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location Buenos Aires, Argentina
age 21
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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 $f_2(a,b)=b$?
May
17
asked Confused about Wikipedia page on differential forms