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swissgirl
 one year ago
Best ResponseYou've already chosen the best response.0Its my weekly question :)

swissgirl
 one year ago
Best ResponseYou've already chosen the best response.0If you dont like this one I can give you another one

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1It's been a little while since I've done actual analysis :P

swissgirl
 one year ago
Best ResponseYou've already chosen the best response.0Ya so let me get you smth easier

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1Well, the first part is pretty easy. It's basically just the definition of "not uniformly continuous"

swissgirl
 one year ago
Best ResponseYou've already chosen the best response.0BTW Congratulations. Where is ur offer from?

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1U of Rochester. I'm still waiting to hear back from most schools I applied to. Unfortunately, I also just recently got a rejection from Princeton :(

swissgirl
 one year ago
Best ResponseYou've already chosen the best response.0awwwwww :( That is sad They are stupiidddddd

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1Ah. This new one I can do. It's just the IVP (intermediate value theorem). To be honest, I wasn't expecting them to accept me.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1Let's see how much I remember. Last time proved this, we started with the assumption that \(f(a)f(b)\le 0\). In this particular case, if \(f(a)=0\) or \(f(b)=0\), we're done. If \(f(a)f(b)<0\), then one of \(f(a),f(b)\) is negative, and the other positive. At this point, we had another theorem that proved there was then some \(c\in[a,b]\) s.t. \(f(c)=0\). Then define \(g(x)=f(x)+y\). Then, from what we just proved, it immediately follows that \(g(c)=y\). If we started with \(g(b)\le y\le g(a)\), take \(f(x)=g(x)y\), so \(f(b)\le 0 \le g(a)\). From this we get that \(f(b)f(a)\le 0\) and we're done.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1However, I cannot immediately remember how we proved that if \(f(a)f(b)\le0\), then there was some \(c\in[a,b]\) s.t. \(f(c)=0\).

swissgirl
 one year ago
Best ResponseYou've already chosen the best response.0hmmm gonna check my notes

swissgirl
 one year ago
Best ResponseYou've already chosen the best response.0hmmm I found a proof in my book using the supremum

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1Well, I'm sorry, but I've got to go now. I need to get up early tomorrow, so I need to head to bed. Have a good night, and good luck with this!
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