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This problem would be easy if I could use the fact that $\exp(x)=e^x$, but I have to use the following definition: $$\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ I can also use the fact that $$\exp(x+y)=\exp(x)\exp(y)$$ So how do I prove, using those two equations, that $$\forall x\in \mathbb{R}:\exp(x)>0 $$ I mean, I can't just use the definition, because if $x<0$ then it isn't so obvious that $\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}>0$. Can someone give me a hint or two?

Thanks!

egreg
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Dunno
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2 Answers2

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$\exp(x) = \exp(x/2+x/2)=\exp(x/2)^2 \ge 0$ for all $x$.

$ \exp(x)=\exp(x_0)\exp(x-x_0) $ implies that if exp is zero for some $x_0$ then it is zero for all $x$.

$\exp(0)=1$ shows that exp is not the zero function.

lhf
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  • ooh, that's clever! I'll accept after I finish working on it, thanks :) – Dunno Jan 04 '14 at 12:24
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    Still not quite out of the woods: because the property $f(x+y) = f(x) f(y)$ is trivially satisfied by the zero function, it alone cannot be used to prove the desired behavior of $\exp$. In order to prove positivity, there also needs to be some statement about the value of $\exp(x)$ at some point. That's where the series definition can be used, or some other property. – heropup Jan 04 '14 at 12:44
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It's quite simple: We can show that $\exp(x) > 0$ for all $x > 0$ using the series definition. Then note that $\exp(0) = 1$, again using the series definition. Then for $x < 0$, consider the second definition $1 = \exp(0) = \exp(x+(-x)) = \exp(x) \exp(-x)$, from which it easily follows that $\exp(x) = 1/\exp(-x) > 0$.

heropup
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