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Math Help - continuety parameter question..

  1. #1
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    continuety parameter question..

    i tried to solve it
    but i didnt get any parameter value
    ??
    Last edited by transgalactic; January 18th 2009 at 12:54 PM.
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  2. #2
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    Here's a thought: what happens when m=0 ?
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  3. #3
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    0^0 *bounded is an undefined form
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    Quote Originally Posted by transgalactic View Post
    0^0 *bounded is an undefined form
    Look carefully... the definition x^m \sin \frac1{x} holds only for x \neq 0. You cant simply substitute 0 in the place of x to check for continuity.

    Thus your problem for m = 0 reads:

    f(x) = \begin{cases} \sin \frac1{x} & x\neq 0 \\ 0 & x = 0 \end{cases}

    Is f(x) continuous at x = 0?

    (This is a very popular problem. I am sure you would have seen it somewhere in your notes/text)
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  5. #5
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    ok so its 1*bounded wich is not defined either

    ??
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    Quote Originally Posted by transgalactic View Post
    ok so its 1*bounded wich is not defined either

    ??
    The function is well defined. The point of my post was to ask if f(x) is continuous at 0. I claim it is not. Can you prove it?

    To prove it is discontinuous at 0, observe that sequence x_n = \frac{2}{(2n+1)\pi} converges to 0, but f(x_n) = (-1)^n
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  7. #7
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    how does it finds the values of the m parameter
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  8. #8
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    First prove the following result:

    (*) \lim_{x \to 0} x^m \sin \frac1{x} exists and equal to 0 iff m > 0.

    Now for part (A) we need to find m for which \lim_{x \to 0} f(x) = f(0) = 0, Thus we need m for which \lim_{x \to 0} x^m \sin \frac1{x} = 0. From the result (*), this happens iff m > 0.

    For part (B), we need values of 'm' for which \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0} exists. But \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0} = x^{m-1} \sin \frac1{x}. Again by the result, the limit exists iff m-1 > 0 \implies m > 1

    Do part (C) yourself, its instructive
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  9. #9
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    its seems you already answered C
    by solving A and B
    it is continues on x=0 when m>0
    it is differentiable on x=0 when m>1
    so its continuously differentiable on m>1

    thats the correct solution?
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  10. #10
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    Quote Originally Posted by transgalactic View Post
    its seems you already answered C
    by solving A and B
    it is continues on x=0 when m>0
    it is differentiable on x=0 when m>1
    so its continuously differentiable on m>1

    thats the correct solution?
    No! A function is continuously differentiable at x = 0 means the derivative at 0 exists and the derivative is continuous at 0 ! I never proved any result about the derivative!

    Ok I will do the last one too... But I will let you prove some statements on your own.

    Since we have already proved that m > 1 iff f(x) is differentiable at x = 0, for continuous differentiability we need m > 1.
    f'(x) = x^{m-2} \left(mx \sin \frac1{x} - \cos \frac1{x}\right)

    Now first prove that if m > 2, then the function f'(x) is continuous at 0, using (*) and making a similar rule for \lim_{x \to 0} x^m \cos \frac1{x}(prove this too!).

    Now the only remaining case is m=2. For this case,  f'(x) = \begin{cases} 2x\sin \frac1{x} - \cos \frac1{x} & x \neq 0 \\ 0 & x = 0 \end{cases}.

    Choose the sequence x_n = \frac1{n \pi}, and observe that x_n \to 0 but f(x_n) = -(-1)^n. Thus f(x) is not continuously differentiable at x = 0.

    So at the end, let me list the exercises you have to do, to complete the proof rigorously:

    (*) exists and equal to 0 iff m > 0

    (**) f'(x) = x^{m-2} \left(mx \sin \frac1{x} - \cos \frac1{x}\right)

    (***) \lim_{x \to 0} x^m \cos \frac1{x} exists and equal to 0 iff m > 0

    (****) Using (**) and (***), prove that if m > 2, then f'(x) is continuous at x = 0

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  11. #11
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    (*) exists and equal to 0 iff m > 0

    the only way to prove it
    is by cheching the cases of m.
    if m<0 then as x goes to zero and m is negative
    we get 0.
    so the whole limit would be
    0*bouded which is bound
    so i think i found a mistake in your theory
    its working for m differs 0
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  12. #12
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    i got a solution to A
    and it says
    some thing about

    if m=0 then
    <br /> <br />
x_n = \frac{1}{(2n)\pi} <br />
    i dont know how they came up with this Xn
    and what i need to do with it
    ??
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  13. #13
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    Quote Originally Posted by transgalactic View Post
    (*) exists and equal to 0 iff m > 0

    the only way to prove it
    is by cheching the cases of m.
    if m<0 then as x goes to zero and m is negative
    we get 0.
    so the whole limit would be
    0*bouded which is bound
    so i think i found a mistake in your theory
    its working for m differs 0
    Read the question again. It explicitly "shouts" m \geq 0.
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  14. #14
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    Quote Originally Posted by transgalactic View Post
    i got a solution to A
    and it says
    some thing about

    if m=0 then
    <br /> <br />
x_n = \frac{1}{(2n)\pi} <br />
    i dont know how they came up with this Xn
    and what i need to do with it
    ??
    can you explain this thing
    ??
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  15. #15
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    Quote Originally Posted by transgalactic View Post
    can you explain this thing
    ??
    That solution you got is of no use to part A. To demonstrate discontinuity at x_0, we have to find a sequence x_n \to x_0, but the sequence f(x_n) \not \to f(x_0).

    Your sequence x_n \to 0, but f(x_n) = \sin 2n \pi = 0. So it is of no use.

    As a side note, I am really sorry transgalactic, that I am failing in explaining this concept to you. I have already given you the answer for this in my previous posts. The sequence you can use is x_n = \frac2{(2n+1)\pi}

    I am telling the following based on our discussion.....If you take it in the right spirit, I suggest you take tutions. Sometimes a live tutor teaching a new concept is an invaluable experience.
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