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Unlimited function of set E

  1. Oct 30, 2008 #1
    Здравствуйте!

    How will I prove that some function f(x) is unlimited of some set E[itex]\subseteq[/itex]Df, where Df is the domain of the function, or the values of x, which can be used in the function.
    For example:

    How will I prove that this function f(x)=x+2 is unlimited for E=R (set of real numbers)?

    Спасибо за помощь!
     
  2. jcsd
  3. Oct 30, 2008 #2

    HallsofIvy

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    By "unlimited" I presume you mean what I would call unbounded "unbounded".
    Pretty much any time you are asked to prove "f is ****", where **** is some word, you prove it by showing that f satisfies the definition of ****.

    A function is unbounded if it is both unbounded above and unbounded below.

    A function is unbounded above if, given any number Y, there exist x such that f(x)> Y.
    A function is unbounded below if, given any number Y, there exist x such that f(x)< Y.


    Given any Y, is there an x such that x+ 2> Y?

    Given any Y, is there an x such that x+ 2< Y?

    For that simple function, it is just a matter of solving the inequalities.
     
  4. Oct 30, 2008 #3
    I think you're right, but also in my book there is |f(x)|>Y, and that means f(x)>Y and -f(x)>Y, or f(x)<-Y.

    So x+2>Y and
    x+2<-Y

    x>Y-2

    Let's say x=Y-1, so:

    [tex]|f(x)|=|x+2|=|Y-1+2|=|Y+1|=Y+1>Y[/tex], for every [itex]Y \geq 0[/itex].

    x<-2-Y

    x=-3-Y

    [tex]|f(x)=|x+2|=|-3-Y+2|=|-Y-1|=-Y-1<-Y[/tex], for every [itex]Y \geq 0[/itex]

    So |f(x)|>K , for [itex]K \in \mathbb{R}[/itex].

    Did I solved it correctly?
     
    Last edited: Oct 30, 2008
  5. Oct 30, 2008 #4

    HallsofIvy

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    Well, yes, but those are equivalent definitions. If you can find a number X so that f(x)> X and a number Z so that f(x)< Z, take Y to be the larger of |X| and |Z|.

    So x+2>Y and
    x+2<-Y

    x>Y-2

    Let's say x=Y-1, so:

    [tex]|f(x)|=|x+2|=|Y-1+2|=|Y+1|=Y+1>Y[/tex], for every [itex]Y \geq 0[/itex].

    x<-2-Y

    x=-3-Y

    [tex]|f(x)=|x+2|=|-3-Y+2|=|-Y-1|=-Y-1<-Y[/tex], for every [itex]Y \leq 0[/itex]

    So |f(x)|>K , for [itex]K \in \mathbb{R}[/itex].

    Did I solved it correctly?[/QUOTE]

    Yes.
     
  6. Oct 30, 2008 #5
    Just to correct, one mistake of mine. It is [itex]Y \geq 0[/itex], down on the second function written with LateX.

    Also I think that in my book, the autors use |f(x)|>K (for unbounded functions), because |f(x)|<K is bounded of it (the opposite one, vice versa).

    But also your statement is also good. f(x)>K and f(x)<K. Also in some cases maybe it is better to use [itex]|f(x)| \leq K [/itex].
     
  7. Oct 31, 2008 #6
    Sorry, for posting in same topic. I did not know if I need to open new topic.
    I want to ask you about this problem.
    The problem is to prove that f(x)=x+sinx ; E=R

    So, |f(x)|>Y

    x+sinx>Y

    x=Y+2

    Y+2+sin(Y+2)>Y

    sin(Y+2)>-2

    Which is correct.

    So |f(x)|=|x+sinx|=|Y+2+sin(Y+2)|=Y+2+sin(Y+2)>Y

    Is this good?
     
  8. Oct 31, 2008 #7

    Office_Shredder

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    While the order in which you listed things makes it slightly confusing to see what you are doing, it looks like the right idea is there
     
  9. Oct 31, 2008 #8

    HallsofIvy

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    do you mean you want to prove that f(x) is unbounded?

    Which is certainly NOT correct! "x+ sin x> Y" does NOT imply x= Y+2! I think what you meant to say was that [itex]-1\le sin x\le 1[/itex] for all x so [itex]x- 1\le x+ sin x\le x+ 1.
    Now, given in Y, if x> Y+1, x+ sin x> Y+1- 1= Y if x< Y+1, x+ sin x< Y+1- 1= Y.


     
  10. Oct 31, 2008 #9
    If [tex]-1\leq sinx \leq 1[/tex], then [tex]-1+x\leq sinx+x\leq 1+x[/tex].

    So if x+sinx > x-1 >Y; so x-1>Y ; x>Y+1

    x=Y+2

    |f(x)|=|x+sinx|=Y+2+sin(Y+2)>Y;

    I think we got the same thing. Why you think it is not correct?
     
  11. Oct 31, 2008 #10

    Office_Shredder

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    I think you got it backwards, reading it I interpreted it as intending to say x=Y+2 implies x+sinx>Y (hence proving the function is unbounded)
     
  12. Nov 1, 2008 #11

    HallsofIvy

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    Either way, there is no need to connect the argument of the sine with Y at all.
     
  13. Nov 2, 2008 #12
    Ok, we got: x+sinx > x-1 >Y; so x-1>Y ; x>Y+1

    But where does this x+ sin x> Y+1- 1= Y if x< Y+1, x+ sin x< Y+1- 1= Y comes from?

    Thanks in advance.
     
  14. Nov 2, 2008 #13

    HallsofIvy

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    That was a typo on my part. What I meant to say was "If x< Y- 1, then x+ sin x< Y-1+ sin x and, since [itex]sin x\le 1[/itex], [itex]x+ sin x\le Y-1+ 1= Y[/itex] so that Y is not a lower bound for x+ sin x.
     
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