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Confused about supremum

  1. Jun 4, 2008 #1
    Let 0 < r < 1. Then [tex]\sup_{x\in[-r,r]}f(x)=f(r)}[/tex], right? However, the text I'm reading says it's [tex]f(-r)[/tex]. How could this be? For example, say r = 0.5, then the least upper bound of [-0.5, 0.5] is 0.5, or r, right? I don't see how it could be -r. Thanks for any help.
  2. jcsd
  3. Jun 4, 2008 #2


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    Hi AKBAR! :smile:

    That will be true if f is increasing in [-r,r], but not in most other cases.

    What is f? What is the relevance of 0 < r < 1? What is the context? :confused:
  4. Jun 4, 2008 #3


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    You are taking the supremum ("least upper bound") of the values of f not x!

    For example, if f(x)= x, then f(-r)= -r, f(r)= r and f takes on all values between -r and r. In that case the sup of f(x) on the interval (not the sup of the interval itself) is f(r).

    But if f(x)= -x, then f(-r)= r, f(r)= -r and now the supremum occurs at -r: f(-r) is the largest value (in fact it is the maximum value).

    And it can get more complicated than that. If f(x)= -x2, then f(-r)= f(r)= -r2< 0. The maximum (and so sup) occurs in the middle of the interval. The sup is f(0)= 0.
  5. Jun 4, 2008 #4
    Ahhhh....that really clears things up. Thank you guys.

    The function in question was [tex]f(x)=\frac{(n-1)!}{(1+x)^n}[/tex]. So [tex]\sup_{-r\leq x\leq r}\frac{(n-1)!}{(1+x)^n}=\frac{(n-1)!}{(1-r)^n}[/tex] The r comes from the radius of convergence of a Taylor expansion (I'm reading about where T(x) = f(x) ).

    Thanks again for the help.
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