Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
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[2-16] Computing polynomial expressions of the roots via the Lagrange interpolation formula
We first restate the polynomial \(P_{\alpha}\) introduced in Section [2-4] for computing the root \(\alpha\).\begin{align} P_\alpha(x) &= V(x)\cdot \Bigl[ \ \sum_{i=1}^6 \sigma_i\!\left(\frac{\alpha }{x-v}\right) \ \Bigr] =\ V(x)\cdot \big( \frac{\alpha }{x-{v_1}}+\frac{\alpha }{x-{v_2}} +\frac{\beta }{x-{v_3}}+\frac{\beta }{x-{v_4}} +\frac{\gamma }{x-{v_5}}+\frac{\gamma }{x-{v_6}} \big) \notag \\ \end{align}
\begin{align} \notag \\ &=\alpha(x-v_2)(x-v_3)(x-v_4)(x-v_5)(x-v_6) +\alpha(x-v_1)(x-v_3)(x-v_4)(x-v_5)(x-v_6) \notag \\ &+\beta(x-v_1)(x-v_2)(x-v_4)(x-v_5)(x-v_6) +\beta(x-v_1)(x-v_2)(x-v_3)(x-v_5)(x-v_6) \notag \\ &+\gamma(x-v_1)(x-v_2)(x-v_3)(x-v_4)(x-v_6) +\gamma(x-v_1)(x-v_2)(x-v_3)(x-v_4)(x-v_5) \\ \end{align}
\begin{align} L_{\alpha}(x) =\Biggl[ \ &\alpha \cdot \frac{(x-v_2)(x-v_3)(x-v_4)(x-v_5)(x-v_6)}{(v_1-v_2)(v_1-v_3)(v_1-v_4)(v_1-v_5)(v_1-v_6)} +\alpha \cdot \frac{(x-v_1)(x-v_3)(x-v_4)(x-v_5)(x-v_6)}{(v_2-v_1)(v_2-v_3)(v_2-v_4)(v_2-v_5)(v_2-v_6)} \notag \\ +&\beta \cdot \frac{(x-v_1)(x-v_2)(x-v_4)(x-v_5)(x-v_6)}{(v_3-v_1)(v_3-v_2)(v_3-v_4)(v_3-v_5)(v_3-v_6)} +\beta \cdot \frac{(x-v_1)(x-v_2)(x-v_3)(x-v_5)(x-v_6)}{(v_4-v_1)(v_4-v_2)(v_4-v_3)(v_4-v_5)(v_4-v_6)} \notag \\ +&\gamma \cdot \frac{(x-v_1)(x-v_2)(x-v_3)(x-v_4)(x-v_6)}{(v_5-v_1)(v_5-v_2)(v_5-v_3)(v_5-v_4)(v_5-v_6)} +\gamma \cdot \frac{(x-v_1)(x-v_2)(x-v_3)(x-v_4)(x-v_5)}{(v_6-v_1)(v_6-v_2)(v_6-v_3)(v_6-v_4)(v_6-v_5)} \quad \Biggr] \\ \notag \\ \end{align}
\begin{align} \alpha&=L_{\alpha}(v_1) \\ \end{align}
First, using determinants, the Lagrange interpolation formula can be written as (16.4).
For this formulation, see the following web page; it gives a very concise explanation.
\(\qquad\) Toshizumi Fukui, "Linear Algebra Lecture Notes" (see p. 55)
\(\qquad\) http://www.rimath.saitama-u.ac.jp/lab.jp/Fukui/lectures/Linear_algebra.pdf
\begin{align} L_{\alpha}(x)&=-\frac{1}{\Delta} \begin{vmatrix} 0 & 1 & x & {{x}^{2}} & {{x}^{3}} & {{x}^{4}} & {{x}^{5}}\\ \alpha & 1 & {v_1} & {{v}_{1}^{2}} & {{v}_{1}^{3}} & {{v}_{1}^{4}} & {{v}_{1}^{5}}\\ \alpha & 1 & {v_2} & {{v}_{2}^{2}} & {{v}_{2}^{3}} & {{v}_{2}^{4}} & {{v}_{2}^{5}}\\ \beta & 1 & {v_3} & {{v}_{3}^{2}} & {{v}_{3}^{3}} & {{v}_{3}^{4}} & {{v}_{3}^{5}}\\ \beta & 1 & {v_4} & {{v}_{4}^{2}} & {{v}_{4}^{3}} & {{v}_{4}^{4}} & {{v}_{4}^{5}}\\ \gamma & 1 & {v_5} & {{v}_{5}^{2}} & {{v}_{5}^{3}} & {{v}_{5}^{4}} & {{v}_{5}^{5}}\\ \gamma & 1 & {v_6} & {{v}_{6}^{2}} & {{v}_{6}^{3}} & {{v}_{6}^{4}} & {{v}_{6}^{5}}\\ \end{vmatrix}\\ \notag \\ \Delta &= \begin{vmatrix} 1 & {v_1} & {{v}_{1}^{2}} & {{v}_{1}^{3}} & {{v}_{1}^{4}} & {{v}_{1}^{5}}\\ 1 & {v_2} & {{v}_{2}^{2}} & {{v}_{2}^{3}} & {{v}_{2}^{4}} & {{v}_{2}^{5}}\\ 1 & {v_3} & {{v}_{3}^{2}} & {{v}_{3}^{3}} & {{v}_{3}^{4}} & {{v}_{3}^{5}}\\ 1 & {v_4} & {{v}_{4}^{2}} & {{v}_{4}^{3}} & {{v}_{4}^{4}} & {{v}_{4}^{5}}\\ 1 & {v_5} & {{v}_{5}^{2}} & {{v}_{5}^{3}} & {{v}_{5}^{4}} & {{v}_{5}^{5}}\\ 1 & {v_6} & {{v}_{6}^{2}} & {{v}_{6}^{3}} & {{v}_{6}^{4}} & {{v}_{6}^{5}}\\ \end{vmatrix} =\displaystyle \prod_{0 \lt i \lt j \leq 6} (v_j-v_i)\\ \end{align}
Substitute (16.6) into (16.4) and (16.5).
