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On an infinite summation

Let \{x_n\}_{n=1}^{\infty} be a sequence of real numbers. Compute:

    \[\mathcal{V} = \sum_{n=1}^{\infty} \sin^2 x_n \prod_{k=1}^{n-1} \cos^2 x_k + \prod_{n=1}^{\infty} \cos^2 x_n\]

Solution

First and foremost we set a_n = \sin^2 x_n and it is obvious that 0 \leq a_n \leq 1. We are making use of probabilistic methods. Suppose than an infinite number of coins are flipped. Let a_n be the probability that the n -th coin toss lands heads and let us consider the first time heads comes up. Then a_n \prod \limits_{k=1}^{n-1} (1 -a_k) is the probability that the first head appears in the n – th flip and \prod \limits_{n=1}^{\infty} (1-a_n) is the probability that all flips come up tails. Thus,

    \[\sum_{n=1}^{\infty} \sin^2 x_n \prod_{k=1}^{n-1} \cos^2 x_k + \prod_{n=1}^{\infty} \cos^2 x_n=1\]

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Trigonometric equality

Prove that in any triangle ABC it holds that

    \[\sum \sqrt{\frac{\sin A}{\sin B \sin C}} = \sqrt{\frac{2R}{r} \sum \sin A}\]

where R denotes the circumradius and r the inradius.

Solution

Using the law of sines we have that

    \[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2 R\]

and if we denote \mathcal{A} the area of the triangle then

    \[\frac{r \left ( a + b + c \right )}{2} = \mathcal{A} = \frac{abc}{4R}\]

Thus,

    \begin{align*} \sum \sqrt{\frac{\sin A}{\sin B \sin C}} &= \sum \sqrt{\frac{a}{2R} \cdot \frac{2R}{b} \cdot \frac{2R}{c}} \\ &=\sum \sqrt{\frac{ a}{bc} \cdot 2 R} \\ &=\sum \sqrt{\frac{a}{bc} \cdot \frac{abc}{2\mathcal{A}}} \\ &=\frac{a+b+c}{\sqrt{2 \mathcal{A}}} \\ &= \sqrt{\frac{a+b+c}{\frac{2\mathcal{A}}{a+b+c}}} \\ &= \sqrt{\frac{a+b+c}{r}} \\ &= \sqrt{\frac{1}{r} \sum a} \\ &= \sqrt{\frac{2R}{r} \sum \frac{a}{2R}} \\ &= \sqrt{\frac{2R}{r} \sum \sin A} \end{align*}

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Non linear system

Solve the system

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

Solution

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 mathematica.gr .

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Finite trigonometric product

Evaluate the product

\Pi = \left( \tan 61^\circ - \sqrt{3} \right) \left( \tan 62^\circ - \sqrt{3} \right) \cdots \left( \tan 89^\circ - \sqrt{3} \right)

Solution

One good way to begin is by invoking the identity

    \[\tan \alpha + \tan \beta = \frac{\sin \left ( \alpha + \beta \right )}{\cos \alpha \cos \beta}\]

Then noting that \sqrt{3}=\tan 60^\circ we have that

    \begin{align*} \Pi &= \prod_{n=61}^{89} \left ( \tan n^\circ - \sqrt{3} \right ) \\ &= \prod_{n=61}^{89} \left ( \tan n^\circ - \tan 60^\circ \right )\\ &= \prod_{n=1}^{29} \frac{\sin n^\circ}{ \cos 60^\circ  \cos \left ( 90^\circ-n^\circ \right )} \\ &=\frac{1}{\left ( \cos 60^\circ \right )^{29}}\prod_{n=1}^{29} \frac{\cancel{\sin n^\circ}}{ \cancel{\sin n^\circ}} \\ &=\frac{1}{ \left( \cos 60^\circ \right)^{29} } \prod_{n=1}^{29} 1  \\ &= 2^{29} \end{align*}

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Arithmetic – Harmonic Progression

Consider the harmonic sequence

\displaystyle 1, \frac{1}{2}, \frac{1}{3} ,  \cdots, \frac{1}{n} , \cdots

Prove that if we pick dinstinct terms of the above sequence we can construct an arithmetic progression sequence of as large (finite) length as we want.

Solution

We just observe that

\displaystyle \frac{1}{n!}, \frac{2}{n!}, \cdots, \frac{n!}{n!}

are all distinct terms of the sequence.

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