Results 1 to 4 of 4

Math Help - HELP!! thankx

  1. #1
    Newbie pc31's Avatar
    Joined
    Jun 2008
    From
    Maine, US
    Posts
    12

    HELP!! thankx

    3 questions:
    1) let f: (a,b) -> (c,d) be continuos surjective monotonic function. prove f is invertible and its inverse is continuous

    2) f: [0,1] -> R
    f(x) = 0 if x is irrational;
    f(x) = 1/n if x is rational with x = m/n and gcd(m,n) = 1
    prove that f is riemann integrable and its integral from 0 to 1 is 0

    3) let f(x) < g(x) < h(x) for all x in [a,b]
    suppose f and h are riemann intagrable and integral of f from 0 to 1 equals integral of h from 0 to 1
    prove g is riemann intagrable and integral of g from 0 to 1 equals integral of f from 0 to 1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by pc31 View Post
    3 questions:

    2) f: [0,1] -> R
    f(x) = 0 if x is irrational;
    f(x) = 1/n if x is rational with x = m/n and gcd(m,n) = 1
    prove that f is riemann integrable and its integral from 0 to 1 is 0
    Before we get started lets note a few things. f is bounded above on [0,1] by 1. There is an irrational and rational point in every interval of postive length.

    Let \epsilon > 0 Pick an N \in \mathbb{N} such that \frac{1}{N} < \frac{\epsilon }{4}.

    Now lets construct a partion of [0,1]. let n > N

    A=\left[ 0,\frac{1}{n}\right]

    B=\bigcup_{i=2}^{n}\left[ \frac{1}{i}-\frac{\epsilon}{2^{i+1}},\frac{1}{i}+\frac{\epsilo  n}{2^{i+1}}\right]

    let C=\left[ 1-\frac{1}{n},1\right]

    Now set P=[A \cup B \cup C]


    |U(f,P)-L(f,P)|
     < \left|1 \cdot \left[ \frac{1}{n}-0 \right]+\sum_{i=2}^{n} 1 \cdot \left( \frac{1}{n} + \frac{ \epsilon}{2^{i+1}} - \frac{1}{n} + \frac{\epsilon}{2^{i+1}} \right) +1 \cdot \left( 1 -1+\frac{1}{n} \right) \right|

    =\left| \frac{2}{n}+\epsilon \sum_{i=2}^{n}\left( \frac{1}{2^i}\right) \right|<2\cdot \frac{\epsilon}{4}+\frac{\epsilon}{2}=\epsilon

    So f is integrable on [0,1]. Since it is integrable the value of the integral is equal to the supreemum of the lower integrals. Since it is always zero the value of the integral is 0.

    Good luck.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    For 3) use the squeeze theorem.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by pc31 View Post
    3 questions:
    1) let f: (a,b) -> (c,d) be continuos surjective monotonic function. prove f is invertible and its inverse is continuous
    First note, that if f is a constant function, we have the result right off the bat, so lets assume it is not constant, that is, it is increasing or decreasing. Now, a function is invertible if it is both 1-1 and onto. we are given that f is onto, thus, to prove it is invertible, it remains to show it is 1-1. we will do this using the definition: A function f is 1-1 if x_1 \ne x_2 \implies f(x_1) \ne f(x_2) for x_1,x_2 \in \text{dom}(f).

    Assume x_1 \ne x_2. Then, we may assume, without loss of generality, that x_2 > x_1, and that f is an increasing function (if this is not the case, we can easily assume otherwise and the proof remains valid). Since f is montone, x_2 > x_1 implies that f(x_2) > f(x_1). But that means f(x_1) \ne f(x_2), as desired.

    Now to prove continuity of the inverse function (i think this works ).

    Let \{ x_n \} be a convergent sequence in \text{dom}(f). and assume \lim_{n \to \infty} f(x_n) = L. this is possible by the continuity of f. by the Heine definition of continuity, we need to show that \lim_{n \to \infty} f^{-1}(f(x_n)) = f^{-1}(L). But that is, \lim_{n \to \infty} x_n = f^{-1}(L). this follows by our original assumption and the continuity of f.




    (Note that the domain of the inverse function is the codomain of the function. thus, for x in the domain of f, it makes sense to write f(x) to represent an element in the domain of f^{-1}(x))
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum