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Forum: Archives Topic: A collection of problems in Analysis 
Tolaso J Kos 
Posted: Sun Dec 23, 2018 5:59 pm


Replies: 4 Views: 1022

Greetings, the current version of the booklet " A collection of problems in Analysis" is out. This is version $11$. Milestone of $500$ exercises is reached , so this new version does not contain any new exercises. However, a new appendix about the Pisot numbers has been added to the bookle... 


Global announcement Topic: GDPR compliance 
Tolaso J Kos 
Posted: Thu May 24, 2018 6:16 pm


Replies: 0 Views: 584

Dear members and guest of mathimatikoi.org community, as many of you are probably aware the European Union launches on May 25, 2018 a new law under the name $2016/679$ regarding the privacy policy of personal data. We would like to seize this moment to inform you how we handle your personal data: &s... 


Forum: Archives Topic: A collection of problems in Analysis 
Tolaso J Kos 
Posted: Fri May 18, 2018 12:03 pm


Replies: 4 Views: 1022

A new version ( version 9 ) is out. Hope you find it entertaining. Awaiting to hear your feedback! 


Forum: Analysis Topic: Analysis 
Tolaso J Kos 
Posted: Wed Feb 07, 2018 7:50 pm


Replies: 2 Views: 251



Forum: Analysis Topic: Analysis 
Tolaso J Kos 
Posted: Wed Feb 07, 2018 7:49 pm


Replies: 2 Views: 251

prove that $(0,1)$ is uncountable Assume that $(0,1)$ is countable. Then you can write $[0,1]=(x_n)_{n \geq 0}$. Do the following steps:  split $[0,1]$ into three equal parts $[0,1/3],[1/3,2/3],[2/3,1]$. Then $x_0$ is not in one of the given intervals. Denote it by $[a_0,b_0]$.  split $[a_0,b_0]$... 


Forum: Real Analysis Topic: Real analysis 
Tolaso J Kos 
Posted: Wed Feb 07, 2018 7:47 pm


Replies: 3 Views: 500

Asis ghosh wrote: what can you say A if LUB A = GLB A ?
I beg your pardon? 


Forum: Real Analysis Topic: Spivak, Michael : Calculus 
Tolaso J Kos 
Posted: Tue Jan 02, 2018 2:20 pm


Replies: 1 Views: 271

Yes Michael Spivak's book is a good one although it is densely written ...! I highly recommend it ...! 


Forum: Real Analysis Topic: On an evaluation of an arctan limit 
Tolaso J Kos 
Posted: Sun Oct 29, 2017 7:37 pm


Replies: 1 Views: 294

Evaluate the limit
$$\Omega = \lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \frac{\frac{1}{n} \arctan \left ( \frac{k}{n} \right )}{1+2\sqrt{1+\frac{1}{n} \arctan \left ( \frac{k}{n} \right )}}$$
Dan Sitaru 


Forum: Real Analysis Topic: Limit of a sequence 
Tolaso J Kos 
Posted: Wed Oct 25, 2017 8:02 pm


Replies: 0 Views: 268

Define the sequence $\{k_n\}_{n \in \mathbb{N}}$ recursively as follows
$$k_0 = \frac{1}{\sqrt{2}} \quad , \quad k_{n+1}={\frac {1{\sqrt {1k_{n}^{2}}}}{1+{\sqrt {1k_{n}^{2}}}}}$$
Evaluate the limit
$$\ell = \lim_{n \rightarrow + \infty} \left(\frac{4}{k_{n+1}}\right)^{2^{n}}$$ 


Forum: General Mathematics Topic: On an operation 
Tolaso J Kos 
Posted: Mon Sep 25, 2017 10:22 pm


Replies: 0 Views: 352

Define
\begin{equation} x* y = \frac{\sqrt{x^2+3xy+y^22x2y+4}}{xy+4} \end{equation}
Evaluate
$$\mathcal{V} = \left ( \left ( \cdots \left ( \left ( 2007 * 2006 \right )*2005 \right )* \cdots \right ) * 1 \right )$$ 


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