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I'm Nina. I have a really tough homework that counts as a test and I couldn't do it. It's really urgent. Help please!

  1. Let $\alpha \in \Bbb{R}^+$. Let $F$ be the function defined in $\Bbb{R}$ by $$F(x) = \frac{1}{\cosh\alpha - \cos x}.$$ What is the decomposition into simple elements of the fraction? $$ -\frac{2X}{X^2 - (e^{\alpha}+ e^{-\alpha} )X+1} $$

  2. Give an other expression for $\displaystyle -\frac{e^{\alpha}}{e^{ix} - e^{\alpha}}$.

  3. And for $\displaystyle -\frac{e^{\alpha}}{e^{ix} - e^{-\alpha}}$.

And big thanks.

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1 Answer 1

use notations for $cosh(x)=cos(i*x)$

$cosh(x)=cos(-i*x)$

$cosh(x)=(1/2)*(e^{-x}+e^x)$

EDITED: for $2$ and $3$ ,you divide both side by $e^a$ because $e^a$ never equal to $0$ also note that $(1/2)*(e^a+e^{-a})=cosh(a)$;if you divide both side by $2$ you get

$X/((x^2)/2-(1/2)*(e^a+e^{-a})*x+1/2)$.which is equal to

$X/(X^2/2-cosh(a)*X+1/2)$

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@Nina i hope this will help –  dato datuashvili Sep 1 '12 at 18:12

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