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Proving Continuity

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine a function which is discontinuous at 1,1/2,1/3...and/not 0, but continuous elsewhere.



    2. Relevant equations



    3. The attempt at a solution
    I figure for the "not zero" part, I would do
    f(x) = {x, x = 1/n where n is a natural number
    {0, x =/= 1/n where n is a natural number

    The thing is, I don't know how to prove it (because I just thought of it based on the function where f(x) = x for x is rational and = 0 for irrational), or whether it even works.

    Then for the "and zero" part, I'm completely stumped. Can someone lead me in the right direction please?

    EDIT: Sorry for the misleading title, but the people who evaluate the work are really strict...and last time I had to show proof for something when the question asked to simply state something...

    EDIT: Am I allowed to just say
    f(x) = {1/0 for x = 1/n, where n is an element of natural numbers
    {0 otherwise ?
     
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 10, 2009 #2

    Mark44

    Staff: Mentor

    I think your idea works, with a little modification.

    f(x) = [1/x, for x [itex]\in[/itex] {1, 2, 3, ... }
    .........[x, otherwise

    You can't say f(x) = 1/0, under any circumstances, if that's indeed what you meant to say.
     
  4. Oct 10, 2009 #3
    How do I begin to proving that? Should I prove that since the limit is 0 for x =/= to 1/x, there is no limit for the other values of x?

    And how do I account for the 0?

    f(x) = {0, x = 0
    .........{1/x, x element of natural numbers
    .........{x, otherwise ?
     
  5. Oct 10, 2009 #4

    Mark44

    Staff: Mentor

    Your problem statement said to "determine" a function that met the given criteria. It didn't say anything about proving anything.
    You need only two cases: one for x in the natural numbers (i.e., 1, 2, 3, ...), and the other for all other numbers. "Otherwise" includes 0.
     
  6. Oct 10, 2009 #5
    I will point out the functions that are being cooked up are discontinuous at the natural numbers, but the original challenge was for the function to be discontinuous at x = 1, 1/2, 1/3, 1/4,.... , but continuous at x = 0 and at all other numbers (I assume for all other real numbers).

    I have such a function in mind, but the real problem is that the function sought has an infinite number of discontinuities near 0, but how do you engineer it so that it is continuous at 0?

    --Elucidus
     
  7. Oct 10, 2009 #6
    I should probably change it to:
    f(x) = x, x=1/n where n is a natural number.
    f(x) = 0, otherwise
    Would this then ascertain that the function is discontinuous at 1,1/2,1/3,1/4...while continuous at the other values?

    I was thinking of this based on the other function:
    f(x) = 0 x is irrational
    f(x) = 1/q where x = p/q in lowest form
    Doesn't this function have a limit of 0 at the irrational numbers while it doesn't have a limit at the other numbers?

    Moreover, I indicated that I needed to find a function where it is discontinuous at zero as well (on top of 1/(natural numbers)). So in this case, would I need to add a third condition to the function?
     
  8. Oct 11, 2009 #7
    [tex]f(x) = \sum_{i = 1}^\infty \frac{1}{x - 1/i} = \frac{1}{x - 1} ~+~ \frac{1}{x - 1/2} ~+~ \frac{1}{x - 1/3} ~+~ ...[/tex]
    ?
     
  9. Oct 11, 2009 #8

    Mark44

    Staff: Mentor

    A detail I overlooked...:eek:

    Here's another stab at it.
    f(x) = [1, for x = 1/n, n in {1, 2, 3, 4, ...}
    .........[0, otherwise
    So f(1) = f(1/2) = f(1/3) = ... = f(1/n) = 1, while f(x) = 0 for other values of x.
     
  10. Oct 11, 2009 #9
    I thought of that as well, but then that would be like the function:
    f(x) = 1 x is rational
    f(x) = 0 x is irrational,
    which, according to the textbook, does not have a limit for any value of a because then around a there would always be a 1 and a zero, so epsilon = 1/4 will not satisfy the definition of a limit.
     
