Solving the Quintic: Correction to x^5 + 20 x + 32 == 0 root example
- To: mathgroup at smc.vnet.net
- Subject: [mg95311] Solving the Quintic: Correction to x^5 + 20 x + 32 == 0 root example
- From: "Q.E.D." <aoe at netzero.net>
- Date: Thu, 15 Jan 2009 06:10:40 -0500 (EST)
- Organization: Adtech Computers, Inc.
There is an error on one of the "Solving the Quintic" pages. It is necessary to change a minus sign to a plus sign to get a correct N[%] value -1.363962165089939` The change is marked below with the comment (*fixed*). The "Solving the Quintic" page in error is http://library.wolfram.com/examples/quintic/main.html The page shows the following: There are polynomials of degree greater than four which do not factor over the rationals, but that can still be solved in radicals. An example is the quintic: x^5 + 20 x + 32 == 0 one of whose roots is 1/5 (-(2500 Sqrt[5] + 250 Sqrt[50 - 10 Sqrt[5]] - 750 Sqrt[50 + 10 Sqrt[5]])^(1/5) - (-2500 Sqrt[5] + 750 Sqrt[50 - 10 Sqrt[5]] + 250 Sqrt[50 + 10 Sqrt[5]])^(1/5) +(*fixed*) (2500 Sqrt[5] + 750 Sqrt[50 - 10 Sqrt[5]] + 250 Sqrt[50 + 10 Sqrt[5]])^(1/5) - (2500 Sqrt[5] - 250 Sqrt[50 - 10 Sqrt[5]] + 750 Sqrt[50 + 10 Sqrt[5]])^(1/5)) The corrected result can be checked using: N[#^5 + 20 # + 32 == 0 &[%], 100] Can anyone who has the "Solving the Quintic" poster see if it is correct on there? 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)] Check the roots using: N[Solve[x^5 + 20 x + 32 == 0, x], 100] == N[%, 100] The "Solving the Quintic" poster notebooks are available at http://library.wolfram.com/infocenter/TechNotes/158/ The Radical.nb notebook is what was used here. Change the simp function to do Expand[Simplify[%]] on its result and discard the Discriminant definition which is built into versions 6 & 7 to get the results in a form like what is shown here. Also change Solve::rad to SolveQuintic::rad in two places before evaluating all cells. Q.E.D.