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From: Chip Eastham on 14 Mar 2010 08:15
On Mar 14, 4:36 am, AP <marc.picher...(a)wanadoo.fr.invalid> wrote:
> On Fri, 12 Mar 2010 11:50:22 -0800 (PST), Chip Eastham
>
>
>
> <hardm...(a)gmail.com> wrote:
> >On Mar 12, 12:25 pm, Dan Cass <dc...(a)sjfc.edu> wrote:
> >> > let p_k , k=1,2,3..., be the sequence of the positive
> >> > roots of tanx=x
> >> > how to prove 1/(p_1)^2+1/(p_2)^2+1/(p_3)^2......=1/10
> >> > thanks
>
> >> > If the sequence p_k is  the positive roots of
> >> > tanx=-x, then I have a
> >> > proof of  1/(p_1)^2+1/(p_2)^2+1/(p_3)^2......=1/3
> >> > and also  1/(p_1)^4+1/(p_2)^4+1/(p_3)^4......=13/180
> >> >   (the p_k are the eigenvalues of a linear operator :
> >> > f->int_a,b K(t,x)f(x)dx
>
> >> I would like to see your proof in the case of the squared reciprocals of the positive roots of tan(x) = -x.
> >> I would be VERY surprised if that sum was exactly 1/3...
>
> >I was also surprised, but apparently we can
> >perhaps imitate what Euler did to find zeta(2),
> >which was to pretty much treat the series for
> >sin(x), having periodic roots at multiples of pi,
> >as if it were x times a polynomial in x^2:
>
> >http://www.math.wpi.edu/IQP/BVCalcHist/calc3.html
>
> in this page  one use a polynomial's result
> for  S=1-u/3!+u^/5!+...but S  is not a polynomial
> because S=sinz/z with z^2=u
> and all roots of sinz/z are not pi, 2*pi, 3*pi....

All roots of sin(z) are 0, +/- pi, +/- 2*pi,... ,
so when we pass to S(u) = sin(z)/z with u = z^2,
the roots are u = pi^2, 4*pi^2, 9*pi^2, etc.

Consider the complex plane if one wishes to
pursue Enrico's point (see upthread) in the
simpler context of sine:

sin(x+iy) = sin(x)*cosh(y) + i cos(x)*sinh(y)

where z has real part x and imaginary part y.
For this to be zero, both real and imaginary
parts of sin(z) must be zero. Now the real
roots of sin(x) are as we've described above,
and cosh(y) has no real roots, so x must be
an integer multiple of pi (happy Pi Day!).
Then cos(x) is +/- 1 depending on parity, so
in addition we need sinh(y) = 0. The only
real root of sinh(y) is y = 0. Ergo the only
complex roots of sin(z) are z the integer
multiples of pi.

Indeed we can write sine in terms of an
infinite product (celebrating the other use
of Pi today):

sin(pi*z)

= pi*z*PRODUCT(1 - (z/n)^2) FOR n = 1,2,3...

_____+oo z^2
= pi*z* | | ( 1 - --- )
| | n = 1 n^2

[Infinite Products -- Wikipedia]
http://en.wikipedia.org/wiki/Infinite_product

Now my point (when I replied to myself above,
adapting Euler's argument to tan(x) = x) was
that if one has a convergent Taylor series to
compare with a convergent product of the form
PRODUCT (1 - (z/p_k)^2) FOR k = 1 to +oo
then the constant term of the Taylor series
should be 1, the z coefficient should be 0,
and the z^2 coefficient -SUM(1/p_k)^2.
Thus the latter can be deduced from Taylor
series coefficients, as Euler did for zeta(2).

regards, chip
From: Chip Eastham on 14 Mar 2010 08:26
On Mar 14, 8:15 am, Chip Eastham <hardm...(a)gmail.com> wrote:
> On Mar 14, 4:36 am, AP <marc.picher...(a)wanadoo.fr.invalid> wrote:
>
>
>
> > On Fri, 12 Mar 2010 11:50:22 -0800 (PST), Chip Eastham
>
> > <hardm...(a)gmail.com> wrote:
> > >On Mar 12, 12:25 pm, Dan Cass <dc...(a)sjfc.edu> wrote:
> > >> > let p_k , k=1,2,3..., be the sequence of the positive
> > >> > roots of tanx=x
> > >> > how to prove 1/(p_1)^2+1/(p_2)^2+1/(p_3)^2......=1/10
> > >> > thanks
>
> > >> > If the sequence p_k is  the positive roots of
> > >> > tanx=-x, then I have a
> > >> > proof of  1/(p_1)^2+1/(p_2)^2+1/(p_3)^2......=1/3
> > >> > and also  1/(p_1)^4+1/(p_2)^4+1/(p_3)^4......=13/180
> > >> >   (the p_k are the eigenvalues of a linear operator :
> > >> > f->int_a,b K(t,x)f(x)dx
>
> > >> I would like to see your proof in the case of the squared reciprocals of the positive roots of tan(x) = -x.
> > >> I would be VERY surprised if that sum was exactly 1/3...
>
> > >I was also surprised, but apparently we can
> > >perhaps imitate what Euler did to find zeta(2),
> > >which was to pretty much treat the series for
> > >sin(x), having periodic roots at multiples of pi,
> > >as if it were x times a polynomial in x^2:
>
> > >http://www.math.wpi.edu/IQP/BVCalcHist/calc3.html
>
> > in this page  one use a polynomial's result
> > for  S=1-u/3!+u^/5!+...but S  is not a polynomial
> > because S=sinz/z with z^2=u
> > and all roots of sinz/z are not pi, 2*pi, 3*pi....
>
> All roots of sin(z) are 0, +/- pi, +/- 2*pi,... ,
> so when we pass to S(u) = sin(z)/z with u = z^2,
> the roots are u = pi^2, 4*pi^2, 9*pi^2, etc.
>

