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Baby rudin, Limit of nth root of p

  1. Jul 4, 2012 #1
    Part (b) of theorem 3.20 is to prove that the limit as n approaches infinity of the nth root of p equals one(for p>0). The proof given in the text uses some inequality derived from the binomial theorem which seems to me to just come out of nowhere and provide a completely unintuitive proof. That's great that this inequality works but how would I know to use this specific inequality in the first place? Could someone either provide a more explanatory proof which perhaps does not use the binomial theorem, or clarify the proof in the text so that it doesn't seem so unnatural?
     
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  3. Jul 4, 2012 #2

    Zondrina

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    Recall that the nth root can be re-written like so :

    [itex]\sqrt[n]{p}[/itex] = p1/n

    Recall further that the lim n-> ∞ 1/n = 0 so we know that lim n-> ∞ [itex]\sqrt[n]{p}[/itex] = 1 since anything to the power of zero is one.

    Something you could do alternatively is use the M, N definition of a limit going to infinity.

    That should help.
     
  4. Jul 4, 2012 #3

    Integral

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    Perhaps you need to spend a little more time staring at that proof.
    [tex] ( 1 + x_n)^n=p [/tex] is a algebraic restatement of [tex] x_n = {^n\sqrt{p}}-1 [/tex]

    The RHS of the inequality is simply the first 2 terms of the exansion of [tex] ( 1 + x_n)^n[/tex] . Since he has dropped many postitive terms what is left is smaller then the full expansion, thus the inequality.

    The different expansions need to be a tool on your mathematical belt, get used to it and learn to use them.

    I observed this fact within a few months of picking up my first scienctic calculator. No matter where I started, repetive square root operations ended up at 1.
     
    Last edited: Jul 4, 2012
  5. Jul 4, 2012 #4

    Bacle2

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    .

    This is a good idea, but it needs the assumption of continuity; I don't have Rudin--

    nor his book ;) --with me, I don't know if he proved continuity of x1/n

    by this point. I don't see how else you can justify f(lim n→∞)=limn→∞

    applied to x1/n
     
  6. Jul 4, 2012 #5
    Integral,

    I think you are right that I need to spend more time staring at this proof(which I have been doing for quite a while now).

    I have no problem understanding this proof in that I can follow the algebra. However I don't understand how one could come up with this proof from scratch. Often in order to truly understand a proof, you have to not just be able to follow the reasoning but also understand how it was thought that this type of reasoning should be applied in the first place. I have stared at this proof for so long that I can easily replicate it and understand why it must be true(because the equation provides the desired result) but if I was faced with a different problem requiring the same technique I would be lost. Is there some sort of "scratch work" which is left out here which would lead one to this proof?
     
  7. Jul 4, 2012 #6

    Bacle2

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    I don't know if this is what you are looking for, but a more general approach I can think

    of is using the fact (check) that the sequence x1/n (for fixed x, of course) is

    monotone and bounded, so that it must converge --to its lub. Re inequalities, I don't

    see how to avoid just learning a bunch of them, as using inequalities is the bread-and-

    butter of Standard Analysis.
     
  8. Jul 4, 2012 #7

    Mute

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    Figuring out how to do a proof 'from scratch' comes from experience and having tried lots of different things (that didn't pan out).
     
  9. Jul 8, 2012 #8
    This proof is easy, use the inf of the set. The proof involveing n^(1/n) is the one that involves the binomial theorem.
     
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