Results 1 to 5 of 5

Math Help - how many times can we differentiate this function??

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    how many times can we differentiate this function??

    how many times can we differentiate this splitted function on point x=0
    ??
    <br />
 f(x) = \{ x^{2n} \sin (\frac{1}{x}) <br />
,x \ne 0<br /> <br /> <br /> <br />
    {0,x=0}


    i did one differentiation
    what to do next??
    <br />
  f(x)\; = \;x^2 \sin ({1 \over x}),x \ne 0<br />
       <br />
  f'(0) = \mathop {\lim }\limits_{x \to 0} {{x^2 \sin ({1 \over x}) - 0} \over {x - 0}} = 0   <br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Abu-Khalil's Avatar
    Joined
    Oct 2008
    From
    Santiago
    Posts
    148
    You asumme n=1 or x^{2n} was an error?

    Anyway, knowing that, you may define the derivate function of f(x) as f'(x)=\begin{cases}2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right), x\not=0\\<br />
0,x=0<br />
\end{cases}.

    And now you have problems because f''(x)=\lim_{x\to0}\frac{2x\sin\left(\frac{1}{x}\r  ight)-\cos\left(\frac{1}{x}\right)-0}{x-0}=\lim_{x\to0}2\sin\left(\frac{1}{x}\right)+\frac  {\cos\left(\frac{1}{x}\right)}{x} doesn't converge.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    what is the correct way of proving it??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,596
    Thanks
    1421
    Proving what? That \lim_{x\rightarrow 0} 2 sin(1/x)+ \frac{cos(1/x)}{x} does not exist? That should be obvious: cos(1/x) is always between -1 and 1 and the denominator goes to 0 so that term does not converge. 2 sin(1/x) is always between -2 and 2 so it cannot "offset" the fact that cos(1/x)/x diverges.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i got a different result.
    i assumed that in point x=0 the function turns to 0 too.
    <br />
f(x) = \ x^{2n} \sin (\frac{1}{x})<br />
    <br />
f'(x)=2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})<br />

    <br />
f''(x)=\lim_{x->0+}\frac{2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})-0}{x-0}=\\<br /> <br />
     <br />
f''(x)=\lim_{x->0+}\frac{2nx^{2n-2}\sin (\frac{1}{x})-x^{2n-1}\frac{1}{x^2}\cos(\frac{1}{x})-0}{1}<br />

    so in the plus side i will get 0
    and on the minus side
    but i cant keep doing this derivatives till i get two different limits
    there must be an easier way
    ??
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differentiate n times
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: November 6th 2011, 06:16 PM
  2. Simplify then differentiate times 3
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 5th 2010, 10:26 AM
  3. Taking derivative of a function times a log
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 13th 2010, 02:43 PM
  4. How do I differentiate a vector times vector?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 13th 2009, 10:12 PM
  5. Replies: 19
    Last Post: February 9th 2009, 08:43 AM

Search Tags


/mathhelpforum @mathhelpforum