a^n+b^n+c^n=

[KY]

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,.......)

[Physics Form]

It equals zero, you spamming mathforum fuckwit.

"KY" <wkfk...@yahoo.co.jp> wrote in message
news:1237185793.46235.12578...@gallium.mathforum.org...
"I cannot possibly shut the fuck up. Musatov live in my behind"

[Dan Cass]

> eqs:={a + b + c = 0,a^3 + b^3 + c^3 = 5,a^5 + b^5 +
> c^5 = 15};
>
> solve(eqs):
>
> > assign(%);
> 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.

This is a re-post without some random interposed stuff.

[Dan Cass]

eqs:={a + b + c = 0,a^3 + b^3 + c^3 = 5,a^5 + b^5 + c^5 = 15};

3 3 3 5 5 5
eqs := {a + b + c = 0, a + b + c = 5, a + b + c = 15}

[Jim Ferry]

a^n + b^n + c^n = x(n), where x(n) is given by
the initial conditions

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,
162/25, 15, 7499/375, 189/5.

x(2009) is the ratio of a 1468- to a 1019-digit
number.

Alternatively, use the values of a, b, and c
given by Dan Cass to compute x(n).

[Robert H. Lewis]

> 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
> the initial conditions
>
> 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,
> 162/25, 15, 7499/375, 189/5.
>
> x(2009) is the ratio of a 1468- to a 1019-digit
> number.

Nice solution, Jim Ferry. Perhaps I can expound on it a bit.

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,
a^3 + b^3 + c^3 = u,
a^5 + b^5 + c^5 = v, the resultant is

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
Fordham University