Results 1 to 3 of 3

Math Help - Values of derivatives

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    62

    Values of derivatives

    Let f: \mathbb{R} \rightarrow \mathbb{R} be infinitely differentiable and suppose that
    f(\frac{1}{n})= \frac{n^{2}}{n^{2}+1}
    for all n=1,2,3,.... Determine the values of
    f^{'}(0), f^{''}(0), f^{'''}(0), f^{(4)}(0),....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Markeur View Post
    Let f: \mathbb{R} \rightarrow \mathbb{R} be infinitely differentiable and suppose that
    f(\frac{1}{n})= \frac{n^{2}}{n^{2}+1}
    for all n=1,2,3,.... Determine the values of
    f^{'}(0), f^{''}(0), f^{'''}(0), f^{(4)}(0),....
    Note first that by continuity \displaystyle f(0)=f\left(\lim \frac{1}{n}\right)=\lim f\left(\frac{1}{n}\right)=1. Note then that since f'(x) exists one has that if x_n\to 0 and x_n\ne 0,\text{ }n\in\mathbb{N} then \displaystyle f'(x)=\lim_{n\to\infty}\frac{f\left(x+x_n\right)-f(x)}{x_n}. In particular \displaystyle f'(0)=\lim_{n\to\infty}\frac{f\left(\frac{1}{n}\ri  ght)-f(0)}{\frac{1}{n}}=\lim_{n\to\infty}\frac{\frac{n^  2}{n^2+1}-1}{\frac{1}{n}}=0. Generalize.
    Last edited by Drexel28; January 20th 2011 at 04:16 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Markeur View Post
    Let f: \mathbb{R} \rightarrow \mathbb{R} be infinitely differentiable and suppose that
    f(\frac{1}{n})= \frac{n^{2}}{n^{2}+1}
    for all n=1,2,3,.... Determine the values of
    f^{'}(0), f^{''}(0), f^{'''}(0), f^{(4)}(0),....
    Since f\bigl(\frac1n\bigr) = \frac{n^{2}}{n^{2}+1} = \frac1{1+(1/n)^2}, it follows that one such function would be f(x) = \frac1{1+x^2} = (1+x^2)^{-1} = 1-x^2+x^4-x^6+\ldots. Comparing that with the Taylor series f(x) = 1+xf'(0)+\frac{x^2}{2!}f''(0)+\ldots, you see that for this particular function f^{(n)}(0) = 0 if n is odd, and f^{(2n)}(0) = (-1)^nn!.

    That tells you what you should expect the answer to be, but of course it isn't a proof. To set about proving the result, you will have to use techniques like those in Drexel28's comment. But knowing what to expect the answer to be can sometimes be a help in establishing it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: September 6th 2010, 03:55 PM
  2. Derivatives involving absolute values
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 13th 2010, 02:40 AM
  3. Derivatives, Integrals, Absolute Values Oh My!
    Posted in the Calculus Forum
    Replies: 11
    Last Post: July 26th 2010, 09:06 AM
  4. Derivatives of Absolute Values?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 24th 2009, 01:45 AM
  5. Local Max/Min Values using Partial Derivatives
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 18th 2009, 07:39 PM

Search Tags


/mathhelpforum @mathhelpforum