Prove that sum 1/a^2 = 1/10, where the sum runs over all
positive real solutions of the equation tan(x)=x.
I'm certain I've seen this problem posed, and possibly
solved, between 5 and 10 years ago on this newsgroup.
If anyone can supply a reference, please e-mail me at
bra...@cs.dal.ca.
Thanks,
David Bradley
I don't have a reference but I know how to solve it. (For some time, I
believed I invented the problem :-()
My approach, to my dismay, uses the theory of trace class integral
operators on L^2(0,1) applied to the Green function of a differential
operator with boundary conditions. The windfall profit is that I can
calculate the sums of 1/a^4, 1/a^6, etc., and prove in advance that they
are all rational numbers.
Does anyone have a solution using, for example, complex variable methods
(Cauchy integral etc.) without excursions into other theories?
And others may be interested in this problem, so I am posting it to the
whole newsgroup instead of e-mailing it.
Cheers, ZVK (Slavek).
> In article <5s293q$65g$1...@news.dal.ca>, David Bradley <bra...@cs.dal.ca=
> wrote:
> >The following problem has been submitted to the American
> >Mathematical Monthly:
> >
> >Prove that sum 1/a^2 =3D 1/10, where the sum runs over all
> >positive real solutions of the equation tan(x)=3Dx.
> >
> >I'm certain I've seen this problem posed, and possibly
> >solved, between 5 and 10 years ago on this newsgroup.
> >If anyone can supply a reference, please e-mail me at
> >bra...@cs.dal.ca.
> =
> I don't have a reference but I know how to solve it. (For some time, I=
> believed I invented the problem :-()
> My approach, to my dismay, uses the theory of trace class integral
> operators on L^2(0,1) applied to the Green function of a differential
> operator with boundary conditions. The windfall profit is that I can
> calculate the sums of 1/a^4, 1/a^6, etc., and prove in advance that the=
y
> are all rational numbers.
> =
> Does anyone have a solution using, for example, complex variable metho=
ds
> (Cauchy integral etc.) without excursions into other theories?
This exact problem was discussed in the Mathematica Journal 3(4):
26-28. There Peter Marksteiner (p...@katz.cc.univie.ac.at) suggested the
following general method for computing sums involving the roots of a
function: =
Invent a function with simple poles at the roots xn having residue
1/xn^2. For our example, the function:
f[z_] :=3D g[z]/h[z]
where:
g[z_] :=3D Sin[z]/z
h[z_]:=3DSin[z]-z Cos[z]
has the required properties: It has simple poles at the roots of
tan(z)-z: =
h[z]=3D=3DExpand[Cos[z] (Tan[z]-z)]
True
At each pole zn , the residue is given by Res(g(z)/ h(z), zn) =3D
g(zn)/h'(zn) since g(zn) * 0 and h'(zn) =3D 0 (see J. Mardsen, Basic
Complex Analysis, W. H. Freeman 1973). Thus, the residue is:
g[z]/h'[z] /. z -> zn
1/zn^2
It can be shown that the integral of f(z) around a circular contour
centered =
at the origin (and not crossing any of its poles) tends to zero as the
radius =
of the contour tends to infinity. Applying the residue theorem (where
the factor of two multiplying the summation comes from the fact that the
contour encloses both the positive xn and negative zeros -xn of
tan(z)-z. The residue at z =3D 0 is:
Residue[f[z],{z,0}]
-1/5
so the exact value for s is given by
Solve[2 s+%=3D=3D0,s]
1/10
Cheers,
Paul =
____________________________________________________________________ =
Paul Abbott Phone: +61-8-9380-2734
Department of Physics Fax: +61-8-9380-1014
The University of Western Australia =
Nedlands WA 6907 mailto:pa...@physics.uwa.edu.au =
AUSTRALIA http://www.pd.uwa.edu.au/Paul
God IS a weakly left-handed dice player
____________________________________________________________________
This seems to be a popular problem. In 1988 it was given to me as a
homework problem, but only to compute the sum numerically. In Jan. 1994
I asked for a proof on sci.math, and got the following response from
"Terry Tao" <t...@math.Princeton.EDU>:
Examine the poles and residues of
sec^2 x - 1
-----------
x^2(tan x - x)
and note that the function decays suitably fast at infinity.