Re: Arctan rational

On Sun, 30 Mar 2008 18:41:54 -0500, Robert Israel
<israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

quasi <quasi@xxxxxxxx> writes:

On Sun, 30 Mar 2008 00:32:18 EDT, León-Sotelo
<francisco.lsotelo@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.

quasi
.

Relevant Pages

  • Re: Arctan rational
    ... If you prove it by iterating the identity for tan, ... But then you also need to verify ... Then again, for sums of this special type, maybe you can. ... we can use this to get other ways of writing pi/4 as sums of distinct ...
    (sci.math)
  • Re: Arctan rational
    ... If you prove it by iterating the identity for tan, ... But then you also need to verify ... Then again, for sums of this special type, maybe you can. ... we can use this to get other ways of writing pi/4 as sums of distinct ...
    (sci.math)
  • Re: Arctan rational
    ... If you prove it by iterating the identity for tan, ... Then again, for sums of this special type, maybe you can. ... we can use this to get other ways of writing pi/4 as sums of distinct ... Oops: also arctan. ...
    (sci.math)
  • Re: Arctan rational
    ... quasi writes: ... If you prove it by iterating the identity for tan, ... Then again, for sums of this special type, maybe you can. ... we can use this to get other ways of writing pi/4 as sums of distinct ...
    (sci.math)