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Little o function

  1. Oct 12, 2011 #1
    "Little o" function

    Claim: et= 1 + t +o(t)

    Proof:
    et = 1 + t + t2/2! + t3/3! +...
    Let g(t)=t2/2! + t3/3! +...
    g(t) is o(t), thus et= 1 + t +o(t).
    =================

    I don't understand why g(t) is o(t).
    Why is it true that
    lim g(t)/t = 0 ?
    t->0

    I know that for example lim[f(x)+g(x)]=lim f(x) + lim g(x), so if lim f(x)=0 and lim g(x)=0, then
    lim[f(x)+g(x)] = 0+0 = 0. I believe this property is true only when computing the limit of a sum of a FINITE number of functions.

    But g(t)/t above is an infinite sum; it has an infinite number of terms. How can we compute and prove that the limit of g(t)/t is 0? I know each term goes to 0, but we are summing an INFINITE number of terms, so how can you be sure that the limit is 0?

    Any help/explanations would be much apprecaited!
     
  2. jcsd
  3. Oct 12, 2011 #2

    Char. Limit

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    Re: "Little o" function

    Well, the thing about our "infinite function" is that it can be rewritten in closed form like so:

    [tex]g(t) = \sum_{n=2}^\infty \frac{t^n}{n!}[/tex]

    Now if this is g(t), then g(t)/t is like so:

    [tex]\frac{g(t)}{t} = \frac{1}{t} \sum_{n=2}^\infty \frac{t^n}{n!} = \sum_{n=2}^\infty \frac{t^{n-1}}{n!}[/tex]

    Now it's easy to see that for all n>1, t^(n-1)/n! tends to zero as t tends to zero. So therefore, every term in the sum tends to zero, and the sum itself thus tends to zero.
     
  4. Oct 13, 2011 #3
    Re: "Little o" function

    I am having some trouble with this. If the upper limit of the sum is, say, 10, (i.e. a finite series where we are summing up a FINITE number of functions) then I can understand that the limit of the sum would be 0. There is a theorem/property of limits that guarantees this, which is lim[f(x)+g(x)]=lim f(x) + lim g(x) when the limits of the RHS both exist.
    so if lim f(x)=0 and lim g(x)=0, then lim[f(x)+g(x)] = 0+0 = 0.

    But for g(t)/t here, we are summing up an INFINITE number of terms that tend to 0, how can we justify or even rigorously prove that
    lim g(t)/t = 0 ?
    t->0

    Thanks for the help!
     
  5. Oct 13, 2011 #4

    disregardthat

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    Re: "Little o" function

    That's not necessarily true for series not converging absolutely, I believe.
     
  6. Oct 13, 2011 #5

    Char. Limit

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    Re: "Little o" function

    Then prove that it converges absolutely. I'm almost certain that this particular series does.
     
  7. Oct 13, 2011 #6

    disregardthat

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    Re: "Little o" function

    Surely this series does, since the terms are always of the same sign for positive t (which proves it for negative t as well). I was pointing out that the general statement does not hold.
     
  8. Oct 14, 2011 #7
    Re: "Little o" function

    OK! Now I have another question.

    Definition: A function f is called o(h) if
    lim [f(h)/h] = 0.
    h->0

    I understand what o(h) means, but in one of the examples from my textbook an o(√h) pops up, but I don't understand the meaning of o(√h).

    Do we say f is o(√h) if
    lim [f(h)/√h] = 0?
    √h->0

    or do we say f is o(√h) if
    lim [f(√h)/√h] = 0?
    √h->0

    Thanks for explaining!
     
  9. Oct 14, 2011 #8

    disregardthat

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    Re: "Little o" function

    The first alternative, but as h --> 0.
     
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