Apunts de Matemàtiques i Física
PREAMBLE
Projecte
TRIGONOMETRY
Cordes en cercles
SPECIAL FUNCTIONS
Nu and related functions
ALGEBRA
Matrius i determinants
Racionalització

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:
\[ \begin{align} x_0 &= - \frac{b}{3\cdot a} + u_0^+ + v_0^- \\ x_1 &= - \frac{b}{3\cdot a} - \frac{ u_0^+ + u_0^-}{2} + i\sqrt{3}\frac{ u_0^+ - u_0^-}{2} \\ x_2 &= - \frac{b}{3\cdot a} - \frac{ u_0^+ + u_0^-}{2} - i\sqrt{3}\frac{ u_0^+ - u_0^-}{2} \end{align} \] \[ \begin{align} u_0^+ &= \sqrt[3]{- \frac{\beta}{2} + \sqrt{\left(\frac{\beta}{2}\right)^2+\left(\frac{\alpha}{3}\right)^3}}\\ u_0^- &= \sqrt[3]{- \frac{\beta}{2} - \sqrt{\left(\frac{\beta}{2}\right)^2+\left(\frac{\alpha}{3}\right)^3}} \end{align} \] \[\beta = \frac{27a^2d - 9abc + 2b^3}{27\cdot a^3}\] \[\alpha = \frac{3\cdot a\cdot c - b^2}{3\cdot a^2}\]

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
\(a=\) \(b=\) \(c=\) \(d=\)

\(\pi=\)
\(\beta=\)
\(\alpha=\)
\(\left(\displaystyle\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3 = \)
\(\Delta = -108a^4\left[\left(\displaystyle\frac{\beta}{2}\right)^2+\left(\displaystyle\frac{\alpha}{3}\right)^3\right] =\)
\(\Delta = 18abcd + b^2c^2 - 27a^2d^2 - 4b^3d - 4ac^3 = \)

\(u_0^+ =\)
\(u_0^- =\)

\(x_0 =\)
\(x_1 =\)
\(x_2 =\)

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 \(x=\frac{a+b+c}{3}\)