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Implicit Hyperbolic Function

by ekkilop
Tags: function, hyperbolic, implicit
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ekkilop
#1
Nov16-12, 09:28 AM
P: 28
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.
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tiny-tim
#2
Nov16-12, 03:35 PM
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hi ekkilop!

have you tried simplifying it by putting x = 2sinhy (and maybe a = 2b) ?
ekkilop
#3
Nov17-12, 02:00 PM
P: 28
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...

tiny-tim
#4
Nov17-12, 06:29 PM
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Implicit Hyperbolic Function

hi ekkilop!
Quote Quote by ekkilop View Post
0 = (4+2a)sinh(n*y)sinh(y) + 2a[sinh(n*y) - cosh(n*y)cosh(y)]
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)
ekkilop
#5
Nov19-12, 02:27 PM
P: 28
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!
tiny-tim
#6
Nov19-12, 02:32 PM
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i don't think we can go any further than that
Mute
#7
Nov19-12, 02:52 PM
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Quote Quote by ekkilop View Post
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!
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.
ekkilop
#8
Nov21-12, 12:51 PM
P: 28
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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