Limit of a sequence

September 28, 2019 / Leave a comment

Let f:[0, 1] \rightarrow (0, +\infty) be a continuous function and A be the set of all positive integers n such that there exists x_n such that

    \[\int_{x_n}^{1} f(t) \, \mathrm{d}t = \frac{1}{n}\]

Prove that \{x_n\}_{n \in A} is infinite and evaluate the limit

    \[\ell = \lim_{n \rightarrow +\infty} n \left( x_n -1 \right)\]

A factorial limit

September 27, 2019 / Leave a comment

Let \cdot! denote the factorial of a real number; that is x!=\Gamma(x+1). Evaluate the limit:

    \[\ell = \lim_{x \rightarrow n} \frac{x!-n!}{x-n}\]

Solution

It holds that

    \begin{align*} \lim_{x\rightarrow n} \frac{x!-n!}{x-n} &= \lim_{x\rightarrow n} \frac{\Gamma(x+1)- \Gamma(n+1)}{x-n} \\ &=\Gamma'(n+1) \\ &=\Gamma(n+1) \psi^{(0)}(n+1) \\ &=n! \left ( \mathcal{H}_n - \gamma \right ) \end{align*}

where \mathcal{H}_n denotes the n-th harmonic number and \gamma the Euler – Mascheroni constant.

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Linear isometry

September 22, 2019 / Leave a comment

Let f:\mathbb{R}^2 \rightarrow \mathbb{R}^2. If:

  • f(\mathbf{0})=\mathbf{0}
  • \left| {f\left( {\bf{u}} \right) - f\left( {\bf{v}} \right)} \right| = \left| {{\bf{u}} - {\bf{v}}} \right| for all {{\bf{u}},{\bf{v}}}

then prove that f is linear.

Solution

For convenience, identify \mathbb{R}^2 with \mathbb{C} here. Then note that for any such function f:\mathbb{C} \to \mathbb{C}, also z_1 \cdot f(z) a solution for any point z_1 on the unit circle. Also \overline{f(z)} is a solution. Note that \vert f(1)\vert=1 and hence we can wlog assume that f(1)=1. So f(i) is a point on the unit circle with distance \sqrt{2} to 1. Hence f(i) =\pm i, so w.l.o.g. assume that f(i)=i. But then for any z \in \mathbb{C}, both z and f(z) have the same distance to 0,1 and i. So supposing z \ne f(z), all 0,1,i lie on the perpendicular bisector between these points and in particular 0,1 and i are collinear which clearly is absurd. Hence f(z)=z for all z which proves the claim.

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Zeta logarithmic series

September 15, 2019 / Leave a comment

Let \zeta denote the zeta function. Prove that

    \[\sum_{k=1}^{\infty} \left( \log\left(1+\frac{1}{k}\right)-\frac{\zeta(k+1)}{k+1} \right)=0\]

Solution

    \begin{align*} \sum_{k=1}^{\infty} \left ( \frac{\zeta(k+1)}{k+1} - \frac{1}{k+1} \right ) &=\sum_{k= 2}^{\infty} \frac{\zeta(k)-1}{k}\\ &=\sum_{k =2}^{\infty} \sum_{n =2}^{\infty} \frac{1}{kn^k}\\ &=\sum_{n=2}^{\infty} \sum_{k= 2}^{\infty} \frac{1}{kn^k}\\ &=-\sum_{n=2}^{\infty} \left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right)\\ &=-\sum_{n= 1}^{\infty} \left(\ln\left(1-\frac{1}{n+1}\right)+\frac{1}{n+1}\right) \\ &-\sum_{n =1}^{\infty} \left(\ln\left(\frac{n}{n+1}\right)+\frac{1}{n+1}\right)\\ &=\sum_{n= 1}^{\infty} \left(\ln\left(\frac{n+1}{n}\right)-\frac{1}{n+1}\right) \\ &=\sum_{n= 1}^{\infty} \left(\ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}\right) \end{align*}

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Mellin transform integral

September 14, 2019 / Leave a comment

Evaluate the integral:

    \[\mathcal{J} = \int_0^\infty x \log x e^{-\sqrt{x}} \, \mathrm{d}x\]

Solution

We are evaluating the Mellin transform of the function f(x)=e^{-\sqrt{x}}.

    \begin{align*} \mathcal{M}\left ( f \right ) &= \int_{0}^{\infty} x^{s-1} e^{-\sqrt{x}} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{u=\sqrt{x}}{=\! =\! =\! =\! =\!} 2\int_{0}^{\infty} u^{2s-1} e^{-u} \, \mathrm{d}u\\ &= 2 \Gamma\left ( 2s \right ) \end{align*}

where \Gamma is the Euler’s Gamma function. Hence,

    \begin{align*} \int_{0}^{\infty} x \log x e^{-\sqrt{x}} \, \mathrm{d}x &= \mathcal{M}'(f) \bigg|_{s=2} \\ &= \left ( 2 \Gamma(2s) \right )'\bigg|_{s=2}\\ &= 4 \Gamma(2s) \psi^{(0)}(2s) \bigg|_{s=2}\\ &=4 \Gamma(4) \psi^{(0)}(4) \\ &= 4 \cdot 6 \cdot \left ( \frac{11}{6} - \gamma \right ) \\ &=44 - 24 \gamma \end{align*}

where \gamma is the Euler – Mascheroni constant and \psi^{(0)} is the digamma.

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