Algebraic Equations in Educational Model of College Physical Education Course Education

Ruffini proved that the general equation does not have the so-called Lagrangian resolution to satisfy an auxiliary equation of degree lower than fifth. This means that the use of Lagrangian pre-solvers does not reduce the degree of higher-order equations. Ruffini examines the rational functions of the roots of general quadratic equations [9]. If p is the number of the process that remains unchanged under the replacement of the original equation roots, of course, p must be a factor of n!. Ruffini delved into the position of the p numbers that make the rational function invariant under the permutation of the original equation roots. He pointed out that the quintic equation n!p {{n!} \over p} could only be 2 or 5 or 6, not 3 or 4. There will be no 3rd or 4th order auxiliary equations that satisfy the Lagrangian resolution formula. If n!p {{n!} \over p} is 2, it must be divisible by 5. This is not possible. If n!p {{n!} \over p} is 5, a quintic equation that satisfies the resolved procedure exists. But we can't reduce it to a two-term equation like B anyway, so the problem remains unsolvable.

Lagrange points the way to the solution of algebraic equations. He thought that the idea of permutation should be followed, but Ruffini declared that there is no radical solution to the general quintic equation. But that doesn't mean Lagrange's theory is wrong. For example, the bisector equation of the form Xⁿ −1 = 0. He solved the bisect equation of a degree less than 11. Gauss became a blockbuster by solving the 17th-degree equation, so how did Gauss deal with it? The Gaussian approach divides the prime-order cyclotomic equation into lower-order equations [10]. Gauss realized Lagrangian's ideal on the equation of the circle. He solves the equation of a process by successfully determining the function of the roots (pre-solution).

Gaussian studies the problem from the root of the modulo p. Any prime number p, ∃g ∈ Z such that, for any k ≡ 0(mod p),, ∃i ∈ Z, such an integer g is called a fundamental root modulo p. Let x₁, x₂, x₃, x₄, x₅, x₆, x₇ be the root of the septate equation| x⁷ −1 = 0. We can quickly get x₁=ζ³¹⁻¹ =ζ, x₂=ζ³²⁻¹ =ζ³, x₃=ζ³³⁻¹ =ζ⁹ =ζ², x₄=ζ³⁴⁻¹ =ζ²⁷ = ζ⁶, x₅=ζ³⁵⁻¹ =ζ⁸¹ = ζ⁴, x₆ = ζ³⁶⁻¹ =ζ²⁴³ = ζ⁵, x₇=ζ³⁷⁻¹ =ζ⁷²⁷ = ζ by taking g = 3, x_i = ζ^gi−1.

The group G = {1,τ, τ²,τ³,τ⁴,τ⁵} can be found, and the group x⁷ −1 = (x −1)(x⁶ + x⁵ + x⁴ + x³ + x² + x +1), so solving equation x⁷ −1 = 0 is solving the equation x⁶ + x⁵ + x⁴ + x³ + x² + x +1 = 0. The equation is hexadecimal. Lagrangians can reduce the degree for equations of degree 6=2×3 or 6=3×2. Therefore, taking the presolution u = x₁, it can only take two values under all permutations of: I(u=x1)={1}∩ r1=3I(β=x1+x3+x5)=A={1,τ2,τ4}∩ r=2G={1,τ,τ2,τ3,τ4,τ5} \eqalign{& I\left({u = {x_1}} \right) = \left\{1 \right\} \cr & \cap \,\,\,{r_1} = 3 \cr & I\left({\beta = {x_1} + {x_3} + {x_5}} \right) = A = \left\{{1,{\tau ^2},{\tau ^4}} \right\} \cr & \cap \,\,\,\,\,\,\,r = 2 \cr & G = \left\{{1,\tau,{\tau ^2},{\tau ^3},{\tau ^4},{\tau ^5}} \right\} \cr}

Where β= x₁ + x₃ + x₅ under all transformations of G, we can only take two values: 1 β=β1=x1+x3+x5τ β=β2=x2+x4+x6 \eqalign{& 1\,\,\,\,\,\,\beta = {\beta _1} = {x_1} + {x_3} + {x_5} \cr & \tau \,\,\,\,\,\beta = {\beta _2} = {x_2} + {x_4} + {x_6} \cr}

