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Bernstein limit

June 17, 2019 / Leave a comment

Let a \neq b be real numbers. Consider the function:

    \[f(x) = \left\{\begin{matrix} a & , & 0 \leq x < c \\ b & , & c \leq x \leq 1 \end{matrix}\right.\]

as well as the polynomial \displaystyle \mathrm{B}_n(x) = \sum_{i=0}^{n} \binom{n}{i} x^i \left ( 1-x \right )^{n-i} f \left ( \frac{i}{n} \right ). Evaluate the limit \lim \limits_{n \rightarrow +\infty} \mathrm{B}_n(c).

Solution

The polynomial \mathrm{B}_n is nothing else than a Bernstein one. f is Riemann integrable on [0, 1] and c is a discontinuity of a first kind. Thus,

    \[\lim_{n \rightarrow +\infty} \sum_{i=0}^{n} \binom{n}{i} \left ( 1-c \right )^{n-i} c^i f \left ( \frac{i}{n} \right ) = \frac{f\left ( c^- \right ) + f \left ( c^+ \right )}{2} = \frac{a+b}{2}\]

where f(c^-) = \lim \limits_{x \rightarrow c^-} f(x) and f(c^+)=\lim \limits_{x \rightarrow c^+} f(x).

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An iteration limit

June 16, 2019 / Leave a comment

Let n \geq 1 be a natural number and let f_n=x^{x^{x^{\dots^{x}}}}  where the number of x‘s in the definition of f_n is n. For example:

    \[f_1 = x \quad , \quad f_2 = x^x \quad , \quad f_3=x^{x^x}\]

Evaluate the limit:

    \[\ell = \lim_{x\rightarrow 1} \frac{f_n(x) - f_{n-1}(x)}{\left ( 1-x \right )^{n+1}}\]

Solution

We easily see by induction that

    \[f_n(x) = 1 +\mathcal{O} \left ( 1-x \right ) \quad \text{for} \quad n \geq 1\]

as well as

    \[f_n(x) = f_{n-1}(x) + (-1)^n \left ( 1-x \right )^n + \mathcal{O} \left ( \left ( 1-x \right )^{n+1} \right ) \quad \text{for} \quad n \geq 2\]

Thus,

    \[\frac{f_n(x) - f_{n-1}(x)}{\left ( 1-x \right )^n} = (-1)^n + \mathcal{O}\left ( \left ( 1-x \right )^{n} \right )\]

and thus \ell does not exist.

 

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Double summation

June 16, 2019 / Leave a comment

Let \alpha>3 be a real number. Evaluate the series:

    \[\mathcal{S} = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{n}{\left ( n+m \right )^\alpha}\]

Solution

Since the summands are all positive , we can sum by triangles. Thus,

    \begin{align*} \mathcal{S}&=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{n}{\left ( n+m \right )^\alpha} \\ &=\sum_{n=2}^{\infty} \frac{\sum \limits_{k=1}^{n-1} k}{ n^\alpha} \\ &=\frac{1}{2} \sum_{n=2}^{\infty} \frac{n\left ( n-1 \right )}{n^\alpha} \\ &= \frac{1}{2} \left ( \sum_{n=2}^{\infty} \frac{1}{n^{\alpha-2}} - \sum_{n=2}^{\infty} \frac{1}{n^{\alpha-1}} \right ) \\ &= \frac{\zeta\left ( \alpha-2 \right ) - \zeta\left ( \alpha-1 \right )}{2} \end{align*}

where \zeta is the Riemann zeta function.

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Inequality of a function

June 9, 2019 / Leave a comment

Let f:[0, 1] \rightarrow \mathbb{R} be a differentiable function with continuous derivative. Prove that:

    \[\left | f\left ( \frac{1}{2} \right ) \right | \leq \int_{0}^{1} \left | f(x) \right | \, \mathrm{d}x + \frac{1}{2} \int_{0}^{1} \left | f'(x) \right | \, \mathrm{d}x\]

Solution

For 0 \leq x \leq \frac{1}{2} it holds that

    \[f\left ( \frac{1}{2} \right ) -f(x) = \int_{x}^{1/2} f'(t)\, \mathrm{d}t\]

Taking absolute values and using basic properties of the integral we get

    \[\left | f\left ( \frac{1}{2} \right ) \right | \leq \left | f(x) \right | + \int_{0}^{1/2} \left | f'(t) \right | \, \mathrm{d}t\]

Integrating we have:

(1)   \begin{equation*} \frac{1}{2}\left | f\left ( \frac{1}{2} \right ) \right | \leq \int_{0}^{1/2}\left | f(x) \right | + \frac{1}{2}\int_{0}^{1/2} \left | f'(t) \right | \, \mathrm{d}t \end{equation*}

Working similarly on \left[\frac{1}{2} , 1 \right] we get

(2)   \begin{equation*} \frac{1}{2}\left | f\left ( \frac{1}{2} \right ) \right | \leq \int_{1/2}^{1}\left | f(x) \right | + \frac{1}{2}\int_{1/2}^{1} \left | f'(t) \right | \, \mathrm{d}t \end{equation*}

Adding equations (1), (2) we get the result.

