# Bound for $e^{-\alpha x}$

For a part of my proof I need to establish that $$e^{-\alpha x} \lt h(x)$$, where $$\alpha,x \gt 0,$$ and $$x,\alpha \in\mathbb{R}$$. I thought for a while and couldn't find a function independent of $$\alpha$$ that fulfills my criteria. Any ideas?

• Are you asking for an example of a function $h$ for which the inequality is true? – kimchi lover Jan 15 at 14:21
• $h(x)=1$ fits the bill. – Adrian Keister Jan 15 at 14:31
• If you don't want any specific $h$, there are a lot (and basically useless) functions that bound $e^{-ax}$ from above. For example given any strictly positive integer $n$, $h_n(x)=n$ is always strictly larger than $e^{-ax}$ and independent of $a$. – James Jan 15 at 14:36
• I'm voting to close the question for lack of context. Without context, we cannot know what kinda of upper bound you want, and the question becomes too open-ended. – Simply Beautiful Art Jan 15 at 14:37
• $$\alpha,x>0\implies e^{-\alpha x}<1.$$ – Julián Aguirre Jan 15 at 15:15

Also note that: $$e^{-\alpha x}=\sum_{n=0}^\infty\frac{(-1)^n\alpha^nx^n}{n!}$$ A polynomial of this expansion could be used, although take care to see that it is consistently greater as this is a series with alternating sign
• But a polynomial (of positive degree) gives a worse bound than the obvious $h(x)=1$. – David C. Ullrich Jan 15 at 17:41
• The higher order used the closer this approximation is surely, especially at smaller values of $x$, since $h(x)=1$ is most accurate for $x\to\infty$ – Henry Lee Jan 15 at 18:53