Discussion:
Follow-up on Leo Wapner's product of tangents problem...
(too old to reply)
David Bernier
2017-06-17 05:11:59 UTC
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Leo Wapner asked in sci.math:

a(n) = tan(1) tan(2) ... tan(n)
does a(n) converge, or does it diverge?

Dec 28, 2006 , sci.math archives:

< http://mathforum.org/kb/message.jspa?messageID=5461260 > .

In 2009, Jim Ferry proved in sci.math that:

for any integers n,k > 0 with n even and k relatively prime to n, we have:

prod(X=1,n-1,tan(X*(k/n)*Pi/2) ) = (-1)^((k-1)/2)

and based on his proof, I sketched a proof that if n,k > 0 with n odd
and k odd and n relatively prime to n, then:

| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1.

So, if n, k > 0 , with k odd and gcd(n,k) = 1, then:

| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1 .

If , in the above, n/k is sufficiently close to Pi/2 ,
in other words if (k/n)*Pi/2 is sufficiently close to 1,
then | a(n) | is of the order of , well it depends how close
k*(Pi/2) is to n.... and k odd so tan( k *Pi/2) undefined (alias +/- oo ).

cf.:
< http://mathforum.org/kb/message.jspa?messageID=6851966 > .

===

This motivates the question:

If xi is an irrational number, for what values of C>0 , if any,
can one be assured that there exist infinitely many co-prime integers
m, n with n>0, n odd, m positive or negative, such that:

| xi - m/n | < C/(n^2) ?

This has a resemblance to the question answered by Hurwitz's theorem in
Diophantine approximation,
< https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) > ,

except here, we require the denominator n to be an *odd* number ...

David Bernier
David Bernier
2017-06-17 06:10:08 UTC
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Post by David Bernier
a(n) = tan(1) tan(2) ... tan(n)
does a(n) converge, or does it diverge?
< http://mathforum.org/kb/message.jspa?messageID=5461260 > .
prod(X=1,n-1,tan(X*(k/n)*Pi/2) ) = (-1)^((k-1)/2)
and based on his proof, I sketched a proof that if n,k > 0 with n odd
| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1.
| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1 .
If , in the above, n/k is sufficiently close to Pi/2 ,
in other words if (k/n)*Pi/2 is sufficiently close to 1,
then | a(n) | is of the order of , well it depends how close
k*(Pi/2) is to n.... and k odd so tan( k *Pi/2) undefined (alias +/- oo ).
< http://mathforum.org/kb/message.jspa?messageID=6851966 > .
===
If xi is an irrational number, for what values of C>0 , if any,
can one be assured that there exist infinitely many co-prime integers
| xi - m/n | < C/(n^2) ?
This has a resemblance to the question answered by Hurwitz's theorem in
Diophantine approximation,
< https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) > ,
except here, we require the denominator n to be an *odd* number ...
I haven't personally seen a solution .

Leo Wapner then asked on January 1, 2007 about another product
series, that one involving sines of the positive integers,

ref.: archives,

Topic: "Sequence: a(n)=(2 sin 1)(2 sin 2)(2 sin 3)...(2 sin n) =
(2^n)(sin 1)(sin 2)(sin 3) ... (sin n) "


< http://mathforum.org/kb/message.jspa?messageID=5464104 > .

David Bernier
David Bernier
2017-06-17 15:14:06 UTC
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Post by David Bernier
a(n) = tan(1) tan(2) ... tan(n)
does a(n) converge, or does it diverge?
< http://mathforum.org/kb/message.jspa?messageID=5461260 > .
prod(X=1,n-1,tan(X*(k/n)*Pi/2) ) = (-1)^((k-1)/2)
and based on his proof, I sketched a proof that if n,k > 0 with n odd
| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1.
| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1 .
If , in the above, n/k is sufficiently close to Pi/2 ,
in other words if (k/n)*Pi/2 is sufficiently close to 1,
then | a(n) | is of the order of , well it depends how close
k*(Pi/2) is to n.... and k odd so tan( k *Pi/2) undefined (alias +/- oo ).
< http://mathforum.org/kb/message.jspa?messageID=6851966 > .
===
If xi is an irrational number, for what values of C>0 , if any,
can one be assured that there exist infinitely many co-prime integers
| xi - m/n | < C/(n^2) ?
This has a resemblance to the question answered by Hurwitz's theorem in
Diophantine approximation,
< https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) > ,
except here, we require the denominator n to be an *odd* number ...
Is it true that for irrational (real) numbers,
an irrational alpha has a periodic continued fraction expansion if and
only if "alpha" is a quadratic irrational, i.e. those irrationals
which are algebraic, real, and with a minimal polynomial
of degree two in Z[x] or Q[x] ?

thanks,

David Bernier
FredJeffries
2017-06-17 16:05:44 UTC
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Post by David Bernier
Is it true that for irrational (real) numbers,
an irrational alpha has a periodic continued fraction expansion if and
only if "alpha" is a quadratic irrational, i.e. those irrationals
which are algebraic, real, and with a minimal polynomial
of degree two in Z[x] or Q[x] ?
Yep.
http://wstein.org/edu/124/lectures/lecture19/lecture19/node2.html
David Bernier
2017-06-17 20:14:39 UTC
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Post by FredJeffries
Post by David Bernier
Is it true that for irrational (real) numbers,
an irrational alpha has a periodic continued fraction expansion if and
only if "alpha" is a quadratic irrational, i.e. those irrationals
which are algebraic, real, and with a minimal polynomial
of degree two in Z[x] or Q[x] ?
Yep.
http://wstein.org/edu/124/lectures/lecture19/lecture19/node2.html
I see Lagrange proved that quadratic irrationals have a
periodic continued fraction expansion, and the proof is
quite intricate.

I'll note a further question on odd denominator
Diophantine approximations, although I have no
idea how to tackle it:

For what values of the constant C > 0 is it true that,
for any quadratic irrational alpha, there exist
infinitely many co-prime m, n with n>0 AND n odd such that:

| alpha - m/n | < C/(n^2) ?

Note: If we should remove the proviso that n must be odd
and instead allow either parity (odd/even) for the
denominator 'n' , then Hurwitz's theorem in number
theory shows C = 1/sqrt(5) is allowable:

< https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) > .

I have no idea if that helps for the "n odd only" problem.

David Bernier
David Bernier
2017-06-18 00:12:58 UTC
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Post by David Bernier
a(n) = tan(1) tan(2) ... tan(n)
does a(n) converge, or does it diverge?
< http://mathforum.org/kb/message.jspa?messageID=5461260 > .
prod(X=1,n-1,tan(X*(k/n)*Pi/2) ) = (-1)^((k-1)/2)
and based on his proof, I sketched a proof that if n,k > 0 with n odd
| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1.
| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1 .
If , in the above, n/k is sufficiently close to Pi/2 ,
in other words if (k/n)*Pi/2 is sufficiently close to 1,
then | a(n) | is of the order of , well it depends how close
k*(Pi/2) is to n.... and k odd so tan( k *Pi/2) undefined (alias +/- oo ).
< http://mathforum.org/kb/message.jspa?messageID=6851966 > .
===
If xi is an irrational number, for what values of C>0 , if any,
can one be assured that there exist infinitely many co-prime integers
| xi - m/n | < C/(n^2) ?
This has a resemblance to the question answered by Hurwitz's theorem in
Diophantine approximation,
< https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) > ,
except here, we require the denominator n to be an *odd* number ...
I wouldn't be surprised if the partial products
of Leonard Wapner's product of tangent are
unbounded.


Heuristics suggest c.f.s. convergents of Pi/2 with even denominator
can be put to use in obtaining (presumably) large partial
products: if the denominator is odd, look at the numerator
and if the deominator is even, look at the numerator minus one:

the heuristics:

For the 10 consecutive convergents in the continued
fraction expansion of pi/2, denoted p_j / q_j for 1 <=j <=10:

namely:

11/7, 344/219, 355/226, 51819/32989, 52174/33215, 260515/165849,
573204/364913, 4846147/3085153, 5419351/3450066, 37362253/23785549

If q_j as above is odd, I take the product tan(1) tan(2)... tan(p_j)
and
if q_j as above is even (226 and 3450066), I take the
product tan(1) tan(2) ... tan( p_j -1) .

The results , in absolute value, are copied below:

abs(prod(X=1,11, tan(X))) = 216.945

abs(prod(X=1,344, tan(X))) = 371.312

abs(prod(X=1,355-1, tan(X))) = 356.710 // 226 even denominator

abs(prod(X=1,51819, tan(X))) = 935.194

abs(prod(X=1,52174, tan(X))) = 476029.121

abs(prod(X=1,260515, tan(X))) = 315387.380

abs(prod(X=1,573204, tan(X))) = 3749154.523

abs(prod(X=1,4846147, tan(X))) = 5417369.011

abs(prod(X=1,5419351-1, tan(X))) = 5820292.563 // 3450066 even denom.

abs(prod(X=1,37362253, tan(X))) = 12055170.596

David Bernier
David Bernier
2017-06-18 00:54:52 UTC
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Post by David Bernier
a(n) = tan(1) tan(2) ... tan(n)
does a(n) converge, or does it diverge?
< http://mathforum.org/kb/message.jspa?messageID=5461260 > .
prod(X=1,n-1,tan(X*(k/n)*Pi/2) ) = (-1)^((k-1)/2)
[snip]

Leo Wapner asked almost exactly the same question in 1998 in the form:

Does tan(1) tan(2) ... tan(n) converge in absolute value as n goes to
infinity?

From Google Groups:

Lim (as n-> inf) of (tan 1)(tan 2)(tan 3)...(tan n) = ?

< https://groups.google.com/forum/#!topic/sci.math/VJ2mLXOISus > .

David Bernier
David Bernier
2017-06-18 07:09:36 UTC
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Post by David Bernier
Post by David Bernier
a(n) = tan(1) tan(2) ... tan(n)
does a(n) converge, or does it diverge?
< http://mathforum.org/kb/message.jspa?messageID=5461260 > .
prod(X=1,n-1,tan(X*(k/n)*Pi/2) ) = (-1)^((k-1)/2)
and based on his proof, I sketched a proof that if n,k > 0 with n odd
| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1.
| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1 .
If , in the above, n/k is sufficiently close to Pi/2 ,
in other words if (k/n)*Pi/2 is sufficiently close to 1,
then | a(n) | is of the order of , well it depends how close
k*(Pi/2) is to n.... and k odd so tan( k *Pi/2) undefined (alias +/- oo ).
< http://mathforum.org/kb/message.jspa?messageID=6851966 > .
===
If xi is an irrational number, for what values of C>0 , if any,
can one be assured that there exist infinitely many co-prime integers
| xi - m/n | < C/(n^2) ?
This has a resemblance to the question answered by Hurwitz's theorem in
Diophantine approximation,
< https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) > ,
except here, we require the denominator n to be an *odd* number ...
I wouldn't be surprised if the partial products
of Leonard Wapner's product of tangent are
unbounded.
Heuristics suggest c.f.s. convergents of Pi/2 with even denominator
can be put to use in obtaining (presumably) large partial
products: if the denominator is odd, look at the numerator
For the 10 consecutive convergents in the continued
11/7, 344/219, 355/226, 51819/32989, 52174/33215, 260515/165849,
573204/364913, 4846147/3085153, 5419351/3450066, 37362253/23785549
If q_j as above is odd, I take the product tan(1) tan(2)... tan(p_j)
and
if q_j as above is even (226 and 3450066), I take the
product tan(1) tan(2) ... tan( p_j -1) .
abs(prod(X=1,11, tan(X))) = 216.945
abs(prod(X=1,344, tan(X))) = 371.312
abs(prod(X=1,355-1, tan(X))) = 356.710 // 226 even denominator
And the PARI/gp calculator gives numerically:

? abs(prod(X=1, 354, tan(X*(226/355)*(Pi/2))))
%3 = 355.000000000000000000000000000000000000000000000000000000000

or in other words,
for beta = (113/355)*pi,

| tan(beta) tan(2 beta) ... tan(354 beta) | ~= 355 .

beta ~= 0.999999915086, very close to 1.

David Bernier
Post by David Bernier
abs(prod(X=1,51819, tan(X))) = 935.194
abs(prod(X=1,52174, tan(X))) = 476029.121
abs(prod(X=1,260515, tan(X))) = 315387.380
abs(prod(X=1,573204, tan(X))) = 3749154.523
abs(prod(X=1,4846147, tan(X))) = 5417369.011
abs(prod(X=1,5419351-1, tan(X))) = 5820292.563 // 3450066 even denom.
abs(prod(X=1,37362253, tan(X))) = 12055170.596
David Bernier
David Bernier
2017-06-18 13:10:35 UTC
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Post by David Bernier
Post by David Bernier
Post by David Bernier
a(n) = tan(1) tan(2) ... tan(n)
does a(n) converge, or does it diverge?
< http://mathforum.org/kb/message.jspa?messageID=5461260 > .
prod(X=1,n-1,tan(X*(k/n)*Pi/2) ) = (-1)^((k-1)/2)
and based on his proof, I sketched a proof that if n,k > 0 with n odd
| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1.
| prod(X=1,n-1,tan(X*(k/n)*Pi/2)) | = 1 .
If , in the above, n/k is sufficiently close to Pi/2 ,
in other words if (k/n)*Pi/2 is sufficiently close to 1,
then | a(n) | is of the order of , well it depends how close
k*(Pi/2) is to n.... and k odd so tan( k *Pi/2) undefined (alias +/- oo ).
< http://mathforum.org/kb/message.jspa?messageID=6851966 > .
===
If xi is an irrational number, for what values of C>0 , if any,
can one be assured that there exist infinitely many co-prime integers
| xi - m/n | < C/(n^2) ?
This has a resemblance to the question answered by Hurwitz's theorem in
Diophantine approximation,
< https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) > ,
except here, we require the denominator n to be an *odd* number ...
I wouldn't be surprised if the partial products
of Leonard Wapner's product of tangent are
unbounded.
Heuristics suggest c.f.s. convergents of Pi/2 with even denominator
can be put to use in obtaining (presumably) large partial
products: if the denominator is odd, look at the numerator
For the 10 consecutive convergents in the continued
11/7, 344/219, 355/226, 51819/32989, 52174/33215, 260515/165849,
573204/364913, 4846147/3085153, 5419351/3450066, 37362253/23785549
If q_j as above is odd, I take the product tan(1) tan(2)... tan(p_j)
and
if q_j as above is even (226 and 3450066), I take the
product tan(1) tan(2) ... tan( p_j -1) .
abs(prod(X=1,11, tan(X))) = 216.945
abs(prod(X=1,344, tan(X))) = 371.312
abs(prod(X=1,355-1, tan(X))) = 356.710 // 226 even denominator
? abs(prod(X=1, 354, tan(X*(226/355)*(Pi/2))))
%3 = 355.000000000000000000000000000000000000000000000000000000000
or in other words,
for beta = (113/355)*pi,
| tan(beta) tan(2 beta) ... tan(354 beta) | ~= 355 .
beta ~= 0.999999915086, very close to 1.
David Bernier
Post by David Bernier
abs(prod(X=1,51819, tan(X))) = 935.194
abs(prod(X=1,52174, tan(X))) = 476029.121
abs(prod(X=1,260515, tan(X))) = 315387.380
abs(prod(X=1,573204, tan(X))) = 3749154.523
abs(prod(X=1,4846147, tan(X))) = 5417369.011
abs(prod(X=1,5419351-1, tan(X))) = 5820292.563 // 3450066 even denom.
Furthermore, the PARI/gp calculator finds that numerically:

? abs(prod(X=1, 5419350 , tan(X*(3450066/5419351)*(Pi/2))))
%4 = 5419351.00000000000000000000000000000000000000000000000000000

or in words:

if beta := (1725033/5419351)*pi ~= 0.99999999999999295109782 ,

then:
| tan(beta) tan(2 beta) ... tan(5419350 beta) | ~= 5419351 .

Even when there are identities, the challenge
is to get a good enough estimate of
| tan(1) tan(2) ... tan(5419350 ) | given that

| tan(beta) tan(2 beta) ... tan(5419350) | ~= 5419351
and that
| beta - 1 | ~= 7.0489 x 10^(-15) .

d/dx ( log( |tan(x) | ) ) = ? +/- 2/(sin(2x)) .

Then if all goes well,

log( | tan(1) tan(2) ... tan(5419350 ) | ) - log( | tan(beta) tan(2
beta) ... tan(5419350) |)

is the sum of 5419350 definite integrals of
d/dx log( | tan(x) tan(2x) ... tan(5419350x ) | ) =
d/dx log( | tan(x)|) + .... d/dx (log( | tan(5419350x)|),
with the j'th interval beginning at
j*beta and ending at j, for 1 <= j <= 5419350 .
These intervals are of course very very tiny,
but on the other hand d/dx log( |tan(x)| ) can
take both large positive values, and "large" negative values.

And the work isn't finished ...

David Bernier
Post by David Bernier
Post by David Bernier
abs(prod(X=1,37362253, tan(X))) = 12055170.596
David Bernier
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