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<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="assets/xml/rss.xsl" media="all"?><rss version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Grupo G</title><link>https://PedroGome.github.io/</link><description>Libre difusion de Conocimiento</description><atom:link href="https://PedroGome.github.io/rss.xml" rel="self" type="application/rss+xml"></atom:link><language>es</language><copyright>Contents © 2017 <a href="mailto:pedro.gomezm@edu.uah.es">Pedro</a> </copyright><lastBuildDate>Mon, 16 Oct 2017 21:39:22 GMT</lastBuildDate><generator>Nikola (getnikola.com)</generator><docs>http://blogs.law.harvard.edu/tech/rss</docs><item><title>Clase 1 Fundamentos Fisicos</title><link>https://PedroGome.github.io/posts/clase-1-fundamentos-fisicos/</link><dc:creator>Pedro</dc:creator><description><div id="outline-container-sec-1" class="outline-2">
<h2 id="sec-1">Posición dada la aceleración</h2>
<div class="outline-text-2" id="text-1">
<p>
Dado que la velocidad viene dada por la derivada de la posición y la aceleración es la derivada de la velocidad se puede expresar de la siguiente manera:
\[ \frac{dx}{dt} x = v \]
\[ \frac{dv}{dt} v = a \]
</p>
</div>
</div></description><category>apuntes</category><category>fundamentos físicos</category><category>mathjax</category><guid>https://PedroGome.github.io/posts/clase-1-fundamentos-fisicos/</guid><pubDate>Mon, 16 Oct 2017 12:48:44 GMT</pubDate></item><item><title>Series y Sucesiones</title><link>https://PedroGome.github.io/posts/series-y-sucesiones/</link><dc:creator>Pedro</dc:creator><description><div id="outline-container-sec-1" class="outline-2">
<h2 id="sec-1">Sucesiones</h2>
<div class="outline-text-2" id="text-1">
<p>
<b>Sucesión:</b> \((a_n)_{n=1} a_1 , a_2 , a_3 , \cdots , a_n\)
</p>
<p>
ej. \(\left\{ \left(-1 \right)^{n} \right\}_{n=1} ^{\infty} \Rightarrow -1,1,-1,1,\cdots\)
</p>
</div>
<div id="outline-container-sec-1-1" class="outline-3">
<h3 id="sec-1-1">Converge hacia L</h3>
<div class="outline-text-3" id="text-1-1">
<p>
Si cuando los términos se acercan hacia L
\[\left(a_n\right)_{n=1}^{\infty}\]
\[\lim_{n\rightarrow\infty}a_n = L\]
</p>
</div>
</div>
</div>
<div id="outline-container-sec-2" class="outline-2">
<h2 id="sec-2">Series</h2>
<div class="outline-text-2" id="text-2">
</div><div id="outline-container-sec-2-1" class="outline-3">
<h3 id="sec-2-1">P-Series</h3>
<div class="outline-text-3" id="text-2-1">
<p>
\(\frac{1}{n^{p}}\) \(p\) determina la convergencia, si \(p&gt;1\) la serie convergerá
</p>
</div>
</div>
<div id="outline-container-sec-2-2" class="outline-3">
<h3 id="sec-2-2">Nth Term Test</h3>
<div class="outline-text-3" id="text-2-2">
<p>
Si \(\sum\limits_{n=1}^{\infty}{a_n}\) y \(\lim\limits_{n\rightarrow \infty}\left[{a_n}\right] \not= 0\) la serie diverge.
</p>
<p>
Nota, este test solo demuestra divergencia cuando el límite es mayor que cero, si es cero, no demuestra nada.
