how do I find all the roots of tanh(z)=z/b?

Discussion in 'Maple' started by Luna Moon, Jan 4, 2008.

  1. Luna Moon

    Luna Moon Guest

    Hi all,

    How do I find out all the roots of tanh(z)=z/b in the complex domain,
    where b is a real number and z is complex valued?

    Thanks a lot!
     
    Luna Moon, Jan 4, 2008
    #1
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  2. Methods for actually finding such solutions are
    a little outside my interests (but I have, at
    times, been interested in the algebraic transcendence
    of the solutions to equations such as this for "most"
    algebraic values of b), but perhaps something in
    these searches will help:

    http://www.google.com/search?q=transcendental-equation+tanh

    http://books.google.com/books?q=transcendental-equation+tanh

    http://scholar.google.com/scholar?q=transcendental-equation+tanh

    Dave L. Renfro
     
    Dave L. Renfro, Jan 4, 2008
    #2
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  3. Solve the equation (*):
    X''(x) = z^2 X(x)
    X(0) = 0
    X'(1) = b X(1)

    All solution must satisfy
    X(x) = c_1 cosh(z x) + c_2 sinh(z x). Then
    c_1=0
    and c_2!=0 if and only if z cosh(z) = b sinh(z).

    Hence the solutions to tanh(z)=z/b are exactly those z for which (*) has
    a non-zero solution.

    Next, if X(x) is a solution, then let bX(x) be its complex conjugate. Then

    z^2 int_0^1 |X(x)|^2 dx =
    int_0^1 X''(x) bX(x) dx =
    int_0^1 |X'(x)|^2 dx

    where the last equality follows by integration by parts (being careful
    to show that the cross terms really are zero).

    Hence z^2 is a real number.

    Hence z is either real (graphically has 1 or three solutions) or pure
    imaginary (graphically has infinitely many solutions, since z=iy, and
    tan(y)=y/b).

    I have to admit that I only came to this solution because I am familiar
    with Robin boundary conditions. Surely there is a more direct way to
    prove this.
     
    Stephen Montgomery-Smith, Jan 4, 2008
    #3
  4. Considering that there are infinitely many, what exactly do you
    mean by "find"?

    (I'm removing the removable singularity at 0, writing tanh(z)/z = 1 at z=0)

    You might note that tanh(z)/z is real if and only if z either real or imaginary.
    This is because if z = x + i y with x and y real,
    Im(tanh(z)/z) = (sin(2y) x - sinh(2x) y)/(2 (x^2+y^2)(sinh(x)^2 + cos(y)^2))
    and sin(2y)/y < 2 < sinh(2x)/x for x, y <> 0.
    For real z, tanh(z)/z is an even function, decreasing on [0,infty) and
    always in the interval (0,1]. So if b > 1, there are two real solutions
    to tanh(z) = z/b (one the negative of the other).
    For imaginary z = it, tanh(z)/z = tan(t)/t. If 0 < b < 1, there is one
    solution to tan(t)/t = 1/b for 0 < t < pi/2 and its negative in -pi/2 < t < 0.
    Then for any real nonzero b, there is one solution to tan(t)/t = 1/b in each
    of the intervals (n-1/2) pi < t < (n+1/2) pi for nonzero integer n.
     
    Robert Israel, Jan 4, 2008
    #4
  5. There are those two, together with the trivial third solution: z = 0.
    Luna Moon: I suspect that Stephen Montgomery-Smith and Robert have already
    answered your question adequately for your needs. But if not, you might
    look at my response

    <http://groups.google.com/group/sci.math/msg/90e9adde033b1056>

    in the sci.math thread "roots of trancendental equation" (2007 Aug. 29).
    There, I considered the equation cot(x) = c*x for c > 0. I could probably
    write up something similar for your problem; if you would like that, let me
    know.

    David
     
    David W. Cantrell, Jan 7, 2008
    #5
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