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Implicit Hyperbolic Function

  1. Nov 16, 2012 #1
    Hi all,

    In studying the eigenvalues of certain tri-diagonal matrices I have encountered a problem of the following form:

    {(1+a/x)*2x*sinh[n*arcsinh(x/2)] - 2a*cosh[(n-1)*arcsinh(x/2)]} = 0

    where a and n are constants. I'm looking to find n complex roots to this problem, but isolating x is troublesome. I attempted to use the implicit derivatives to obtain an expression for x in terms of a and n but it didn't lead me anywhere.

    Is there a general approach to finding the roots of equations of this type? If not, can one find any general properties of the roots, e.g. if they belong to a certain half-plane etc.

    The problem may be simplified somewhat if we choose a=-2 and try to find x as a function of n but even here the roots are hard to find.

    Any advice would be much appreciated.
    Thank you.
     
  2. jcsd
  3. Nov 16, 2012 #2

    tiny-tim

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    hi ekkilop! :smile:

    have you tried simplifying it by putting x = 2sinhy (and maybe a = 2b) ?
     
  4. Nov 17, 2012 #3
    Hi tiny-tim,
    Thanks for your reply.
    I did try this and it cleans things up a bit. In particular it becomes clear that a=-2 is a convenient choice since we get

    0 = (4+2a)sinh(n*y)sinh(y) + 2a[sinh(n*y) - cosh(n*y)cosh(y)]

    after expanding cosh((n-1)y). However, it is not clear to me how to proceed from here. Is there perhaps some other substitution that would make life easier? Or maybe there's a standard form to rewrite sinh(n*y) and cosh(n*y). My attempts from here just seems to make things more complicated...
     
  5. Nov 17, 2012 #4

    tiny-tim

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    hi ekkilop! :smile:
    the only thing i can suggest is to divide throughout by cosh(n*y)cosh(y),

    and get tanh(ny) = a function of tanh(y) and sech(y)
     
  6. Nov 19, 2012 #5
    This is not a bad idea. All in all I can boil things down to

    tanh(ny) = cosh(y)

    which has the expected roots (found numerically). Solving for n is straight forward but inverting seems impossible, at least in terms of standard functions.
    If n is a positive integer, what can be said about y?
    Thank you for all your help!
     
  7. Nov 19, 2012 #6

    tiny-tim

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    i don't think we can go any further than that :smile:
     
  8. Nov 19, 2012 #7

    Mute

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    In this particular case that you have simplified to, perhaps the best you could do now is rewrite this equation as a polynomial. If you set ##x = e^y##, you can rearrange the equation into a degree 2n+2 polynomial, which you could then perhaps study to see if you can say anything useful about the roots for different n. (Remember that since x = exp(y), only positive valued x's are valid solutions to the polynomial, the rest are erroneous).

    You should be able to write down a polynomial for the case ##a \neq -2## as well, but I'm not sure how helpful that will ultimately be, as you will have to vary both a and n.
     
    Last edited: Nov 19, 2012
  9. Nov 21, 2012 #8
    Hi Mute!
    Thanks for your reply. The problem is actually a result of a polynomial of degree n, which has been rewritten in it's present form. The coefficients of all n+1 terms are non-zero integers dependent on a, except for the leading term. I could probably do a more thorough analysis on the bounds of the solutions though, so thank you for the inspiration!
     
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