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| master1729... |
 Posted: Wed Nov 04, 2009 7:37 am |
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what do you know about cos(pi/19) ?
nice expressions or properties ? |
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| master1729... |
| Posted: Wed Nov 04, 2009 7:56 am |
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[quote]what do you know about cos(pi/19) ?
nice expressions or properties ?
[/quote]
algebraic order is 9. |
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| Jim Ferry... |
| Posted: Wed Nov 04, 2009 12:30 pm |
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On Nov 4, 4:02 pm, Robert Israel
<isr... at (no spam) math.MyUniversitysInitials.ca> wrote:
[quote]master1729 <tommy1... at (no spam) gmail.com> writes:
what do you know about cos(pi/19) ?
nice expressions or properties ?
algebraic order is 9.
Indeed, its minimal polynomial over the rationals is
-1+10*x+40*x^2-160*x^3-240*x^4+672*x^5+448*x^6-1024*x^7-256*x^8+512*x^9
which might look a bit nicer as
-1+5*z+10*z^2-20*z^3-15*z^4+21*z^5+7*z^6-8*z^7-z^8+z^9
where z = 2*x. Thus 2*cos(pi/19) is an algebraic integer (which
is also obvious from the fact that it is exp(i pi/19) + exp(-i pi/19),
the sum of two roots of unity).
Its conjugates are cos(3 pi/19), cos(5 pi/19), ..., cos(17 pi/19).
--
Robert Israel isr... at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
[/quote]
Let M(z,a) be the minimal (monic) polynomial in z for the algebraic
integer a. Then, re-writing Robert's expression above we have
M(z,2*cos(pi/19)) = M(z,0) M(z,1) M(z,2*cos(pi/9)) M(z,2*sin(pi/5)) -
1
where
M(z,0) = z,
M(z,1) = z - 1,
M(z,2*cos(pi/9)) = z^3 - 3*z - 1, and
M(z,2*sin(pi/5)) = z^4 - 5*z^2 + 5. |
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| Dann Corbit... |
| Posted: Wed Nov 04, 2009 1:59 pm |
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In article
<1721388518.10802.1257356286507.JavaMail.root at (no spam) gallium.mathforum.org>,
tommy1729 at (no spam) gmail.com says...
[quote]
what do you know about cos(pi/19) ?
nice expressions or properties ?
[/quote]
http://oldweb.cecm.sfu.ca/cgi-bin/isc/lookup?
number=.9863613034027223736025091948190671107284815032028763167436651350
63454066494&lookup_type=simple |
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| Robert Israel... |
| Posted: Wed Nov 04, 2009 4:02 pm |
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master1729 <tommy1729 at (no spam) gmail.com> writes:
[quote]what do you know about cos(pi/19) ?
nice expressions or properties ?
algebraic order is 9.
[/quote]
Indeed, its minimal polynomial over the rationals is
-1+10*x+40*x^2-160*x^3-240*x^4+672*x^5+448*x^6-1024*x^7-256*x^8+512*x^9
which might look a bit nicer as
-1+5*z+10*z^2-20*z^3-15*z^4+21*z^5+7*z^6-8*z^7-z^8+z^9
where z = 2*x. Thus 2*cos(pi/19) is an algebraic integer (which
is also obvious from the fact that it is exp(i pi/19) + exp(-i pi/19),
the sum of two roots of unity).
Its conjugates are cos(3 pi/19), cos(5 pi/19), ..., cos(17 pi/19).
--
Robert Israel israel at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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| Jim Ferry... |
| Posted: Thu Nov 05, 2009 6:36 am |
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On Nov 4, 5:30 pm, Jim Ferry <corkleb... at (no spam) hotmail.com> wrote:
[quote]On Nov 4, 4:02 pm, Robert Israel
isr... at (no spam) math.MyUniversitysInitials.ca> wrote:
master1729 <tommy1... at (no spam) gmail.com> writes:
what do you know about cos(pi/19) ?
nice expressions or properties ?
algebraic order is 9.
Indeed, its minimal polynomial over the rationals is
-1+10*x+40*x^2-160*x^3-240*x^4+672*x^5+448*x^6-1024*x^7-256*x^8+512*x^9
which might look a bit nicer as
-1+5*z+10*z^2-20*z^3-15*z^4+21*z^5+7*z^6-8*z^7-z^8+z^9
where z = 2*x. Thus 2*cos(pi/19) is an algebraic integer (which
is also obvious from the fact that it is exp(i pi/19) + exp(-i pi/19),
the sum of two roots of unity).
Its conjugates are cos(3 pi/19), cos(5 pi/19), ..., cos(17 pi/19).
--
Robert Israel isr... at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Let M(z,a) be the minimal (monic) polynomial in z for the algebraic
integer a. Then, re-writing Robert's expression above we have
M(z,2*cos(pi/19)) = M(z,0) M(z,1) M(z,2*cos(pi/9)) M(z,2*sin(pi/5)) -
1
where
M(z,0) = z,
M(z,1) = z - 1,
M(z,2*cos(pi/9)) = z^3 - 3*z - 1, and
M(z,2*sin(pi/5)) = z^4 - 5*z^2 + 5.
