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Math Help - Problems with Spivak Calculus Chapter 6 Problem 6

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    Problems with Spivak Calculus Chapter 6 Problem 6

    I have the solutions book and see the answers, but they make no sense to me.
    Any/All help is appreciated.
    a) Find a function that is discontinuous at 1, 1/2, 1/4, 1/4... but continuous at all other points.
    b) Find a function that is discontinuous at 1, 1/2, 1/4, 1/4... and at 0 but continuous at all other points.

    Problem 7 doesn't seem to have enough explanation in the solution either. How is f being continuous 0 used?
    7) Suppose f satisfies f(x+y) = f(x) + f(y) and f is continuous at 0. Prove it is continuous at A for all values of A.
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    Lightbulb Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Attached Thumbnails Attached Thumbnails Problems with Spivak Calculus Chapter 6 Problem 6-discontinous-function.png  
    Last edited by MaxJasper; October 3rd 2012 at 09:39 PM.
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Well that certainly didn't help at all. Thanks, anyway.
    Thanks from Nervous
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    For 6a, you can also consider

    f(x)=\begin{cases}\frac{1}{k}& \frac{1}{2^k}<x\le \frac{1}{2^{k-1}}, k=1,2,\dots\\0&\text{otherwise}\end{cases}
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Problem #7: Ignoring issues of the domain, f continuous at a means: \lim_{x\to a} f(x) = f(a).

    Now observe that, always, \lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h).
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Quote Originally Posted by johnsomeone View Post
    Problem #7: Ignoring issues of the domain, f continuous at a means: \lim_{x\to a} f(x) = f(a).

    Now observe that, always, \lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h).
    I think that one also has to prove that f(0) = 0.
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Thanks, I managed something with 7, but I still don't understand 6. Is there any way to show how you construct an answer?
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Quote Originally Posted by patrickmanning View Post
    I still don't understand 6. Is there any way to show how you construct an answer?
    Can you construct a function that is discontinuous at x = 0? For example, f(x) = 0 for x <= 0 and f(x) = 1 for x > 0. If this is clear, then what is difficult about problem 6? I suggested a function whose graph looks like a staircase.
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Quote Originally Posted by emakarov View Post
    Can you construct a function that is discontinuous at x = 0? For example, f(x) = 0 for x <= 0 and f(x) = 1 for x > 0. If this is clear, then what is difficult about problem 6? I suggested a function whose graph looks like a staircase.
    In your example in this post, f(x) is really discontinuous at x = 0? It seems I don't even understand continuity properly. At the point where there is a jump, that point is discontinuous? Because it is different when approached from left and right sides, right?
    Wouldn't that mean in your first example, f(x) was also discontinuous at x 0?
    Last edited by patrickmanning; October 4th 2012 at 11:34 AM.
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Quote Originally Posted by patrickmanning View Post
    At the point where there is a jump, that point is discontinuous? Because it is different when approached from left and right sides, right?
    Yes. Would you agree that a person's spacial position as a function of time is a continuous function? If f(t) = 0 for t <= 0 and f(1) = 1 for t > 0 described my spacial position, then I would be at position 0 before time 0 and then suddenly move to position 1 (say, 1 mile from position 0). If you know how to construct such transportation, I would certainly like to hear!

    Formally, f(t) is not continuous at 0 because any interval containing t = 0 is mapped by f to a set that contains some y_1 and y_2 such that |y_1 - y_2| = 1. In other words, however small you make an interval around 0, the image of that interval contains numbers that are sufficiently far apart, i.e., not arbitrarily close. It does not matter how precisely you approximate t = 0: if you change t just a little bit, f(t) can change by 1.
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Thank you. I hate to be a bother, but can you continue helping by answering my question about your first example? Isn't f(x) discontinuous at x = 0 in your example with f(x) = 1/k?
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Quote Originally Posted by patrickmanning View Post
    sn't f(x) discontinuous at x = 0 in your example with f(x) = 1/k?
    As defined in post #4, f(x) = 0 for x\le 0. As far as x > 0, f(x) decreases as x decreases. In fact, f(x) tends to 0 as x tends to 0 from the right. Since \lim_{x\to0}f(x)=f(0), f is continuous at 0.
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Thank you. I understand well now. Your example actually is not discontinuous at 1/3 then. Changing it a bit to:
    f(x) = 1/b , 1/a < x <= 1/b where a = b+1 and b = 1, 2, 3... will fix that.
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Quote Originally Posted by patrickmanning View Post
    Your example actually is not discontinuous at 1/3 then. Changing it a bit to:
    f(x) = 1/b , 1/a < x <= 1/b where a = b+1 and b = 1, 2, 3... will fix that.
    I though that "1, 1/2, 1/4, 1/4..." from post #1 was supposed to mean "1, 1/2, 1/4, 1/8...," but you are right if it should be "1, 1/2, 1/3, 1/4..."
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    Re: Problems with Spivak Calculus Chapter 6 Problem 6

    Haha, you are right. I didn't realise I typed 1/4 twice. Thank you so much for everything. I understand this all now.
    For 6b, you can just change it to "otherwise, x = -1" and now x = 0 will be discontinuous as from the left, f(x) approaches -1 but from the right, f(x) still approaches 0.
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