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From: AP on 12 Mar 2010 11:15

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
From: Dan Cass on 12 Mar 2010 02:25
>
> 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...
From: Chip Eastham on 12 Mar 2010 14:50
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
From: AP on 12 Mar 2010 16:37
On Fri, 12 Mar 2010 12:25:56 EST, Dan Cass <dcass(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...
can you prove your affirmation ?

From: Rob Johnson on 12 Mar 2010 19:33
In article <85qkp5p3ai0r76v0o81gf5u20q89hptmm3(a)4ax.com>,
AP <marc.pichereau(a)wanadoo.fr.invalid> 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

Here's something to try. Consider the function

1
f(z) = --------
tan(z)-z

This function has residue -6/5 at 0 and 1/z^2 at every root of
tan(z) = z. I haven't crossed all the t's and dotted all the i's,
but it looks as if you can get a proof by contour integration that
the sum is (-1 + 6/5)/2 = 1/10.

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
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