Prove that $\exp(x)>0$ using only formal definition of exp
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!
$\endgroup$ 12 Answers
$\begingroup$$\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.
$\endgroup$ 2 $\begingroup$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 $\exp(x) = 1/\exp(-x) > 0$.
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