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swissgirl

  • 3 years ago

@KingGeorge

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  1. swissgirl
    • 3 years ago
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  2. swissgirl
    • 3 years ago
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    Its my weekly question :)

  3. swissgirl
    • 3 years ago
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    If you dont like this one I can give you another one

  4. KingGeorge
    • 3 years ago
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    It's been a little while since I've done actual analysis :P

  5. swissgirl
    • 3 years ago
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    Ya so let me get you smth easier

  6. KingGeorge
    • 3 years ago
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    Well, the first part is pretty easy. It's basically just the definition of "not uniformly continuous"

  7. swissgirl
    • 3 years ago
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    BTW Congratulations. Where is ur offer from?

  8. swissgirl
    • 3 years ago
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  9. KingGeorge
    • 3 years ago
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    U 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 :(

  10. swissgirl
    • 3 years ago
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    awwwwww :( That is sad They are stupiidddddd

  11. KingGeorge
    • 3 years ago
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    Ah. 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.

  12. KingGeorge
    • 3 years ago
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    Let'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.

  13. KingGeorge
    • 3 years ago
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    However, I cannot immediately remember how we proved that if \(f(a)f(b)\le0\), then there was some \(c\in[a,b]\) s.t. \(f(c)=0\).

  14. swissgirl
    • 3 years ago
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    hmmm gonna check my notes

  15. swissgirl
    • 3 years ago
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    hmmm I found a proof in my book using the supremum

  16. KingGeorge
    • 3 years ago
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    Well, 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!

  17. swissgirl
    • 3 years ago
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    Thanks @KG :)

  18. swissgirl
    • 3 years ago
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    Good night

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