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Equació cúbica
\[a\cdot x^3+b\cdot x^2+c\cdot x+d = 0\]
\[x^3+\frac{b}{a}\cdot x^2+\frac{c}{a}\cdot x+\frac{d}{a} = 0\]
\[(x+A)^3=x^3+3\cdot A\cdot x^2+3\cdot A^2\cdot x+A^3\]
Si fem \(3\cdot A = \displaystyle\frac{b}{a}\) que implica \(A = \displaystyle\frac{b}{3\cdot a}\)
podem eliminar a potència quadrada utilitzant
\[\left(x+\frac{b}{3\cdot a}\right)^3 = x^3+\frac{b}{a}\cdot x^2 + \frac{b^2}{3\cdot a^2}\cdot x+\frac{b^3}{27\cdot a^3}\]
\[x^3+\frac{b}{a}\cdot x^2 = \left(x+\frac{b}{3\cdot a}\right)^3 - \frac{b^2}{3\cdot a^2}\cdot x - \frac{b^3}{27\cdot a^3}\]
Així, tornant a l'equació original
\[\left(x+\frac{b}{3\cdot a}\right)^3 - \frac{b^2}{3\cdot a^2}\cdot x - \frac{b^3}{27\cdot a^3} +\frac{c}{a}\cdot x+\frac{d}{a} = 0\]
\[\left(x+\frac{b}{3\cdot a}\right)^3 + \left[\frac{c}{a}- \frac{b^2}{3\cdot a^2}\right]\cdot x
+ \left[\frac{d}{a} - \frac{b^3}{27\cdot a^3}\right] = 0\]
Si volem fer el canvi \( t = x + \displaystyle\frac{b}{3\cdot a}\) cal
\[ \left(x+\frac{b}{3\cdot a}\right)^3
+ \left[\frac{c}{a}- \frac{b^2}{3\cdot a^2}\right]\cdot \left(x+\frac{b}{3\cdot a}\right)
- \left[\frac{c}{a}- \frac{b^2}{3\cdot a^2}\right]\cdot \left(\frac{b}{3\cdot a}\right)
+ \left[\frac{d}{a} - \frac{b^3}{27\cdot a^3}\right] = 0\]
\[ \left(x+\frac{b}{3\cdot a}\right)^3
+ \left[\frac{c}{a}- \frac{b^2}{3\cdot a^2}\right]\cdot \left(x+\frac{b}{3\cdot a}\right)
+ \left[-\frac{bc}{3a^2}+ \frac{b^3}{9\cdot a^3} + \frac{d}{a} - \frac{b^3}{27\cdot a^3}\right] = 0\]
\[ \left(x+\frac{b}{3\cdot a}\right)^3
+ \left[\frac{c}{a}- \frac{b^2}{3\cdot a^2}\right]\cdot \left(x+\frac{b}{3\cdot a}\right)
+ \left[\frac{d}{a} -\frac{bc}{3a^2} + \frac{2b^3}{27\cdot a^3}\right] = 0\]
\[ \left(x+\frac{b}{3\cdot a}\right)^3
+ \left[\frac{3ac-b^2}{3\cdot a^2}\right]\cdot \left(x+\frac{b}{3\cdot a}\right)
+ \left[\frac{27a^2d - 9abc + 2b^3}{27\cdot a^3}\right] = 0\]
\[ t = x+\frac{b}{3\cdot a}\]
\[\alpha = \frac{3\cdot a\cdot c - b^2}{3\cdot a^2}\]
\[\beta = \frac{27a^2d - 9abc + 2b^3}{27\cdot a^3}\]
Així les nostres solucions seran
\[ x = - \displaystyle\frac{b}{3\cdot a}+t\]
on \(t\) satisfa l'equació reduïda
\[t^3+\alpha\cdot t + \beta = 0\]
Solució de l'equació reduïda
La contribució històrica de Cardano i Vietta és fer un canvi
\[t = u + v\]
\[(u + v)^3+\alpha\cdot(u + v) + \beta = 0\]
\[
\begin{align}
(u + v)^3 &= u^3 + 3\cdot u^2\cdot v+3\cdot u\cdot v^2 +v^3\\
&= u^3 + 3\cdot u \cdot v\cdot(u+v) +v^3
\end{align}
\]
\[u^3 + (\alpha + 3\cdot u \cdot v)\cdot(u+v) + v^3 + \beta = 0\]
ara cal imposar una relació entre u i v
\[\alpha + 3\cdot u \cdot v= 0\]
amb la que
\[v = -\frac{\alpha}{3\cdot u} \]
i la nostra equació es transforma en
\[u^3 - \frac{\alpha^3}{27\cdot u^3} + \beta = 0\]
\[27u^6 - \alpha^3 + 27\beta\cdot u^3 = 0\]
\[27(u^3)^2 + 27\beta\cdot u^3 - \alpha^3 = 0\]
que és una equacio bicúbica que és pot resoldre
\[u^3 = \frac{- 27\beta \pm \sqrt{(27\beta)^2-4\cdot27\cdot(-\alpha^3)}}{2\cdot27}
= \frac{-\beta\pm \sqrt{\beta^2+\displaystyle\frac{4}{27}\alpha^3}}{2} \]
Aquesta equació té tres solucions (una real i dues complexes)
\[
\begin{align}
u_0 = u_0^+ &= \sqrt[3]{- \frac{\beta}{2} + \sqrt{\left(\frac{\beta}{2}\right)^2+\left(\frac{\alpha}{3}\right)^3}}\\