</p>
</div>
</div>
<div id="outline-container-sec-2-3" class="outline-3">
<h3 id="sec-2-3">The Integral Test</h3>
<div class="outline-text-3" id="text-2-3">
<p>
If \(\sum\limits_{n=1}^{\infty}{a_n}\) where \(a_n = f\left(n\right)\) for \(f\) is continious, decreasing and $f\left(n\right)≥ 0$\\ Then:
</p>
\begin{equation*}
\sum\limits_{n=1}^{\infty}{a_n} \ \text {and} \ \int\limits_{1}^{\infty}{f\left(n\right)}\ dn
\end{equation*}
<p>
Either both converge or both diverge
</p>
</div>
</div>
<div id="outline-container-sec-2-4" class="outline-3">
<h3 id="sec-2-4">Direct Comparison Test</h3>
<div class="outline-text-3" id="text-2-4">
\begin{align*}
&amp;\sum\limits_{n=1}^{\infty}{\frac{1}{n^{2}+1}} &amp;\frac{1}{n^{2}+1}\frac{1}{n}\ &amp;\text{Therefore it diverges}\\
&amp;\sum\limits_{n=1}^{\infty}{\frac{1}{\sqrt{n-11}}} &amp;\frac{1}{\sqrt{n-11}}&gt;\frac{1}{\sqrt{n}}\ &amp;\text{Therefore it diverges}\\
&amp;\sum\limits_{n=1}^{\infty}{\left(\frac{4}{5+n}\right)^n} &amp;\left(\frac{4}{5+n}\right)^n\frac{1}{n^3}\ &amp;\text{Needs another test}\\
&amp;\sum\limits_{n=1}^{\infty}{\frac{n}{n^2+2}} &amp;\dfrac{1}{n+\frac{2}{n}}
</div>
<div id="outline-container-sec-2-5" class="outline-3">
<h3 id="sec-2-5">Limit Comparison Test</h3>
<div class="outline-text-3" id="text-2-5">
<p>
Given \(\sum\limits_{n=1}^{\infty}{a_n}\) and \(\sum\limits_{n=1}^{\infty}{b_n}\) where \(a_{n}&gt;0\) and \(b_{n}&gt;0\)
\\
</p>
\begin{multicols}{2}
\begin{equation*}
\lim\limits_{n\rightarrow\infty}\left[\frac{a_n}{b_n}\right]=L
\end{equation*}
\columnbreak
\begin{itemize}
\item If \(L&gt;0\) and finite, either both converge or both diverge
\item If \(b_n\) converges and \(L=0\) both converge
\item If \(b_n\) diverges and \(L=\infty\) both diverge
\end{itemize}
\end{multicols}
\begin{flalign*}
\text{Example:}\ \ \ &amp;\sum\limits_{n=1}^{\infty}{\frac{n^2}{n^5-4}}\ \ \ \sum\limits_{n=1}^{\infty}{\frac{1}{n^3}}\\
&amp;\lim\limits_{n\rightarrow\infty}\left[\frac{n^2}{n^5-4}:\frac{1}{n^3}\right]=\lim\limits_{n\rightarrow\infty}{\frac{n^5}{n^5-4}}=1\\
\text{Example:}\ \ \ &amp;\sum\limits_{n=1}^{\infty}{\frac{2n}{n^2+2}}\ \ \ \sum\limits_{n=1}^{\infty}{\frac{1}{n}}\\
&amp;\lim\limits_{n\rightarrow\infty}\left[\frac{2n}{n^2+2}:\frac{1}{n}\right]=\lim\limits_{n\rightarrow\infty}{\frac{2n^2}{n^2+2}}=2
\end{flalign*}
<p>
\section{Alternating Series Test}
\(a_{n}&gt;0\) then<br>
\(\sum\limits_{n=1}^{\infty}{\left(-1\right)^{n}a_n}\) and $∑\limits<sub>n=1</sub><sup>∞</sup>{\left(-1\right)<sup>n+1</sup>a<sub>n</sub>}$<br>
converge if:<br>
{\setlength{\abovedisplayskip}{0pt}
</p>
\begin{align*}
1.\ &amp;\lim\limits_{n\rightarrow\infty}{a_n}=0\ &amp;\Rightarrow \text{nth term test}\\
2.\ &amp;a_{n+1}<a_n all terms decrease>
}
<p>
\section{Alternating Series Estimation Theorem}
</p>
\begin{align*}
\text{If}\ &amp;\sum\limits_{n=1}^{\infty}{\left|a_{n}\right|}\ \text{converges, then}\ \sum\limits_{n=1}^{\infty}{a_n}\ \text{also converges}\\
