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Convergence of series

October 28, 2020 / Leave a comment

For v =\langle x_1, x_2, \dots, x_n \rangle \in \mathbb{R}^n we define \left \| v \right \|_p = \left ( \sum \limits_{i=1}^{n} \left | x_i \right |^p \right )^{1/p} and \left \| v \right \|_{\infty} = \max \limits_{1 \leq i \leq n} \left | x_i \right |. For which p does the series

    \[\sum_{p=1}^{\infty} \left ( \left \| v \right \|_p - \left \| v \right \|_{\infty} \right)\]

converge?

Solution

We may assume that v is not the zero vector and n>1 otherwise the series is trivially convergent. Then, we show that the series is convergent if and only if there is exactly one component of maximal absolute value.

(a) If the above condition is satisfied then, without loss of generality, let x_1 be the component of maximal absolute value and let \displaystyle t = \frac{\max \limits_{2 \leq i \leq n} \left | x_i \right |}{\left | x_i \right |} \in [0, 1]. Hence , as p \rightarrow +\infty

    \begin{align*} 0 &\leq \left \| v \right \|_p - \left \| v \right \|_{\infty} \\ &= \left \| v \right \|_{\infty} \left ( \left ( 1 + \left ( n-1 \right )t^p \right )^{1/p} - 1 \right ) \\ &=\left \| v \right \|_{\infty} \left ( \exp \left ( \frac{\ln \left ( 1+ (n-1) t^p \right )}{p} \right ) - 1\right ) \\ &\sim \left \| v \right \|_{\infty} \left ( n-1 \right ) \frac{t^p}{p} \end{align*}

and the given series is convergent because \displaystyle \sum_{p=1}^{\infty} \frac{t^p}{p} < + \infty.

(b) If the above condition is not satisfied, then there are at least 2 components of maximal absolute value and therefore

    \begin{align*} \left \| v \right \|_p - \left \| v \right \|_{\infty} & \geq \left \| v \right \|_{\infty} \left ( 2^{1/p} -1 \right )\\ &= \left \| v \right \|_{\infty} \left ( \exp \left ( \frac{\ln 2}{p} \right ) -1 \right ) \\ &\sim \left \| v \right \|_{\infty} \frac{\ln 2}{p} \end{align*}

and the given series is not convergent because \displaystyle \sum_{p=1}^{\infty} \frac{1}{p} = +\infty.

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Huygen’s Inequality

October 28, 2020 / Leave a comment

Let \alpha \in \left[0, \frac{\pi}{2} \right). Prove that

    \[2 \sin \alpha + \tan \alpha \geq 3\alpha\]

Solution

Let f(\alpha) = 2 \sin \alpha + \tan \alpha. f is twice differentiable with

    \[f''(\alpha) = -2 \sin \alpha +2 \tan \alpha + 2 \tan^3 \alpha \geq 0\]

Hence f is convex. The tangent at (0, 0) has equation y=3x. The result follows.

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Sigma divisor sum

October 27, 2020 / Leave a comment

Let \sigma denote the divisor function. Prove that

    \[\lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{k=1}^{n} \sigma(k) = \frac{\pi^2}{12}\]

Solution

We have successively:

    \begin{align*} \frac{1}{n^2} \sum_{k\leq n}\sigma(k) &= \frac {1}{n^2}\sum_{k \leq n}\sum_{q \mid k}q \\ &= \frac {1}{n^2}\sum_{q,d \mid qd \leq n}q \\ &= \frac {1}{n^2} \sum_{d\leq n} \sum_{q \leq n/d}q \\ &= \frac {1}{2n^2}\sum_{d \leq n} \left ( \frac {n^2}{d^2}+ \mathcal{O} \left ( \frac {n}{d}\right) \right ) \\ &= \frac{1}{2} \sum_{d \leq n} \frac{1}{d^2} + \mathcal{O} \left ( \frac{1}{n} \sum_{d \mid n} \frac{1}{d} \right ) \\ & = \frac{1}{2} \sum_{d \leq n} \frac{1}{d^2} + \mathcal{O} \left ( \frac{\ln n}{n} \right ) \\ & \rightarrow \frac{\pi^2}{12} \end{align*}

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Derivative at 0

October 11, 2020 / Leave a comment

Let

    \[f(x) = \frac{\sqrt{1+2x} \cdot \sqrt[4]{1+4x} \cdot \sqrt[6]{1+6x} \cdot \cdots \cdot \sqrt[100]{1+100x}}{\sqrt[3]{1+3x} \cdot \sqrt[5]{1+5x} \cdot \sqrt[7]{1+7x} \cdot \cdots \cdot \sqrt[101]{1+101x}}\]

Evaluate f'(0).

Solution

The domain of f is \mathcal{A}_f = \left ( - \frac{1}{101} , +\infty \right ). Let us consider the logarithmic of f

    \[g(x) = \ln f(x) = \sum_{k=2}^{101} \frac{(-1)^k}{k} \ln \left ( 1 + k x \right )\]

and differentiate it; Hence,

    \[g'(x) = \frac{f'(x)}{f(x)} = \sum_{k=2}^{101} \frac{(-1)^k}{1+kx}\]

For x=0 we have

    \[g'(0) = \sum_{k=2}^{101} (-1)^k =0\]

Thus, f'(0)=0 since f(0)=1.

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Integral function

September 22, 2020 / Leave a comment

Let f, g:\mathbb{R} \rightarrow \mathbb{R} be two continuous functions such that

    \[\int_{\int_{x}^{1} f(t) \, \mathrm{d}t}^{\int_{0}^{x}f(t) \, \mathrm{d}t} g(t) \, \mathrm{d}t>0 \quad \text{for all} \;\; x \in \mathbb{R} \setminus \{0, 1 \}\]

If g(x) + g(2-x)=2 and g(x) \neq 0 for all x \in \mathbb{R}, then prove that:

  1. \displaystyle \int_{0}^{x} f(t) \, \mathrm{d}t > \int_{x}^{1} f(t) \, \mathrm{d}t
  2. \displaystyle \int_{0}^{1} f(t)\, \mathrm{d}t = 0
  3. f(1)=f(0)=0
  4. the equation \displaystyle f(x) \int_{x}^{1} f(t) \, \mathrm{d}t = f(x) f'(x) has at least a root in (0, 1) if f is differentiable.
  5. the equation \displaystyle xf(x) = \int_{x}^{1} f(t) \, \mathrm{d}t has at least a root in (0, 1).
  6. the area of g bounded by the axis x'x and the lines x=0 , x=2 is 2 square meters.

Solution

  1. Since g(x) \neq 0 for all x \in \mathbb{R} it follows that the sign of g is constant. For x=1 we get that g(1)=1>0 hence g(x)>0 for all x \in \mathbb{R}. Therefore,

        \[\int_{\int_{x}^{1} f(t) \, \mathrm{d}t}^{\int_{0}^{x}f(t) \, \mathrm{d}t} g(t) \, \mathrm{d}t>0 \overset{g(x)>0}{\Leftarrow \! =\! =\! =\! =\! \Rightarrow } \int_{0}^{x} f(t) \, \mathrm{d}t > \int_{x}^{1} f(t) \, \mathrm{d}t\]

  2. Taking limits we have

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