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location Buenos Aires, Argentina
age 20
visits member for 3 years, 5 months
seen 3 hours ago

(my about me is currently blank)


Jul
25
comment Why are $\sin$ and $\cos$ (and perhaps $\tan$) “more important” than their reciprocals?
Fun fact: In some languages, in particular Spanish, sine still has the same not safe for work meaning. This has lead to uncountable repressed giggles in high school math class.
Jul
22
comment Compute integral $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$
@Matt: Isn't it simpler to just say $\mathrm{d}u = e^x\ \mathrm{d}x = u\ \mathrm{d}x \implies \frac{\mathrm{d}u}{u} = \mathrm{d}x$?
Jul
21
comment Compute integral $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$
After doing the change of variables, shouldn't it be $u^2$ instead of $e^{2u}$ in the numerator?
Jul
21
comment Motivation for definition of logarithm in Feynman's Lectures on Physics
I just saw in the feynmanlectures.info site that this has been added to the errata. Awesome!
Jul
20
comment Find all solutions, other than $2$ for $12x^3-23x^2-3x+2=0$
@Austin: You don't need to ask. You could just plug in those values and check whether they satisfy your equation.
Jul
19
comment Help solving differential equation: $y' = x\sqrt{4+y^{2}}/{y(9+x^{2})}$
The right side of the equality can be written as $\frac{\sqrt{4+y^2}}{y} \frac{x}{(9+x^2)}$. Does this help?
Jul
15
comment Motivation for definition of logarithm in Feynman's Lectures on Physics
I get it now, thank you! This paragraph is really weird, considering that the rest of the lectures is very well written and clear.
Jul
15
comment Motivation for definition of logarithm in Feynman's Lectures on Physics
@joriki: That really helps. Thanks!
Jul
15
comment Is the function $y=\ln x^2$ the same as $y=2\ln |x|$?
Short answer: yes.
Jul
14
comment Problems regarding exponents
@Rick: Yeah, I think I'll change it. Thanks for the advice!
Jul
7
comment I have to show $(1+\frac1n)^n$ is monotonically increasing sequence
Unless I'm missing something, that sequence is increasing, not decreasing.
Jun
30
comment Does this weird sequence have a limit?
That's interesting. Does this change if instead of picking a number from ${1,2,3,4,5,6}$ we choose a random real number, or maybe one from the interval $[0,1]$?
Jun
30
comment Does this weird sequence have a limit?
@anon: Making a needlessly complicated definition was sort of the point. Also, you don't necessarily have to choose $k$ randomly. You can start from $1$ and work your way up if you want; the point is not so much in what order the terms are calculated, but that you can calculate $a_k$ for any $k$ you want.
Jun
30
comment Does this weird sequence have a limit?
@AndréNicolas: What I mean if that $a_k$ has already been calculated, there is no need to roll the die again. We just look at the list and check what was the value of $a_k$.
Jun
23
comment If a function has a finite limit at infinity, does that imply its derivative goes to zero?
Also, a nitpick: shouldn't it be $x > 0$ instead of $x \ge 0$?
Jun
23
comment If a function has a finite limit at infinity, does that imply its derivative goes to zero?
This is the answer I like more, simply because you provided a function for which it is easy to check that it's a counterexample (just differentiate and take limits). Thanks!
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
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
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!