  11. Oct 11, 2009 #10
    Consider the function g defined by g(x) = 1/[1/x], where [y] denotes the greatest integer function. Of course you'll have to modify this a little, but this basically works.
     
  12. Oct 11, 2009 #11

    Mark44

    Staff: Mentor

    No, I think the two functions are fundamentally different. For any number a that someone picks, it's possible to find an epsilon small enough that all values of x in (a - epsilon, a + epsilon), f(x) = 0.
     
  13. Oct 11, 2009 #12
    Would
    f(x) = x, x=1/n, n is natural
    f(x) = 0, otherwise
    also work though?
     
  14. Oct 11, 2009 #13
    Ohh, could the function be:
    f(x) = 1/x, x is a whole number (assuming whole = natural + 0)
    f(x) = 0 otherwise ?
    I don't understand why you can't define a function to equal something over 0.
     
  15. Oct 11, 2009 #14

    Mark44

    Staff: Mentor

    I don't think so, but I could be wrong, as I haven't put a whole lot of thought into it. The problem is that as n gets large, 1/n gets close to 0, so the points where the function is discontinuous are tending to smooth out. You don't have that problem if f(x) = 1, for x = 1/n.
     
  16. Oct 11, 2009 #15

    Mark44

    Staff: Mentor

    Division by zero is undefined! Period. Haven't any of your math teachers told you this numerous times as you were learning mathematics?

    That's why you can't define a function value to be 1/0 or anything else over zero.
     
  17. Oct 11, 2009 #16
    If a function isn't defined at 0, then the function isn't continuous at zero...
     
  18. Oct 11, 2009 #17

    Mark44

    Staff: Mentor

    True, but you can't define a function value by giving it a value that violates the rules of arithmetic. As an analogy, suppose you looked up the word "xerpy" in the dictionary, and found that it said "there is no definition for this word."
     
  19. Oct 11, 2009 #18
    The function

    [tex]f(x) = \left\{ \begin{array}{rl} 1, & x \text{ rational} \\
    0, & x \text{ irrational} \end{array}[/tex]

    (aka the rational comb function) fails since it is not continuous anywhere. The function sought needs to be continuous everywhere except at 1, 1/2, 1/3, 1/4, etc.

    The function

    [tex]f(x) = \left\{ \begin{array}{rl} 1/q, & x = p/q \text{ a ratio of integers} \\
    0, & x \text{ irrational} \end{array}[/tex]

    (aka the denominator function) fails since it is discontinuous at all rationals and continuous at all irrationals.

    The function

    [tex]f(x) = \left\{ \begin{array}{rl} x, & x = 1/n, n \text{ a natural number} \\
    0, & \text{ otherwise} \end{array}[/tex]

    has merit. Can you prove it satisfies the requirement?

    The functions involving 1/[1/x] or 1/x for x whole run into the problem I pointed out earlier in that they have discontinuities at the whole numbers, not 1, 1/2, 1/3, 1/4, etc.

    --Elucidus
     
  20. Oct 12, 2009 #19
    I realize that the last function satisfies the criteria, because you can choose delta to be the shortest distance from any 1/n to a, then x would not be any of the 1/n if we took epsilon to be greater than or equal to 1/n for a very large n, and so the function would not be continuous at x = 1/n but continuous with a limit of 0 otherwise.
     
  21. Oct 12, 2009 #20
    It's just harder to find a function which is discontinuous at 1,1/2,1/3...AND 0 as well while being continuous elsewhere.

    EDIT: Would the following work:

    f(x) = {sin 1/x, x=/= 1/n, n is natural
    .........{2, x = 1/n, n is natural

    That way, the function is discontinuous at 0 because it is not defined, and for x = 1,1/2,1/3... the limit from the sin function will not be equal to the value of the function at that x value.
     
    Last edited: Oct 12, 2009
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