Re: Arctan rational

On Mar 30, 2008 8:45 PM CT, quasi wrote:

On Sun, 30 Mar 2008 17:37:04 -0700 (PDT), gaew
<bennett@xxxxxxxxxxx>
wrote:

On Mar 30, 5:36 pm, gaew <benn...@xxxxxxxxxxx> wrote:
On Mar 30, 5:45 pm, quasi <qu...@xxxxxxxx> wrote:



On Sun, 30 Mar 2008 19:43:44 -0500, quasi <qu...@xxxxxxxx> wrote:
On Sun, 30 Mar 2008 18:41:54 -0500, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

quasi <qu...@xxxxxxxx> writes:

On Sun, 30 Mar 2008 00:32:18 EDT, León-Sotelo
<francisco.lsot...@xxxxxxxxx> wrote:

How can I get this equality:

Arctan (1) = arctan (1 / 2) + arctan (1 / 5) + arctan (1 / 8)?

If you prove it by iterating the identity for tan(u+v), then what you
get is

 tan ( arctan(1/2) + arctan(1/5) + arctan(1/8) ) = 1

But then you also need to verify

   -5*Pi/4 < arctan(1/2) + arctan(1/5) + arctan(1/8) < 5*Pi/4

Of course, that's easy since arctan(1/2), arctan(1/5), arctan(1/8) are
all positive and less than arctan(1) = Pi/4.

But in general, you can't just ignore the issue.

Hmmm ...

Then again, for sums of this special type, maybe you can.

Ok, here's a challenge problem (for which I don't know the answer):

Do there exist _distinct_ positive integers d_1, ..., d_n, such that,
letting a_k denote arctan(1/d_k),

   tan(a_1 + ... + a_n) = 1

but

   a_1 + ... + a_n > Pi/4

?

Given d, we can write arctan(1/d) = arctan(1/(q+d)) + arctan(1/(p+d)) if
d^2 + 1 = p q.

Starting from arctan(1/2) + arctan(1/5) + arctan(1/8) = pi/4
we can use this to get other ways of writing pi/4 as sums of distinct
arctan(1/d_k).  And then
5 pi/4
= arctan(1)
+ arctan(1/2) + arctan(1/5) + arctan(1/8)
+ arctan(1/3)+arctan(1/7)+arctan(1/6)+arctan(1/31)+arctan(1/9)+arctan(1/73)
+
arctan(1/4)+arctan(1/10)+arctan(1/12)+arctan(1/13)+arctan(1/14)

 +arctan(1/17)+arctan(1/21)+arctan(1/31)+arctan(1/43)+
arctan(1/57)
 +arctan(1/73)+arctan(1/91)+arctan(1/183)
+arctan(1/11)+arctan(1/15)+arctan(1/18)+arctan(1/19)+arctan(1/22)

+arctan(1/23)+arctan(1/24)+arctan(1/25)+arctan(1/27)+arctan(1/28)

+arctan(1/30)+arctan(1/32)+arctan(1/41)+arctan(1/44)+arctan(1/46)

+arctan(1/47)+arctan(1/58)+arctan(1/74)+arctan(1/75)+arctan(1/83)

+arctan(1/92)+arctan(1/111)+arctan(1/119)+arctan(1/157)+arctan(1/162)

+arctan(1/184)+arctan(1/211)+arctan(1/242)+arctan(1/288)+arctan(1/463)

+arctan(1/553)+arctan(1/757)+arctan(1/993)+arctan(1/1893)+arctan(1/3307)

+arctan(1/5403)+arctan(1/8373)+arctan(1/33673)

Not quite.

You have arctan(1/73) used twice.

My current feeling is that it can't be done.

quasi

So take Robert's representation and replace arctan(1/73) with
arctan(1/78)+arctan(1/1139).

Gaew

and similarly with 31....Robert beat me to it!

Thus, based on the latest data, my current feeling is
that it _can_ be done!

quasi

...Robert never ceases to amaze me ^_^
.

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