Here we need to set its root as β₁, β₂ to get the quadratic auxiliary equation about β, then β1+β2=x1+x2+x3+x4+x5+x6=−1β1×β2=(x1+x3+x5)(x2+x4+x6)=(ζ+ζ2+ζ4)(ζ3+ζ6+ζ5)=ζ4+1+ζ6+ζ5+1+1+ζ3+ζ2)=3−1=2 \eqalign{& {\beta _1} + {\beta _2} = {x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} = - 1 \cr & {\beta _1} \times {\beta _2} = \left({{x_1} + {x_3} + {x_5}} \right)\left({{x_2} + {x_4} + {x_6}} \right) = \cr & \left({\zeta + {\zeta ^2} + {\zeta ^4}} \right)\left({{\zeta ^3} + {\zeta ^6} + {\zeta ^5}} \right) = \cr & {\zeta ^4} + 1 + {\zeta ^6} + {\zeta ^5} + 1 + 1 + {\zeta ^3} + {\zeta ^2}) = 3 - 1 = 2 \cr}

From this we can obtain the quadratic auxiliary equation for β as y² + y + 2 = 0. We can quickly solve its root β₁, β₂.

Where u = x₁ can only take three values under all permutations of A 1 u=u1=x1τ2 u=u2=x3τ4 u=u3=x5 \eqalign{& 1\,\,u = {u_1} = {x_1} \cr & {\tau ^2}\,\,u = {u_2} = {x_3} \cr & {\tau ^4}\,\,u = {u_3} = {x_5} \cr}

Here we need to set its root as u₁, u₂, u₃ to get the cubic auxiliary equation about u, then u1+u2+u3=x1+x3+x5=β1u1u2+u1u3+u2u3=x1x3+x3x5+x1x5=ζ3+ζ5+ζ6=x2+x4+x6=β2u1u2u3=x1x3x5=ζζ2ζ4=1 \eqalign{& {u_1} + {u_2} + {u_3} = {x_1} + {x_3} + {x_5} = {\beta _1} \cr & {u_1}{u_2} + {u_1}{u_3} + {u_2}{u_3} = {x_1}{x_3} + {x_3}{x_5} + {x_1}{x_5} = {\zeta ^3} + {\zeta ^5} + {\zeta ^6} = {x_2} + {x_4} + {x_6} = {\beta _2} \cr & {u_1}{u_2}{u_3} = {x_1}{x_3}{x_5} = \zeta {\zeta ^2}{\zeta ^4} = 1 \cr}

At this point we get the cubic auxiliary equation u³ −β₁u² + β₂u −1 = 0 about u, and its root u is easy to get. At this point we also get the origins of the original equation.

Of course, we can also use the second option. It's just that you need to solve a cubic equation first and then a quadratic equation. Its method is I(u=x1)={1}∩ r1=2I(β=x1+x4)=A={1,τ3}∩ r=3G={1,τ,τ2,τ3,τ4,τ5} \eqalign{& I\left({u = {x_1}} \right) = \left\{1 \right\} \cr & \cap \,\,\,\,{r_1} = 2 \cr & I\left({\beta = {x_1} + {x_4}} \right) = A = \left\{{1,{\tau ^3}} \right\} \cr & \cap \,\,r = 3 \cr & G = \left\{{1,\tau,{\tau ^2},{\tau ^3},{\tau ^4},{\tau ^5}} \right\} \cr}

Finally, the roots of the original secret equation can also be found.

