Re: Solving the Quintic: Correction to x^5 + 20 x + 32 == 0 root example

• To: mathgroup at smc.vnet.net
• Subject: [mg95490] Re: Solving the Quintic: Correction to x^5 + 20 x + 32 == 0 root example
• From: "Q.E.D." <aoe at netzero.net>
• Date: Tue, 20 Jan 2009 05:44:55 -0500 (EST)
• References: <gkn5j8\$igv\$1@smc.vnet.net>

```> All of the roots are given by evaluating this:
>
> Module[{prim1, prim2, prim3, prim4, r1, r2, r3, r4},
>  r1 = -(2500*5^(1/2) + 250* (50 - 10*5^(1/2))^(1/2) -
>       750 *(50 + 10*5^(1/2))^(1/2))^(1/5);
>  r2 = (2500*5^(1/2) + 750 *(50 - 10*5^(1/2))^(1/2) +
>      250 *(50 + 10*5^(1/2))^(1/2))^(1/5);
>  r3 = -(-2500*5^(1/2) + 750 *(50 - 10*5^(1/2))^(1/2) +
>       250* (50 + 10*5^(1/2))^(1/2))^(1/5);
>  r4 = -(2500*5^(1/2) - 250* (50 - 10*5^(1/2))^(1/2) +
>       750 *(50 + 10*5^(1/2))^(1/2))^(1/5);
>  prim1 = (-1 - 5^(1/2) + 2^(1/2)*I*(5 - 5^(1/2))^(1/2))/4;
>  prim2 = (-1 + 5^(1/2) - 2^(1/2)*I*(5 + 5^(1/2))^(1/2))/4;
>  prim3 = (-1 + 5^(1/2) + 2^(1/2)*I*(5 + 5^(1/2))^(1/2))/4;
>  prim4 = (-1 - 5^(1/2) - 2^(1/2)*I*(5 - 5^(1/2))^(1/2))/4;
>  {x -> #} & /@ ({{r1, r2, r3, r4}.{1, 1, 1, 1},
>      {r1, r2, r3, r4}.{prim2, prim4, prim1, prim3},
>      {r1, r2, r3, r4}.{prim3, prim1, prim4, prim2},
>      {r1, r2, r3, r4}.{prim1, prim2, prim3, prim4},
>      {r1, r2, r3, r4}.{prim4, prim3, prim2, prim1}}/5)]

Further simplification is possible:
750 *(50 + 10*5^(1/2))^(1/2) - 250* (50 - 10*5^(1/2))^(1/2) ->
500*(125 - 10*5^(1/2))^(1/2)

750 *(50 - 10*5^(1/2))^(1/2) + 250 *(50 + 10*5^(1/2))^(1/2) ->
500*(125 + 10*5^(1/2))^(1/2)

Also (5^(1/2))^5 can be factored out resulting in all of the roots
given by evaluating this:

Module[{prim1, prim2, prim3, prim4, r1, r2, r3, r4},
r1 = -5^(1/2)*(100 - 20*(25 - 2*5^(1/2))^(1/2))^(1/5);
r2 = 5^(1/2)*(100 + 20*(25 + 2*5^(1/2))^(1/2))^(1/5);
r3 = -5^(1/2)*(-100 + 20*(25 + 2*5^(1/2))^(1/2))^(1/5);
r4 = -5^(1/2)*(100 + 20*(25 - 2*5^(1/2))^(1/2))^(1/5);
prim1 = (-1 - 5^(1/2) + 2^(1/2)*I*(5 - 5^(1/2))^(1/2))/4;
prim2 = (-1 + 5^(1/2) - 2^(1/2)*I*(5 + 5^(1/2))^(1/2))/4;
prim3 = (-1 + 5^(1/2) + 2^(1/2)*I*(5 + 5^(1/2))^(1/2))/4;
prim4 = (-1 - 5^(1/2) - 2^(1/2)*I*(5 - 5^(1/2))^(1/2))/4;
{x -> #} & /@ ({{r1, r2, r3, r4}.{1, 1, 1, 1},
{r1, r2, r3, r4}.{prim2, prim4, prim1, prim3},
{r1, r2, r3, r4}.{prim3, prim1, prim4, prim2},
{r1, r2, r3, r4}.{prim1, prim2, prim3, prim4},
{r1, r2, r3, r4}.{prim4, prim3, prim2, prim1}}/5)]

Q.E.D.

```

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