Trascendental equation
rjfmv
anybody give me a clue of how to do it? or give me an indication of
where to look for a solution
Thank you for your help
Pubkeybreaker
Look up "Newton-Raphson"
Me
1. Let y = sin x -x
2. Solve numerically using whatever flavor you desire.
"rjfmv" <rjfm....@gmail.com> wrote in message
news:1138627091.6...@g43g2000cwa.googlegroups.com...
Chip Eastham
Look near x = 0.
-- c
David W. Cantrell
> I am trying to solve a trascendental equation of type sin(x)=x, can
> anybody give me a clue of how to do it? or give me an indication of
> where to look for a solution
Unfortunately, saying "of type sin(x)=x" is not being specific enough.
Chip may have thought that you wanted to solve sin(x) = x _itself_ and
so suggested that you "Look near x = 0." Please clarify. Are you perhaps
wanting the nonzero real solutions of equations of the form sin(x) = c*x
where c is a constant, |c| < 1?
David
Michael Varney
Plot sin(x) and x as a guide to your thinking.
G. A. Edgar
considered a function of x is not an elementary function (in the
technical meaning of that term).
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
ldier...@skynet.be
rjfmv schreef:
if you are looking in the reals you have several roots x = 0 +k.pi
>From analysis you know dat in the interval ]0,pi/2[ sinx/x < 1 so sin
x < x.
I don't think you'll find further real roots.
rjfmv
dealing with:
I am seeking a solution for the biharmonic equation
(1) f,xxxx+2*f,xxyy+f,yyyy=0
that governs problems in plane elasticity.
I am seeking for solutions in the form
(2) g(x,y)=exp(-k*x)*F(y)
where k is a constant that may be complex.
Substituting into (1) f(x,y) by g(x,y) yields the following fourth
order equation
F,yyyy+2*k^2*F,yy+k^4*F=0
The domain for F is [-h/2,h/2]
Assuming for boundary conditions
F(h/2)=0
F(-h/2)=0
F,y(h/2)=0
F,y(-h/2)=0
Which correspond admitting a traction free condition on the edge of a
membrane [0,L]x[-h/2,h/2]
An eigenvalue problem associated with the differential equation is
obtained and the corresponding complex eigenvalues are the nonzero
roots of the transcendental equations
sin(2*k)-2*k=0
sin(2*k)+2*k=0
rjfmv
rjfmv
rjfmv
Dave L. Renfro
> An eigenvalue problem associated with the differential
> equation is obtained and the corresponding complex
> eigenvalues are the nonzero roots of the transcendental
> equations
> sin(2*k)-2*k=0
> sin(2*k)+2*k=0
Timothy Y. Chow, "What is a closed-form number",
American Mathematical Monthly 106 #5 (May 1999),
440-448.
http://arxiv.org/PS_cache/math/pdf/9805/9805045.pdf
Chow mentions that Ferng-Ching Lin proved that
if Schanuel's conjecture is true, then none of
the nonzero (complex) solutions to sin(x) = x
will be a Liouvillian number. A Liouvillian number
(to be distinguished from a Liouville number,
although many web page authors don't seem to)
belongs to the countable collection of complex
numbers that can be obtained from the rational
numbers in finitely many applications of exponential,
logarithmic, and algebraic (explicit or implicit)
operations. This means anything you can write down
explicitly using trig. functions, their inverses,
exponential functions, and logarithm functions
(the latter two can be to any base that is also
a Liouvillian number), and you can also -- from
time to time -- use the numbers you've obtained
as coefficients to any degree polynomial and
include the zeros of that polynomial in the
collection (regardless of whether that polynomial's
roots can be expressed in terms of radicals).
Chow actually mentions cos(x) = x in his article,
but since every trig. function is quadratically
related to every other trig. function, the same
holds true for the (nontrivial) solutions to
sin(x) = x, tan(x) = x, etc. In fact, pretty
much any solution to a trig. function equal
to any rational function of x -- see Theorem 1
on p. 443 of Chow's paper -- which takes care
of many of the solutions to transcendental
equations that arise in physics (e.g. Kepler's
equation, zeros of the Bessel function J_n(x)
where n = 3/2, 5/2, 7/2, ..., etc.). You might
also be interested in my recent sci.math thread
"Regarding tan(x) = x" <http://tinyurl.com/dgn25>.
Dave L. Renfro
Virgil
ldier...@skynet.be wrote:
Actually, for all x <> 0, |sin(x)| < |x|, so ther can only be the one
root.
