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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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)


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.


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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A binomial squared sum

Prove that

\displaystyle \binom{n}{0}^2 + \binom{n}{1} ^2 + \cdots + \binom{n}{n}^2 = \binom{2n}{n}


Let us begin by recalling the binomial expansion

\displaystyle \left ( 1+x \right )^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} x^k


    \begin{align*} \left ( 1+x \right )^{2n} &= \left ( 1+x \right )^n \left ( 1+x \right )^n \\ &= \left [ \sum_{i=0}^{n} \binom{n}{i} x^i \right ] \left [ \sum_{j=0}^{n} \binom{n}{j} x^j \right ]\\ &=\sum_{i, j} \binom{n}{i} \binom{n}{j} x^{i+j} \\ &= \sum_{k=0}^{2n} x^k \sum_{i+j=k}\binom{n}{i} \binom{n}{j} \end{align*}

Equatating the coefficients of x^n we have that

    \begin{align*} \binom{2n}{n} &= \sum_{i+j=n} \binom{n}{i} \binom{n}{j} \\ &= \sum_{i=0}^{n} \binom{n}{i} \binom{n}{n-i}\\ &= \sum_{i=0}^{n} \binom{n}{i}^2 \end{align*}

yielding the result.

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A trigonometric sum


\displaystyle \mathcal{S} = \frac{1}{\cos 0^\circ \cos 1^\circ}+\frac{1}{\cos 1^\circ \cos 2^\circ}+\cdots+ \frac{1}{\cos 88^\circ \cos 89^\circ}


We note that

\begin{aligned} \frac{1}{\cos k \cos (k+1)} &=\frac{1}{\sin 1}\frac{\sin 1}{\cos k \cos (k+1)} \\ &= \frac{1}{\sin 1}\frac{\sin (k+1)\cos k - \sin k \cos (k+1)}{\cos k \cos (k+1)}\\ &= \frac{1}{\sin 1}\left [ \tan (k+1)- \tan k \right ] \end{aligned}

Thus summing telescopically we have that

\displaystyle \sum_{k=0}^{88}\frac{1}{\cos k \cos (k+1)}=\frac{1}{\sin 1} [\tan 89-\tan 0]=\frac{\cos 1}{\sin^2 1}

since \cot 1 = \tan 89 .

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