Symmetrical equation

A symmetric equation is a polynomial (completely rational) equation whose coefficient sequence is symmetric. Because every solution also has a solution, they are also called reciprocal equations. If the coefficients are symmetrical in terms of their absolute value , but differ in terms of their sign , one speaks of an antisymmetrical equation. ${\displaystyle \textstyle x}$ ${\displaystyle \textstyle {\frac {1}{x}}}$

definition

A polynomial equation -th degree ${\displaystyle n}$

${\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{1}x+a_{0}=0}$

is called symmetric if applies to all . If on the other hand , the equation is called antisymmetric . The polynomial ${\displaystyle a_{k}=a_{n-k}}$ ${\displaystyle k=0,1,\dotsc ,n}$ ${\displaystyle a_{k}=-a_{n-k}}$

${\displaystyle p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{1}x+a_{0}}$

is then also palindromic ( ger .: palindromic polynomial ) or antipalindromisch called.

properties

Let it be a polynomial of degree with real or complex coefficients. ${\displaystyle p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{1}x+a_{0}}$ ${\displaystyle n}$

1. When is palindromic or antipalindromic, is . ${\displaystyle p(x)}$ ${\displaystyle a_{0}\neq 0}$
2. When is anti-palindromic and straight, then applies . ${\displaystyle p(x)}$ ${\displaystyle n}$ ${\displaystyle a_{n/2}=0}$
3. ${\displaystyle p(x)}$is palindromic if and only if and antipalindromic if and only if . ${\displaystyle p(x)=x^{n}p(1/x)}$ ${\displaystyle p(x)=-x^{n}p(1/x)}$
4. If is palindromic and odd, then . When is antipalindromic, then applies . ${\displaystyle p(x)}$ ${\displaystyle n}$ ${\displaystyle p(-1)=0}$ ${\displaystyle p(x)}$ ${\displaystyle p(1)=0}$
5. If and is palindromic or antipalindromic , then is and . and are then solutions of the same multiplicity of the (anti-) symmetric equation . ${\displaystyle p(x)}$ ${\displaystyle p(x_{0})=0}$ ${\displaystyle x_{0}\neq 0}$ ${\displaystyle p(1/x_{0})=0}$ ${\displaystyle x_{0}}$ ${\displaystyle 1/x_{0}}$ ${\displaystyle p(x)=0}$
6. If and are palindromic polynomials, so is the product . If both factors are antipalindromic, the product is also palindromic. If one factor is palindromic and the other is antipalindromic, the product is antipalindromic. ${\displaystyle p(x)}$ ${\displaystyle q(x)}$ ${\displaystyle p(x)\cdot q(x)}$
7. If and are palindromic or antipalindromic polynomials, then is also palindromic or antipalindromic. ${\displaystyle p(x)}$ ${\displaystyle p(x)\cdot q(x)}$ ${\displaystyle q(x)}$
8. If with every solution of the equation the reciprocal is also a solution of the equation with the same multiplicity as , then the equation is symmetric or antisymmetric. ${\displaystyle x_{0}}$ ${\displaystyle p(x)=0}$ ${\displaystyle 1/x_{0}}$ ${\displaystyle x_{0}}$
9. If a polynomial is of degree , then a palindromic and an antipalindromic polynomial is of degree . ${\displaystyle r(x)}$ ${\displaystyle m}$ ${\displaystyle x^{m}\cdot r\left(x+{\tfrac {1}{x}}\right)}$ ${\displaystyle x^{m}\cdot r\left(x-{\tfrac {1}{x}}\right)}$ ${\displaystyle 2m}$
10. If a palindromic (or antipalindromic) polynomial is of degree , then there is exactly one polynomial of degree with (or ). ${\displaystyle p(x)}$ ${\displaystyle n=2m}$ ${\displaystyle r(x)}$ ${\displaystyle m}$ ${\displaystyle p(x)=x^{m}\cdot r\left(x+{\tfrac {1}{x}}\right)}$ ${\displaystyle p(x)=x^{m}\cdot r\left(x-{\tfrac {1}{x}}\right)}$
11. If all coefficients are real and all complex zeros have the magnitude 1, then it is palindromic or antipalindromic. ${\displaystyle a_{i}}$ ${\displaystyle p(x)}$ ${\displaystyle p(x)}$

General solution strategies

Special solution strategies help with equations from the 5th degree, since there is no longer a general solution formula for determining the zeros.