Chip Eastham
Thanks for the clarification. As I understand it, your interest
is specifically in the non-zero complex roots of:
sin(z) = z and sin(x) = -z
where z = 2k in your notation above. I suspect, but haven't
verified, that these are isolated roots.
Dave Renfro contributed a nice post below on the essential
impossibility of finding a "closed form" for these values. I
will therefore make a brief comment about approximations.
One approach is to use a Newton or related root-finding method
in the complex numbers C. As an alternative the single complex
equation can be rewritten as a system of 2 equations in two real
unknowns by separating the real and imaginary parts of z = x+iy,
and using the Euler equation to separate the real and imaginary
parts of sin(z). Newton's method on a system of equations is
often referred to as Newton-Raphson. The inverse of the Jacobi
matrix (of first partials) replaces division by an ordinary derivative
when translating from the scalar to the system formulation. In
using exact derivatives it will all come down to equivalent ways
of doing the iterations.
regards, chip
David W. Cantrell
> Thank you for your comment, I will specify the problem that I am
> dealing with:
> An eigenvalue problem associated with the differential equation is
> obtained and the corresponding complex eigenvalues are the nonzero
> roots of the transcendental equations
> sin(2*k)-2*k=0
> sin(2*k)+2*k=0
Thanks for the clarification.
I have just now found a nice simple approximation for the nonzero complex
solutions of sin(z) = +/- z.
The positive real parts of the solutions are approximately
x = p - log(2p)/p where p = (2n + 1) Pi/2 and n is a positive integer,
and the corresponding positive imaginary parts are approximately
y = acosh((-1)^n x/sin(x)).
For each solution x + i y with positive real and imaginary parts, there are
three other solutions: -x + i y, x - i y, -x - i y.
The approximation is excellent for large n, but not for small n; see below.
If it's not adequately accurate for your purpose, it will at least give you
an excellent starting point for some iterative numerical procedure...
The following is merely copied from a Mathematica notebook. The first list
gives my approximations of the first nine solutions; the second list gives
solutions obtained using Mathematica's FindRoot; the third list shows the
errors in my approximations.
Regards,
David W. Cantrell
----------------------------------------------
approxs = Table[N[x + I*ArcCosh[((-1)^n*x)/Sin[x]] /.
x -> (1/2)*(1 + 2*n)*Pi - (2*Log[(1 + 2*n)*Pi])/((1 + 2*n)*Pi)], {n, 1, 9}]
{4.236336985556712 + 2.243525772899499*I,
7.50331009861341 + 2.7673400884586408*I,
10.714493922494109 + 3.1027628431366585*I,
13.900772020085618 + 3.352071951995091*I,
17.07373184728186 + 3.551032364850773*I,
20.238686467146476 + 3.716744977128801*I,
23.398427836460947 + 3.858800105661019*I,
26.554570654718322 + 3.9831390088093053*I,
29.708117274786243 + 4.093705190711631*I}
Table[z /. FindRoot[Sin[z] == (-1)^n*z, {z, approxs[[n]]}], {n, 1, 9}]
{4.2123922304906625 + 2.2507286116018617*I,
7.497676277776385 + 2.7686782829873215*I,
10.712537397279261 + 3.1031487458252496*I,
13.899959713976465 + 3.352209884853505*I,
17.07336485315183 + 3.5510873470220825*I,
20.23851770783002 + 3.716767679752499*I,
23.398355225651308 + 3.8588089931055745*I,
26.55454726549156 + 3.9831416403399618*I,
29.70811982527604 + 4.093704924765334*I}
approxs - %
{0.023944755066049872 - 0.007202838702362602*I,
0.005633820837024572 - 0.0013381945286807806*I,
0.00195652521484746 - 0.0003859026885910666*I,
0.0008123061091538375 - 0.00013793285841412484*I,
0.00036699413002949655 - 0.00005498217130961436*I,
0.00016875931645543574 - 0.000022702623698300783*I,
0.00007261080963871791 - 8.887444555405466*10^-6*I,
0.000023389226761594273 - 2.631530656493908*10^-6*I,
-2.550489796249167*10^-6 + 2.6594629698450944*10^-7*I}
David W. Cantrell
> "rjfmv" <rjfm....@gmail.com> wrote:
> > Thank you for your comment, I will specify the problem that I am
> > dealing with:
> [snip]
> > An eigenvalue problem associated with the differential equation is
> > obtained and the corresponding complex eigenvalues are the nonzero
> > roots of the transcendental equations
> > sin(2*k)-2*k=0
> > sin(2*k)+2*k=0
>
> Thanks for the clarification.
