Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

Demonstrating a most-intriguing summation

68 views
Skip to first unread message

xadrezus

unread,
Apr 13, 2007, 8:43:58 AM4/13/07
to
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.

Thanks in advance and best regards.

David W. Cantrell

unread,
Apr 13, 2007, 11:35:31 AM4/13/07
to
"xadrezus" <xadr...@yahoo.com> wrote:
> 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'm glad that today isn't April 1; otherwise I might have thought you
were joking.

Based on some numerical work of my own, I agree that the sum _appears_ to
be exactly 1/10. But I have no idea how your conjecture might be proven.

> 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.

I hesitate to mention this because I doubt it will help, but you might look
at items (2) and (3) at <http://mathworld.wolfram.com/TancFunction.html>.

> 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

How? In other words, what caused you to consider that sum?

> and I would like to know how to prove it correct, which it is.

You seem to be _completely_ convinced, in absence of a proof. Why?

David W. Cantrell

Robert Israel

unread,
Apr 13, 2007, 12:43:36 PM4/13/07
to
"xadrezus" <xadr...@yahoo.com> writes:

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...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

xadrezus

unread,
Apr 14, 2007, 8:25:17 AM4/14/07
to

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...@math.MyUniversitysInitials.ca> wrote:

> University of British Columbia Vancouver, BC, Canada- Hide quoted text -
>
> - Show quoted text -


xadrezus

unread,
Apr 14, 2007, 8:50:44 AM4/14/07
to

Hi, Mr. Cantrell:

Thanks for answering my request for help, I'll comment your
questions interspersed with your original post:

On Apr 13, 5:35 pm, David W. Cantrell <DWCantr...@sigmaxi.net> wrote:
> [...]


> Based on some numerical work of my own, I agree that the sum _appears_ to
> be exactly 1/10. But I have no idea how your conjecture might be proven.


Mr. Israel's reply below is enough to concoct a demonstration, be
it rigorous or non-rigorous.


> I hesitate to mention this because I doubt it will help, but you might look
> at items (2) and (3) at <http://mathworld.wolfram.com/TancFunction.html>.


Very interesting link, thanks a lot for it.


> How? In other words, what caused you to consider that sum?


Well, actually this sum is akin to a particular case of a much
more general infinite summation, also involving the roots of
tan(x)=x. This was about the simplest case of all, and I set up
to the task of trying and demonstrating it first, before
attempting the more general case. I was more or less properly
oriented, but the essential step of first converting the
tan(x)=x equation to its cos(x)-sin(x)/x=0 form (as pointed
below by Mr. Israel) was sorely missing.


> You seem to be _completely_ convinced, in absence of a proof. Why?


Because there *is* at least one correct, already published
proof of it which I'm aware of. Most regrettably, it seems not
to be freely available but only as a pay-per-article or pay-per-
subscription item, so it wasn't an option for me to get it in
order to learn the details and further, it would spoil the fun.
I'd love to claim to be the first in bringing this amazing sum
to public light but alas, it was already well-known.

By the way, see if you can find the closed-form value of
this same sum, but with 1/(r^2+1) instead of 1/r^2. It
may very pleasantly surprise you as well.

Thanks again and best regards.

0 new messages