# Dystopian

less info
reputation
5
bio website location age member for 1 year seen Jul 23 '12 at 12:21 profile views 3

# 24 Actions

 Jul7 comment Quick question about comparing functions.I can sure have $h(x)=f(x)-g(x)$ but the problems is, I wouldn't know in what interval would it be nonnegative or nonpositive. It's more interesting and convenient if I'd know if it is nonnegative or nonpositive when I $\underline{already}$ have an interval. Jul7 awarded Supporter Jul7 awarded Scholar Jul7 awarded Commentator Jul7 accepted Quick question about comparing functions. Jul7 comment Quick question about comparing functions.Oh yeah, I've forgotten that stipulation. Many thanks, you just saved me from all that strenuous graphing work. Jul7 comment Quick question about comparing functions.Oh, yeah I'm getting what you're trying to say. This is similar to the first derivative test. From what I understood: If the derivative of f is greater than g for all values of x in the given interval, then we could say that the values of f is greater than g in the given interval. Is that it? Jul7 asked Quick question about comparing functions. May26 comment Conventions for antiderivatives notation.Oh thanks, I guess it's just the notation. For some time, I've wondered how could mathematicians come up with such brilliant notations. I just came around asking; maybe I got it all wrong in my head. May26 asked Conventions for antiderivatives notation. May13 comment Proving a theorem on limits that approach infinity.Yeah, I was too focused on getting a value for delta. I'm not very comfortable with assuming a value; but it works anyways. May13 revised Proving a theorem on limits that approach infinity.There shouldn't be a zero in -x<(delta) May13 asked Proving a theorem on limits that approach infinity. May13 comment About limit theoremsI mean $|(f(x)-K)|+|(g(x)-L)|<\epsilon$ Latex is making me dizzy. May13 comment About limit theoremsDoes that mean $|f(x)+g(x)-(K+L)|<|(f(x)-K)|+|g(x)-K)|$ Thus $|f(x)+g(x)-(K+L)|<\epsilon$ since $|(f(x)-K)|+|g(x)-K)|=\epsilon$ May13 comment About limit theoremsIt's ambiguous. May12 comment About limit theoremsRight. Last one though, for those who actually looked at the link, how did we get $<\frac{1}{2}\epsilon + \frac{1}{2}\epsilon$ then $=\epsilon$? May12 revised About limit theoremsadded 73 characters in body May12 comment About limit theoremsWow, that's pretty cool. I still can't think of it in general terms though, since the example was quite specific but that certainly did help. Thanks. May12 awarded Student