>
> I have just now found a nice simple approximation for the nonzero complex
> solutions of sin(z) = +/- z.
>
> The positive real parts of the solutions are approximately
>
> x = p - log(2p)/p where p = (2n + 1) Pi/2 and n is a positive integer,
In my haste to get to bed last night, and wanting to post before I did so,
I overlooked an extremely simple improvement to the approximation for the
real parts. Well, I say it's an improvement, despite the fact that it makes
matters slightly worse for large n, because it substantially reduces the
error for small n, where the original approximation was poor. For the
improvement, just add 1 to the argument of the logarithm, so that the
approximation then becomes
x = p - log(2p + 1)/p
and keep everything else as it was.
In my previous post, I had not mentioned |relative error|. It's worst when
n = 1, whether you use my original approximation (in which case it's about
0.52%) or today's improvement (in which case it's about 0.06%).
> and the corresponding positive imaginary parts are approximately
>
> y = acosh((-1)^n x/sin(x)).
>
> For each solution x + i y with positive real and imaginary parts, there
> are three other solutions: -x + i y, x - i y, -x - i y.
>
> The approximation is excellent for large n, but not for small n; see
> below. If it's not adequately accurate for your purpose, it will at least
> give you an excellent starting point for some iterative numerical
> procedure...
>
> The following is merely copied from a Mathematica notebook. The first
> list gives my approximations of the first nine solutions; the second list
> gives solutions obtained using Mathematica's FindRoot; the third list
> shows the errors in my approximations.
The four new lists shown below my signature give some data for the improved
approximation. Twenty approximations are shown, beginning with n = 1 and
incrementing by 5.
Regards,
David W. Cantrell
-----------------------------------------------------
approxs = Table[N[x + I*ArcCosh[(-1)^n*x/Sin[x]] /. x -> p - Log[2*p + 1]/p
/. p -> (2*n + 1)*Pi/2], {n, 1, 96, 5}]
{4.21493737388774 + 2.2499288649969973*I,
20.237501844379725 + 3.716904961801436*I,
36.009462237116956 + 4.2838185238070565*I,
51.74655712428149 + 4.643442854620518*I,
67.47150002201913 + 4.907445874946697*I,
83.1906933197758 + 5.116139731349149*I,
98.90668338787728 + 5.288729526232592*I,
114.62069461109014 + 5.43588213146465*I,
130.33339252414683 + 5.564140248458534*I,
146.04517156002584 + 5.67780772863048*I,
161.7562810563726 + 5.779867233807703*I,
177.4668867241893 + 5.872471134688547*I,
193.17710319773096 + 5.957223671265011*I,
208.88701242285936 + 6.035352586572378*I,
224.59667460437774 + 6.10781843358829*I,
240.3061350126397 + 6.175386885367933*I,
256.01542837330373 + 6.238678126711618*I,
271.7245817883028 + 6.298201526706402*I,
287.4336167327473 + 6.35438056095421*I,
303.14255045273615 + 6.40757109922307*I}
sols = Table[z /. FindRoot[Sin[z] == (-1)^n*z, {z, approxs[[n]]}],
{n, 1, 20}]
{4.212392230490661 + 2.2507286116018608*I,
20.23851770783002 + 3.716767679752499*I,
36.0098660163716 + 4.283781587775027*I,
51.746768302821785 + 4.643427957051896*I,
67.47162863497536 + 4.907438416522555*I,
83.19077943783753 + 5.116135465966929*I,
98.90674489376762 + 5.28872685705572*I,
114.62074063827144 + 5.435880348977437*I,
130.3334282071958 + 5.564138998156587*I,
146.04520000024652 + 5.67780681724206*I,
161.75630423436908 + 5.7798665486047*I,
177.46690596251636 + 5.872470606283898*I,
193.1771194123405 + 5.957223255029243*I,
208.88702626769256 + 6.035352252729705*I,
224.59668655848978 + 6.107818161648313*I,
240.30614543480496 + 6.1753866608509185*I,
256.01543753741043 + 6.238677939147297*I,
271.72458990697197 + 6.298201368369828*I,