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An arccotangent integral

June 6, 2019 / 1 Comment on An arccotangent integral

Let \zeta denote the Riemann zeta function. Prove that

    \[\int_0^\infty \left ( \arccot x \right )^3\; \mathrm{d}x = \frac{3 \pi^2 \ln 2}{4} - \frac{21\zeta(3)}{8}\]

Solution

Background: I have always been of the opinion that integrals of this form diverged when integrated from something to infinity. Apparently, this is not the case since the above not only converges but it has a nice closed form involving the Riemann zeta function. The technique to break it down , is pretty simple actually. Integration by parts! Yep! We combine IBP along with Fourier series and voilà. However, somewhere in the middle of this evaluation process we will come across a famous constant , \mathcal{G} , coming literally out of nowhere. But this is what to expect when reducing trigonometric integrals to logarithmic ones.

We begin our evaluation by stating two lemmata:

Lemma 1: It holds that \displaystyle \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{1+x^2} \, \mathrm{d}x = \frac{\pi \ln 2}{2} - \mathcal{G}.

Proof: Successively, we have that:

    \begin{align*} \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{1+x^2} \, \mathrm{d}x &\overset{x=\tan \theta}{=\! =\! =\! =\! =\!} \int_{0}^{\pi/4} \frac{\ln \left ( 1+\tan^2 \theta \right )}{1+\tan^2 \theta} \cdot \sec^2 \theta \, \mathrm{d}\theta\\ &=\int_{0}^{\pi/4} \ln \left ( 1+ \tan^2 \theta \right ) \, \mathrm{d}\theta \\ &= \int_{0}^{\pi/4} \ln \sec^2 \theta \, \mathrm{d} \theta \\ &=2 \int_{0}^{\pi/4} \ln \sec \theta \, \mathrm{d} \theta \\ &=-2 \int_{0}^{\pi/4} \ln \cos \theta \, \mathrm{d} \theta \\ &=-2 \left ( -\int_{0}^{\pi/4} \left (\sum_{n=1}^{\infty} (-1)^n \frac{\cos 2n \theta}{n} - \ln 2 \right ) \, \mathrm{d}\theta \right )\\ &=2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \int_{0}^{\pi/4} \cos 2n\theta \, \mathrm{d} \theta +2 \int_{0}^{\pi/4} \ln 2 \, \mathrm{d}\theta \\ &= 2 \sum_{n=1}^{\infty} \frac{(-1)^n \sin \frac{n \pi}{2}}{2n^2} +\frac{\pi \ln 2}{2} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^n \sin \frac{n \pi}{2}}{2n^2} + \frac{\pi \ln 2}{2} \\ &=\frac{\pi \ln 2}{2} + \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}\\ &= \frac{\pi \ln 2}{2} - \mathcal{G} \end{align*}

where \mathcal{G} is the Catalan’s constant. Also,

    \[\ln 2\cos x = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\cos 2n \theta}{n}\]

which is the known Fourier series expansion of the \ln \cos function.

Lemma 2: It holds that \displaystyle \int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x = \frac{7 \zeta(3)}{16} - \frac{\pi \mathcal{G}}{4}.

Proof: Successively, we have that:

    \begin{align*} \int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x &\overset{x=\tan \theta}{=\! =\! =\! =\!} \int_{0}^{\pi/4} \theta \ln \tan \theta \, \mathrm{d} \theta \\ &=\int_{0}^{\pi/4} \theta \sum_{n \; \text{odd}} \frac{\cos 2n \theta}{n} \, \mathrm{d} \theta \\ &=\sum_{n \; \text{odd}} \frac{1}{n} \int_{0}^{\pi/4} \theta \cos 2 n \theta \, \mathrm{d} \theta \\ &= \frac{7 \zeta(3)}{16} - \frac{\pi \mathcal{G}}{4} \end{align*}

Hence,

    \begin{align*} \int_{0}^{\infty} \left ( \mathrm{arccot}x \right )^3 \, \mathrm{d}x &= \int_{0}^{\infty} \left ( x \right ) ' \left ( \mathrm{arccot}x \right )^3 \, \mathrm{d}x \\ &=\left [ x \left ( \mathrm{arccot}x \right )^3 \right ]_0^{\infty} +3 \int_{0}^{\infty} \frac{x\left( \mathrm{arccot }x \right )^2}{ x^2+1} \, \mathrm{d}x \\ &=3 \left [ \frac{\ln \left ( 1+x^2 \right ) \, \left (\mathrm{arccot}(x) \right )^2}{2} \right ]_0^{\infty} + 3 \int_{0}^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x \\ &=3 \left ( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x + \int_1^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x \right ) \\ &=3 \bigg( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x + \\ &\quad \quad \quad + \int_{0}^{1} \frac{\ln \left ( 1+\frac{1}{x^2} \right ) \arctan x}{1+\frac{1}{x^2}} \cdot \frac{1}{x^2} \, \mathrm{d}x \bigg ) \\ &=3 \bigg ( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x + \\ &\quad \quad \quad + \int_{0}^{1} \frac{\left (\ln \left ( 1+x^2 \right ) - 2 \ln x \right ) \arctan x}{1+x^2} \, \mathrm{d}x \bigg )\\ &=3\bigg ( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )\left ( \mathrm{arccot} x + \arctan x \right )}{x^2+1} \, \mathrm{d}x - \\ &\quad \quad \quad - 2 \int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x \bigg ) \\ &=3\left ( \frac{\pi}{2} \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{1+x^2} \, \mathrm{d}x - 2 \int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x \right )\\ &=\frac{3 \pi^2 \ln 2}{4} - \frac{21\zeta(3)}{8} \end{align*}

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