4 problems found
Show that, provided \(q^2\ne 4p^3\), the polynomial \[ \hphantom{(p\ne0, \ q\ne0)\hspace{2cm}} x^3-3px +q \hspace {2cm} (p\ne0, \ q\ne0) \] can be written in the form \[ a(x-\alpha)^3 + b(x-\beta)^3\,, \] where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(pt^2 -qt +p^2=0\), and \(a\) and \(b\) are constants which you should express in terms of \(\alpha\) and \(\beta\). Hence show that one solution of the equation \(x^3-24x+48=0\,\) is \[ x= \frac{2 (2-2^{\frac13})}{1-2^{\frac13}} \] and obtain similar expressions for the other two solutions in terms of \(\omega\), where \(\omega = \mathrm{e}^{2\pi\mathrm{i}/3}\,\). Find also the roots of \(x^3-3px +q=0\) when \(p=r^2\) and \(q= 2r^3\) for some non-zero constant \(r\).
In this question, you may use without proof the results \[ 4 \cosh^3 y - 3 \cosh y = \cosh (3y) \ \ \ \ \text{and} \ \ \ \ \mathrm{arcosh} \, y = \ln ( y+\sqrt{y^2-1}). \] \noindent[ {\bf Note: } \(\mathrm{arcosh}y\) is another notation for \(\cosh^{-1}y\,\)] Show that the equation \(x^3 - 3a^2x = 2a^3 \cosh T\) is satisfied by \( 2a \cosh \l \frac13 T \r\) and hence that, if \(c^2\ge b^3>0\), one of the roots of the equation \(x^3-3bx=2c\) is \(\ds u+\frac{b}{u}\), where \(u = (c+\sqrt{c^2-b^3})^{\frac13}\;\). Show that the other two roots of the equation \(x^3-3bx=2c\) are the roots of the quadratic equation \[\ds x^2 + \Big( u+\frac{b}{u}\Big) x + u^2+\frac{b^2}{u^2}-b=0\, ,\] and find these roots in terms of \(u\), \(b\) and \(\omega\), where \(\omega = \frac{1}{2}(-1 + \mathrm{i}\sqrt{3})\). Solve completely the equation \(x^3-6x=6\,\).
Given that \(\alpha = \e^{\mathrm{i} \pi/3}\) , prove that \(1 + \alpha^2 = \alpha\). A triangle in the Argand plane has vertices \(A\), \(B\), and \(C\) represented by the complex numbers \(p\), \(q\alpha^2\) and \(- r\alpha\) respectively, where \(p\), \(q\) and \(r\) are positive real numbers. Sketch the triangle~\(ABC\). Three equilateral triangles \(ABL\), \(BCM\) and \(CAN\) (each lettered clockwise) are erected on sides \(AB\), \(BC\) and \(CA\) respectively. Show that the complex number representing \(N\) is \mbox{\(( 1 - \alpha) p- \alpha^2 r\)} and find similar expressions for the complex numbers representing \(L\) and \(M\). Show that lines \(LC\), \(MA\) and \(NB\) all meet at the origin, and that these three line segments have the common length \(p+q+r\).
Let \(\omega=\mathrm{e}^{2\pi\mathrm{i}/3}.\) Show that \(1+\omega+\omega^{2}=0\) and calculate the modulus and argument of \(1+\omega^{2}.\) Let \(n\) be a positive integer. By evaluating \((1+\omega^{r})^{n}\) in two ways, taking \(r=1,2\) and \(3\), or otherwise, prove that \[ \binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots+\binom{n}{k}=\frac{1}{3}\left(2^{n}+2\cos\left(\frac{n\pi}{3}\right)\right), \] where \(k\) is the largest multiple of \(3\) less than or equal to \(n\). Without using a calculator, evaluate \[ \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} \] and \[ \binom{24}{2}+\binom{24}{5}+\cdots+\binom{24}{23}\,. \] {[}\(2^{25}=33554432.\){]}
Solution: Since \(\omega^3 = 1\) and \(\omega \neq 1\) we must have that \((\omega-1)(1 + \omega + \omega^2) = 0\) but by dividing by \(\omega - 1\) we obtain the desired result. \(1+\omega^2 = -\omega\) so \(|1 + \omega^2| = |-\omega| = 1\) and \(\arg ( 1 + \omega^2) = \arg(-\omega) = \pi - \frac{2\pi}{3} = \frac{\pi}{3}\) \begin{align*} && (1 + 1)^n &= \sum_{k=0}^n \binom{n}{k}\\ && (1+ \omega)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{k} \\ && (1+ \omega^2)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{2k} \\ \Rightarrow && 2^n+(-\omega^2)^n + (-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n (1+1+1)\binom{n}{k} + \sum_{k=0, k \equiv 1 \pmod{3}}^n (1 + \omega + \omega^2) \binom{n}{k} + \sum_{k=0, k \equiv 2 \pmod{3}}^n (1 + \omega^2 + \omega) \binom{n}{k} \\ \Rightarrow && 2^n +((-\omega)^n)^{-1}+(-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n \binom{n}{k} \end{align*} \(2^n +((-\omega)^n)^{-1}+(-\omega)^n = 2^n + 2 \textrm{Re}(-\omega^n) = 2^n + 2 \cos \frac{n\pi}{3}\) Therefore our answer follows. \begin{align*} \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} &= \frac13 \l 2^{25} + 2\cos (\frac{25 \pi}{3}) \r \\ &= \frac13 \l 2^{25} + 2 \cos \frac{\pi}{3} \r \\ &= \frac13 \l 2^{25} + 1 \r \\ &= \frac13 \l (4096 \cdot 4096 \cdot 2) + 1 \r \\ &= 11\,184\,811 \end{align*} Notice that \(S_2 = \binom{24}{2} + \cdots +\binom{24}{23} = \binom{24}{1} + \cdots + \binom{24}{22} = S_1\) and \(S_0 = \binom{24}0 + \cdots + \binom{24}{21} = \frac13 \l 2^{24} + 2 \r\) Therefore since \(S_0 + 2 \cdot S_2 = 2^{24}\) we must have \begin{align*} S_2 &= \frac12 \l 2^{24} - \frac13 \l 2^{24} + 2 \r \r \\ &= \frac13 \l 2^{24} - 1 \r \\ &= \frac13 \l 16777216- 1 \r \\ &= \frac13 \cdot 16777215 \\ &= 5\,592\,405 \end{align*}