287.4336239734992 + 6.354380426043789*I,
303.14255694930637 + 6.407570983312318*I}
errors = approxs - %
{0.0025451433970795634 - 0.0007997466048634827*I,
-0.0010158634502950292 + 0.00013728204893670437*I,
-0.0004037792546469632 + 0.00003693603202936657*I,
-0.00021117854029739647 + 0.000014897568621385915*I,
-0.00012861295623167734 + 7.458424142292586*^-6*I,
-0.00008611806173064451 + 4.265382219870162*^-6*I,
-0.00006150589034348286 + 2.6691768715281228*^-6*I,
-0.000046027181298313735 + 1.7824872129423852*^-6*I,
-0.00003568304896361951 + 1.2503019464205067*^-6*I,
-0.000028440220688707996 + 9.113884198441724*^-7*I,
-0.000023177996496315245 + 6.852030036696988*^-7*I,
-0.000019238327070070227 + 5.284046498132966*^-7*I,
-0.00001621460953060705 + 4.162357676307238*^-7*I,
-0.000013844833205212126 + 3.338426726173793*^-7*I,
-0.000011954112039802567 + 2.7193997631513867*^-7*I,
-0.000010422165274803774 + 2.2451701475745267*^-7*I,
-9.164106700154662*^-6 + 1.875643214077627*^-7*I,
-8.118669143186708*^-6 + 1.5833657407426927*^-7*I,
-7.240751926929079*^-6 + 1.3491042150093335*^-7*I,
-6.496570222225273*^-6 + 1.1591075210759527*^-7*I}
relerrors = Abs[errors/sols]
{0.0005585939689007047, 0.000049817693760930954, 0.000011180996290910405,
4.074769303828492*^-6, 1.90435036260852*^-6, 1.0345020961582238*^-6,
6.215547375783586*^-7, 4.014105227186867*^-7, 2.7370149379777627*^-7,
1.9468863535027375*^-7, 1.4326077611037997*^-7, 1.0838671078997641*^-7,
8.392424608005709*^-8, 6.62706582293883*^-8, 5.3218887390979525*^-8,
4.336611055690016*^-8, 3.5792004544991767*^-8, 2.9875958568523044*^-8,
2.5189257595371955*^-8, 2.1429367174462924*^-8}
rjfmv
Let me ask you something that to me is not obvious, why do you say that
x = p - log(2p)/p where p = (2n + 1) Pi/2 is an approximation for the
real solutions of
sin(x+iy)=sin(x)ch(y)+i*cos(x)*sh(y)=x+i*y
considering that the real part is
sin(x)ch(y)=x
If dealing with tan(x)=x, can you derive a similar solution?
David W. Cantrell
> Thank you for your answer!
You're welcome.
> Let me ask you something that to me is not obvious, why do you say
> that
> x = p - log(2p)/p
Recall that I actually prefer x = p - log(2p + 1)/p.
> where p = (2n + 1) Pi/2 is an approximation for the
> real solutions of
> sin(x+iy)=sin(x)ch(y)+i*cos(x)*sh(y)=x+i*y
> considering that the real part is
> sin(x)ch(y)=x
The derivation of the approximation was rather messy. But if you _really_
want to see the derivation, I suppose I could outline it here for you.
> If dealing with tan(x)=x, can you derive a similar solution?
I'm not sure you'd call it "similar". But in any event, please see my
result given as (2) and (3) at
<http://mathworld.wolfram.com/TancFunction.html> which gives the real
solutions as a series. [And BTW, unless I've made a mistake, tan(z) = z has
no (nonreal) complex solutions.]
David Cantrell
Dave L. Renfro
> And BTW, unless I've made a mistake, tan(z) = z
> has no (nonreal) complex solutions.
More generally, tan(z) = az has no nonreal complex
solutions if (a is real) and (a = 1 or a > 1), and
exactly two nonreal solutions (both of which are
purely imaginary) if a is real and 0 < a < 1.
This can be proved using Rouché's theorem -- see
Greenleaf [1] (Chapter 6.7, Example 6.27, pp. 414-416)
and Hille [3] (Section 9.2, pp. 255-256).
The result for a = 1 or a > 1 also follows from the
fact that the eigenvalues for certain Sturm-Liouville
problems are real.
Finally, I've come across a third way in Hardy [2]
(p. 480) that is more elementary. Since p. 480 of
Hardy's book is one of google's book searches that
says "page's content is restricted" pages, here's
what Hardy says in the top third of p. 480. ("\leq and
"\geq" stand for "less than or equal to" and "greater
than or equal to", respectively.)
