quartic equation
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From: GL(2,R) on 16 May 2010 05:20 On May 16, 8:15 am, kinor <ki...(a)excite.com> wrote: > How does one find all complex solutions to the equation > > x^4 - x^3 + x^2 - x + 1 = 0. > > Thanks. > > k Use the geometric series formula, followed by De Moivre's theorem to find the answer.
From: GL(2,R) on 16 May 2010 05:31 On May 16, 8:15 am, kinor <ki...(a)excite.com> wrote: > How does one find all complex solutions to the equation > > x^4 - x^3 + x^2 - x + 1 = 0. > > Thanks. > > k Another way to solve this quartic would be to note that its coefficients are symmetric; you can solve it by first assuming non- zero x, and then dividing through by x^2 to get x^2 - x + 1 - (1/x) + 1/x^2 = 0. Now use the substitution y = (x + 1/ x) to turn this into an equation in y, that is y^2 - y - 1 = 0, which implies that {{y -> 1/2 (1 - Sqrt[5])}, {y -> 1/2 (1 + Sqrt[5])}}. Now solve back for the original equation in x to get the four solutions (I used mathematica) {{x -> 1/4 (1 + Sqrt[5] - I Sqrt[16 - (-1 - Sqrt[5])^2])}, {{x -> 1/4 (1 + Sqrt[5] + I Sqrt[16 - (-1 - Sqrt[5])^2]), {{x -> 1/4 (1 - Sqrt[5] - I Sqrt[16 - (-1 + Sqrt[5])^2])}, {x -> 1/4 (1 - Sqrt[5] + I Sqrt[16 - (-1 + Sqrt[5])^2])}} However, it would just be easier to use De Moivre's; this is just an alternative method.
From: A N Niel on 16 May 2010 06:19 In article <710fbf54-8398-450f-8959-b7c4fb7cfbd6(a)h37g2000pra.googlegroups.com>, 2,R <charlescalculus_robertobaggio(a)hotmail.com> wrote: > On May 16, 8:15�am, kinor <ki...(a)excite.com> wrote: > > How does one find all complex solutions to the equation > > > > x^4 - x^3 + x^2 - x + 1 = 0. > > > > Thanks. > > > > k > > Another way to solve this quartic would be to note that its > coefficients are symmetric; you can solve it by first assuming non- > zero x, and then dividing through by x^2 to get > > x^2 - x + 1 - (1/x) + 1/x^2 = 0. Now use the substitution y = (x + 1/ > x) to turn this into an equation in y, that is > > y^2 - y - 1 = 0, which implies that > > {{y -> 1/2 (1 - Sqrt[5])}, {y -> 1/2 (1 + Sqrt[5])}}. > > Now solve back for the original equation in x to get the four > solutions (I used mathematica) > {{x -> 1/4 (1 + Sqrt[5] - I Sqrt[16 - (-1 - Sqrt[5])^2])}, > {{x -> 1/4 (1 + Sqrt[5] + I Sqrt[16 - (-1 - Sqrt[5])^2]), > {{x -> 1/4 (1 - Sqrt[5] - I Sqrt[16 - (-1 + Sqrt[5])^2])}, > {x -> 1/4 (1 - Sqrt[5] + I Sqrt[16 - (-1 + Sqrt[5])^2])}} > > However, it would just be easier to use De Moivre's; this is just an > alternative method. "easier" or "better" ?? This method gives you the answers in radicals. The "geometric series" method gives you the answer in terms of trigonometry. Which do you want?
From: Ray Vickson on 16 May 2010 16:36 On May 16, 2:31 am, "GL(2,R)" <charlescalculus_robertobag...(a)hotmail.com> wrote: > On May 16, 8:15 am, kinor <ki...(a)excite.com> wrote: > > > How does one find all complex solutions to the equation > > > x^4 - x^3 + x^2 - x + 1 = 0. > > > Thanks. > > > k > > Another way to solve this quartic would be to note that its > coefficients are symmetric; you can solve it by first assuming non- > zero x, and then dividing through by x^2 to get > > x^2 - x + 1 - (1/x) + 1/x^2 = 0. Now use the substitution y = (x + 1/ > x) to turn this into an equation in y, that is > > y^2 - y - 1 = 0, which implies that > > {{y -> 1/2 (1 - Sqrt[5])}, {y -> 1/2 (1 + Sqrt[5])}}. > > Now solve back for the original equation in x to get the four > solutions (I used mathematica) > {{x -> 1/4 (1 + Sqrt[5] - I Sqrt[16 - (-1 - Sqrt[5])^2])}, > {{x -> 1/4 (1 + Sqrt[5] + I Sqrt[16 - (-1 - Sqrt[5])^2]), > {{x -> 1/4 (1 - Sqrt[5] - I Sqrt[16 - (-1 + Sqrt[5])^2])}, > {x -> 1/4 (1 - Sqrt[5] + I Sqrt[16 - (-1 + Sqrt[5])^2])}} Using the command 'solve(x^4-x^3+x^2-x+1=0,x)', Maple gets 1/4+1/4*5^(1/2)+1/4*I*2^(1/2)*(5-5^(1/2))^(1/2), -1/4*5^(1/2)+1/4+1/4*I*2^(1/2)*(5+5^(1/2))^(1/2), -1/4*5^(1/2)+1/4-1/4*I*2^(1/2)*(5+5^(1/2))^(1/2), 1/4+1/4*5^(1/2)-1/4*I*2^(1/2)*(5-5^(1/2))^(1/2) Numerically, these agree with yours. R.G. Vickson > > However, it would just be easier to use De Moivre's; this is just an > alternative method.
From: cwldoc on 19 May 2010 14:27
> How does one find all complex solutions to the > equation > > x^4 - x^3 + x^2 - x + 1 = 0. > > Thanks. > > k Let f(x) = x^4 - x^3 + x^2 - x + 1 and g(x) = (x + 1)f(x) = x^5 + 1 The roots of g(x)=0 are w, w^3, w^5, w^7, w^9, where w = exp(i pi/5) = cos(pi/5) + i sin(pi/5) But the roots of g(x)=0 are just the roots of f(x)=0 with -1 appended. Since w^5 = -1 is not a root of f(x)=0, that means the roots of f(x)=0 are w, w^3, w^7, w^9 |