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symmetric quartic equation
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(Topic)
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Besides the biquadratic equation, there are other types of quartic equations
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(1) |
which can be reduced to quadratic equations. If the left hand side of (1) is $P(z)$ , one may write the identity
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(2) |
If we assume first that $a_4 = a_0$ and $a_3 = a_1$ , the identity is $$\frac{P(z)}{z^2} = a_0\left(z^2+\frac{1}{z^2}\right)+a_1\left(z+\frac{1}{z}\right)+a_2.$$ We set $\displaystyle z+\frac{1}{z} := x$ , whence $\displaystyle z^2+\frac{1}{z^2} = x^2-2$ ; hence the identity is simplified to $$\frac{P(z)}{z^2} = a_0x^2+a_1x+a_2-2a_0.$$ Accordingly, we obtain the roots of the so-called symmetric quartic equation
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(3) |
if we first determine the roots $x_1$ and $x_2$ of the quadratic $$a_0x^2+a_1x+a_2-2a_0 = 0$$ and then solve the equations $z+\frac{1}{z} = x_1$ and $z+\frac{1}{z} = x_2$ which can be written
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(4) |
Note, that the roots of either equations (4) are inverse numbers of each other (see properties of quadratic equations). Therefore, as well the inverse number of any root of the symmetric quartic (3) is a root of (3); this fact is, by the way, clear also because of the identity $$z^4P\!\left(\frac{1}{z}\right) = P(z).\\$$
Example. The equation $$2z^4-5z^3+4z^2-5z+2 = 0$$ is symmetric. Thus we solve first $$2x^2-5x+4-2\cdot2 = 0,$$ which yields $x_1 = 0$ , $x_2 = \frac{5}{2}$ . Secondly we solve $$z^2+1 = 0, \quad z^2-\frac{5}{2}z+1 = 0$$ which yield all the four roots $z = \pm i$ , $z = \frac{1}{2}$ , $z = 2$ of the quartic.
There is still the quartic equation
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(5) |
which reduces to quadratics -- the identity (2) for it reads $$\frac{P(z)}{z^2} = a_0\left(z^2+\frac{1}{z^2}\right)+a_1\left(z-\frac{1}{z}\right)+a_2.$$ The substitution $\displaystyle z-\frac{1}{z} := x$ converts it to $$\frac{P(z)}{z^2} = a_0x^2+a_1x+a_2+2a_0.$$ Thus, (5) can be solved by determining first the roots $x_1$ and $x_2$ of $$a_0x^2+a_1x+a_2+2a_0 = 0,$$ then the roots $z$ of $z-\frac{1}{z} = x_1$ and $z-\frac{1}{z} = x_2$ which may written $$z^2-x_1z-1 = 0, \quad z^2-x_2z-1 = 0.$$ Hence one infers, that if $z$ is a root of (5), so is also its opposite inverse $-\frac{1}{z}$ ; this is apparent also due to the identity $$z^4P\!\left(-\frac{1}{z}\right) = P(z).$$
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- ERNST LINDELÖF: Johdatus korkeampaan analyysiin. Fourth edition. Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).
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"symmetric quartic equation" is owned by pahio.
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Cross-references: clear, propertiess of quadratic equation, inverse numbers, equations, left hand side, quadratic equations, reduced, quartic equations, biquadratic equation
This is version 11 of symmetric quartic equation, born on 2008-05-26, modified 2008-05-27.
Object id is 10629, canonical name is SymmetricQuarticEquation.
Accessed 888 times total.
Classification:
| AMS MSC: | 12D99 (Field theory and polynomials :: Real and complex fields :: Miscellaneous) | | | 30-00 (Functions of a complex variable :: General reference works ) |
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Pending Errata and Addenda
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