Results 1 to 8 of 8

Thread: derivative problem

  1. #1
    Member
    Joined
    Feb 2009
    From
    Chennai
    Posts
    148

    derivative problem

    If $\displaystyle f(x-y)=f(x)g(y)-g(x)f(y)$ and $\displaystyle g(x-y)=g(x)g(y)+f(x)f(y)$ for all $\displaystyle x,y \in \mathbb{R}$ and the right hand derivative of $\displaystyle f(x)$ at zero exists. Find $\displaystyle g'(x)$ at $\displaystyle x=0$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    $\displaystyle g(-x) = g(0-x) = g(x)$

    Assuming $\displaystyle g$ is differentiable at 0, the limit

    $\displaystyle \lim_{d \rightarrow 0} \frac{g(d)-g(-d)}{2d}$ converges and is equal to $\displaystyle g'(0)$; which yields $\displaystyle g'(0)=0$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by Bruno J. View Post
    $\displaystyle g(-x) = g(0-x) = g(x)$

    Assuming $\displaystyle g$ is differentiable at 0, the limit

    $\displaystyle \lim_{d \rightarrow 0} \frac{g(d)-g(-d)}{2d}$ converges and is equal to $\displaystyle g'(0)$; which yields $\displaystyle g'(0)=0$.
    You need to provide the explanation of why g is symmetric

    CB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    I thought I had!

    $\displaystyle g(x-0) = g(x)g(0)+f(x)f(0)$
    $\displaystyle g(0-x) = g(0)g(x)+f(0)f(x)$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2009
    From
    Chennai
    Posts
    148

    query

    Its not given that g is differentiable
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2009
    From
    Chennai
    Posts
    148

    query

    Its not given that g is differentiable, so u cant assume g' exists
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Chandru1 View Post
    Its not given that g is differentiable, so u cant assume g' exists
    Here we go then!

    You have $\displaystyle g(x) = g(x)g(0)+f(x)f(0) = g(x)g(0)$. Hence either $\displaystyle g(0) = 1$ or $\displaystyle g(x)$ is identically zero which instantly yields $\displaystyle g'(0)=0$. Hence suppose $\displaystyle g(0)=1$.

    $\displaystyle f(0) = 0$ is easy; set x=y.
    $\displaystyle f(x)=-f(-x)$ is also easy.

    $\displaystyle f(x+y)=f(x)g(-y)-g(x)f(-y) = f(x)g(y)+g(x)f(y)$ by the above.

    $\displaystyle g(x+y)=g(x)g(-y)+f(x)f(-y) = g(x)g(y)-f(x)f(y)$

    Now consider the complex numbers

    $\displaystyle m(x) = g(x)+if(x)$.

    We have $\displaystyle m(x)m(y) = g(x+y)+if(x+y) = m(x+y)$.

    In particular $\displaystyle m(0)=1$, and $\displaystyle m(x)m(-x)=1$.


    Hence the function $\displaystyle m : \mathbb{R} \rightarrow \mathbb{C}$ is a morphism of groups $\displaystyle m : \mathbb{R}^+ \rightarrow \mathbb{C}^\times$. Hence we have $\displaystyle m(x)=\exp(tx)$ for some $\displaystyle t=a+bi \in \mathbb{C}$. Clearly $\displaystyle a=0$ since $\displaystyle m(0)=1$. Hence $\displaystyle m(x)=\exp(xbi)$

    Hence $\displaystyle g(x) = \Re \exp(xbi) = \cos(bx)$,
    so that $\displaystyle g'(0) = 0$.
    Last edited by Bruno J.; Jun 16th 2009 at 03:02 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Just a little extra :
    In case you are wondering why expressing m as an exponential map is possible, consider

    $\displaystyle k=m(1)$

    Then for all integers n:
    $\displaystyle m(n)=m(1+...+1)=k^n$
    $\displaystyle k=m(1)=m(1/n+...+1/n)=m(1/n)^n$
    so that $\displaystyle m(1/n) = k^{1/n}$

    from this it follows easily that for all rational x
    $\displaystyle m(x) = k^{x}$.

    Hence if $\displaystyle m$ is continuous on any interval, it coincides everywhere on that interval with $\displaystyle k^{x}$ (irrationals included).
    But $\displaystyle f$ is certainly continuous at 0 (it's differentiable there!). To see that g is continuous at 0, we have

    $\displaystyle g(x/2 + x/2) = g(x/2)^2-f(x/2)^2$

    i.e.
    $\displaystyle g(x/2)^2-g(x) = f(x/2)^2$

    and from this it is easy to show (with sequences for example) that $\displaystyle g$ is continuous at 0 (using the fact that $\displaystyle f(x/2)^2$ is continuous at 0 and is equal to 0)
    Last edited by Bruno J.; Jun 16th 2009 at 05:19 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivative Problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 21st 2010, 06:22 PM
  2. Replies: 1
    Last Post: Oct 7th 2010, 08:23 AM
  3. Derivative Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Sep 28th 2010, 08:37 PM
  4. Derivative problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Sep 27th 2010, 06:13 AM
  5. Derivative Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 15th 2009, 02:36 PM

Search Tags


/mathhelpforum @mathhelpforum