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From: Chip Eastham on 12 Mar 2010 20:23
On Mar 12, 2:50 pm, 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
>
> It's trickier since the series for tangent "blows
> up" at pi/2, and also there's a double counting
> of positive and negative fixed points tan(x) = x
> that requires taking a square root of the series.
>
> I don't have time to give the details right now,
> but I'll check back later.
>
> regards, chip

Here's a sketch for tan x = x that mimics Euler's
analysis of sin x = 0 leading to zeta(2) = pi^2/6:

tan x = x

sin x
----- = cos x
x

0 = (1 - x^2/2 + x^4/4! - x^6/6! + ... )

- (1 - x^2/6 + x^4/5! - x^6/7! + ... )

= 0 - x^2/3 + x^4/30 - x^6/(5!*7) + ...

Factoring out -x^2/3 yields:

0 = 1 - x^2/10 + 3x^4/(5!*7) - ...

Now Euler's technique amounts to equating the
series to a product formula:

0 = PRODUCT (1 - (x/p_k)^2) FOR k = 1 to +oo

where the p_k are the positive fixed points of
tan x = x described by the OP. Thus if we
equate the x^2 terms in the last two formulas:

1/10 = SUM (1/p_k)^2 FOR k = 1 to +oo

as was to be shown.

Furthermore we can give similar treatment to
the "negative fixed points", i.e. positive
roots of tan x = -x, which leads to:

sin x
- ----- = cos x
x

0 = (1 - x^2/2 + x^4/4! - x^6/6! + ... )

+ (1 - x^2/6 + x^4/5! - x^6/7! + ... )

= 2 - 2x^2/3 + 5x^4/(4!*6) - 7x^6/(6!*8) + ...

0 = 1 - x^2/3 + 5x^4/(4!*6*2) - 7x^6/(6!*8*2) + ...

Calling the positive fixed points q_k for the sake
of clarity, we have by the same argument as before:

1/3 = SUM (1/q_k)^2 FOR k = 1 to +oo

regards, chip
From: Enrico on 13 Mar 2010 08:47
> "Chip Eastham" <hardmath(a)gmail.com> ha scritto nel messaggio
> news:fbc0e9ae-3e2c-4979-aef7-93fa30d0a285(a)q21g2000yqm.googlegroups.com...
> On Mar 12, 2:50 pm, Chip Eastham <hardm...(a)gmail.com> wrote:
> ...
> Now Euler's technique amounts to equating the
> series to a product formula:

> 0 = PRODUCT (1 - (x/p_k)^2) FOR k = 1 to +oo
>...

nice demonstration, but there is just one point I don't understand:
why is not possible to have other factor in the product? for example

0 =(1+(x/a)^2) PRODUCT (1 - (x/p_k)^2) FOR k = 1 to +oo

One should demonstrate that there are no more solutions in the whole complex
plain.

regards, Enrico.

From: Chip Eastham on 13 Mar 2010 10:54
On Mar 13, 8:47 am, "Enrico" <a...(a)b.nomail> wrote:
> > "Chip Eastham" <hardm...(a)gmail.com> ha scritto nel messaggio
> >news:fbc0e9ae-3e2c-4979-aef7-93fa30d0a285(a)q21g2000yqm.googlegroups.com....
> > On Mar 12, 2:50 pm, Chip Eastham <hardm...(a)gmail.com> wrote:
> > ...
> > Now Euler's technique amounts to equating the
> > series to a product formula:
> > 0 = PRODUCT (1 - (x/p_k)^2)  FOR k = 1 to +oo
> >...
>
> nice demonstration, but there is just one point I don't understand:
> why is not possible to have other factor in the product? for example
>
> 0 =(1+(x/a)^2)  PRODUCT (1 - (x/p_k)^2)  FOR k = 1 to +oo
>
> One should demonstrate that there are no more solutions in the whole complex
> plain.

Hi, Enrico:

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.

regards, chip


From: AP on 14 Mar 2010 04:36
On Fri, 12 Mar 2010 11:50:22 -0800 (PST), Chip Eastham
<hardmath(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....
>It's trickier since the series for tangent "blows
>up" at pi/2, and also there's a double counting
>of positive and negative fixed points tan(x) = x
>that requires taking a square root of the series.
>
>I don't have time to give the details right now,
>but I'll check back later.
>
>regards, chip

From: Gerry on 14 Mar 2010 07:31
On Mar 14, 2:54 am, Chip Eastham <hardm...(a)gmail.com> wrote:
> On Mar 13, 8:47 am, "Enrico" <a...(a)b.nomail> 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.

I'm pretty sure the tan x = x problem was an American Math Monthly
problem, maybe as long as 30 years ago.
--
GM
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