a(n) = (tan 1)(tan 2)(tan 3) . . . (tan n): Diverges?
Leonard M. Wapner
diverges but I can't show this. Help! Hint?
Thanks,
LW
PhysicsHowie
also equals L, but \lim a(n+1) = \lim \tan(n+1) a(n) = L \lim \tan(n+1)
(assuming the limits exist). But \lim \tan(n+1) does not exist, so one
_sees_ the answer. Sorry I can't prove it, but at least you have a
supporter in your assumption that it diverges!
William Elliot
On Thu, 28 Dec 2006, Leonard M. Wapner wrote:
> The nth term of a sequence is defined in the subject line. I think it
> diverges but I can't show this. Help! Hint?
>
What? You want us to be like Fermat and put the answer in the subject
line? Ok, I've a wonderful solution but it's too long for the subject
line. Sorry.
David Bernier
> The nth term of a sequence is defined in the subject line. I think it
> diverges but I can't show this. Help! Hint?
>
If we define s(n) = ln( | (tan 1)(tan2)(tan3) ... (tan n) | ) ,
then
s(n) = sum_{k=1..n} ln(|tan(k)|).
Using standard double precision arithmetic, this is what I
get for s(n) [ which shows up as 'sum'] up to n=100,000:
n=10000 term = -1.1364040832 sum = -54.7215526357
n=20000 term = -0.3345324360 sum = -78.3926419001
n=30000 term = 0.2969768906 sum = -79.9014561211
n=40000 term = 1.0764385327 sum = -60.0180530064
n=50000 term = 4.0240661668 sum = -14.9794426904
n=60000 term = 1.1994447135 sum = 58.8954361712
n=70000 term = 0.3725486002 sum = 88.0657002481
n=80000 term = -0.2598239960 sum = 93.2285200073
n=90000 term = -1.0191921173 sum = 77.4192066964
n=100000 term = -3.3305992366 sum = 40.8990882576
I don't know how precise the sum values are. Conceivably,
any of them could be far off.
David Bernier
Proginoskes
David Bernier wrote:
> Leonard M. Wapner wrote:
> > The nth term of a sequence is defined in the subject line. I think it
> > diverges but I can't show this. Help! Hint?
> >
>
> If we define s(n) = ln( | (tan 1)(tan2)(tan3) ... (tan n) | ) ,
> then
>
> s(n) = sum_{k=1..n} ln(|tan(k)|).
>
> Using standard double precision arithmetic, this is what I
Precision, schmecision. The sequence ln(|tan k|) clearly does not
converge to 0, so s(n) (the series) cannot possibly converge.
--- Christopher Heckman
Denis Feldmann
True (and strange nobody pointed it yet). But a(n) could still converge
towards 0. This, of course, would implie that huge terms tan(n)
corrsponding to n near pi/2+ kpi (which is possible, as pi is
irrational) are compensated by even smaller terms (with n nearer to k
pi). At a glance, as n mod pi/2 is evenly distributed, it looks liuke
such a compensation is imopssible, but it also looks quite hard to prove
...
>
> --- Christopher Heckman
>
David W. Cantrell
> Proginoskes a écrit :
> > David Bernier wrote:
> >> Leonard M. Wapner wrote:
> >>> The nth term of a sequence is defined in the subject line. I think
> >>> it diverges but I can't show this. Help! Hint?
> >>>
> >> If we define s(n) = ln( | (tan 1)(tan2)(tan3) ... (tan n) | ) ,
> >> then
> >>
> >> s(n) = sum_{k=1..n} ln(|tan(k)|).
> >>
> >> Using standard double precision arithmetic, this is what I
> >> get for s(n) [ which shows up as 'sum'] up to n=100,000: [...]
> >
> > Precision, schmecision. The sequence ln(|tan k|) clearly does not
> > converge to 0, so s(n) (the series) cannot possibly converge.
>
> True (and strange nobody pointed it yet).
Yes. ;-) Before I saw your response, I was going to respond to Christopher
by saying "Series, schemeries."
> But a(n) could still converge towards 0.
Indeed.
> This, of course, would implie that huge terms tan(n)
> corrsponding to n near pi/2+ kpi (which is possible, as pi is
> irrational) are compensated by even smaller terms (with n nearer to k
> pi). At a glance, as n mod pi/2 is evenly distributed, it looks liuke
> such a compensation is imopssible, but it also looks quite hard to prove
> ...
Despite my intuition to the contrary, I'm going to suggest, based
on (admittedly weak) numerical evidence, that a(n) _does_ converge to 0.
David W. Cantrell
Kira Yamato
Not sure if the following is useful at all. Just a wild hunch.
Let
F_n(x) = a(x+n).
Can we say anything about the function
F(x) = lim_{n->oo} F_n(x)
that maybe useful?
Whenever F(x) exists, we have
F(x) = F(x + pi).
F(x) has zeros at all negative integers.
F has poles whenever
x = u + v pi + pi/2
for some integer u and v.
