Results 1 to 4 of 4

Thread: Find all entire functions f

  1. #1
    Junior Member
    Joined
    Dec 2011
    Posts
    57
    Thanks
    1

    Find all entire functions f

    Find all the entire functions f such that
    $\displaystyle f(0) = i $ and $\displaystyle \left |f(z)- \cos z \right |\geq{\sqrt{2}}$
    for all $\displaystyle {z\in{\mathbb{C}}}$.


    My attempt at this question...
    Let $\displaystyle g(z) = \frac{1}{f(z)-\cos z}$.
    Since $\displaystyle f(z)$ and $\displaystyle \cos z$ are entre and $\displaystyle f(z) - \cos z \neq 0, \forall {z\in{\mathbb{C}}}$ (since $\displaystyle \left |f(z)- cos z \right |\geq{\sqrt{2}}$).
    $\displaystyle \Rightarrow g(z)$ is entire & $\displaystyle \left |g(z) \right |= \frac{1}{\left |f(z)-\cos z\right |} \leq \frac{1}{\sqrt{2}}, \forall {z\in{\mathbb{C}}}$.
    By Liouville's Theorem, $\displaystyle g(z) \equiv k$ for some $\displaystyle {k\in{\mathbb{C}}$.
    $\displaystyle \Rightarrow \frac{1}{f(z)-\cos z} \equiv k$ for some $\displaystyle {k\in{\mathbb{C}}, k \neq 0$, since $\displaystyle f(z) - \cos z \neq 0, \forall {z\in{\mathbb{C}}}$.
    Hence $\displaystyle f(z) = \frac{1}{k} + \cos z$
    Now it is given that $\displaystyle f(0) = i$, and $\displaystyle cos(0) = 1$, hence $\displaystyle \frac{1}{k} = i-1$.
    $\displaystyle \Rightarrow f(z) = i - 1 + \cos z$

    Is this correct? Are there any other solutions? Thanks in advance!.
    Last edited by alphabeta89; Oct 17th 2012 at 12:44 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    1,061
    Thanks
    410

    Re: Find all entire functions f

    Quote Originally Posted by alphabeta89 View Post
    Find all the entire functions f such that
    $\displaystyle f(0) = i $ and $\displaystyle \left |f(z)- \cos z \right |\geq{2}$
    for all $\displaystyle {z\in{\mathbb{C}}}$.
    $\displaystyle \text{Umm... } f(0) = i \text{ and } \cos(0) = \frac{e^{i(0)} + e^{-i(0)}}{2} = \frac{1 + 1}{2} = 1,$

    $\displaystyle \text{so } 2 \le \lVert f(0)- \cos(0) \rVert = \lVert -1 + i \rVert = \sqrt{2} < 2.$

    The answer to this problem is that there are no such functions.
    Last edited by johnsomeone; Oct 16th 2012 at 11:35 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2011
    Posts
    57
    Thanks
    1

    Re: Find all entire functions f

    Quote Originally Posted by johnsomeone View Post
    $\displaystyle \text{Umm... } f(0) = i \text{ and } \cos(0) = \frac{e^{i(0)} + e^{-i(0)}}{2} = \frac{1 + 1}{2} = 1,$

    $\displaystyle \text{so } 2 \le \lVert f(0)- \cos(0) \rVert = \lVert -1 + i \rVert = \sqrt{2} < 2.$

    The answer to this problem is that there are no such functions.
    Hi, there is a typo in my question. I have made the necessary changes. Sorry for that. Is my solution still correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    1,061
    Thanks
    410

    Re: Find all entire functions f

    Yes - that looks absolutely correct to me. And because your reasoning works backwards from assuming such an entire f, there are no other solutions. Well done.
    Last edited by johnsomeone; Oct 17th 2012 at 09:12 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Family of Entire Functions
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Jul 2nd 2010, 07:25 AM
  2. Entire functions and polynomials. Poles and singularities.
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 27th 2010, 08:07 AM
  3. Entire function
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Oct 31st 2009, 01:42 AM
  4. entire function
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Nov 12th 2008, 05:44 AM
  5. Entire Functions
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Nov 26th 2007, 06:21 PM

Search Tags


/mathhelpforum @mathhelpforum