"We know already (Ex. XVII.4) that the equation has
infinitely many real roots. Now let z = x + iy, and
equate real and imaginary parts. We obtain
(sin 2x) / [cos(2x) + cosh(2y)] = ax,
(sinh 2y) / [cos(2x) + cosh(2y)] = ay,
so that, unless x or y is zero, we have
(sin 2x)/(2x) = (sinh 2y)/(2y).
This is impossible, the left-hand side being
numerically less, and the right-hand side
numerically greater than unity. Thus x = 0
or y = 0. If y = 0 we come back to the real
roots of the equation. If x = 0 then tanh(y) = ay.
It is easy to see that this equation has no
real root other than zero if a \leq 0 or
a \geq 1, and two such roots if 0 < a < 1.
Thus there are two purely imaginary roots
if 0 < a < 1; otherwise all the roots are real."
Arguments similar to the one Hardy used can also be
found in [4], [5], and [6] (p. 662).
[1] Frederick P. Greenleaf, INTRODUCTION TO COMPLEX
VARIABLES, W. B. Saunders Company, 1972, xii + 588 pages.
[2] Godfrey H. Hardy, A COURSE OF PURE MATHEMATICS,
10'th edition, Cambridge University Press,
1952/1996, xii + 509 pages.
[3] Einar Hille, ANALYTIC FUNCTION THEORY, Volume I,
Ginn and Company, 1959, xii + 308 pages.
http://tinyurl.com/8d6tf [p. 255]
http://tinyurl.com/a5h7w [p. 256]
[4] "Solution to Monthly Problem #E1295", American
Mathematical Monthly 65 #6 (June/July 1958), 450.
[5] "Solution to Monthly Problem #E1857", American
Mathematical Monthly 74 #6 (June/July 1967), 722-723.
[6] "Solution to Monthly Problem #6488", American
Mathematical Monthly 93 #8 (October 1986), 660-664.
Dave L. Renfro
rjfmv
tan(x)=x;
However, in what concerns the solution you kindly gave for
sin(2*k)-2*k=0
sin(2*k)+2*k=0
I still couldn't derive it, I guess I haven't enough math skills.
Could you please give me an outline of how to deduce your solution to
the problem,
x = p - log(2p + 1)/p.
where p = (2n + 1) Pi/2
Best Regards,
Ricardo
David W. Cantrell
> Thank you for your contributions, I have no longer doubts about the
> tan(x)=x;
> However, in what concerns the solution you kindly gave for
> sin(2*k)-2*k=0
> sin(2*k)+2*k=0
> I still couldn't derive it, I guess I haven't enough math skills.
> Could you please give me an outline of how to deduce your solution to
> the problem,
To give some context for those who may just now be joining us:
Ricardo wanted the nonzero complex solutions of sin(z) = +/- z.
Presumably, those solutions cannot be given in closed form in terms of well
known functions. I had stated an approximation, given above, for the real
part of such solutions, and said that the imaginary part could then be
approximated using y = acosh((-1)^n x/sin(x)).
Due to Ricardo's request, I'll now indicate how the approximation was
derived. It's not pretty at all (which is one reason I was reluctant to
post it). If anyone has an elegant way to get this approximation (or a
better one), please post it.
-------------------------------------
Begin with
(0) sin(x + iy) = +/- (x + iy)
where x and y are resp. the real and imaginary parts of z. If x + iy is a
solution, then so are -x + iy, -x - iy and x - iy. Knowing that, it
suffices now to search only for first quadrant solutions, and so we shall
assume henceforth that both x and y are positive.
Equating real parts and equating imaginary parts of (0), we get resp.
(1) cosh(y) sin(x) = +/- x
and
(2) sinh(y) cos(x) = +/- y.
Solving (1) for y gives
(3) y = acosh(+/- x/sin(x)).
[Trivial note: To avoid losing the solution z = 0 already, please think of
x/sin(x) here as actually being 1/sinc(x), which has the value 1 at x = 0.]
Then using (3) to substitute for y in (2), we get an equation in x alone:
(4) sinh(acosh(+/- x/sin(x))) cos(x) = +/- acosh(+/- x/sin(x)).
This can be rewritten, without hyperbolic functions, as
(5)
sqrt((x /sin(x))^2 - 1) cos(x) = +/- log(+/- x/sin(x) + sqrt((x/sin(x))^2 - 1)).
Solving this for x exactly in terms of familiar functions seems hopeless,
and so it's time to start approximating. Initially, I was concerned with
getting an approximation which is good when x is large. Accordingly, I then
ignored the constant terms under the square roots, giving
(6) x cot(x) = log(+/- 2x/sin(x))
once signs had been taken care of. This is still too messy, so we must
continue to approximate.