&amp;\sum\limits_{n=1}^{\infty}{a_n}\ \text{is \underline{absolutely convergent} if}\ \sum\limits_{n=1}^{\infty}{\left|a_{n}\right|}\ \text{also converges}\\
&amp;\sum\limits_{n=1}^{\infty}{a_n}\ \text{is \underline{conditionally convergent} if}\ \sum\limits_{n=1}^{\infty}{\left|a_{n}\right|}\ \text{diverges}
\end{align*}
<p>
example:<br>
{\setlength{\abovedisplayskip}{0pt}
</p>
\begin{align*}
\sum\limits_{n=1}^{\infty}{\left(-1\right)^{n}\frac{1}{n}}\ \text{meets the alternating series test but}\ \sum\limits_{n=1}^{\infty}{\left|-1\right|^{n}\frac{1}{n}}\Rightarrow\sum\limits_{n=1}^{\infty}{\frac{1}{n}}\ \text{diverges}
\end{align*}
<p>
}
</p>
<p>
\pagebreak
\section{Ratio Test}
{\setlength{\abovedisplayskip}{0pt}
</p>
\begin{flalign*}
&amp;\lim\limits_{n\rightarrow\infty}{\left[\frac{a_{n+1}}{a_n}\right]}\\
&amp;
\end{flalign*}
<p>
}{
\setlength{\columnseprule}{.3pt}
</p>
<p>
\begin{multicols}{2}
{\centering \textbf{Converges}
</p>
\begin{align*}
&amp;\sum\limits_{n=1}^{\infty}{\frac{2}{n^2+1}}\ \ \ \lim\limits_{n\rightarrow\infty}{\left[\frac{a_{n+1}}{a_n}\right]}\\
&amp;\lim\limits_{n\rightarrow\infty}{\left[\frac{2}{(n+1)^2+1}\cdot\frac{n^2+1}{2}\right]}\\
&amp;\lim\limits_{n\rightarrow\infty}{\left[\frac{n^2+1}{(n+1)^2+1}\right]}\\
&amp;\lim\limits_{n\rightarrow\infty}{\left[\frac{n^2+1}{n^2+2n+2}\right]}=1
\end{align*}
<p>
}
\vfill\null
\columnbreak
{
\centering \textbf{Diverges}
</p>
\begin{align*}
&amp;\sum\limits_{n=1}^{\infty}{\frac{1}{n}\left(\frac{3}{2}\right)^{n}}\ \ \ \lim\limits_{n\rightarrow\infty}{\left[\frac{a_{n+1}}{a_n}\right]}\\
&amp;\lim\limits_{n\rightarrow\infty}{\left[\frac{3^{n+1}}{\left(n+1\right)2^{n+1}}\cdot\frac{2^n n}{3^n}\right]}\\
&amp;\lim\limits_{n\rightarrow\infty}{\left[\frac{3^{n}3n2^n}{(n+1)2^{n}2\cdot 3^{n}}\right]}\\
&amp;\lim\limits_{n\rightarrow\infty}{\left[\frac{3n}{2n+2}\right]}=\frac{2}{3}
\end{align*}
<p>
}
\end{multicols}}
For any series \(\sum\limits_{n=1}^{\infty}{a_n}\) find \(\lim\limits_{n\rightarrow\infty}{\left|\frac{a_{n+1}}{a_n}\right|=L}\)
</p>
\begin{enumerate}
\item If \(L&lt;1\), the series converges absolutely
\item If \(L&gt;1\) (or \(\infty\)), the series diverges
\item If \(L=1\), no conclusion can be made
\end{enumerate}
<p>
\section{Root Test}
For any series \(\sum\limits_{n=1}^{\infty}{a_n}\) find \(\lim\limits_{n\rightarrow\infty}{\left[\sqrt[\leftroot{-2}\uproot{2}n]{\left|a_n\right|}\right]=L}\)
</p>
\begin{enumerate}
\item If \(L&lt;1\), the series converges absolutely
\item If \(L&gt;1\) (or \(\infty\)), the series diverges
\item If \(L=1\), no conclusion can be made
\end{enumerate}
</a_n></div>
</div>
</div></div></description><category>apuntes</category><category>cálculo</category><category>mathjax</category><guid>https://PedroGome.github.io/posts/series-y-sucesiones/</guid><pubDate>Sun, 15 Oct 2017 20:38:12 GMT</pubDate></item></channel></rss>