Sorry, Ed. I discovered a couple of typos in the message I sent. Please
ignore that message. Here's the corrected message.
That is a cute result. It seems clear that there should be some u_0
where product(n=1...N, (u + sin n)) ->0 if uu_0, but who would have thought that u_0 would be rational!
We have
product(n=1...N, (5/4 + sin n)) = exp sum(n=1...N, ln (5/4 + sin n))
The common difference of the value of n, viz 1, is incommensurate with
the periodicity of the summand, viz 2 pi. Thus, the law of large numbers
applies. Thus, for any -1<=y_0oo, the proportion of
values of n in the partial sum
sum(n=1...N ln (5/4 + sin n))
where y_0oo [sum(n=1...N, ln (5/4 + sin n))] / N = int(x=-pi/2...pi/2, ln (5/4
+ sin x) dx) / pi ___ [1]
We have the following indefinite integral equation:
d x 2 p tan (ax/2) + q
int(------------) = ------------- arctan ---------------- (if p>q)
p + q sin ax a sqrt(p²-q²) sqrt(p²-q²)
therefore, setting p=u, q=1, a=1,
d x 2 u tan (x/2) + 1
int(---------) = ---------- arctan --------------- ___ [2]
u + sin x sqrt(u²-1) sqrt(u²-1)
Define, for u>1,
I(u) = int(x=-pi/2...pi/2, ln (u + sin x) dx)
Differentiating under the integral sign,
d I(u) dx
------ = int(x=-pi/2...pi/2, --------- )
d u u + sin x
[ 2 u tan (x/2) + 1 ]x=pi/2
= [ ---------- arctan --------------- ] from eqn [2]
[ sqrt(u²-1) sqrt(u²-1) ]x=-pi/2
2 [ u tan (x/2) + 1 ]x=pi/2
= ---------- [ arctan --------------- ]
sqrt(u²-1) [ sqrt(u²-1) ]x=-pi/2
2 [ u+1 -u+1 ]
= ---------- [ arctan ---------- - arctan ---------- ] ___ [3]
sqrt(u²-1) [ sqrt(u²-1) sqrt(u²-1) ]
Now,
u+1 -u+1 1-u²
---------- * ---------- = ---- = -1
sqrt(u²-1) sqrt(u²-1) u²-1
and if tan x_0 = y_0, tan x_1 = y_1, and y_0 y_1 = -1, then x_0 and
x_1 differ by pi/2. Thus, from equation 3,
d I(u) pi
------ = ----------
d u sqrt(u²-1)
Integrate with respect to u.
I(u) = pi ln(u+sqrt(u²-1)) + c ___ [4]
where c is the constant of integration. Going back to the definition of I,
I(u) = int(x=-pi/2...pi/2, ln (u + sin x) dx)
= int(x=-pi/2...pi/2, ln (u (1 + (sin x)/u)) dx)
= int(x=-pi/2...pi/2, ln u + ln (1 + (sin x)/u) dx)
= pi ln u + int(x=-pi/2...pi/2, ln (1 + (sin x)/u) dx) ___ [5]
As u->oo,
(sin x)/u -> 0
=> 1 + (sin x)/u -> 1
=> ln (1 + (sin x)/u) -> 0
convergence being uniform in all cases, thus the integral in equation
5 approaches 0. Thus, putting equations 4 and 5 together, as u->oo
pi ln u -> pi ln(u+sqrt(u²-1)) + c
=> c = lim_u->oo [pi ln u - pi ln(u+sqrt(u²-1))]
= pi lim_u->oo [ln u - ln(u+sqrt(u²-1))]
= pi lim_u->oo [ln u - ln(u+sqrt(u²))]
= pi lim_u->oo [ln u - ln(2 u)]
= -pi ln 2.
Therefore
I(u) = pi ln(u+sqrt(u²-1)) - pi ln 2
In particular,
I(5/4) = pi ln[5/4+sqrt(25/16-1)] - pi ln 2
= pi ln[5/4+sqrt(9/16)] - pi ln 2
= pi ln(5/4+3/4) - pi ln 2
= 0.
Thus, going back to equation 1,
lim_N->oo [sum(n=1...N, ln (5/4 + sin n))] / N = 0
=> sum(n=1...N, ln (5/4 + sin n)) = o(N)
As stated above, the law of large numbers applies. Moreover,
ln (5/4 + sin n) is bounded by ln 1/4 and ln 9/4, therefore, any
short-term effects whereby large terms dominate the sum are always
smoothed out in the long run as N->oo. Thus
sum(n=1...N, ln (5/4 + sin n)) = O(1)
and sum(n=1...oo, ln (5/4 + sin n)) does not diverge. Thus
product(n=1...oo, (5/4 + sin n)) doesn't diverge or have 0 as a limit.
