{b}}}\right)$$ I get division by zero whenever $e^{\pm\frac{2i\pi{x}}{b}}=1$. Although, I know both expressions give singularities when $e^{\pm\frac{2i\pi{x}}{b}}=1$, the first one cancels out the singularity and outputs zero. Is there a mathematical way to simplify an expression such as this and go around the division-by-zero problem? Perhaps we can say that the $\log (\frac {n} {0}) = 0$? Sin embargo no satisfacerme porque no aceptan aplicaciones de cómputo.
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$$\frac{1}{2}i\log\left(\frac{1-e^{\frac{2i\pi{x}}{b}}}{1-e^{-\frac{2i\pi{x}}{b}}}\right)$ $ Entonces que $y = (\pi x)/b$ para obtener\begin{align} \ln\left( \frac{1 - e^{2i y}}{1- e^{-2i y}} \right) &= \ln\left( \frac{e^{i y} \, (e^{-i y} - e^{i y})}{e^{- i y} \, (e^{i y} - e^{-i y}) } \right) \\ &= \ln\left( - e^{2 i y} \right) = \ln\left( e^{i (\pi + 2y)}\right) \\ &= i \, (\pi + 2y) = i \, \pi \, \left( 1 + \frac{2 x}{b}\right). \end {Alinee el} ahora, $$\frac{i}{2} \, \ln\left( \frac{1 - e^{2i y}}{1- e^{-2i y}} \right) = - \frac{\pi}{2} \, \left(1 + \frac{2 x}{b}\right).$ $
Desde $1 = e^{\pm 2 \, n \, \pi \, i}$ $n \geq 0$, entonces el $e^{\pm \frac{2 \pi \, i \, x}{b}} = 1 = e^{2 \, n \, \pi \, i}$ conduce a $\frac{x}{b} = n$ y la expresión para el logaritmo se convierte en $$\frac{i}{2} \, \ln\left( \frac{1 - e^{2i y}}{1- e^{-2i y}} \right) = - \frac{\pi}{2} \, \left(1 + \frac{2 x}{b}\right) = - \frac{\pi \, (1 + 2n)}{2}.$ $
Utilizar $ \frac{1-e^{ik}}{1-e^{-ik}}\frac{1-e^{ik}}{1-e^{ik}}=\frac{1-\sin^2$ (k) + \cos^2 (k)-2 \cos(k)-i \sin(k) (\cos(k) 2-2)} {2-2\cos(k)} = \\ \frac{-\cos(k) (2-2 \cos(k))-i \sin(k) (\cos(k) 2-2)} {2-2\cos(k)}=-\cos(k)-i\sin(k) =- e ^ {ik} $$... ninguna singularidad en la fracción y una expresión general bastante simple para usted!
