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Principal equation form

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In mathematics and, more specifically, in theory of equations, the principal form of an irreducible polynomial of degree at least three is a polynomial of the same degree n without terms of degrees n−1 and n−2, such that each roots of either polynomial is a rational function of a root of the other polynomial.

The principal form of a polynomial can be found by applying a suitable Tschirnhaus transformation to the given polynomial.

Definition

Let

f ( x ) = x n + a 1 x n − 1 + ⋯ + a n − 1 x + a n {\displaystyle f(x)=x^{n}+a_{1}x^{n-1}+\cdots +a_{n-1}x+a_{n}}

be an irreducible polynomial of degree at least three.

Its principal form is a polynomial

g ( y ) = y n + b 3 y n − 3 + ⋯ + b n − 1 y + b n , {\displaystyle g(y)=y^{n}+b_{3}y^{n-3}+\cdots +b_{n-1}y+b_{n},}

together with a Tschirnhaus transformation of degree two

φ ( x ) = x 2 + α x + β {\displaystyle \varphi (x)=x^{2}+\alpha x+\beta }

such that, if r is a root of f, ϕ ( r ) {\displaystyle \phi (r)} is a root of ⁠ g {\displaystyle g} ⁠.

Expressing that ⁠ g {\displaystyle g} ⁠ does not has terms in ⁠ y n − 1 {\displaystyle y^{n-1}} ⁠ and ⁠ y n − 2 {\displaystyle y^{n-2}} ⁠ leads to a system of two equations in ⁠ α {\displaystyle \alpha } ⁠ and ⁠ β {\displaystyle \beta } ⁠, one of degree one and one of degree two. In general, this system has two solutions, giving two principal forms involving a square root. One passes from one principal form to the secong by changing the sign of the square root.

Quintic case

Synthesis advice for the quadratic Tschirnhaus key

This is the given quintic equation:

x 5 − a x 4 + b x 3 − c x 2 + d x − e = 0 {\displaystyle x^{5}-ax^{4}+bx^{3}-cx^{2}+dx-e=0}

That quadratic equation system leads to the coefficients of the quadratic Tschirnhaus key:

F i r s t c l u e {\displaystyle \mathrm {First\,clue} } a s + 5 t + a 2 − 2 b = 0 {\displaystyle as+5t+a^{2}-2b=0}
S e c o n d c l u e {\displaystyle \mathrm {Second\,clue} } b s 2 − 10 t 2 + a b s − 3 c s − 2 a c + b 2 + 2 d = 0 {\displaystyle bs^{2}-10t^{2}+abs-3cs-2ac+b^{2}+2d=0}

By polynomial division that Tschirnhaus transformation can be made:

( x 2 + s x + t ) 5 − u ( x 2 + s x + t ) 2 + v ( x 2 + s x + t ) − w = 0 {\displaystyle (x^{2}+sx+t)^{5}-u(x^{2}+sx+t)^{2}+v(x^{2}+sx+t)-w=0}

Calculation examples

This is the first example:

x 5 − x 4 − x 2 − 1 = 0 {\displaystyle x^{5}-x^{4}-x^{2}-1=0}
y = x 2 − 1 4 ( 19 − 265 ) x − 1 20 ( 265 − 15 ) {\displaystyle y=x^{2}-{\tfrac {1}{4}}(19-{\sqrt {265}})x-{\tfrac {1}{20}}({\sqrt {265}}-15)}
y 5 + 1 80 ( 24455 − 1501 265 ) y 2 − 1 160 ( 5789 265 − 93879 ) y − 1 4000 ( 5393003 265 − 87785025 ) = 0 {\displaystyle y^{5}+{\tfrac {1}{80}}(24455-1501{\sqrt {265}})y^{2}-{\tfrac {1}{160}}(5789{\sqrt {265}}-93879)y-{\tfrac {1}{4000}}(5393003{\sqrt {265}}-87785025)=0}

And this is the second example:

x 5 + x 4 + x 3 + x 2 − 1 = 0 {\displaystyle x^{5}+x^{4}+x^{3}+x^{2}-1=0}
y = x 2 + 1 3 ( 30 − 3 ) x + 1 15 30 {\displaystyle y=x^{2}+{\tfrac {1}{3}}({\sqrt {30}}-3)x+{\tfrac {1}{15}}{\sqrt {30}}}
y 5 − 1 45 ( 465 − 61 30 ) y 2 + 2 45 ( 1616 − 289 30 ) y − 1 1125 ( 33758 30 − 183825 ) = 0 {\displaystyle y^{5}-{\tfrac {1}{45}}(465-61{\sqrt {30}})y^{2}+{\tfrac {2}{45}}(1616-289{\sqrt {30}})y-{\tfrac {1}{1125}}(33758{\sqrt {30}}-183825)=0}

