Re: Demonstrating a most-intriguing summation


Hi, Mr. Israel:

Thank you very much for your prompt and helpful reply.

Thanks to it I'm now able to complete the demonstration. My original
attempts were also oriented to get some polynomial form of the
problem,
then change variables and/or use symmetric polynomials to deal with
the sum of the inverse of the square of the roots, but I wasn't
having any success because I missed converting the tan(x)=x original
equation to cos(x)-sin(x)/x, which is much more amenable to series
expansion treatment and thus is the key to success.

If I recall correctly this is about the third time you help me
here, so thanks again for your kindness, you're a real asset to
this newsgroup.

Best regards.


On Apr 13, 6:43 pm, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
"xadrezus" <xadre...@xxxxxxxxx> writes:
Hi all, best regards:

I've been trying to demonstrate the following correct closed form of
the infinite sum:

n=Inf
--------
\
> 1 / (root[n])^2) = 1/10
/
--------
n=1

where root[n] is the n-th positive root of the equation: tan(x) = x

I've checked it numerically, by summing up the first 1,000,000 roots
and indeed it
evaluates to 0.099999..., agreeing with the sum's final value of 0.1,
but all my
attempts to demonstrate its correctness have been unsuccessful, and my
web
searches have also failed to produce any freely accessible results.

Thus I would be very obligued if some of you would give me some
ideas,
techniques, hints, or pointers to freely accessible documents
featuring
such demonstration, or even the complete demonstration itself, if at
all
possible. BTW, this is no homework assignment or such, it's just that
this
sum caught my attention and I would like to know how to prove it
correct,
which it is.

Write the equation as f(x) = cos(x) - sin(x)/x = 0.
Note that
f(x) = sum_{k=1}^infty (1/(2k)! - 1/(2k+1)!) (-1)^k x^(2*k)
= sum_{k=1}^infty (-1)^k c_k x^(2k)
where c_k = 2k/(2k+1)!.

Let S_n(x) = sum_{k=1}^n (-1)^k c_k x^(2k).
Then the sum of 1/r^2 over the positive roots of S_n(x) is the sum of
the roots of the polynomial
P_n(s) = sum_{j=0}^{n-1} (-1)^j c_{n-j} t^j
But that sum is c_2/c_1 = 1/10.

So all that remains is to show that the positive roots of f(x) are
sufficiently well approximated by the positive roots of S_n(x)...
--
Robert Israel isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada- Hide quoted text -

- Show quoted text -


.

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