
#1
Nov1612, 09:28 AM

P: 25

Hi all,
In studying the eigenvalues of certain tridiagonal matrices I have encountered a problem of the following form: {(1+a/x)*2x*sinh[n*arcsinh(x/2)]  2a*cosh[(n1)*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 halfplane 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
Nov1612, 03:35 PM

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hi ekkilop!
have you tried simplifying it by putting x = 2sinhy (and maybe a = 2b) ? 



#3
Nov1712, 02:00 PM

P: 25

Hi tinytim,
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((n1)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... 



#4
Nov1712, 06:29 PM

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Implicit Hyperbolic Function
hi ekkilop!
and get tanh(ny) = a function of tanh(y) and sech(y) 



#5
Nov1912, 02:27 PM

P: 25

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! 



#6
Nov1912, 02:32 PM

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i don't think we can go any further than that




#7
Nov1912, 02:52 PM

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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. 



#8
Nov2112, 12:51 PM

P: 25

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 nonzero 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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