Re: Arctan rational

On Sun, 30 Mar 2008 21:49:04 -0700 (PDT), gaew <bennett@xxxxxxxxxxx>
wrote:

On Mar 30, 8:59 pm, quasi <qu...@xxxxxxxx> wrote:
On Sun, 30 Mar 2008 19:32:18 -0500, Robert Israel



<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
quasi <qu...@xxxxxxxx> writes:

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.

Oops: also arctan(1/31).  
But that's easily rectified.

arctan(1)+arctan(1/2)+arctan(1/3)+arctan(1/4)+arctan(1/5)+arctan(1/6)
+arctan(1/7)+arctan(1/8)+arctan(1/9)+arctan(1/10)+arctan(1/11)+arctan(1/12 )
+arctan(1/13)+arctan(1/14)+arctan(1/15)+arctan(1/17)+arctan(1/18)
+arctan(1/19)+arctan(1/21)+arctan(1/22)+arctan(1/23)+arctan(1/24)
+arctan(1/25)+arctan(1/27)+arctan(1/28)+arctan(1/30)+arctan(1/31)
+arctan(1/32)+arctan(1/33)+arctan(1/41)+arctan(1/43)+arctan(1/44)
+arctan(1/46)+arctan(1/47)+arctan(1/57)+arctan(1/58)+arctan(1/73)
+arctan(1/74)+arctan(1/75)+arctan(1/78)+arctan(1/83)+arctan(1/91)
+arctan(1/92)+arctan(1/111)+arctan(1/119)+arctan(1/157)+arctan(1/162)
+arctan(1/183)+arctan(1/184)+arctan(1/211)+arctan(1/242)+arctan(1/288)
+arctan(1/463)+arctan(1/512)+arctan(1/553)+arctan(1/757)+arctan(1/993)
+arctan(1/1139)+arctan(1/1893)+arctan(1/3307)+arctan(1/5403)+arctan(1/8373 )
+arctan(1/33673)

Not bad.

So you think you're a true arctan master?

(hehe)

Try this ...

Do there exist _distinct_ positive integers d_1, ..., d_n, such that,
for S = {1,..., n}, and letting a_k denote arctan(1/d_k),

   (1) tan(sum({a_k | k in S})) = 1

   (2) for every nonempty proper subset S' of S,

            tan(sum({a_k | k in S'})) =/= 1

   (3) sum({a_k | k in S}) > Pi/4

?

quasi

Ahhh, the plot thickens!

I believe the answer is still "yes". Start with the obvious identity

5 Pi/4 = 44 arctan (1/11) - 9 arctan (1/239) - 16 arctan (1/682) - 12
arctan (12943).

Yep, totally obvious -- in fact, if I recall correctly, that identity
was taught in elementary school.

Then expand 9 of our arctan(1/11)'s until we cancel off the
arctan(1/239)'s, 16 to cancel the arctan(1/682)'s, etc. We'll be left
with a sum of arctan(1/n)'s which total 5 Pi/4. By suitable expansion,
I suspect we can again ensure that they are distinct and (I think)
have no subsums equal to Pi/4.

Hmmm ...

Looks like a job for Maple.

But I think Maple may balk since it knows the deal -- if it succeeds
at this problem (really just a warmup), it gets the real problem
(truly formidable, trust me).

quasi
.

Relevant Pages

  • Re: Arctan rational
    ... result: Pi/4 ... Next, a warmup ... ... Although Maple had to be asked to simplify it, ...
    (sci.math)
  • Re: Arctan rational
    ... On Mar 30, 2008 8:45 PM CT, quasi wrote: ... letting a_k denote arctan, ... we can use this to get other ways of writing pi/4 as sums of distinct ... ...Robert never ceases to amaze me ^_^ ...
    (sci.math)
  • Re: Arctan rational
    ... result: Pi/4 ... Note -- Maple simplifies it automatically. ... Next, a warmup ... ... Although Maple had to be asked to simplify it, ...
    (sci.math)
  • Re: Arctan rational
    ... letting a_k denote arctan, ... we can use this to get other ways of writing pi/4 as sums of distinct ...
    (sci.math)