zodiac
Reputation
300
Top tag
Next privilege 500 Rep.
Access review queues
 Feb20 comment What's your favorite proof accessible to a general audience? Another reason I like it - the more biased the coin is, the longer it takes to get one random bit out Sep30 awarded Explainer Jan19 answered Mathematically, why does $[ma]\mathrm{d}x = [mv]\mathrm{d}v$? Sep25 revised Why is integral of $(\tan x)^3$ not $((\sec x)^2)/2 - \ln(\sec x)$? add latex formatting Sep25 suggested approved edit on Why is integral of $(\tan x)^3$ not $((\sec x)^2)/2 - \ln(\sec x)$? Sep24 comment L'Hopsital Rule Understanding Sep24 comment L'Hopsital Rule Understanding That's not true, $f'/g'$ might be undefined while $\lim f'/g'$ exists. Sep24 comment L'Hopsital Rule Understanding Fair enough. But why doesn't your answer point out this limitation of your proof? I can comment on your answer instead if you want. Sep24 comment L'Hopsital Rule Understanding I think any proof or proof intuition for l'Hopital's rule must be roughly as complex as mine. Your proof shows that $\lim f/g = f'/g'$, but l'Hopital's rule is that $\lim f/g = \lim f'/g'$. The later is much harder to prove than the former. In fact, I think you should point out this limitation in your answer Sep24 comment L'Hopsital Rule Understanding Your proof doesn't work for applying l'Hopital's rule twice, your final expression is a ratio of derivatives without a limit in front. Sep24 comment L'Hopsital Rule Understanding Because if you apply l'Hopital's rule twice, you are eliminating the zeroth and first order terms, and focusing on the quadratic terms of the functions Sep24 comment L'Hopsital Rule Understanding then you need to talk about quadratic, cubic, quartic...etc approximations. I know the proof of taylor's theorem isn't very simple, but I find the fact that any nice function has a first, second, etc order approximation to be self-evident Sep24 comment L'Hopsital Rule Understanding I know, this is how I express the intuition that "the first-order term determines the limit" which I think of when using l'Hopital's rule. Sep24 answered L'Hopsital Rule Understanding Sep24 revised L'Hopsital Rule Understanding add latex formatting Sep24 suggested approved edit on L'Hopsital Rule Understanding Sep23 comment Show that $\gcd(a,b)=\operatorname{lcm}(a,b)$ if and only if $a=b$. I'm sure this will work, but sometimes these results about gcd and lcm are used as lemmas to prove the fundamental theorem of arithmetic Sep23 awarded Critic Sep23 comment Show that $\gcd(a,b)=\operatorname{lcm}(a,b)$ if and only if $a=b$. What are you allowed to assume? For example, if you are allowed to use the fact that every number has a unique prime factorization, then the proof is almost trivial (if it still isn't, I can write it as an answer) Sep23 revised To what extent are divisibility by different primes independent? replace \mod by \pmod and remove brackets