Javier
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 Jun 8 awarded Caucus Jun 4 accepted How do we know that $\exp(x)$ agrees with raising a number to a rational power? Jun 4 comment How do we know that $\exp(x)$ agrees with raising a number to a rational power? Yeah, sorry about that, fixed it. I think I liked Jim Belk's answer a little more, but thanks for your help! Jun 4 revised How do we know that $\exp(x)$ agrees with raising a number to a rational power? deleted 168 characters in body Jun 4 comment How do we know that $\exp(x)$ agrees with raising a number to a rational power? Yes, we used that property to define rational powers, but we also knew how to define integer powers and roots, and we used $a^{b^c}=a^{bc}$ to define rational powers in terms of those. Jun 4 comment How do we know that $\exp(x)$ agrees with raising a number to a rational power? @Limitless: The advanced part is showing that such a function is unique. Wikipedia doesn't give such a proof, and makes some statements that I don't know about, such as the fact that it must be Lebesgue measurable. Jun 4 asked How do we know that $\exp(x)$ agrees with raising a number to a rational power? Jun 1 accepted Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ Jun 1 comment Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ That works, then. Don't get me wrong, the rest of your answer is great too, it's just not precisely what I had in mind when I thought of the question. Thanks! Jun 1 comment Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ I guess I was looking for answers that use mostly the definition of a limit together with some of $e^x$'s properties, but I'm not sure that's practical. Jun 1 comment Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ I'm seeing that I haven't really thought this through. I would say that the only definition of $e^x$ that makes this problem an interesting problem is the compound interest one, since the ones you suggest are related to the fact that $(e^x)' = e^x$. But then it seems that the problem reduces to showing that the derivative of $\lim_{n \to \infty}(1+\frac{x}{n})^n$ is equal to itself. Jun 1 comment Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ I hadn't really thought about how to define $e^x$, but I guess it would have to be $e^x = \lim_{n\to \infty}(1+\frac{x}{n})^n$, because if you use the other ones that were suggested (power series, differential equation), then the proof becomes almost trivial. I guess that complicates things, doesn't it? Jun 1 comment Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ @Thomas: Isn't using the power series representation of $e^x$ pretty much equivalent to saying that $(e^x)' = e^x$? Jun 1 asked Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ May 28 comment How do you explain paradoxes to non-mathematicians? Those are not paradoxes. May 26 comment How to show $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$ @Derrick: the idea is that we want to be left only with the circular and hyperbolic sines, so we use the identities I mentioned to replace the cosines with sines. I'm on my iPod so I can't really type out all the TeX, but that is the gist of it. May 26 revised How to show $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$ added 372 characters in body May 26 answered How to show $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$ May 20 accepted Confused about Wikipedia page on differential forms May 19 comment Confused about Wikipedia page on differential forms So, just to make sure I understand: Suppose we're working in $\mathbb{R}^2$, and instead of $x,y$ I call them $f_1, f_2$ for clarity, so they're functions from $\mathbb{R}^2$ to $\mathbb{R}$. Then we can say that $f_1(a, b) = a$ and $f_2(a,b)=b$?