Symmetrical equations

Symmetrical equations can therefore be reduced to 4th degree equations (or less) up to degree 9, so that all zeros can also be determined here.

The general solution strategy for symmetric equations with even degree and real coefficients is based on the following steps: ${\displaystyle n}$

1. Division of all terms of the polynomial by ${\displaystyle \textstyle x^{\frac {n}{2}}}$
2. Combining the terms with the same coefficient and factoring out the coefficient ${\displaystyle a_{i}=a_{n-i}}$
3. Substitution , , etc. ${\displaystyle \textstyle (x+{\frac {1}{x}})=u}$ ${\displaystyle \textstyle (x^{2}+{\frac {1}{x^{2}}})=u^{2}-2}$ ${\displaystyle \textstyle (x^{3}+{\frac {1}{x^{3}}})=u^{3}-3\cdot u}$
4. Multiplying out leads to a polynomial in degrees ${\displaystyle u}$ ${\displaystyle \textstyle {\frac {n}{2}}}$
5. Solutions for calculate ${\displaystyle u}$
6. Substituting each solution of in the substitution equation and solving for so that two solutions from the equation can be determined with each . ${\displaystyle u}$ ${\displaystyle \textstyle (x+{\frac {1}{x}})=u}$ ${\displaystyle x}$ ${\displaystyle u}$ ${\displaystyle x}$ ${\displaystyle x^{2}-u\cdot x+1=0}$

In symmetric odd- degree equations with real coefficients, a zero is either 1 or −1. This is determined by inserting, then the corresponding linear factor or using the polynomial division is subdivided, so that a symmetrical equation of an even degree is created. ${\displaystyle \textstyle (x-1)}$ ${\displaystyle \textstyle (x+1)}$

Calculation of the substitutions

Further substitutions for higher potencies can be determined in the following way: If one searches, for example, the substitution for , the approach can be used ${\displaystyle \textstyle (x^{4}+{\frac {1}{x^{4}}})}$

${\displaystyle \left(x+{\frac {1}{x}}\right)^{4}=u^{4}}$

by multiplying the expression

${\displaystyle \left(x^{4}+{\frac {1}{x^{4}}}\right)+4\cdot \left(x^{2}+{\frac {1}{x^{2}}}\right)+6=u^{4}}$

win. Inserting the already known substitution and arranging leads to ${\displaystyle \textstyle (x^{2}+{\frac {1}{x^{2}}})=u^{2}-2}$

${\displaystyle x^{4}+{\frac {1}{x^{4}}}=u^{4}-4\cdot u^{2}+2}$

In this way, more and more substitutions for higher powers of can be constructed recursively from the already known substitutions for smaller powers. ${\displaystyle x}$

Resubstitution to calculate the solutions

Once a solution has been found to solve the substitution equation according to. This results in two solutions for each of the quadratic equations ${\displaystyle u_{i}}$ ${\displaystyle \textstyle (x+{\frac {1}{x}})=u_{i}}$ ${\displaystyle x}$ ${\displaystyle u_{i}}$ ${\displaystyle x}$

${\displaystyle x^{2}-u_{i}\cdot x+1=0}$,

and therefore for everyone ${\displaystyle u_{i}}$

${\displaystyle x_{1/2}={\frac {u_{i}}{2}}\pm {\sqrt {{\frac {u_{i}^{2}}{4}}-1}}}$.

