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I am trying to show that this function $s:\mathbb R→\mathbb R$ defined by $s(x)=e^x - e^{-x}$ is one to one.

My approach is using the natural $\log$ ($\ln$) to cancel out the $e$'s: $$e^x - e^{-x} = e^y - e^{-y}\\ \ln(e^x - e^{-x}) = \ln(e^y - e^{-y})\\ x + x = y + y\\ 2x = 2y\\ x = y$$ I just wanted to know A. if my algebra is correct and you can do the above operations and B. if it isn't correct, would the next best way to prove its one to one by just plugging in zero for the original functions?

Thanks.

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    $\begingroup$ The algebra is not right. The ln of a sum is not the sum of the ln's. $\endgroup$ Mar 14 '13 at 21:18
  • $\begingroup$ What sort of tools do you have? I assume you're not supposed to use calculus? $\endgroup$ Mar 14 '13 at 21:22
  • $\begingroup$ Yes I think it should be done using basic algebra, but if there is a simple way to do prove x = y using calculus I'd be enlightened to see it as well. $\endgroup$
    – Goose
    Mar 14 '13 at 21:23
  • $\begingroup$ A simple calculus-proof would be: Let $f(x) = e^x - e^{-x}$. Then $f'(x) > 0$ for all $x$, so $f$ is strictly increasing and thus $f(x) = f(y)$ implies $x = y$. $\endgroup$
    – TMM
    Mar 14 '13 at 21:32
  • $\begingroup$ @TMM thank you that is pretty simple and makes sense, I'd still like an algebraic one if possible. Gerry Myerson, thanks I just edited my question. $\endgroup$
    – Goose
    Mar 14 '13 at 21:39
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$$\begin{align} e^x - e^{-x} &= e^y - e^{-y} \\ e^x + e^{-y} &= e^y + e^{-x} \\ e^x(1 + e^{-x-y}) &= e^y(1 + e^{-x-y})\end{align}$$ Since $e^{-x-y} \geq 0$ for all $x,y$ we have $1 + e^{-x-y} \neq 0$, so we can cancel the terms on both sides to get $$e^x = e^y$$ Since $e^x$ is injective, the result follows.

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Perhaps it is enough to observe that since $e^x$ is strictly increasing $x>y \Rightarrow e^x > e^y$

so

$e^{-x} < e^{-y}$

$-e^{-x} > -e^{-y}$

$e^x -e^{-x} > e^y -e^{-y}$

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If $e^x-e^{-x}=w$, then $e^{2x}-we^x-1=0$, a quadratic equation in $e^x$. By the quadratic formula, $e^x=(w\pm\sqrt{w^2+4})/2$. We reject the solution with the minus sign, since it makes $e^x$ negative, which is impossible. So $e^x=(w+\sqrt{w^2+4})/2$, and $x=\log((w+\sqrt{w^2+4})/2)$, a unique solution. So $f(x)=e^x-e^{-x}$ is one to one.

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