The solution process of Gauss's famous 17th-degree circle equation is roughly similar to the above. We can simply express it as I(u=x1)={1}∩ r1=2I(α=x1+x9)=A={1,τ8}∩ r2=2I(β=x1+x5+x9+x13)=B={1,τ4,τ8,τ12}∩ r3=2I(γ=x1+x3+x5+x7+x9+x11+x13+x15)=C={1,τ2,τ4,τ6,τ8,τ10,τ12,τ14}∩ r4=2G={1,τ,τ2,τ3,τ4,⋯,τ14,τ15} \eqalign{& I\left({u = {x_1}} \right) = \left\{1 \right\} \cr & \cap \,\,\,\,{r_1} = 2 \cr & I\left({\alpha = {x_1} + {x_9}} \right) = A = \left\{{1,{\tau ^8}} \right\} \cr & \cap \,\,{r_2} = 2 \cr & I\left({\beta = {x_1} + {x_5} + {x_9} + {x_{13}}} \right) = B = \left\{{1,{\tau ^4},{\tau ^8},{\tau ^{12}}} \right\} \cr & \cap \,\,{r_3} = 2 \cr & I\left({\gamma = {x_1} + {x_3} + {x_5} + {x_7} + {x_9} + {x_{11}} + {x_{13}} + {x_{15}}} \right) = C = \cr & \left\{{1,{\tau ^2},{\tau ^4},{\tau ^6},{\tau ^8},{\tau ^{10}},{\tau ^{12}},{\tau ^{14}}} \right\} \cr & \cap \,\,{r_4} = 2 \cr & G = \left\{{1,\tau,{\tau ^2},{\tau ^3},{\tau ^4}, \cdots,{\tau ^{14}},{\tau ^{15}}} \right\} \cr}

Here g still takes 3. We decompose the equation x¹⁶ + x¹⁵ +⋯+ x³ + x² + x¹ + x +1 = 0 of degree 16 into 2×2×2×2. The process of solving the equation is roughly similar to the above. So in the end, Gauss got the roots of the 17th-degree bisector equation. It is easy to see that Gauss's method of solving the circle equation is Lagrangian's method. The general cycloid equation is expressed as xⁿ −1 = 0 (n is a prime number). Assuming n −1 = 2× r₁ × r₂ ×⋯× r_m we can obtain its roots by the above method, Gauss declares that any degree of the circular equation is solvable.

Gauss is very clear about the connotation of the theory of Lagrangian algebraic equations, including that the pre-solved formula should be selected and the degree of the equation should be reduced. His procedure for solving the circle equation is an implementation of the Lagrangian method. Lagrange's theory of solving algebraic equations is validated here by Gauss. Therefore, without Lagrange's algebraic equation solving theoretical Gaussian, it is impossible to obtain the success of solving the 17th-degree centroid equation.

The most significant victory of the Lagrangian algebraic equation solving theory directly led to the emergence of the Galois theory. Its influence on Galois was enormous. The first is spiritual motivation [11]. Lagrange's solving algebraic equations stimulated Galois's idea to the greatest extent. Lagrange's study of the solution of equations of the fifth degree and above led Galois to be the first to realize that the different representations of numerical values must be taken into account when studying particular equations. So permutation doesn't change the value of numbers, only their indication.

Second, many algebraists say that Galois changed the face of solving algebraic equations. This has led to pure algebra in mathematics since him. Lagrange divided algebra into two parts: mathematics and pure algebra. Galois continued Lagrange's work. He further developed algebra based on Lagrange.

Finally, Lagrange gave Galois great help in solving the algebraic equations. Some of the fundamental theories of Lagrange still appear in Galois's view. Some of Galois's ideas have emerged in Lagrange's work. Such as presolution. Lagrange requires three conditions when discussing the pre-solution formula selection: 1) The pre-solution procedure should be a rational expression of the original equation roots and known numbers. 2) Related numbers and known numbers can rationally express each root of the equation. 3) must be the root of a solvable equation. 1) is the criterion for describing the choice of presolution formula. 2) and 3) describe the properties that the pre-solution should satisfy. It is undeniable that Lagrange did select the pre-solution recipe and solve the algebraic equation according to the above three points. This may be one of the reasons why he didn't get the final victory. Galois realized that choosing the pre-solved formula is enough to satisfy 1) and 2) for an equation. And there is no pre-solution formula that needs to satisfy the three conditions simultaneously. Therefore, Galois referred to Lagrange's research theory on the problem of pre-solution selection.

In addition, it was said in the previous article that Lagrange pointed out the direction for solving algebraic equations, but he did not point out specific measures. That is to say, according to Lagrange's idea, it should be possible to find the root of the equation, and Galois did it. Under the guidance of Lagrange, he found this specific path to the end. According to the idea of solving Lagrange's theoretical algebraic equations, a subgroup sequence is finally obtained [12]. The core of Galois's theory is to find such a subgroup sequence. In any respect, Lagrangian has the most significant influence on Galois. Lagrange established the framework of the idea, and it is within this framework Galois won the victory of solving algebraic equations.