u_1 = u_1^+ &= \sqrt[3]{- \frac{\beta}{2} + \sqrt{\left(\frac{\beta}{2}\right)^2+\left(\frac{\alpha}{3}\right)^3}}
\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\\
u_2 = u_2^+ &= \sqrt[3]{- \frac{\beta}{2} + \sqrt{\left(\frac{\beta}{2}\right)^2+\left(\frac{\alpha}{3}\right)^3}}
\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)
\end{align}
\]
També podriem haver partit de
\[
u^3 = \frac{1}{2}\left(-\beta + \sqrt{\beta^2+\displaystyle\frac{4}{27}\alpha^3}\right)
= - \frac{\beta}{2} \pm \frac{1}{2}\sqrt{\beta^2+\frac{4}{27}\alpha^3}
= - \frac{\beta}{2} \pm \frac{\beta}{2}\sqrt{1+\frac{4\alpha^3}{27\beta^2}} =\ ...
\]
Semblaria que també tenim aquestes noves solucions
\[
\begin{align}
u_0^- &= \sqrt[3]{- \frac{\beta}{2} - \sqrt{\left(\frac{\beta}{2}\right)^2+\left(\frac{\alpha}{3}\right)^3}}\\
u_1^- &= \sqrt[3]{- \frac{\beta}{2} - \sqrt{\left(\frac{\beta}{2}\right)^2+\left(\frac{\alpha}{3}\right)^3}}
\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\\
u_2^- &= \sqrt[3]{- \frac{\beta}{2} - \sqrt{\left(\frac{\beta}{2}\right)^2+\left(\frac{\alpha}{3}\right)^3}}
\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)
\end{align}
\]
és fàcil comprovar que
\[
\begin{array}{rlcrl}
u_0^+\cdot u_0^- &= - \frac{\alpha}{3} & & \frac{1}{u_0^+} &= - \frac{3\cdot u_0^-}{\alpha}\\
u_1^+\cdot u_1^- &= - \frac{\alpha}{3}\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)
& & \frac{1}{u_1^+} &= - \frac{3\cdot u_1^-}{\alpha}
\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)
= - \frac{3\cdot u_2^-}{\alpha}\\
u_2^+\cdot u_2^- &= - \frac{\alpha}{3}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)
& & \frac{1}{u_2^+} &= - \frac{3\cdot u_2^-}{\alpha}
\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)
= - \frac{3\cdot u_1^-}{\alpha}\\
\end{array}
\]
Com que
\[ x = - \frac{b}{3\cdot a} + t= - \frac{b}{3\cdot a} + u + v\]
les solucions seran
\[
\begin{align}
x_0 &= - \frac{b}{3\cdot a} + u_0 + v_0 = - \frac{b}{3\cdot a} + u_0 - \frac{\alpha}{3u_0}\\
x_1 &= - \frac{b}{3\cdot a} + u_1 + v_1 = - \frac{b}{3\cdot a} + u_1 - \frac{\alpha}{3u_1}\\
x_2 &= - \frac{b}{3\cdot a} + u_2 + v_2 = - \frac{b}{3\cdot a} + u_2 - \frac{\alpha}{3u_2}
\end{align}
\]
Si utilitzem les primeres solucions per \(u\) (super-índex positiu)
\[
\begin{align}
x_0 &= - \frac{b}{3\cdot a} + u_0^+ + v_0^+
= - \frac{b}{3\cdot a} + u_0^+ - \frac{\alpha}{3u_0^+}
= - \frac{b}{3\cdot a} + u_0^+ + u_0^- \\
x_1 &= - \frac{b}{3\cdot a} + u_1^+ + v_1^+
= - \frac{b}{3\cdot a} + u_1^+ - \frac{\alpha}{3u_1^+}
= - \frac{b}{3\cdot a} + u_1^+ + u_2^- \\
x_2 &= - \frac{b}{3\cdot a} + u_2^+ + v_2^+
= - \frac{b}{3\cdot a} + u_2^+ - \frac{\alpha}{3u_2^+}
= - \frac{b}{3\cdot a} + u_2^+ + u_1^-
\end{align}
\]
Si utilitzem les segones solucions per \(u\) (super-índex negatiu)
\[
\begin{align}
x_0 &= - \frac{b}{3\cdot a} + u_0^- + v_0^-
= - \frac{b}{3\cdot a} + u_0^- - \frac{\alpha}{3u_0^-}
= - \frac{b}{3\cdot a} + u_0^- + u_0^+ \\
x_1 &= - \frac{b}{3\cdot a} + u_1^- + v_1^-
= - \frac{b}{3\cdot a} + u_1^- - \frac{\alpha}{3u_1^-}
= - \frac{b}{3\cdot a} + u_1^- + u_2^+ \\
x_2 &= - \frac{b}{3\cdot a} + u_2^- + v_2^-
= - \frac{b}{3\cdot a} + u_2^- - \frac{\alpha}{3u_2^-}
= - \frac{b}{3\cdot a} + u_2^- + u_1^+
\end{align}
\]
La primera seria exactament la mateixa i les altres dues s'intercanvien.
