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Math Help - differentiable function

  1. #1
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    differentiable function

    Hello. I have this problem,
    If r>0 is a rational number, let f: R to R be defined by f(x)=(abs(x)^r)*sin(1/x) for x not equalling zero and f(0)=0. Determine the values of r for which f'(0) (first derivative) exists.

    I could really really use some help. Thank you.
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  2. #2
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    What we have is,
    f(x)=\left\{ \begin{array}{c}|x|^{a/b}\sin (1/x) \mbox{ for }x\not = 0\\ 0 \mbox{ for }x=0\end{array}\right\}

    If the function fails to be continous at x=0 (it is continous everywhere else because the product of two continous functions at a point is continous at that point).

    So what is the limit,
    \lim_{x\to 0}|x|^{a/b}\sin \frac{1}{x} equal to?
    If, a=0 then,
    |x|^{a/b}=1.
    And we have discontinuity be the limit
    \lim_{x\to 0}\sin \frac{1}{x}
    Does not exist.
    So we need to consider the cases when a\not = 0.

    We require the limit from the right to conincide with the limit from the left. Since |x| is always positive and \sin \frac{1}{x} does change from positive and negative. We can only have for limit existence to be 0. Which is persicely what the functional value at the point is, f(0)=0.
    Thus, we can consider cases when the right-handed limits are zero.
    In that case,
    \lim_{x\to 0^+} x^{a/b}\sin \frac{1}{x}
    We can write,
    \lim_{x\to 0^+} x^{a/b-1}  x\sin \frac{1}{x}
    (But that limit does not exist, ever.
    If you appraoch it a even or odd multiplies of \pi)*

    *)I did not prove that, but it makes sense to me.
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  3. #3
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    What are the actual values of a/b though that make it continuous?
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  4. #4
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    Quote Originally Posted by dave007rules View Post
    What are the actual values of a/b though that make it continuous?
    Look at what I posted (assuming my last conjecture statement was true). None.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Look at what I posted (assuming my last conjecture statement was true). None.
    I think that you need to rethink that last statement.
    \begin{array}{l}<br />
  - 1 \le \sin \left( {\frac{1}{x}} \right) \le 1 \\ <br />
  - \left| x \right|^r  \le \left| x \right|^r \sin \left( {\frac{1}{x}} \right) \le \left| x \right|^r ,\quad r \ge 1 \\ <br />
 \end{array}
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  6. #6
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    Quote Originally Posted by Plato View Post
    I think that you need to rethink that last statement.
    \begin{array}{l}<br />
  - 1 \le \sin \left( {\frac{1}{x}} \right) \le 1 \\ <br />
  - \left| x \right|^r  \le \left| x \right|^r \sin \left( {\frac{1}{x}} \right) \le \left| x \right|^r ,\quad r \ge 1 \\ <br />
 \end{array}
    See, I was thinking what happens when we approach zero as,
    \frac{1}{\pi (2k)}
    And,
    \frac{1}{\pi (2k+1)}
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  7. #7
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    Quote Originally Posted by dave007rules View Post
    Hello. I have this problem,
    If r>0 is a rational number, let f: R to R be defined by f(x)=(abs(x)^r)*sin(1/x) for x not equalling zero and f(0)=0. Determine the values of r for which f'(0) (first derivative) exists.

    I could really really use some help. Thank you.
    Other posters have shown that the function is continuous at x=0 as long as
    r!=0. However that does not answer your question which was about
    differentiability at x=0.

    As x approaches zero sin(1/x) has zeros closer and closer together with the
    interval between them going to zero, so there are always points inside and
    interval [-epsilon, epsilon] such that:

    [f(0)-f(x)]/x=0, x!=0,

    Also if the derivative of d(0) of |x|^r is not zero, there are other points in the same interval for which:

    [f(0)-f(x)]/x ~= d(0)

    which means that f(x) is not differentiable at 0. I will leave it to the reader
    to show that of d(0)=0 then f is differentiable at 0, and its derivative is 0.

    That the derivative of |x|^r is zero at x=0 means that r>1.

    RonL
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