The absolute term of the quadratic equation is 1, so it follows from the Vietnamese Root Law that the two solutions of this quadratic equation must be reciprocal. So it is also determined for everyone . ${\displaystyle x}$ ${\displaystyle \textstyle {\frac {1}{x}}}$

Other reciprocal equations

For reciprocal equations, for which there is always a solution next to each , substitutions can be constructed and thus solutions calculated. Substitution is suitable for this ${\displaystyle \textstyle x_{i}}$ ${\displaystyle \textstyle -{\frac {1}{x_{i}}}}$

${\displaystyle x-{\frac {1}{x}}=u}$

With the methods already described, substitutions of higher potencies can be determined:

${\displaystyle x^{2}+{\frac {1}{x^{2}}}=u^{2}+2;\qquad x^{3}-{\frac {1}{x^{3}}}=u^{3}+3\cdot u;\qquad x^{4}+{\frac {1}{x^{4}}}=u^{4}+4\cdot u^{2}+2}$

As can be seen here, for the even powers of there is a sum, not a difference. ${\displaystyle \textstyle x^{2\cdot n}+{\frac {1}{x^{2\cdot n}}}}$ ${\displaystyle x}$

This allows, for example, the following special types of equations to be solved:

${\displaystyle x^{4}\pm a\cdot x^{3}+b\cdot x^{2}\mp a\cdot x+1=0}$

${\displaystyle x^{6}\pm a\cdot x^{5}+b\cdot x^{4}+c\cdot x^{3}+b\cdot x^{2}\mp a\cdot x+1=0}$

${\displaystyle x^{8}\pm a\cdot x^{7}+b\cdot x^{6}\pm c\cdot x^{5}+d\cdot x^{4}\mp c\cdot x^{3}+b\cdot x^{2}\mp a\cdot x+1=0}$

As can be easily seen from the examples, these equations do not have the structure of an antisymmetric equation in the sense of the definition given above. The solution by substitution is only possible if unequal signs always occur and only for the coefficients of odd powers.

Solution formulas for special equations

The following examples show how the substitution leads to an equation in . ${\displaystyle \textstyle u}$

4th degree symmetric equation

For a quartic equation in normal form

${\displaystyle x^{4}+a\cdot x^{3}+b\cdot x^{2}+a\cdot x+1=0}$

results after dividing by and combining the terms: ${\displaystyle x^{2}}$

${\displaystyle \left(x^{2}+{\frac {1}{x^{2}}}\right)+a\cdot \left(x+{\frac {1}{x}}\right)+b=0}$

After the substitution with and the quadratic equation results in : ${\displaystyle \textstyle (x+{\frac {1}{x}})=u}$ ${\displaystyle \textstyle (x^{2}+{\frac {1}{x^{2}}})=u^{2}-2}$ ${\displaystyle u}$

${\displaystyle u^{2}+a\cdot u+b-2=0}$

From this one determines the solutions and . ${\displaystyle u_{1}}$ ${\displaystyle u_{2}}$

Example: The equation becomes a quadratic equation with the solutions 10/3 and 5/2 for , from which the equations and with the solutions −2; −1/2; 1/3 and 3 also result in the original equation. ${\displaystyle 6x^{4}-5x^{3}-38x^{2}-5x+6=0}$ ${\displaystyle 6u^{2}-5u-50=0}$ ${\displaystyle u}$ ${\displaystyle 3x^{2}-10x+3=0}$ ${\displaystyle 2x^{2}+5x+2=0}$

6th degree symmetric equation

For a 6th degree equation in normal form

${\displaystyle x^{6}+a\cdot x^{5}+b\cdot x^{4}+c\cdot x^{3}+b\cdot x^{2}+a\cdot x+1=0}$

results after dividing by and combining the terms: ${\displaystyle x^{2}}$

${\displaystyle \left(x^{3}+{\frac {1}{x^{3}}}\right)+a\cdot \left(x^{2}+{\frac {1}{x^{2}}}\right)+b\cdot \left(x+{\frac {1}{x}}\right)+c=0}$

