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Continuety parameter question

  1. Jan 18, 2009 #1
    i tried to solve it
    but i didnt get any parameter value
  2. jcsd
  3. Jan 18, 2009 #2
    Re: part A: I think this is a "setup" to make you think the answer is easy. Yes, it appears that: for m =1,2,3,4 ... etc, the function is continuous at x=0. Your logic looks good to me.

    However, the sin(1/x) function oscillates infinitely fast as x --> 0, so I think you'll find the answers to B and C to be surprising.
  4. Jan 18, 2009 #3
    how to solve it?
  5. Jan 18, 2009 #4
    Can you differentiate f(x) "formally" ? I get this:

    f'(x)= mxm-1sin(1/x) + xm-1cos(1/x)(-1)x-2

    For the case of m=1, I get:

    f'(x) = sin(1/x) - cos(1/x)/x2

    The first term doesn't approach anything as x--> 0. It oscillates infinitely fast between +1 and -1. No limit, therefore not continuous.

    Also, in second term, cos(1/x) -->1, but 1/x2 --> infinity, so again no limit.

    Your assignment, Mr. Phelps, should you choose to accept it ... :) ... is to find an m > 0 where f'(0) is continuous. See if you can find one.
  6. Jan 19, 2009 #5
    i got a solution to A
    and it says
    some thing about

    if m=0 then

    x_n = \frac{1}{(2n)\pi}
    i dont know how they came up with this Xn
    and what i need to do with it
  7. Jan 19, 2009 #6
    Oops, I missed the m > OR = 0 part.
    For f(x) = xmsin(1/x)
    If m=0, then f(x) = sin(1/x)
    So, for m=0, your logic for testing continuity of f(x) at zero breaks down. Sin(1/x) is not continuous at 0.

    Now, regarding that "solution" you found that says:
    [itex]x_n = \frac{1}{(2n)\pi} [/itex]
    Hmm. Well, let's look at when sin(1/x) = 0
    then 1/x = {[itex]\pi[/itex], 2[itex]\pi[/itex], 3[itex]\pi[/itex], ... } = n [itex]\pi[/itex]
    So I get: x = [itex]\frac{1}{(n)\pi} [/itex]
    Different than your solution, but the same idea: as n gets larger, x gets closer to 0, but ... between every point where sin(1/x) = 0, it oscillates up to +1 (or down to -1):

    when sin(1/x) = 1 --> 1/x = {[itex]\pi[/itex]/2, [itex]2\pi[/itex]+[itex]\pi[/itex]/2, ...} = [itex]2n\pi[/itex]+[itex]\pi[/itex]/2 --> x = [itex]\frac{1}{(2n\pi + \pi/2)}[/itex]
    and when sin(1/x) = -1 --> 1/x = {[itex]3\pi[/itex]/2, [itex]2\pi[/itex]+[itex]3\pi[/itex]/2, ...} = [itex]2n\pi[/itex]+[itex]3\pi[/itex]/2 --> x = [itex]\frac{1}{(2n\pi + 3\pi/2)}[/itex]

    So ... as x ->0, it passes through a full oscillation between each successive [itex]x_n = \frac{1}{(2n)\pi} [/itex], taking on values 0, 1, 0, -1, 0 successively between xn and xn+1

    That's a long-winded answer. Does it help?

    Regards, BobM
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