[/quote]
More information along these lines is at
http://mathworld.wolfram.com/TrigonometryAngles.html |
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| KY... |
| Posted: Fri Nov 06, 2009 5:06 am |
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Guest
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Table[N[{Cos[Pi/k], FunctionExpand[
Cos[Pi/k]]}], {k, 17, 21}]
=
{{0.9829730996839018, 0.9829730996839018},
{0.984807753012208, 0.9848077530122081 + 0.*I},
{0.9863613034027223, 1.126478970802505 + 0.375464157076925*I},
<-----?
{0.9876883405951378, 0.9876883405951377},
{0.9888308262251285, 0.9888308262251286 + 9.614813431917819*^-17*I}} |
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| Jim Ferry... |
| Posted: Fri Nov 06, 2009 5:40 am |
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On Nov 6, 10:06 am, KY <wkfkh... at (no spam) yahoo.co.jp> wrote:
[quote]Table[N[{Cos[Pi/k], FunctionExpand[
Cos[Pi/k]]}], {k, 17, 21}]
{{0.9829730996839018, 0.9829730996839018},
{0.984807753012208, 0.9848077530122081 + 0.*I},
{0.9863613034027223, 1.126478970802505 + 0.375464157076925*I},
-----?
{0.9876883405951378, 0.9876883405951377},
{0.9888308262251285, 0.9888308262251286 + 9.614813431917819*^-17*I}}
[/quote]
I don't understand why you get this. Apparently
FunctionExpand simplifies Cos[Pi/k] iff it is
constructable with compass and straightedge (i.e.,
iff k is a product of a power of 2 and distinct
Fermat primes).
In particular, FunctionExpand[Cos[Pi/19]] returns
Cos[Pi/19] (in Mathematica 6 and 7). What does
it return on your system that evaluates numerically
to 1.126478970802505 + 0.375464157076925*I ? |
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| AP... |
| Posted: Sat Nov 07, 2009 9:27 am |
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Guest
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[quote]what do you know about cos(pi/19) ?
algebraic order is 9.
Indeed, its minimal polynomial over the rationals is
-1+10*x+40*x^2-160*x^3-240*x^4+672*x^5+448*x^6-1024*x^7-256*x^8+512*x^9
which might look a bit nicer as
-1+5*z+10*z^2-20*z^3-15*z^4+21*z^5+7*z^6-8*z^7-z^8+z^9
where z = 2*x. Thus 2*cos(pi/19) is an algebraic integer (which
is also obvious from the fact that it is exp(i pi/19) + exp(-i pi/19),
the sum of two roots of unity).
[/quote]
the fact that 2cos(pi/19)= exp(i pi/19) + exp(-i pi/19),
=sum of two roots of unity
prove only 2*cos(pi/19) is algebraic
so, there exists k in N^* such k*2*cos(pi/19) is algebraic integer
but, why (without calculation) k=1?
thanks |
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| Robert Israel... |
| Posted: Sun Nov 08, 2009 4:43 am |
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Guest
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AP <marc.pichereau at (no spam) wanadoo.fr.invalid> writes:
[quote]what do you know about cos(pi/19) ?
algebraic order is 9.
Indeed, its minimal polynomial over the rationals is
-1+10*x+40*x^2-160*x^3-240*x^4+672*x^5+448*x^6-1024*x^7-256*x^8+512*x^9
which might look a bit nicer as
-1+5*z+10*z^2-20*z^3-15*z^4+21*z^5+7*z^6-8*z^7-z^8+z^9
where z = 2*x. Thus 2*cos(pi/19) is an algebraic integer (which
is also obvious from the fact that it is exp(i pi/19) + exp(-i pi/19),
the sum of two roots of unity).
the fact that 2cos(pi/19)= exp(i pi/19) + exp(-i pi/19),
=sum of two roots of unity
prove only 2*cos(pi/19) is algebraic
so, there exists k in N^* such k*2*cos(pi/19) is algebraic integer
but, why (without calculation) k=1?
thanks
[/quote]
exp(i pi/19) and exp(-i pi/19) are 38'th roots of unity, i.e. solutions of
z^38 - 1 = 0. This is a monic polynomial with integer coefficients, so they
are algebraic integers. The sum of two algebraic integers is an algebraic
integer.
--
Robert Israel israel at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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| AP... |
| Posted: Sun Nov 08, 2009 4:58 am |
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Guest
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On Sun, 08 Nov 2009 03:43:11 -0600, Robert Israel
<israel at (no spam) math.MyUniversitysInitials.ca> wrote:
[quote]AP <marc.pichereau at (no spam) wanadoo.fr.invalid> writes:
the fact that 2cos(pi/19)= exp(i pi/19) + exp(-i pi/19),
=sum of two roots of unity
prove only 2*cos(pi/19) is algebraic
so, there exists k in N^* such k*2*cos(pi/19) is algebraic integer
but, why (without calculation) k=1?
thanks
exp(i pi/19) and exp(-i pi/19) are 38'th roots of unity, i.e. solutions of
z^38 - 1 = 0. This is a monic polynomial with integer coefficients, so they
are algebraic integers. The sum of two algebraic integers is an algebraic
integer.
Thanks[/quote] |
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