• Jeena@piefed.jeena.net
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    2 months ago

    I feel I should understand it, but it’s just outside of my reach. It’s now 10 years after university.

  • davidagain@lemmy.world
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    2 months ago

    I don’t think you can use the x0 plus minus delta in the bracket (or anywhere), because then the function that’s 1 on the rationals and 0 on the irrationals is continuous, because no matter what positive number epsilon is, you can pick delta=7 and x0 plus minus delta is exactly as rational as x0 is so the distance to L is zero, so under epsilon.

    You have to say that
    whenever |x - 0x|<delta,
    |f(x) - L|<epsilon.

    But I think this is one of my favourite memes.

    • affiliate@lemmy.world
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      2 months ago

      unless f(x0 ± δ) is some kind of funky shorthand for the set f(x) : x ∈ ℝ, x - x0 | < δ . in that case, the definition would be “correct”.

      it’s much more likely that it’s a typo, but analysts have been known to cook up some pretty bizarre notation from time to time, so it’s not totally out of the question.

      • davidagain@lemmy.world
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        2 months ago

        There’s notation for that - (x0 - δ, x0 + δ), so you could say
        f(x0 - δ, x0 + δ) ⊂ (L - ε, L + ε)

        • affiliate@lemmy.world
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          2 months ago

          that would be a lot clearer. i’ve just been burned in the past by notation in analysis.

          my two most painful memories are:

          • in the (baby) rudin textbook, he uses f(x+) to denote the limit of _f _from the right, and f(x-) to denote the limit of f from the left.
          • in friedman analysis textbook, he writes the direct sum of vector spaces as M + N instead of using the standard notation M ⊕ N. to make matters worse, he uses M ⊕ N to mean M is orthogonal to N.

          there’s the usual “null spaces” instead of “kernel” nonsense. ive also seen lots of analysis books use the → symbol to define functions when they really should have been using the ↦ symbol.

          at this point, i wouldn’t put anything past them.

  • affiliate@lemmy.world
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    2 months ago

    i still feel like this whole ε-δ thing could have been avoided if we had just put more effort into the “infinitesimals” approach, which is a bit more intuitive anyways.

    but on the other hand, you need a lot of heavy tools to make infinitesimals work in a rigorous setting, and shortcuts can be nice sometimes

    • someacnt_@lemmy.world
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      2 months ago

      Infinitesimal approach is often more convoluted when you perform various operations, like exponentials.

      Instead, epsilon-delta can be encapsulated as a ball business, then later to inverse image check for topology.

      • affiliate@lemmy.world
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        2 months ago

        i think the ε-δ approach leads to way more cumbersome and long proofs, and it leads to a good amount of separation between the “idea being proved” and the proof itself.

        it’s especially rough when you’re chasing around multiple “limit variables” that depend on different things. i still have flashbacks to my second measure theory course where we would spend an entire two hour lecture on one theorem, chasing around ε and η throughout different parts of the proof.

        best to nip it in the bud id say

    • Binette@lemmy.ml
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      2 months ago

      Not an advanced mathematician, but I think it’s just saying that f(x-delta) between f(x + delta) is going to give a value between L - epsilon and L + epsilon.

        • Binette@lemmy.ml
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          2 months ago

          Imagine you have a simple function: y = 2x

          If you have two different x values (let’s say 2 and 4), there exist a y value for every number in between them.

          In this example, the y is going to be in between 4 and 8, for every x in between 2 and 4.

  • emergencyfood@sh.itjust.works
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    2 months ago

    Not a mathematician, but I’m pretty sure this isn’t necessarily true. What if L is -1 and f(x) = x^2? Also I think your function has to be continuous.

    • davidagain@lemmy.world
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      2 months ago

      You’re right on all three counts. It’s not always true, f(x0) has to be L, and the function has to be continuous.