Solució l'equació cúbica
Així, finalment:
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\[
\begin{align}
x_0 &= - \frac{b}{3\cdot a} + u_0^+ + v_0^- \\
x_1 &= - \frac{b}{3\cdot a} - \frac{ u_0^+ + u_0^-}{2} + i\sqrt{3}\frac{ u_0^+ - u_0^-}{2} \\
x_2 &= - \frac{b}{3\cdot a} - \frac{ u_0^+ + u_0^-}{2} - i\sqrt{3}\frac{ u_0^+ - u_0^-}{2}
\end{align}
\]
\[
\begin{align}
u_0^+ &= \sqrt[3]{- \frac{\beta}{2} + \sqrt{\left(\frac{\beta}{2}\right)^2+\left(\frac{\alpha}{3}\right)^3}}\\
u_0^- &= \sqrt[3]{- \frac{\beta}{2} - \sqrt{\left(\frac{\beta}{2}\right)^2+\left(\frac{\alpha}{3}\right)^3}}
\end{align}
\]
\[\beta = \frac{27a^2d - 9abc + 2b^3}{27\cdot a^3}\]
\[\alpha = \frac{3\cdot a\cdot c - b^2}{3\cdot a^2}\]
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Discussió de les solucions l'equació cúbica
Depenent dels valors de \(\left(\displaystyle\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3\)
la situació pot ser molt diferent.
Si \(\left(\displaystyle\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3<0\)
llavors \((u_0^+)^* = u_0^-\) son complexes i totes les solucions son reals:
\[
\begin{align}
x_0 &= - \frac{b}{3\cdot a} + 2\Re[u_0^+] \\
x_1 &= - \frac{b}{3\cdot a} - \Re[ u_0^+] - \sqrt{3}\Im[u_0^+] \\
x_2 &= - \frac{b}{3\cdot a} - \Re[ u_0^+] + \sqrt{3}\Im[u_0^+] \\
\end{align}
\]
Si \(\left(\displaystyle\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3=0\)
llavors \(u_0^+ = u_0^-\) i totes les solucions son reals:
\[
\begin{align}
x_0 &= - \frac{b}{3\cdot a} + 2u_0^+ \\
x_1 &= - \frac{b}{3\cdot a} - u_0^+ \\
x_2 &= - \frac{b}{3\cdot a} - u_0^+
\end{align}
\]
Si \(\left(\displaystyle\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3>0\)
llavors \(u_0^+ \neq u_0^-\) però reals i una solució és real i les altres imaginàries (complex conjugades entre elles).
