Results 1 to 10 of 10

Math Help - Can 2 functions be equal for a continuous range?

  1. #1
    JvJ
    JvJ is offline
    Newbie
    Joined
    Jul 2010
    Posts
    3

    Can 2 functions be equal for a continuous range?

    I've been wondering this for a while. Given two C-infinity functions f(x) and g(x) such that f(x) != g(x), is it possible for f(x) = g(x) for some continuous range (a, b)?

    Basically, I first wanted to know if it was possible for any functions, then realized that there are cheap ways out like absolute value functions and defining the function to be different for different ranges. That's why I'm asking specifically about C-infinity functions (all derivatives, and derivatives of derivatives, and so-on are continuous).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jun 2009
    Posts
    68
    Well, what about two functions that have equal asymptotes as they approach infinity?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Mathelogician's Avatar
    Joined
    Jun 2010
    From
    Iran
    Posts
    89
    Thanks
    1
    Quote Originally Posted by 1005 View Post
    Well, what about two functions that have equal asymptotes as they approach infinity?
    No; for example f(x)=1/x and g(x)=1/x^2.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    there are smooth functions where it is identically 1 in [-1, 1] and 0 in (- \infinite, -2] and [ 2, +\infinite)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    JvJ
    JvJ is offline
    Newbie
    Joined
    Jul 2010
    Posts
    3
    Quote Originally Posted by xxp9 View Post
    there are smooth functions where it is identically 1 in [-1, 1] and 0 in (- \infinite, -2] and [ 2, +\infinite)
    Which functions are those?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by JvJ View Post
    I've been wondering this for a while. Given two C-infinity functions f(x) and g(x) such that f(x) != g(x), is it possible for f(x) = g(x) for some continuous range (a, b)?

    Basically, I first wanted to know if it was possible for any functions, then realized that there are cheap ways out like absolute value functions and defining the function to be different for different ranges. That's why I'm asking specifically about C-infinity functions (all derivatives, and derivatives of derivatives, and so-on are continuous).
    f(x)= \{ \begin{array}{cc} 0 \ \mbox{if} \ x\leq0 \\ e^{-\frac{1}{x^2}} \ \mbox{if} \ x>0 \end{array}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    JvJ
    JvJ is offline
    Newbie
    Joined
    Jul 2010
    Posts
    3
    Quote Originally Posted by Jose27 View Post
    f(x)= \{ \begin{array}{cc} 0 \ \mbox{if} \ x\leq0 \\ e^{-\frac{1}{x^2}} \ \mbox{if} \ x>0 \end{array}
    This actually looks promising. It's still a piecewise function, but it looks like all derivatives approach 0 from the right, and it will never be discontinuous.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by JvJ View Post
    This actually looks promising. It's still a piecewise function, but it looks like all derivatives approach 0 from the right, and it will never be discontinuous.
    It is C^{\infty} and this can be proven by noting that \lim_{x\rightarrow \infty} p(x)e^{-x}=0 where p(x) is any polynomial.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by Jose27 View Post
    f(x)= \{ \begin{array}{cc} 0 \ \mbox{if} \ x\leq0 \\ e^{-\frac{1}{x^2}} \ \mbox{if} \ x>0 \end{array}
    It is famous Cauchy function... f^{(n)}(0)=0 for every n\in \mathbb{N}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,453
    Thanks
    1868
    Which, by the way, shows that the Taylor's series of a function does NOT necessarily converge to that function! The Taylor's series, about x= 0, for this function. is just "0" and converges to 0 for all x.

    Of course, this function is not "analytic" in any neighborhood of 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: February 20th 2011, 09:24 AM
  2. finding the range for the functions
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 12th 2009, 07:37 AM
  3. Algebraically disproving equal functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 25th 2009, 10:59 PM
  4. Range of Composite Functions
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: June 17th 2009, 09:02 AM
  5. Domain and Range Functions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 4th 2008, 05:35 PM

Search Tags


/mathhelpforum @mathhelpforum