tanx=x
AP
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
Dan Cass
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...
Chip Eastham
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
AP
can you prove your affirmation ?
Rob Johnson
AP <marc.pi...@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 <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
Chip Eastham
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
Enrico
> news:fbc0e9ae-3e2c-4979...@q21g2000yqm.googlegroups.com...
> On Mar 12, 2:50 pm, Chip Eastham <hardm...@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.
Chip Eastham
> > "Chip Eastham" <hardm...@gmail.com> ha scritto nel messaggio
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
AP
<hard...@gmail.com> wrote:
>On Mar 12, 12:25 pm, Dan Cass <dc...@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....
Gerry
I'm pretty sure the tan x = x problem was an American Math Monthly
problem, maybe as long as 30 years ago.
--
GM
Chip Eastham
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
Chip Eastham
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
Robert Israel
See
<http://groups.google.ca/group/sci.math.symbolic/browse_thread/thread/c5b555272d9aca60/36a1b0fe634037cc>
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Dave L. Renfro
> 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
AP
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
Chip Eastham
Thanks, Dave, I missed your post for about a week. But
your elementary proof (from Hardy) that tan x = x has
only real roots is satisfying, scarcely more difficult
than a similar proof for sin x = 0.
regards, chip
Dave L. Renfro
> My pb is : how find Green's function to obtain
> the 1/h^2 with h>0 and tanh-h=0
> for eigenvalues
Unfortunately, I haven't done anything with Green's functions
since 1983, so I'd have to review some stuff (which I don't
have with me now anyway) before I could begin thinking about
what you're asking. Maybe someone else who has studied this
more recently can help.
Dave L. Renfro