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Trivial solution for cosh(x)=0 and sinh(x)=0

  1. Apr 12, 2015 #1
    I'm doing Sturm-Loiuville problems and I need to find the eigenvalues for λ

    I'm having difficulty understanding the trivial solutions for the hyperbolic sin and cos when they equal 0.

    I know that cos(x)=0 when ##x= π/2 + πn = (2n+1)π/2##
    sin(x) = 0 when ##x=πn##

    What about cosh(x) and sinh(x)? Please help
     
  2. jcsd
  3. Apr 12, 2015 #2

    Mark44

    Staff: Mentor

    ##sinh(x) = \frac{e^x - e^{-x}}{2}##
    ##cosh(x) = \frac{e^x + e^{-x}}{2}##
    Clearly cosh(x) is never zero. It's pretty easy to find the zeroes of sinh(x).
     
  4. Apr 12, 2015 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    sinh(x)=0 when x=0.
     
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