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Intro To real analysis problem

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data
    a) Find f ([0,3]) for the following function:
    f(x)=1/3 x^3 − x + 1

    b) Consider the following function :
    f(x) = e^(−ax) (e raised to the power of '-a' times 'x') a, x ∈ [0,∞)
    Find values of a for which f is a contraction .

    c) Prove that for all x,y ≤ 0 | 2^x −2^y | ≤ |x−y|
     
    Last edited: Apr 23, 2012
  2. jcsd
  3. Apr 23, 2012 #2

    HallsofIvy

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    Confuse about what? Do you know the definition of "f(A)" for f a function and A a set? It is: { f(x)| x in set A}. You might find it simplest to graph the function. The draw vertical lines at x= 0 and x= 3. Where those vertical lines cross the graph, draw horizontal lines to make a rectangle. f(A) is the set of all y values inside that rectangle.
     
  4. Apr 23, 2012 #3
    I think I have done part A properly. When you are taking f of a set, you are simply mapping each value in the set to another set right?
    for a, I got the set {1, 1/3, 5/3, 7}
    is this correct?

    For part B I am confused because, well to be honest im terrible at proofs ( you can imagine how this class has been going for me ). I know what a contraction is, I simply just do not know how to approach part B.

    For part C, I was able to get a little attempt going, but I seem to have gone astray. Once again the problem is Approaching the proof. My mind becomes all jumbled trying to approach this stuff.
     
  5. Apr 23, 2012 #4

    HallsofIvy

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    No, isn't! Those {f(0), f(1), f(2), f(3)}. That would be correct if were {0, 1, 2, 3}. It is not. "[0, 3]" means "the set of all real numbers from 0 to 3, inclusive".

    Part B does not ask for a proof! I'm glad you know what a contraction is. What is the precise definition of "contraction"? Typically in both problems and proofs, you can use the exact words of a degfinition. Specifically, to show that something is a "contraction" you show that it satisfies the definition.

    In (C) did you notice the comdition that x and y are both negative? What can you say about 2x and 2y when x and y are negative?
     
  6. Apr 23, 2012 #5
    Thank you for your reply!

    Well for part a, I did something similar in my notes ( I think ). Unless I took notes wrong, the professor took the min and max of the interval ( so 0 and 3 respectively ), solved f at those points, then took f prime, set it equal to zero, then solved for x?
    Is this a step in the right direction? Could be very wrong.


    Sorry you are right for part B, I do not know what I was thinking. Ok, so I know a contraction is defined as
    | f(x) - f(y) | <= n | x - y | For some 0 < n < 1
    So would I set
    f(x) = e^(-a_1x)
    f(y) = e^(-a_2x)
    Then use mean value theorum? If so how would I apply it. I seem to not be able to get past the definition.

    For part c I did infact recognize that x and y are negative. This means that 2^x and 2^y will be 0 < 2^x, 2^y <=1
    This also means that -1 < 2^x - 2^y < 1
    => 0 <= | 2^x - 2^y | < 1
    also | x - y | >= 0 <= | 2^x - 2^y |... which yeilds | x - y | >= | 2^x - 2^y |
    oh.... awkward.... :)
    figured it out while typing it in the forum :) I'll keep the results though. they are right? right?
     
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