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I would like to understand the behaviour of

$$\exp(-ax)\,\cosh(bx)$$

or

$$\exp(-ax)\,\sinh(bx)$$

for large $x$, provided that $a,b>0$ and $a>b$ or $a<b$.

JFNJr
  • 1,055
  • The Maple command $$simplify(asympt(exp(-ax)cosh(bx), x, 2)) ,assuming a > 0, b > 0, a > b $$ produces $$1/2,{{\rm e}^{bx-{\it ax}}}+1/2,{{\rm e}^{-bx-{\it ax}}} $$ and the command $$simplify(asympt(exp(-ax)cosh(bx), x, 2)) ,assuming a > 0, b > 0, a < b $$ produces $$1/2,{{\rm e}^{bx-{\it ax}}}+1/2,{{\rm e}^{-bx-{\it ax}}}. $$ – user64494 Aug 02 '13 at 09:42

1 Answers1

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Expand the hyperbolic function as exponentials, i.e. write the first exoression as $$\exp(-ax)\,\cosh(bx) = \frac{1}{2}\exp((b-a)x) - \frac{1}{2}exp(-(a+b)x)$$ From this you can read off the expansion depending on signs and order of $a,b$. The second expression is handled the same way.

gammatester
  • 18,827