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Rational function and polynomial

Prove that there does not exist a rational function f with real coefficients such that

    \[f \left ( \frac{x^2}{x+1} \right ) = \mathrm{P}(x)\]

where \mathrm{P}(x) \in \mathbb{R}[x] is a non constant polynomial.


Since polynomials are defined on x=-1 we have that

    \begin{align*} \mathbb{R} \ni \mathrm{P}(-1) & =\lim_{x \rightarrow -1^+} \mathrm{P}(x) \\ &=\lim_{x \rightarrow -1^+}f \left(\frac{x^2}{x+1} \right) \\ &=\lim_{x \rightarrow \infty} f\left(\frac{x^2}{x+1}\right) \\ &=\lim_{x \rightarrow \infty} \mathrm{P}(x) \end{align*}

Since \mathrm{P} tends to a finite value as x \rightarrow \infty it must be a constant polynomial. In particular, f must be constant in the range of \frac{x^2}{x+1} which is an infinite set, implying that f must also be constant. This proves what we wanted.

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Sum over all positive rationals

For a rational number x that equals \frac{a}{b} in lowest terms , let f(x)=ab. Prove that:

    \[\sum_{x \in \mathbb{Q}^+} \frac{1}{f^2(x)} = \frac{5}{2}\]


First of all we note that

    \[F(s) = \sum_{x \in \mathbb{Q}^+} \frac{1}{f^s(x)} = \sum_{\substack{a,b=1 \\ \gcd(a, b)=1}}^{\infty} \frac{1}{\left ( ab \right )^s}\]

Moreover for s>1 we have that

    \begin{align*} \zeta^2(s) &= \left ( \sum_{a=1}^{\infty} \frac{1}{a^s} \right )^2 \\ &=\sum_{a, b=1}^{\infty} \frac{1}{(ab)^s} \\ &=\sum_{d=1}^{\infty} \sum_{\substack{a, b=1 \\\gcd(a, b)=d}}^{\infty} \frac{1}{\left ( ab \right )^s} \\ &= \sum_{d=1}^{\infty} \frac{1}{d^{2s}} \sum_{\substack{a, b=1 \\\gcd(a, b)=1}}^{\infty} \frac{1}{\left ( ab \right )^s} \\ &= \zeta(2s) F(s) \end{align*}

Hence for s=2 we have that

    \[F(2) = \frac{5}{2}\]

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Infinite Product

Evaluate the infinite product:

    \[\Pi = \prod_{n=3}^{\infty} \frac{\left ( n^3+3n \right )^2}{n^6-64}\]

(IMC 2019 / Day 1 / Problem 1)


    \begin{align*} \prod_{n=3}^{\infty} \frac{\left ( n^3+3n \right )^2}{n^6-64} &= \prod_{n=3}^{\infty} \frac{n^2 \left ( n^2+3 \right )^2}{n^6-69} \\ &=\prod_{n=3}^{\infty} \frac{n^2\left ( n^2+3 \right )^2}{\left ( n^3-8 \right )\left ( n^3+8 \right )} \\ &= \prod_{n=3}^{\infty} \frac{n^2\left ( n^2+3 \right )^2}{\left ( n-2 \right )\left ( n^2+2n+4 \right )\left ( n+2 \right )\left ( n^2-2n+4 \right )} \\ &=\prod_{n=3}^{\infty} \frac{n}{n-2} \cdot \frac{n}{n+2} \cdot \frac{n^2+3}{\left ( n-1 \right )^2+3} \cdot \frac{n^2+3}{\left ( n+1 \right )^2+3} \\ &= \lim_{N \rightarrow +\infty} \frac{N\left ( N-1 \right )}{1 \cdot 2} \cdot \frac{3 \cdot 4}{\left ( N+1 \right ) \left ( N+2 \right )} \cdot \frac{N^2+3}{2^2+3} \cdot \\ & \quad \quad \quad \quad \quad \cdot\frac{3^2+3}{\left ( N+1 \right )^2+3} \\ &= \frac{72}{7} \end{align*}

since the product telescopes.

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On an infinite summation

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

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


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.


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

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


    \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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