Moreover, if there were a non-zero limit L such that
lim_N->oo product(n=1...N, (5/4 + sin n)) = L
then
lim_N->oo sum(n=1...N, ln (5/4 + sin n)) = ln L
=> lim_N->oo sum(n=N...oo, ln (5/4 + sin n)) = 0
=> lim_n->oo ln (5/4 + sin n) = 0
=> lim_n->oo 5/4 + sin n = 1
which is false. Thus product(n=1...N, (5/4 + sin n)) has no non-zero
limit as N->oo. To summarise, product(n=1...oo, (5/4 + sin n)) does
not diverge or have a limit.
--
Richard Sabey
----------------------------------------------------------------------
Dear Mr Pegg
Here is a short TeX-file with a few remarks concerning Tommy's problem.
\hsize=12cm \parindent=0pt
The natural numbers are equidistributed mod $2\pi$. This
implies that for any continuous $2\pi$-periodic function $f$ we
have
$$\lim_{N\to\infty}{1\over N}\sum_{n=1}^N f(n) ={1\over 2\pi}
\int_0^{2\pi} f(x)\,dx\ .$$ In particular,
$$\lim_{N\to\infty}{1\over N}\sum_{n=1}^N \log\Bigl({5\over 4}
+
\sin(n)\Bigr) = {1\over 2\pi}\int_0^{2\pi}\log\Bigl({5\over4}
+\sin(x)\Bigr)\,dx =:A\ .$$ The integral $A$ seems to be
nonelementary, but anyway, it computes to 0. (This has to do
with the Pythagorean (3,4,5)-triangle!) It follows that in order
to solve the original problem one would have to use finer
results about equidistribution.
\end
Sincerely yours
Christian Blatter
---------------------------------------
Ed,
I’ll be interested to see a proof of that, since it appears for all intents and
purposes to go to infinity. To be convergent, one would think the increments of
the final value as n increases would diminish. But this equation has a repeating
pattern of geometric increments and decrements, of which the majority are increments.
It would seem to me that the equation steadily increases geometrically, thus
approaching infinity. That’s my initial take on the problem, but I never did any
math past Calculus 3 in college so I’m probably missing something.
Looking forward to the next update,
Clint Weaver
---------------------------------------
Tommy wants proof that the product, for n=1 to infinity, of (5/4 +
sin(n)) goes to neither zero nor infinity. I can offer a start.
If we take the logarithm of this expression, we need that the
logarithm goes to neither -infinity nor +infinity, but instead stays
finite. The log becomes the sum, for n=1 to infinity, of ln(5/4 +
sin(n)).
Now this clearly never converges to a value, but we can find the mean
value of the terms in the sum by collapsing all the points into the
single sine wave from -pi to +pi, and considering that, due to the
irrationality of pi, sin(n) over the integers hits all parts of this
sine curve equally often. In other words, the mean value is the
integral from x = -pi to +pi of ln(5/4 + sin(x)), divided by the 2pi
length of the interval.
Now I don't know how to integrate this (maybe fool around with
integration by parts, where in the first step dv = dx and you get a
still messy integral involving only x and sin(x) and cos(x), or just
let Mathematica grind away on it for a while). For the proof to work,
we need the mean value of the logarithmic terms to be 0. If it is
non-zero, then the sum will diverge, and the infinite product will
correspondingly head to zero or infinity.
Joseph DeVincentis
---------------------------------------
the integral i mentioned earlier today exp (int ( sin -5/4)) is indeed -1 .
also interesting is the product taken over the primes , lineairy variants etc
eg sin(p)+5/4 , sin(2n)+5/4 (behaves mainly the same ) sin ((f(n)) -(f ' (n))^-a +5/4)
especiallly sin(mn)+5/4. that sin(355n) +5/4 diverges is logical . 355 is a
approximate denomitor ( if im correct srr for my english ) for a rational approcing pi .
most cases where sec (m ) is high, sin(mn) +5/4 diverges. but why does m =142 diverges
and 143 does not !! ?
also sin (n+m) +5/4 should behave the same for all positive real m according to
the integral
( integral is taken over period so ) if m is rational of course.
Secondly id like to wunder if that "integral estimate " only works for products with
a flat upper bound and lower bound with attrackts it strongly ( for the product itself
and the function which is " producted " (mulitplied ? ) , I fear that at first sight.
Have you seen A098066 yet ?
also sec(n+1) is never bigger than (n+1) (n+1) i believe strongly. I even expect a
connection with sin+5/4.
greetz
Tommy
>An item to look at:
>
>http://www.math.niu.edu/~rusin/known-math/99/contour_int
>
>--Ed Pegg Jr
------------------------------------------------------------------