After the substitution with and and the cubic equation results in : ${\displaystyle \textstyle (x+{\frac {1}{x}})=u}$ ${\displaystyle \textstyle (x^{2}+{\frac {1}{x^{2}}})=u^{2}-2}$ ${\displaystyle \textstyle (x^{3}+{\frac {1}{x^{3}}})=u^{3}-3\cdot u}$ ${\displaystyle u}$

${\displaystyle u^{3}+a\cdot u^{2}+(b-3)\cdot u-2\cdot a+c=0}$

The solutions are determined from this , and with the help of the solution formulas for the cubic equation . ${\displaystyle u_{1}}$ ${\displaystyle u_{2}}$ ${\displaystyle u_{3}}$

8th degree symmetric equation

For an 8th degree equation in normal form

${\displaystyle x^{8}+a\cdot x^{7}+b\cdot x^{6}+c\cdot x^{5}+d\cdot x^{4}+c\cdot x^{3}+b\cdot x^{2}+a\cdot x+1=0}$

results after dividing by and combining the terms: ${\displaystyle x^{2}}$

${\displaystyle \left(x^{4}+{\frac {1}{x^{4}}}\right)+a\cdot \left(x^{3}+{\frac {1}{x^{3}}}\right)+b\cdot \left(x^{2}+{\frac {1}{x^{2}}}\right)+c\cdot \left(x+{\frac {1}{x}}\right)+d=0}$

After the substitution with and , and the quartic equation results in : ${\displaystyle \textstyle (x+{\frac {1}{x}})=u}$ ${\displaystyle \textstyle (x^{2}+{\frac {1}{x^{2}}})=u^{2}-2}$ ${\displaystyle \textstyle (x^{3}+{\frac {1}{x^{3}}})=u^{3}-3\cdot u}$ ${\displaystyle \textstyle (x^{4}+{\frac {1}{x^{4}}})=u^{4}-4\cdot u^{2}+2}$ ${\displaystyle u}$

${\displaystyle u^{4}+a\cdot u^{3}+(b-4)\cdot u^{2}+(c-3\cdot a)\cdot u-2\cdot b+d+2=0}$

From this we determined the solutions , , and with the help of the solution formulas for the quartic equation. ${\displaystyle u_{1}}$ ${\displaystyle u_{2}}$ ${\displaystyle u_{3}}$ ${\displaystyle u_{4}}$

Further examples

• The quickest way to determine the zeros of the quadratic equation is with the known solution formulas .
• For cubic equations with real coefficients, +1 or −1 is a solution, after which a polynomial division leads to a quadratic (symmetric) equation.

Examples:

• The left side of the balanced equation 3rd degree with the solutions -1/3, -1 and -3 are calculated by dividing by to , resulting in the further solutions result. ${\displaystyle 3x^{3}+13x^{2}+13x+3=0}$ ${\displaystyle (x+1)}$ ${\displaystyle 3x^{2}+10x+3}$
• The left side of equation antisymmetric 3rd degree with the solutions -1/3, 1 and -3 are calculated by dividing by also increasing , resulting in the further solutions result. ${\displaystyle 3x^{3}+7x^{2}-7x-3=0}$ ${\displaystyle (x-1)}$ ${\displaystyle 3x^{2}+10x+3}$

• In the Quintic equation (equation 5th degree) +1 or −1 is a solution again. A polynomial division by or leads to a symmetric equation of the fourth degree. ${\displaystyle (x+1)}$ ${\displaystyle (x-1)}$

Individual evidence

1. ↑ Frank Celler: Constructive recognition algorithms of classical groups in GAP (dissertation) [1] ( GZIP ; 233 kB)
2. ↑ The Fundamental Theorem for Palindromic Polynomials [2]
3. ↑ Meyers Big computing Duden, Bibliographical Institute AG Mannheim (Dudenverlag), 1961; Without specifying an ISBN and edition