Non linear system

Solve the system

    \[\left\{\begin{matrix} a+b & = &4 \\ a^4+b^4& = & 82 \end{matrix}\right.\]


We set a=2+k and b=2-k where k \in [-2, 2]. The second equation of course is written as

    \[\left ( 2+k \right )^4 + \left ( 2-k \right )^4 = 82 \Leftrightarrow k^4 +24k^2 -25 =0 \Leftrightarrow k = \pm 1\]

Now a, b‘s follow.

Historical note: In the Babylonian signs , tracing back in 1700 BC , there are a lot of geometrical problems that are equivalent to the solution of such systems. In order to be solved the following ( modern ) formulae were used

    \[\alpha = \frac{\alpha + \beta}{2} + \frac{\alpha-\beta}{2} \quad , \quad \beta = \frac{\alpha + \beta}{2} - \frac{\alpha-\beta}{2}\]

    \[\alpha \beta = \left ( \frac{\alpha + \beta}{2} \right )^2 - \left ( \frac{\alpha - \beta}{2} \right )^2\]

The above procedure actually led to the discriminant in order for us to solve a second order equation. For example in a book of that age we see the equation x(6+x)=16. Try to solve this using the above technique.

The above exercise , along with the historical note , can be found at .

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The series converges

Prove that the series

    \[\mathcal{S}  = \sum_{n=1}^{\infty} \frac{\sin ( \sin n)}{n}\]

converges. Examine if the convergence is absolute.


Lemma: Let \alpha \in \mathbb{R} such that \alpha / \pi \notin \mathbb{Q} then the sequence

    \[\omega_n = \sin ( \sin \alpha ) + \sin ( \sin ( 2 \alpha)) + \cdots + \sin ( \sin (n \alpha))\]

is bounded.

With the above lemma in mind the series converges as a consequence of Dirichlet’s theorem. The fact that the series does not converge absolutely can be seen by applying the Jordan inequality

(1)   \begin{equation*} \left | \sin a \right | \geq \frac{2 \left | a \right |}{\pi} \quad , \quad a \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \end{equation*}


\begin{aligned} \left | \frac{\sin \left ( \sin n \right )}{n} \right | + \left | \frac{\sin \left ( \sin \left ( n+1 \right ) \right )}{n+1} \right | &\geq \left | \frac{\sin \left ( \sin n \right )}{n+1} \right | + \left | \frac{\sin \left ( \sin \left ( n+1 \right ) \right )}{n+1} \right | \\ &\geq \frac{2}{\pi} \left | \frac{\sin n}{n+1} \right | + \frac{2}{\pi} \left | \frac{\sin (n+1)}{n+1} \right |\\ &\geq \frac{2}{\pi \left ( n+1 \right )} \left | \sin n \cos (n+1) - \cos n \sin (n+1) \right | \\ &= \frac{2}{\pi (n+1)} \left | \sin 1 \right | \end{aligned}

hence the series diverges absolutely.




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Range of convergence

For what values of x \in \mathbb{R} does the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \cos (2^n x)\]


For none. Since for the sequence a_n=\cos 2^n x it holds that a_{n+1} = 2 a_n^2 -1 we deduce that the potential limits are 1 or -\frac{1}{2}. Hence, the sequence cannot tend to zero and the conclusion follows.

Note: One interesting question is the following. For what values of x \in \mathbb{R} does the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \sin (2^n x)\]

converge? The answer may be found at this link.

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Irreducible fraction

Find all positive integers \alpha such that it holds that

(1)   \begin{equation*} \frac{1}{\alpha} = 0.\bar{\alpha} \end{equation*}

where \overline{ \cdot } stands for the period.



    \begin{align*} \frac{1}{\alpha} = 0.\bar{\alpha} &\Leftrightarrow \frac{1}{\alpha} = 0.\alpha \alpha \alpha \dots \\ &\Leftrightarrow \frac{1}{\alpha} = \frac{\alpha}{10} \sum_{i=0}^{\infty} \frac{1}{10^i} \\ &\Leftrightarrow \frac{1}{\alpha} = \frac{\alpha}{9}\\ &\!\!\!\!\!\!\!\overset{\alpha>0}{\Leftarrow \! =\! =\! \Rightarrow } \alpha = 3 \end{align*}

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On the supremum and infimum of a sine sequence

Let \{x_n\}_{n \in \mathbb{N}} be a sequence defined as

    \[x_n = \sin 1 + \sin 3 + \sin 5 + \cdots + \sin (2n -1)\]

Find the supremum as well as the infimum of the sequence x_n.


Background: This problem was on the shortlist of the 2014 Olimpiada Nationala de Matematica de Romania and was suggested by Leo Giugiuc.

We begin by the very well known manipulation.

    \begin{align*} \sum_{k=1}^n\sin (2k-1) &= {\rm Im}\left[\sum_{k=1}^ne^{i(2k-1)}\right]\\ &= {\rm Im}\left[\frac{e^{i}(1-e^{2ni})}{1-e^{2i}}\right]\\ &= {\rm Im}\left[\frac{e^{i}}{1-e^{2i}}(1-e^{2ni})\right]\\ &= {\rm Im}\left[\frac{e^{i}}{1-e^{2i}}(1-\cos 2n+i\sin 2n)\right]\\ &= {\rm Im}\left[\frac{i}{2\sin 1}(1-\cos 2n+i\sin 2n)\right]\\ &=\frac{1-\cos 2n}{2\sin 1} \end{align*}

Thus \displaystyle x_n= \frac{1-\cos 2n}{2\sin 1} and we have to find the supremum and infimum of \cos 2n. Since the values n \mod 2\pi are dense on the unit circle , the same shall hold for 2 n \mod 2\pi implying that \inf \cos 2n =-1 and \sup \cos 2n = 1. Thus,

\displaystyle \inf\{x_n\}=\frac{1}{2\sin 1}(1-1)=0 \quad , \quad \sup\{x_n\}=\frac{1}{2\sin 1}(1+1)=\frac{1}{\sin 1}

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