Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limited strings

  1. Sep 12, 2008 #1
    To prove that some function is limited we use |an|[itex]\leq K (n \in \mathbb{N})[/itex] where K = max { |m| , |M| }

    What this last part mean K = max { |m| , |M| } ?

    Thank you.
    Last edited: Sep 12, 2008
  2. jcsd
  3. Sep 12, 2008 #2
    well, i think that |m| and |M| are the absolute minimum and max, that the function reaches in a certian interval. So if it is limited we choose the max of these two, so we can assure ourselves that our function will fall between these two.

    It is obvious that m<M but since we are talking about abs values, then it is possible that

    |m|>|M| that's why we choose the max of their abs values, since this way we are saying that

    -max(|M|,|m|)< a_n<max(|m|,|M|)
  4. Sep 12, 2008 #3
    but why abs? is { |m|, |M| } some kind of point?
  5. Sep 12, 2008 #4
    i don't know what you exactly mean when you say is { |m|, |M| } some kind of point?

    We say that a function f is bounded on an interval, call it I, if there exists a K>0 such that

    |f(x)|<K for every x in I. This means that f(x) will lay between -K and K that is -K<f(x)<K for all x's in I. Now, asuume that f is continusous in a closed interval [a,b], so it means that in this interval the function reaches it's max and min values. That is there are two points c, d, on the interva. (a,b) such that f(c)=m, and f(d)=M. What this is telling us is that

    [tex] m\leq f(x)\leq M[/tex] for all x in [a,b]. This tells us that the function is bounded. Now let K=max{|m|,|M|} we take abs in here since our function might as well be negative, so

    [tex] |f(x)|\leq K=>-K\leq f(x)\leq K[/tex] But K is either |m| or |M| whichever of these is greater in abs value.

    Say for example that both m and M are positive, then what this is telling us is that for all x's in [a,b] -M<f(x)<M, but if say both m and M are negative then |m|>|M| so this is telling us that

    -|m|<f(x)<|m| for all x
    s in [a,b]
  6. Sep 12, 2008 #5
    so K = max{ |m|, |M| } , means the maximum value of this two in abs ?

    I think I understand you know. And what about this string:


    [tex]|a_n| \leq K [/tex]

    [tex]|a_n|=|\frac{2n+1}{n}|=|2+\frac{1}{n}| > 2[/tex]

    What should I do with this one (this is example, not homework question)

    Thank you.
  7. Sep 12, 2008 #6
    well, now

    [tex]|a_n|=|\frac{2n+1}{n}|=|2+\frac{1}{n}|\leq 3[/tex]

    because when n=1 then a_n=3, and for any n greater than 1 we will get sth that is smaller than 3, so it means that a_n is bounded by 3 for any n.

    P.S. these are sequences, not functions. Well, in a sense they are, but still.
  8. Sep 13, 2008 #7
    So [tex]-3 \leq a_n \leq 3[/tex], like this?
  9. Sep 13, 2008 #8
    And, what about:



    [tex]|a_n|=|\frac{(-1)^n}{n}| \leq K [/tex]

    How will I prove for this one?
  10. Sep 13, 2008 #9

    [tex]|a_n|=|\frac{(-1)^n}{n}|=\frac{|(-1)^n|}{|n|}=\frac{1}{|n|}\leq 1[/tex]

    [tex]|(-1)^n|=1[/tex] because if we take n even, then it is obviously poz. but also if we take n odd, then (-1)^n=-1 but the abs value of -1 is still 1, so that makes it valid.

    NOte , we can do this only when the sequence a_n converges, other wise we would not be able to do this all the time.
  11. Sep 13, 2008 #10
    And what about this:


    [tex]|a_n|=|\frac{(-1)^n+1}{n}|=|\frac{(-1)^n}{n}+\frac{1}{n}|=|\frac{(-1)^n}{n}|+|\frac{1}{n}|=\frac{1}{|n|}+|\frac{1}{n}|=\frac{2}{|n|} \leq 1[/tex]

    Like this?
  12. Sep 13, 2008 #11
    Well u did a couple of mistakes in there, compare it to this

    [tex]|a_n|=|\frac{(-1)^n+1}{n}|=|\frac{(-1)^n}{n}+\frac{1}{n}|\leq|\frac{(-1)^n}{n}|+|\frac{1}{n}|=\frac{1}{|n|}+|\frac{1}{n}|=\frac{2}{|n|} \leq 2[/tex]
  13. Sep 14, 2008 #12
    Ok, thank you very much. In this case [tex]\frac{1}{|n|}=|\frac{1}{n}|[/tex], right?
  14. Sep 14, 2008 #13
    Yup, these are some of the properties of the abs value, and they can be proved.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook