Re: tan n > n (corrected)

Tim Little wrote:
On 2009-03-07, lwapner2@xxxxxxxxx <lwapner2@xxxxxxxxx> wrote:
Let n be a natural number. How many solutions are there to tan n > n?

I conjecture infinitely many. Suitable n are numerators in "good"
rational approximations to pi/2 from below with odd denominator.
E.g. your value 260515 corresponds to 260515/165849.

Using continued fraction approximations, the first few suitable values
for n are 1, 260515, 37362253, 122925461, 534483448, 3083975227,
902209779836, 74357078147863, 214112296674652, 18190586279576483,
248319196091979065, and 1108341089274117551.

The reason why I can only conjecture that there are infinitely many
values is that not all continued fraction approximants are suitable.
For example, every second one gives very negative tan(n) values
instead of positive, and an even denominator yields a very large
1/tan(n) value instead of tan(n).

It may be possible that past some point all the denominators in odd
positions in the sequence are even, or that the convergents are not
"close enough". The set of real numbers with either property of their
convergents has measure zero, but I cannot prove that pi/2 does not
belong to either set.

In the article
" Tangent Sequences, World Records, pi, and the Meaning of Life"
by Ira Rosenholtz,
http://www.jstor.org/pss/2690793

Result 1. on the first page is:
lim_{ n -> oo} tan(n)/n does not exist.

So if A = liminf_{ n -> oo } tan(n)/n (in the extended reals, allowing -oo and +oo)
and
B = limsup_{n -> oo } tan(n)/n , then

A < B by Result 1. in the Rosenholtz article .

If B > 1, then tan(n)/n > 1 for infinitely many n.

Concerning Result 1 , n can be close to k*pi, for some integer k, so then tan(n) ~= 0.

I'm not sure what can be said about A and B besides A < B ...

David Bernier

.

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