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Newsgroups: sci.math
From:
KY <wkfkh... @yahoo.co.jp> Date: Tue, 10 Nov 2009 11:24:04 EST
Local: Tues, Nov 10 2009 11:24 am
Subject: a^n+b^n+c^n=
a + b + c = 0, a^n+b^n+c^n=
(n=2,3,4,6,7,....,14,16,...,2009,..)
a^n+b^n+c^n=
(n=-1,-2,-3,.......)
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Newsgroups: sci.math
From:
"Physics Form" <p... @wako.net> Date: Tue, 10 Nov 2009 12:49:48 -0800
Local: Tues, Nov 10 2009 3:49 pm
Subject: Re: a^n+b^n+c^n=
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Newsgroups: sci.math
From:
Dan Cass <dc... @sjfc.edu> Date: Fri, 13 Nov 2009 12:04:46 EST
Local: Fri, Nov 13 2009 12:04 pm
Subject: Re: a^n+b^n+c^n=
> > a + b + c = 0,
> > a^3 + b^3 + c^3 = 5,
> > a^5 + b^5 + c^5 = 15.
> > -----------------------------------------------
> > a^n+b^n+c^n=
> > (n=2,3,4,6,7,....,14,16,...,2009,..)
> > a^n+b^n+c^n=
> > (n=-1,-2,-3,.......)
> eqs:={a + b + c = 0,a^3 + b^3 + c^3 = 5,a^5 + b^5 +
> solve(eqs):
> > assign(%);
> -0.8361555950 - 0.5454067628 I
> > evalf(b);
> -0.8361555950 + 0.5454067628 I
> > evalf(c);
> 1.672311190
> Then you can plug these a,b,c into a^n + b^n + c^n
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Newsgroups: sci.math
From:
Dan Cass <dc... @sjfc.edu> Date: Fri, 13 Nov 2009 12:02:33 EST
Local: Fri, Nov 13 2009 12:02 pm
Subject: Re: a^n+b^n+c^n=
> a + b + c = 0,
> a^3 + b^3 + c^3 = 5,
> a^5 + b^5 + c^5 = 15.
> -----------------------------------------------
> a^n+b^n+c^n=
> (n=2,3,4,6,7,....,14,16,...,2009,..)
> a^n+b^n+c^n=
> (n=-1,-2,-3,.......)
3 3 3 5 5 5
> solve(eqs):
Now a,b,c are solutions, and order doesn't matter. In one of the orders:
> evalf(a);
-0.8361555950 - 0.5454067628 I
> evalf(b);
-0.8361555950 + 0.5454067628 I
> evalf(c);
1.672311190 Then you can plug these a,b,c into a^n + b^n + c^n and see what you get.
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Newsgroups: sci.math
From:
Jim Ferry <corkleb... @hotmail.com> Date: Fri, 13 Nov 2009 13:48:28 -0800 (PST)
Local: Fri, Nov 13 2009 4:48 pm
Subject: Re: a^n+b^n+c^n=
On Nov 10, 11:24 am, KY <wkfkh... @yahoo.co.jp> wrote:
> a + b + c = 0,
> a^3 + b^3 + c^3 = 5,
> a^5 + b^5 + c^5 = 15.
> -----------------------------------------------
> a^n+b^n+c^n=
> (n=2,3,4,6,7,....,14,16,...,2009,..)
> a^n+b^n+c^n=
> (n=-1,-2,-3,.......)
x(0) = 3, x(1) = 0, x(2) = 18/5,
the forward recursion
x(n) = (9/5) x(n-2) + (5/3) x(n-3) for n > 2,
and the backward recursion
x(n) = (-27/25) x(n+1) + (3/5) x(n+3) for n < 0.
The values of x(n) for n = -3 through 7 are
8442/15625, 729/625, -27/25, 3, 0, 18/5, 5,
x(2009) is the ratio of a 1468- to a 1019-digit
Alternatively, use the values of a, b, and c
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Newsgroups: sci.math
From:
"Robert H. Lewis" <rle... @fordham.edu> Date: Sat, 14 Nov 2009 00:21:27 EST
Local: Sat, Nov 14 2009 12:21 am
Subject: Re: a^n+b^n+c^n=
> On Nov 10, 11:24 am, KY <wkfkh
... @yahoo.co.jp> wrote:
> > a + b + c = 0,
> > a^3 + b^3 + c^3 = 5,
> > a^5 + b^5 + c^5 = 15.
> > -----------------------------------------------
> > a^n+b^n+c^n=
> > (n=2,3,4,6,7,....,14,16,...,2009,..)
> > a^n+b^n+c^n=
> > (n=-1,-2,-3,.......)
> a^n + b^n + c^n = x(n), where x(n) is given by
> x(0) = 3, x(1) = 0, x(2) = 18/5,
> the forward recursion
> x(n) = (9/5) x(n-2) + (5/3) x(n-3) for n > 2,
> and the backward recursion
> x(n) = (-27/25) x(n+1) + (3/5) x(n+3) for n < 0.
> The values of x(n) for n = -3 through 7 are
> 8442/15625, 729/625, -27/25, 3, 0, 18/5, 5,
> x(2009) is the ratio of a 1468- to a 1019-digit
The resultant for a of the original three equations in a, b, and c is 15*a^3 - 27*a - 25. Obviously the resultant for b or c is the same. It is easy to see that a, b, and c must be the three distinct roots of this polynomial. Solving for a^3 shows where the recurrence relation comes from:
> x(n) = (9/5) x(n-2) + (5/3) x(n-3) for n > 2.
I wondered why there is no a^2 term in 15*a^3 - 27*a - 25. If we start over with a + b + c = t,
15*u*a^3 - 15*t^3*a^3 - 15*t*u*a^2 + 15*t^4*a^2 - 9*v*a + 15*t^2*u*a - 6*t^5*a + 9*t*v - 5*u^2 - 5*t^3*u + t^6
So if, for example, t = 0, there is no quadratic term.
Regards,
Robert H. Lewis
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