Polynomial equation

Let \Phi denote the golden ratio. Solve the equation

    \[x^3 +x^2-\Phi^5 x +\Phi^5=0\]

Solution

First of all we note that

    \[\left\{ \begin{array}{l} \Phi ^2 = \Phi + 1\\ \Phi ^3 = \Phi ^2 + \Phi = 2\Phi + 1 \end{array} \right. \Rightarrow \Phi ^5 = 2 \Phi ^2 + 3\Phi + 1 \Rightarrow \bold{\Phi ^5 = 5\Phi + 3}\]

We easily note that \Phi is one root of the equation, hence using Horner we get that

    \begin{align*} x^2 + \Phi ^2x - (2\Phi ^2 + \Phi ) = 0 &\Leftrightarrow x = \frac{ - \Phi ^2 \pm \sqrt {9\Phi ^2 + 6\Phi + 1}}{2} \\ &\Leftrightarrow x = \frac{ - (\Phi + 1) \pm (3\Phi + 1)}{2} \end{align*}

Hence \Phi is a double root and the other root is x=-2-\sqrt{5}.

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Similarity implies equality?

Let A\in \mathbb{C}^{n\times n} be similar to A^2. Does A=A^2 hold?

Solution

No! Take A=\begin{pmatrix}1 &0 \\1&1 \end{pmatrix} then A^2=\begin{pmatrix}1 &0\\ 2&1 \end{pmatrix}. The matrices are similar but not equal.

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On tensors

Let V_1, V_2, W_1, W_2, U_1, U_2 \in \; \mathbb{K} -Vect, V_1 \xrightarrow{\;\; \alpha_1 \;\; }W_1 \xrightarrow{\;\; \beta_1 \;\; }U_1, V_2 \xrightarrow{\alpha_2}W_2 \xrightarrow{\beta_2}U_2 \;\; \mathbb{K}-linear. Prove that

    \[(\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2)=(\beta_1 \alpha_1)\otimes(\beta_2 \alpha_2)\]

Solution

Recall the general definition of the tensor product of linear maps, we have successively:

    \begin{align*} ((\beta_1 \alpha_1) \otimes (\beta_2 \alpha_2))(v_1 \otimes v_2) &= (\beta_1 \alpha_1)(v_1) \otimes (\beta_2 \alpha_2)(v_2) \\ &=\beta_1(\alpha_1(v_1)) \otimes \beta_2(\alpha_2(v_2)) \\ &= (\beta_1 \otimes \beta_2)(\alpha_1(v_1) \otimes \alpha_2(v_2)) \\ &=((\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2))(v_1 \otimes v_2) \end{align*}

Thus, the two linear maps V_1 \otimes V_2 \rightarrow U_1 \otimes U_2 are equal when composed with the canonical bilinear map V_1 \times V_2 \to V_1 \otimes V_2, hence equal (by the universal property).

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Fibonacci series

Let \{F_n\}_{n \in \mathbb{N}} denote the Fibonacci sequence such that F_1=F_2=1 and F_3=2. Prove that

    \[\sum_{n=2}^{\infty} {\rm arctanh} \;  \frac{1}{F_{2n}}  =\frac{\log 3}{2}\]

Solution

It holds that

    \[{\rm arctanh}\; x - {\rm arctanh} \; y= {\rm arctanh} \left ( \frac{x-y}{1-xy} \right )\]

It follows from Cassini’s identity that

(1)   \begin{equation*} {F_{n-1} F_{n+1} - F_n^2 = (-1)^n \end{equation*}

Setting n \mapsto 2n back at (1) we get

(2)   \begin{equation*} F^2_{2n}=F_{2n-1} F_{2n+1} -1 \end{equation*}

Since F_n =F_{n-1} + F_{n-2} we get

(3)   \begin{equation*} F_{2n+1} =F_{2n} + F_{2n-1} \Leftrightarrow F_{2n} = F_{2n+1} - F_{2n-1} \end{equation*}

Hence,

    \begin{align*} \sum_{n=2}^{\infty} {\rm arctanh}\;   \frac{1}{F_{2n}}  &= \sum_{n=2}^{\infty} {\rm arctanh} \; \frac{F_{2n}}{F_{2n}^2}  \\ &=\sum_{n=2}^{\infty} {\rm arctanh} \frac{F_{2n-1} - F_{2n+1}}{1-F_{2n-1}F_{2n+1}} \\ &= \sum_{n=2}^{\infty} {\rm arctanh} \;  \frac{\frac{1}{F_{2n-1}} - \frac{1}{F_{2n+1}}}{1- \frac{1}{F_{2n-1} F_{2n+1}}} \\ &=\lim_{m \rightarrow +\infty} \sum_{n=2}^{m} \bigg [ {\rm arctanh} \left ( \frac{1}{F_{2n-1}} \right ) - \\ &\quad \quad \quad - {\rm arctanh} \left ( \frac{1}{F_{2n+1}} \right ) \bigg ] \\ &= \lim_{m \rightarrow +\infty} \left [ {\rm arctanh} \left ( \frac{1}{F_3} \right ) - {\rm arctanh} \left ( \frac{1}{F_{2m+1}} \right )\right ]\\ &= {\rm arctanh} \left ( \frac{1}{2} \right ) \\ &= \frac{\log 3}{2} \end{align*}

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Logarithmic integral

Evaluate the integral

    \[\mathcal{J} = \int_{0}^{1} \ln^2 \left | \sqrt{x} - \sqrt{1-x} \right |\, \mathrm{d}x\]

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