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Math Help - real analysis question

  1. #1
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    real analysis question

    this is from ross's book elementary analysis: the theory of calculus section 21 #10. i tried reading the section and stuff but it seems like the problems are still hard for me. here are the problems:

    show that there exist continuous functions
    (a) mapping (0,1) onto [0,1]
    (b) mapping (0,1) onto R
    (c) mapping [0,1]U[2,3] onto [0,1]

    also, can someone give me tips on how i should study for this class? this class is the 2nd quarter of analysis i'm taking, and we are 4 weeks into the class. so i'd read the book and think i understand the material, only to get stumped on the questions... the midterm exam is on next week friday, so i'm getting really anxious... help!
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  2. #2
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    Quote Originally Posted by squarerootof2 View Post
    this is from ross's book elementary analysis: the theory of calculus section 21 #10. i tried reading the section and stuff but it seems like the problems are still hard for me. here are the problems:

    show that there exist continuous functions
    (a) mapping (0,1) onto [0,1]
    (b) mapping (0,1) onto R
    (c) mapping [0,1]U[2,3] onto [0,1]
    (a) Look at \sin(2\pi x) this takes the interval (0,1) to [-1,1], so can you do something to modify this function so the image of (0,1) is [0,1]?

    RonL
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  3. #3
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    Quote Originally Posted by squarerootof2 View Post
    this is from ross's book elementary analysis: the theory of calculus section 21 #10. i tried reading the section and stuff but it seems like the problems are still hard for me. here are the problems:

    show that there exist continuous functions
    (a) mapping (0,1) onto [0,1]
    (b) mapping (0,1) onto R
    (c) mapping [0,1]U[2,3] onto [0,1]

    also, can someone give me tips on how i should study for this class? this class is the 2nd quarter of analysis i'm taking, and we are 4 weeks into the class. so i'd read the book and think i understand the material, only to get stumped on the questions... the midterm exam is on next week friday, so i'm getting really anxious... help!
    For (b) consider somthing like \tan( \pi x -\pi/2)

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    (a) Look at \sin(2\pi x) this takes the interval (0,1) to [-1,1], so can you do something to modify this function so the image of (0,1) is [0,1]?

    RonL
    isn't the image of the sin (2pi*x) function already [0,1]? it seems like all the pts in [0,1] are hit already by putting in (0,1) for x? i dont see how we need to modify the function. hmm...
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by squarerootof2 View Post
    isn't the image of the sin (2pi*x) function already [0,1]? it seems like all the pts in [0,1] are hit already by putting in (0,1) for x? i dont see how we need to modify the function. hmm...
    try doing something like shrinking the period and shifting the graph
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    Quote Originally Posted by squarerootof2 View Post

    show that there exist continuous functions
    (a) mapping (0,1) onto [0,1]
    (b) mapping (0,1) onto R
    (c) mapping [0,1]U[2,3] onto [0,1]
    Actually we can construct ugly functions to satisfy requirements....
    For (c) , we can get onto mapping with \sin(\frac{\pi x}{2}) from [0,1] to [0,1]. Now define f([2,3]) = 1. This will make the function continuous, but the function does not look natural
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  7. #7
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    Quote Originally Posted by squarerootof2 View Post
    isn't the image of the sin (2pi*x) function already [0,1]? it seems like all the pts in [0,1] are hit already by putting in (0,1) for x? i dont see how we need to modify the function. hmm...
    \sin(2\pi x) achives a maximum value of 1 and a minimum of -1 in (0,1) .

    RonL
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    Quote Originally Posted by Jhevon View Post
    try doing something like shrinking the period and shifting the graph
    I would leave the period alone and shift the function upwards by 1 and shrink it by a factor of 2.

    RonL
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  9. #9
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    Quote Originally Posted by CaptainBlack View Post
    \sin(2\pi x) achives a maximum value of 1 and a minimum of -1 in (0,1) .

    RonL
    so i've been thinking , and would it work if i take the absolute value of sin (2pi*x) (just to make the function onto [0,1]. would that work?
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  10. #10
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    Quote Originally Posted by squarerootof2 View Post
    so i've been thinking , and would it work if i take the absolute value of sin (2pi*x) (just to make the function onto [0,1]. would that work?
    1. It would leave the function continuous, so that's OK

    2. It would map (0,1) to [0,1] so that's Ok

    so yes it would

    RonL
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