\begin{align} &\left\{ \begin{array}{l} v_{1}=\alpha+2\beta+3\gamma \qquad v_{2}=\alpha+2\gamma+3\beta \qquad v_{3}=\beta+2\alpha+3\gamma \\ v_{4}=\beta+2\gamma+3\alpha \qquad v_{5}=\gamma+2\alpha+3\beta \qquad v_{6}=\gamma+2\beta+3\alpha \\ \end{array} \right. \\ \end{align}
\begin{align} &L_{\alpha}(x)= (-1) \cdot \frac {nL_{\alpha}}{dL_{\alpha}}\\ \notag \\ &r_1=\alpha^3+3 \alpha +1=0. \quad r_2=\beta^2+\alpha \beta +\alpha^2+3=0, \quad r_3=\alpha+\beta+\gamma=0 \\ \end{align}
Finally, \(\alpha\) is obtained simply by substituting \(x=v\) into \(L_{\alpha}(x)\).
\begin{align} \notag \\ &\left\{ \begin{array}{l} nL_{\alpha}=-131220(x^4+15x^2-9x+36)(2\alpha^2\beta+2\beta-2\alpha-1)\\ dL_{\alpha}=2361960(2\alpha^2\beta+2\beta-2\alpha-1)\\ \end{array} \right. \\ \notag\\ &\qquad \therefore \quad L_{\alpha}(x)=\frac{x^4+15x^2-9x+36}{18} \quad \rightarrow \quad \alpha=L_{\alpha}(v)\\ \end{align}
\begin{align} &L_{\beta}(x)=-\frac{1}{\Delta} \begin{vmatrix} 0 & 1 & x & {{x}^{2}} & {{x}^{3}} & {{x}^{4}} & {{x}^{5}}\\ \beta & 1 & {v_1} & {{v}_{1}^{2}} & {{v}_{1}^{3}} & {{v}_{1}^{4}} & {{v}_{1}^{5}}\\ \gamma & 1 & {v_2} & {{v}_{2}^{2}} & {{v}_{2}^{3}} & {{v}_{2}^{4}} & {{v}_{2}^{5}}\\ \alpha & 1 & {v_3} & {{v}_{3}^{2}} & {{v}_{3}^{3}} & {{v}_{3}^{4}} & {{v}_{3}^{5}}\\ \gamma & 1 & {v_4} & {{v}_{4}^{2}} & {{v}_{4}^{3}} & {{v}_{4}^{4}} & {{v}_{4}^{5}}\\ \alpha & 1 & {v_5} & {{v}_{5}^{2}} & {{v}_{5}^{3}} & {{v}_{5}^{4}} & {{v}_{5}^{5}}\\ \beta & 1 & {v_6} & {{v}_{6}^{2}} & {{v}_{6}^{3}} & {{v}_{6}^{4}} & {{v}_{6}^{5}} \end{vmatrix}\\ \notag \\ &L_{\gamma}(x)=-\frac{1}{\Delta} \begin{vmatrix} 0 & 1 & x & {{x}^{2}} & {{x}^{3}} & {{x}^{4}} & {{x}^{5}}\\ \gamma & 1 & {v_1} & {{v}_{1}^{2}} & {{v}_{1}^{3}} & {{v}_{1}^{4}} & {{v}_{1}^{5}}\\ \beta & 1 & {v_2} & {{v}_{2}^{2}} & {{v}_{2}^{3}} & {{v}_{2}^{4}} & {{v}_{2}^{5}}\\ \gamma & 1 & {v_3} & {{v}_{3}^{2}} & {{v}_{3}^{3}} & {{v}_{3}^{4}} & {{v}_{3}^{5}}\\ \alpha & 1 & {v_4} & {{v}_{4}^{2}} & {{v}_{4}^{3}} & {{v}_{4}^{4}} & {{v}_{4}^{5}}\\ \beta & 1 & {v_5} & {{v}_{5}^{2}} & {{v}_{5}^{3}} & {{v}_{5}^{4}} & {{v}_{5}^{5}}\\ \alpha & 1 & {v_6} & {{v}_{6}^{2}} & {{v}_{6}^{3}} & {{v}_{6}^{4}} & {{v}_{6}^{5}} \end{vmatrix}\\ \notag \\ &L_{\beta}(x)=-\frac{x^4+15x^2+36}{9} \qquad \rightarrow \quad \beta=L_{\beta}(v)\\ &L_{\gamma}(x)=\frac{x^4+15x^2+9x+36}{18} \quad \rightarrow \quad \gamma=L_{\gamma}(v)\\ \end{align}
This completes the method for obtaining polynomial expressions in \(v\) for the three roots of the equation \(f(x)\) using Lagrange interpolation.
I hope this is of some help.