If a(n) converges at all, then
F(0) = lim_n a(n) = 0.
So,
F(m + n pi) = 0
for all integer m and n.
--
-kira
Kira Yamato
> On 2006-12-28 23:04:22 -0500, "Leonard M. Wapner" <lwa...@adelphia.net> said:
>
>> The nth term of a sequence is defined in the subject line. I think it
>> diverges but I can't show this. Help! Hint?
>>
>>
>> Thanks,
>>
>> LW
>
> Not sure if the following is useful at all. Just a wild hunch.
>
> Let
> F_n(x) = a(x+n).
Sorry, I meant
F_n(x) = prod_{k=1}^n tan(x+k)
instead.
David W. Cantrell
My intuition is forcing me to retract that suggestion.
BTW, a(347) is approximately 740.6 .
What is the smallest n for which |a(n)| > a(347), or does no such n exist?
Cheers,
David
Leonard M. Wapner
is
bounded. I can't formally resolve any of these issues; but calculating and
plotting the first 200 terms (Mathematica) may yield misleading results.
The plot "suggests" the
sequence is bounded, perhaps converging to zero. Extending the calculation
to 400 terms makes me believe otherwise.
"Leonard M. Wapner" <lwa...@adelphia.net> wrote in message
news:DIqdnf--69snDgnY...@adelphia.com...
David Bernier
[...]
>> Despite my intuition to the contrary, I'm going to suggest, based
>> on (admittedly weak) numerical evidence, that a(n) _does_ converge to 0.
>
> My intuition is forcing me to retract that suggestion.
>
> BTW, a(347) is approximately 740.6 .
> What is the smallest n for which |a(n)| > a(347), or does no such n exist?
Taking absolute values of the tangent values, I get:
n=347:
|tan(n)| = 6.8011356481
|a(n)| = 740.60335501
n=51819:
|tan(n)| = 40589.613531
|a(n)| = 935.19441678
By the way, isn't it true that for 0<x<pi/2, one has
tan(x)tan(pi/2 -x) = 1?
I've been thinking about the probability density function
of log(tan(X)), where X is uniformly distributed on [0, pi/2].
I'm not sure I have it worked out right, so I won't speculate.
David Bernier
Herman Rubin
David W. Cantrell <DWCan...@sigmaxi.net> wrote:
>David W. Cantrell <DWCan...@sigmaxi.net> wrote:
>> Denis Feldmann <denis.feldma...@club-internet.fr> wrote:
>> Indeed.
The terms ln(|tan(k)|) are asymptotically distributed with
mean 0 and variance pi^2/4. Assuming that the terms are
independent, which is of course false, the a(n)/sqrt(n)
should be approximately normal with mean 0 and standard
deviation pi/2.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
David C. Ullrich
My speculation is that when you do get it worked out right
it's really not going to help much. The problem being that
log(tan(X)) is unbounded; something like the fact that
k is asymptotically uniformly distributed mod pi/2 isn't
going to help unless we have very precise information on
how close to 0 it gets, how often.
>David Bernier
>
David C. Ullrich
Robert Israel
Note that those numbers are dense in the reals.
Clearly F_n can't converge in any nice way.
Maybe more interesting would be
G(z) = prod_{n=1}^infty i tan(nz)
For Im(z) > 0, tan(nz) = (1 + O(exp(-2 n Im(z))))/i
so G(z) converges, uniformly on compact sets, to an
analytic function in the open upper half plane. Of
course that doesn't say much about z = 1.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
David Bernier
I've used contfrac at the WIMS site in France to get good
diophantine approximations to pi/2.
n=344:
|tan(n)| ~= 227.50041
|a(n)| ~= 371.31258
n=688:
|tan(n)| ~= 0.00879
|(a(n)| ~= 0.01544
344/219 ~= 1.570776256
pi/2 ~= 1.570796327
And 219pi/2 ~= 344.00439
pi/2 == 219pi/2 (mod pi)
so tan(344) should be positive [true].
If a "good" approximant to pi/2 is m/n with n even, say n=2r, then
pi/2 ~= m/2r, so pi ~= m/r and m ~= r.pi, so tan(m) ~= 0.
So to get |tan(m)| to be large, we want a continued fraction
approximant m/n with n odd.
I don't know whether the largish value of |a(344)| can
be lower bounded (based on pi/2 ~= 344/219, as a part
of the convergents using the continued fraction
approximations to pi/2 with odd denominator)
in some systematic way ...
David Bernier
toni.l...@gmail.com
> BTW, a(347) is approximately 740.6 .
> What is the smallest n for which |a(n)| > a(347), or does no such n exist?
How much do you get for the value of a(599455)?
David Bernier
The PARI-gp program gives this:
gp > exp(sum(X=1,599455,log(abs(tan(X)))))
= 3560687483133952802806124334479966400662813.5528
or |a(599455)| ~= 3.56 x 10^42
David Bernier