It happens that the solutions for x are close to odd multiples of Pi/2.
[This can be seen, for example, by plotting equations (1) and (2)
implicitly.] Using that fact, we may approximate x cot(x) by p(p - x) where
p = (2n + 1) Pi/2 and n is a positive integer. [That's merely using the
first term of the Taylor expansion of x cot(x) at p.] Again using that
fact, we may approximate sin(x) by -/+ 1. Equation (6) then becomes
(7) p(p - x) = log(2x).
At this point, despite the fact that I had initially sought an
approximation which would work well for large x, I decided to see if the
approximation could be improved for small x by modifying equation (7). This
can be done by using log(2x + 1), instead of log(2x). Of course it's also
important that this modification is not significant when x is large. [How
did I decide to use log(2x + 1) in particular? Well, looking at the
implicit plot of equations (1) and (2) helped. Furthermore, note that if
the right side of, say, equation (5) were replaced by log(2x + 1), then we
would have recovered the trivial solution z = 0 precisely. Thus, in some
sense, adding 1 to the argument of the logarithm is exactly what should be
done.] With this modification, equation (7) becomes
(8) p(p - x) = log(2x + 1).
The Lambert W function allows us to solve this, and we get
(9) x = -1/2 + W(p/2 * e^(p^2 + p/2)) / p.
Now even if p itself is not particularly large, the argument of W will be
quite large. This allows us to approximate the Lambert W function well here
using just two terms of the appropriate series; see item (14) at
<http://mathworld.wolfram.com/LambertW-Function.html>. In other words, we
will now approximate W(q) by log(q) - log(log(q)). Applying that
approximation to (9) and simplifying, using properties of logarithms, we
can get
(10) x = p + (log(p/2) - log(p^2 + p/2 + log(p/2))) / p.
In the argument of the second logarithm, log(p/2) is small compared to the
other terms, and so I ignored it, giving
(11) x = p + (log(p/2) - log(p^2 + p/2)) / p
which can be further simplified, using properties of logarithms, to give my
approximation
(12) x = p - log(2p + 1)/p.
Whew! Some maestros could conduct a Mahler symphony with less handwaving!
Hope this answered your question.
Ciao,
David
Robert Israel
>> Thank you for your contributions, I have no longer doubts about the
>> tan(x)=x;
>> However, in what concerns the solution you kindly gave for
>> sin(2*k)-2*k=0
>> sin(2*k)+2*k=0
>> I still couldn't derive it, I guess I haven't enough math skills.
>> Could you please give me an outline of how to deduce your solution to
>> the problem,
>> x = p - log(2p + 1)/p where p = (2n + 1) Pi/2
>
>To give some context for those who may just now be joining us:
>Ricardo wanted the nonzero complex solutions of sin(z) = +/- z.
>Presumably, those solutions cannot be given in closed form in terms of well
>known functions. I had stated an approximation, given above, for the real
>part of such solutions, and said that the imaginary part could then be
>approximated using y = acosh((-1)^n x/sin(x)).
>Due to Ricardo's request, I'll now indicate how the approximation was
>derived. It's not pretty at all (which is one reason I was reluctant to
>post it). If anyone has an elegant way to get this approximation (or a
>better one), please post it.
...
>(4) sinh(acosh(+/- x/sin(x))) cos(x) = +/- acosh(+/- x/sin(x)).
For the + case [I think the - is similar], this can be written as
cos(x) sqrt(x^2 - sin^2 x) = sin(x) acosh(x/sin(x))
The Intermediate Value Theorem can be used to show there is a
solution in each interval 2 n pi < x < (2 n + 1) pi for positive
integers n, and plotting shows the graphs of the two sides intersect
near x = (2 n + 1/2) pi. With q = (2 n + 1/2) pi and x = q + t,
the equation says
- sin(t) sqrt((q + t)^2 - cos^2 t) = cos(t) acosh((q + t)/cos(t))
Expanding this in a Taylor series to first order in t gives the
approximation
t = -arccosh(q)/((q^2-1)^(1/2) + (q^2-1)^(-1/2)
~ - ln(2 q)/q
Actually Maple will give as many terms as desired of the asymptotics
of a solution:
> asympt(RootOf(-sin(t)*sqrt((q+t)^2 - cos(t)^2) -
cos(t)*arccosh((q+t)/cos(t)), t), q);
But the terms get rather complicated.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
rjfmv
Thanks a lot.