arctan(2) problem
Martin Gasevski
Can you help me out with this problem: What is arctan(2) and
arctan(0.5)? The solution should be in terms of pi, like arctan(1) =
pi/4.
Martin
Robert Israel
Martin Gasevski <m_g...@hotmail.com> wrote:
>Can you help me out with this problem: What is arctan(2) and
>arctan(0.5)? The solution should be in terms of pi, like arctan(1) =
>pi/4.
arctan(2) is not a rational multiple of pi. I don't think there's any
simpler way to express it than arctan(2). Of course arctan(1/2),
arctan(2) and pi are related...
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Martin Gasevski
how about this?
arctan(2) = pi/4 + arctan(1/3)
arctan(2) = pi/2 - arctan(1/2)
do you know any solution for arctan(1/2) or arctan(1/3) in terms of pi?
Martin
Jan Kristian Haugland
Martin Gasevski wrote:
> isr...@math.ubc.ca (Robert Israel) wrote in message news:<9irdh0$1ti$1...@nntp.itservices.ubc.ca>...
> > In article <d3e9a11f.01071...@posting.google.com>,
> > Martin Gasevski <m_g...@hotmail.com> wrote:
> >
> > >Can you help me out with this problem: What is arctan(2) and
> > >arctan(0.5)? The solution should be in terms of pi, like arctan(1) =
> > >pi/4.
> >
> > arctan(2) is not a rational multiple of pi. I don't think there's any
> > simpler way to express it than arctan(2). Of course arctan(1/2),
> > arctan(2) and pi are related...
>
> how about this?
> arctan(2) = pi/4 + arctan(1/3)
> arctan(2) = pi/2 - arctan(1/2)
>
> do you know any solution for arctan(1/2) or arctan(1/3) in terms of pi?
If he did, would he have written what he wrote?
--
J K Haugland
http://home.hia.no/~jkhaug00
Stephen Montgomery-Smith
OK, I spent a bit of time with this problem. This is what I came up
with so far.
We have the formula atan(u*v) = atan(u) + atan(v), where
u*v = (u+v)/(1-uv).
Notice that * satisfies the commutative and associative law. We know
atan(1)=pi/4 and atan(1/sqrt(3))=pi/6. Also, we can solve the equation
u*u=v
using the quadratic formula - let us write u=v^*(1/2), and in general
write u*^n for u*u*u*...*u. So the question is the following: can we
write 2 as
1^*(m/2^k) * (1/sqrt(3))^*(n/2^l)
for some integers m,n,k,l?
I experimented a bit, and couldn't do it. That doesn't mean that it is
impossible though.
--
Stephen Montgomery-Smith
ste...@math.missouri.edu
http://www.math.missouri.edu/~stephen
Stephen Montgomery-Smith
tabulated 2^*n for n up to 100, and it doesn't look like this ever
happens. There should be a simple proof of this.
Stephen Montgomery-Smith
integer k?
Stephen Montgomery-Smith
OK, I don't think that atan(2) can be a rational multiple of pi. For
suppose that it is, say m/n pi. Then (1+2i) = sqrt(5) exp(i m/n pi), so
(1+2i)^{2n} = (-3+4i)^n = 5^n. That is, if x=(-3+4i)/5, then x^n=1.
But 5x^2+6x+5=0, which means that 5x^2+6x+5 must divide into x^n-1. But
that is not possible, because polynomials that divide into x^n-1 must
have their leading and trailing coefficients equal to plus or minus 1.
Robert Israel
Stephen Montgomery-Smith <ste...@math.missouri.edu> wrote:
>OK, I don't think that atan(2) can be a rational multiple of pi.
Note that if tan(r pi) = t, then cos(2 r pi) = (1-t^2)/(1+t^2).
If r is rational, 2 cos(2 r pi) = exp(2 r pi i) + exp(-2 r pi i)
is an algebraic integer, and therefore can't be rational unless it
is an integer. So the only cases where r is rational and tan(r pi)
is either rational or the square root of a rational are when
2 cos(2 r pi) is an integer; these are the "obvious" cases where
tan(r pi) is 0, (+/-) 1/sqrt(3), (+/-) 1 or (+/-) sqrt(3). The arctan
of any other rational or square root of a rational is not a rational
multiple of pi.
Robert Israel isr...@math.ubc.ca
Leroy Quet
>Martin
I don't know, but MAYBE this will help.
If arctan(x) and arctan(y) are known, and w = (x + y) /(1 - x y),
then arctan(w) = arctan(x) + arctan(y).
So, by solving for y, if arctan(x) and arctan((w-x)/(1+wx)) both have
closed forms, then arctan(w) has a closed form:
arctan(w) = arctan(x) + arctan((w-x)/(1+wx)).
Thanks,
Leroy Quet
(Spam Block: Extra Q added.)
Robin Chapman
<ste...@math.missouri.edu> =====
>OK, I was a bit stupid. Suppose atan(2) = m/n pi. Then 2^*n = 0. I
>tabulated 2^*n for n up to 100, and it doesn't look like this ever
>happens. There should be a simple proof of this.
>
If x is a rational muliple of pi, then e^(ix) = (1+2i)/sqrt(5)
is a root of unity. But (1+2i)/sqrt(5) lies in the field
generated by the 20-th roots of unity. By standard theory of
cylotomic fields, the only roots of unity in this field are the
20th roots of unity, and this isn't one of them. Contradiction!
------------------------------------------------------------
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"His mind has been corrupted by colours, sounds and shapes."
The League of Gentlemen
Robert Israel
Robert Israel <isr...@math.ubc.ca> wrote:
>In article <3B524B8D...@math.missouri.edu>,
>Stephen Montgomery-Smith <ste...@math.missouri.edu> wrote:
>>OK, I don't think that atan(2) can be a rational multiple of pi.
>Note that if tan(r pi) = t, then cos(2 r pi) = (1-t^2)/(1+t^2).
>If r is rational, 2 cos(2 r pi) = exp(2 r pi i) + exp(-2 r pi i)
>is an algebraic integer, and therefore can't be rational unless it
>is an integer. So the only cases where r is rational and tan(r pi)
>is either rational or the square root of a rational are when
>2 cos(2 r pi) is an integer; these are the "obvious" cases where
>tan(r pi) is 0, (+/-) 1/sqrt(3), (+/-) 1 or (+/-) sqrt(3). The arctan
>of any other rational or square root of a rational is not a rational
>multiple of pi.
Actually, more is true: by the Gelfond-Schneider Theorem,
if x is algebraic and arctan(x) is not a rational multiple of pi, then
arctan(x), pi and 1 are linearly independent over the field of
algebraic numbers. Thus there do not exist algebraic numbers a and
b such that arctan(x) = a pi + b.