Actually Wikipedia has a pretty nice writeup
of details related to Euler's solution here:

[Basel problem -- Wikipedia]
http://en.wikipedia.org/wiki/Basel_problem

regards, chip
From: Robert Israel on 14 Mar 2010 17:35
AP <marc.pichereau(a)wanadoo.fr.invalid> writes:

>
> let p_k , k=1,2,3..., be the sequence of the positive roots of tanx=x
> how to prove 1/(p_1)^2+1/(p_2)^2+1/(p_3)^2......=1/10
> thanks
>
> If the sequence p_k is the positive roots of tanx=-x, then I have a
> proof of 1/(p_1)^2+1/(p_2)^2+1/(p_3)^2......=1/3
> and also 1/(p_1)^4+1/(p_2)^4+1/(p_3)^4......=13/180
> (the p_k are the eigenvalues of a linear operator :
> f->int_a,b K(t,x)f(x)dx

See
<http://groups.google.ca/group/sci.math.symbolic/browse_thread/thread/c5b555272d9aca60/36a1b0fe634037cc>
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Dave L. Renfro on 18 Mar 2010 13:18
Chip Eastham wrote:

> Yes, that's a good point. If we could demonstrate that
> the only roots of x = tan x are on the real line, then
> we'd have something close to a rigorous proof.

For a "precalculus level" proof taken from Hardy's book
"A Course of Pure Mathematics", see pp. 9-10 of the .pdf
file "tan(x) = x" I just archived in the following Math
Forum sci.math post:

http://mathforum.org/kb/message.jspa?messageID=7014308

Note that I also give 2 references where the result can
be proved using Rouché's theorem and 2 references where
the result can be proved using theorems about eigenvalues
for certain Sturm-Liouville problems.

Dave L. Renfro
From: AP on 26 Mar 2010 13:07
On Thu, 18 Mar 2010 10:18:37 -0700 (PDT), "Dave L. Renfro"
<renfr1dl(a)cmich.edu> wrote:

>Chip Eastham wrote:
>
>> Yes, that's a good point. If we could demonstrate that
>> the only roots of x = tan x are on the real line, then
>> we'd have something close to a rigorous proof.
>
>For a "precalculus level" proof taken from Hardy's book
>"A Course of Pure Mathematics", see pp. 9-10 of the .pdf
>file "tan(x) = x" I just archived in the following Math
>Forum sci.math post:
>
>http://mathforum.org/kb/message.jspa?messageID=7014308
>
>Note that I also give 2 references where the result can
>be proved using Rouch�'s theorem and 2 references where
>the result can be proved using theorems about eigenvalues
>for certain Sturm-Liouville problems.
>
>Dave L. Renfro

I read in
>http://mathforum.org/kb/message.jspa?messageID=7014308

:
-------------------------
It can be found in ODE and PDE books, because it affords
a nice illustration of the general behavior of eigenvalues in
a Sturm-Liouville problem. For example, y" + (b^2)y = 0
with the boundary conditions y(0) + y'(0) = 0 and y(1) = 0
gives rise to tan(b) = b when solving for the eigenvalues b^2.
-------------------------

but to obtain the Green's function it is necessary to find
two independent functions u_1 , u_2

solutions of u''=0

with u_1(0)+u_1'(0)=0
u_2(1)=0

and it is not possible, because
u_1(x)=a(x-1)
u_2(x)=b(x-1)


but in the case
u_1(0)=0
u_2(1)+u_2'(1)=0
then u_1=x, u_2=x-2

Green's function is
K(x,t)=-x(t-2)/2 if x<=t
K(x,t)=-t(x-2)/2 if x>=t

and the eigenvalues are k with

y"+(1/k)y=0 ( y :R->R)
y(0)=0
y(1)+y'(1)=0
so k is >0
1/k=h^2
and y=bsin(hx) with
tanh+h=0

(the 1/h^2 with h>0 and tanh+h=0 are the eigenvalue)


My pb is : how find Green's function to obtain
the 1/h^2 with h>0 and tanh-h=0
for eigenvalues

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