Examples of solving the principal form

Along with the Abel Ruffini theorem the following equation is an example that can not be solved in an elementary way, but can be reduced to the Bring Jerrard form by only using cubic radical elements. This shall be demonstrated here:

y 5 − 5 y 2 + 5 y − 5 = 0 {\displaystyle y^{5}-5y^{2}+5y-5=0}

For this given equation the quartic Tschirnhaus key now shall be synthesized. For that example the values u = v = w = 5 are the combination. Therefore, this is the mentioned equation system for that example:

20 α − 15 β = 25 {\displaystyle 20{\color {crimson}\alpha }-15{\color {green}\beta }=25}
15 α + 5 δ = 20 {\displaystyle 15{\color {crimson}\alpha }+5{\color {blue}\delta }=20}
75 α β − 15 α δ − 30 β 2 − 5 δ = 25 {\displaystyle 75{\color {crimson}\alpha }{\color {green}\beta }-15{\color {crimson}\alpha }{\color {blue}\delta }-30{\color {green}\beta }^{2}-5{\color {blue}\delta }=25}

So these are the coefficients of the cubic, quadratic and absolute term of the Tschirnhaus transformation key:

α = − 1 , β = − 3 , δ = 7 {\displaystyle {\color {crimson}\alpha }=-1,\,{\color {green}\beta }=-3,\,{\color {blue}\delta }=7}

That cubic equation leads to the coefficient of the linear term of the key:

5 γ 3 + 110 γ 2 + 1100 γ + 3080 = 0 {\displaystyle 5{\color {orange}\gamma }^{3}+110{\color {orange}\gamma }^{2}+1100{\color {orange}\gamma }+3080=0}
γ = 8 3 11 sinh ⁡ [ 1 3 arsinh ⁡ ( 4 11 11 ) ] − 22 3 {\displaystyle {\color {orange}\gamma }={\frac {8}{3}}{\sqrt {11}}\sinh {\bigl }-{\frac {22}{3}}}

By doing a polynomial division on the fifth power of the quartic Tschirnhaus transformation key and analyzing the remainder rest the coefficients of the mold can be determined. This is the result:

y 5 − 5 y 2 + 5 y − 5 = 0 {\displaystyle y^{5}-5y^{2}+5y-5=0}
z = y 4 − y 3 − 3 y 2 + { 8 3 11 sinh ⁡ [ 1 3 arsinh ⁡ ( 4 11 11 ) ] − 22 3 } y + 7 {\displaystyle z=y^{4}-y^{3}-3y^{2}+{\bigl \{}{\frac {8}{3}}{\sqrt {11}}\sinh {\bigl }-{\frac {22}{3}}{\bigr \}}y+7}
z 5 + 14080 3 { 11 − 2 22 cosh ⁡ [ 1 3 arcosh ⁡ ( 7 22 22 ) ] } z + 11264 3 { 11 − 110 2 sinh ⁡ [ 1 3 arsinh ⁡ ( 1 22 2 ) ] } = 0 {\displaystyle z^{5}+{\frac {14080}{3}}{\bigl \{}11-2{\sqrt {22}}\cosh {\bigl }{\bigr \}}z+{\frac {11264}{3}}{\bigl \{}11-110{\sqrt {2}}\sinh {\bigl }{\bigr \}}=0}

These are the approximations of the solution:

z ≈ − 4.87187987090341241739191116705958390845844658170795795268900739402026742 {\displaystyle z\approx -4.87187987090341241739191116705958390845844658170795795268900739402026742}
y ≈ 1.56670895425072582758152133323240667646412076995364965189840377191745 {\displaystyle y\approx 1.56670895425072582758152133323240667646412076995364965189840377191745}

Literature

References

  1. Weisstein, Eric W. "Principal Quintic Form". mathworld.wolfram.com.
  2. "The solution to the principal quintic via the Brioschi and Rogers-Ramanujan cfrac $R(q)$". Mathematics Stack Exchange.
  3. Jerrard, George Birch (1859). An essay on the resolution of equations. London, UK: Taylor & Francis.
  4. Adamchik, Victor (2003). "Polynomial Transformations of Tschirnhaus, Bring, and Jerrard" (PDF). ACM SIGSAM Bulletin. 37 (3): 91. CiteSeerX 10.1.1.10.9463. doi:10.1145/990353.990371. S2CID 53229404. Archived from the original (PDF) on 2009-02-26.
  5. Klein, Felix (December 28, 1888). "Lectures on the ikosahedron and the solution of equations of the fifth degree". London : Trübner & Co. – via Internet Archive.
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