A la rercerca d'un Discriminant
\[\beta = \frac{27a^2d - 9abc + 2b^3}{27\cdot a^3}\]
\[\alpha = \frac{3\cdot a\cdot c - b^2}{3\cdot a^2}\]
\[
\begin{align}
108\left[\left(\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3\right]
= 27\beta^2+4\alpha^3
&= 27\left(\frac{27a^2d - 9abc + 2b^3}{27\cdot a^3}\right)^2
+4\left(\frac{3\cdot a\cdot c - b^2}{3\cdot a^2}\right)^3 \\
&= \frac{\left(27a^2d - 9abc + 2b^3\right)^2}{27\cdot a^6}
+\frac{4\left(3\cdot a\cdot c - b^2\right)^3}{27\cdot a^6} \\
&= \frac{\left(27a^2d - 9abc + 2b^3\right)^2 + 4\left(3\cdot a\cdot c - b^2\right)^3}{27\cdot a^6}
\\
\end{align}
\]
\[
\begin{align}
108\cdot 27\cdot a^6\left[\left(\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3\right]
&= \left(27a^2d - 9abc + 2b^3\right)^2 + 4\left(3\cdot a\cdot c - b^2\right)^3
\\
\end{align}
\]
\[
\begin{align}
2916a^6\left[\left(\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3\right]
&= \left(27a^2d - 9abc + 2b^3\right)^2 + 4\left(3\cdot a\cdot c - b^2\right)^3
\\
&= 729a^4d^2 + 81a^2b^2c^2 + 4b^6 -486a^3bcd + 108a^2b^3d - 36ab^4c
+ 4\left(27a^3c^3 - 27a^2c^2b^2 + 9ab^4c-b^6\right)
\\
&= 729a^4d^2 + 81a^2b^2c^2 + 4b^6 -486a^3bcd + 108a^2b^3d - 36ab^4c
+ 108a^3c^3 - 108a^2c^2b^2 + 36ab^4c-4b^6
\\
&= 729a^4d^2 -27a^2b^2c^2 -486a^3bcd + 108a^2b^3d + 108a^3c^3
\\
\end{align}
\]
\[
\begin{align}
\frac{2916a^6\left[\left(\displaystyle\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3\right]}{27a^2}
&= 27a^2d^2 -b^2c^2 -18abcd + 4b^3d + 4ac^3
&\\
\\
108a^4\left[\left(\displaystyle\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3\right]
&= 27a^2d^2 -b^2c^2 -18abcd + 4b^3d + 4ac^3
\\
\end{align}
\]
Si volem tenir una quantitat que faci el mateix paper que el discriminant de l'equació de segon grau
caldria
\[
\Delta
= -a^4\left[27\beta^2+4\alpha^3\right]
= -108a^4\left[\left(\displaystyle\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3\right]
= -27a^2d^2 +b^2c^2 +18abcd - 4b^3d - 4ac^3
\]
\[a\cdot x^3+b\cdot x^2+c\cdot x+d = 0\]
\[
\Delta
= 18abcd + b^2c^2 - 27a^2d^2 - 4b^3d - 4ac^3
\]
\(\Delta>0\) arrels reals i les tres diferents
\(\Delta=0\) dues arrels iguals com a mínim
\(\Delta<0\) una arrel real i dues arrels complexes conjugades
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Arrels: valors enters
\[
\begin{align}
f(x)&=(x-a)\cdot(x-b)\cdot(x-c)\\
&=x^3-(a+b+c)x^2+(a\cdot b+ a\cdot c + b\cdot c)x-a\cdot b\cdot c
\end{align}
\]
\[f^\prime(x)=3x^2-2(a+b+c)x+(a\cdot b+ a\cdot c + b\cdot c)\]
\(f^\prime(x)=0\) es tradueix com \(3x^2-2(a+b+c)x+(a\cdot b+ a\cdot c + b\cdot c)=0\)
\[
\begin{align}
\Delta &= 4(a+b+c)^2 - 4\cdot3\cdot(a\cdot b+ a\cdot c + b\cdot c)\\
&= 4\left(a^2+b^2+c^2 - a\cdot b - a\cdot c - b\cdot c\right)\\
&= 4\left(\frac{3(a^2+b^2+c^2) - (a+b+c)^2}{2}\right)
\end{align}
\]
\[(a+b+c)^2=a^2+b^2+c^2 +2\cdot a\cdot b+2\cdot a\cdot c + 2\cdot b\cdot c\]
\[-\left(\cdot a\cdot b+\cdot a\cdot c + \cdot b\cdot c\right)=\frac{a^2+b^2+c^2 - (a+b+c)^2}{2}\]
i per tant
\[
\begin{align}
x &= \frac{2(a+b+c)\pm\sqrt{4\left(a^2+b^2+c^2 - a\cdot b - a\cdot c - b\cdot c\right)}}{2\cdot3}\\
&= \frac{(a+b+c)\pm\sqrt{a^2+b^2+c^2 - a\cdot b - a\cdot c - b\cdot c}}{3}\\
\end{align}
\]
\[f^{\prime\prime}(x)=6x-2(a+b+c)\]
\(f^{\prime\prime}(x)=0\) es tradueix com \(6x-2(a+b+c)=0\)
i llavors el punt de canvi de curvatura és \(x=\frac{a+b+c}{3}\)
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