Search found 597 matches
- Fri May 18, 2018 12:03 pm
- Forum: Archives
- Topic: A collection of problems in Analysis
- Replies: 4
- Views: 7945
Re: A collection of problems in Analysis
A new version ( version 9 ) is out. Hope you find it entertaining. Awaiting to hear your feedback!
Re: Analysis
Also worth noting Cantor's diagonal argument
Re: Analysis
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 $[a_0,b_0]$...
- Wed Feb 07, 2018 7:47 pm
- Forum: Real Analysis
- Topic: Real analysis
- Replies: 3
- Views: 6286
Re: Real analysis
I beg your pardon?Asis ghosh wrote:what can you say A if LUB A = GLB A ?
- Tue Jan 02, 2018 2:20 pm
- Forum: Real Analysis
- Topic: Spivak, Michael : Calculus
- Replies: 1
- Views: 3382
Re: Spivak, Michael : Calculus
Yes Michael Spivak's book is a good one although it is densely written ...! I highly recommend it ...!
- Sun Oct 29, 2017 7:37 pm
- Forum: Real Analysis
- Topic: On an evaluation of an arctan limit
- Replies: 1
- Views: 4335
On an evaluation of an arctan limit
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
$$\Omega = \lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \frac{\frac{1}{n} \arctan \left ( \frac{k}{n} \right )}{1+2\sqrt{1+\frac{1}{n} \arctan \left ( \frac{k}{n} \right )}}$$
Dan Sitaru
- Wed Oct 25, 2017 8:02 pm
- Forum: Real Analysis
- Topic: Limit of a sequence
- Replies: 0
- Views: 2539
Limit of a sequence
Define the sequence $\{k_n\}_{n \in \mathbb{N}}$ recursively as follows
$$k_0 = \frac{1}{\sqrt{2}} \quad , \quad k_{n+1}={\frac {1-{\sqrt {1-k_{n}^{2}}}}{1+{\sqrt {1-k_{n}^{2}}}}}$$
Evaluate the limit
$$\ell = \lim_{n \rightarrow + \infty} \left(\frac{4}{k_{n+1}}\right)^{2^{-n}}$$
$$k_0 = \frac{1}{\sqrt{2}} \quad , \quad k_{n+1}={\frac {1-{\sqrt {1-k_{n}^{2}}}}{1+{\sqrt {1-k_{n}^{2}}}}}$$
Evaluate the limit
$$\ell = \lim_{n \rightarrow + \infty} \left(\frac{4}{k_{n+1}}\right)^{2^{-n}}$$
- Mon Sep 25, 2017 10:22 pm
- Forum: General Mathematics
- Topic: On an operation
- Replies: 0
- Views: 3777
On an operation
Define
\begin{equation} x* y = \frac{\sqrt{x^2+3xy+y^2-2x-2y+4}}{xy+4} \end{equation}
Evaluate
$$\mathcal{V} = \left ( \left ( \cdots \left ( \left ( 2007 * 2006 \right )*2005 \right )* \cdots \right ) * 1 \right )$$
\begin{equation} x* y = \frac{\sqrt{x^2+3xy+y^2-2x-2y+4}}{xy+4} \end{equation}
Evaluate
$$\mathcal{V} = \left ( \left ( \cdots \left ( \left ( 2007 * 2006 \right )*2005 \right )* \cdots \right ) * 1 \right )$$
- Sat Sep 09, 2017 12:16 am
- Forum: Meta
- Topic: MathJaX Upgrade
- Replies: 1
- Views: 4960
MathJaX Upgrade
Greetings, today we upgraded to the latest stable release of MathJaX. This upgrade brings new features: A whole new set of macros thanks to mediawiki MathJaX extension . The revival of xypic.js which had been lost to the previous upgrade to 2.7.0 . Now we have once again the ability to draw commutat...
- Wed Aug 30, 2017 6:50 pm
- Forum: Number theory
- Topic: Is it an integer?
- Replies: 0
- Views: 3432
Is it an integer?
Investigating something on the serieses with Harmonic numbers the following occured. I do not have a solution , yet I would like to see one. Let $n \geq 1$ and denote as $\mathcal{H}_n$ the $n$ - th harmonic number. Let us also denote as ${\rm lcm} ( \cdot , \cdot )$ the least common multiple. Is i...