There are simple rules for computing the derivative of complex (in the sense of not simple, not in the sense of complex numbers) functions such as the product rule or the chain rule. However, when it comes to integration of these function there aren't simple rules to use, and usually you have to find a trick in order to integrate. There is a saying: &quot;differentiation is mechanics, integration is art&quot; (in my language it even rhymes) I have several questions: 1. Is this intuition true? Can we be sure that such rules does not exist for integration? 2. Is there a formal version of this intuition and saying? In example, maybe the computational complexity of computing derivative is lower the complexity of computing integrals? (3. Is there a deep reason for this situation?) Edit: question 3 was already asked here -  but questions 1 and 2 were not asked so I think the question is relevant. Thanks in advance for the answers.
If a function is a combination of other functions whose derivatives are known via composition, addition, etc., the derivative can be calculated using the chain rule and the like. But even the product of integrals can't be expressed in general in terms of the integral of the products, and forget about composition! Why is this?
If $f$ is bounded and $\lim_{x \to c} g(x) = 0$, then $\lim_{x \to c}g(x)f(x) = 0$ I would like to have multiple bibliographies (reference lists) as described in   ().   However, I'd like to use with a single .tex file, which would contain the bibliography database embedded, as in  ...   Now, I want to use letter documentclass, for which the bibliography needs to be defined ()   So what I end up is with this example (code below, compiled with just pdflatex, no bibtex(which I'd like to avoid in this case)):     ... obviously, something works (multibib recognized primary and secondary cites, assigning them number "1" as start) -- however I don't get two, separate primary and secondary list of references; would that be possible in this context? (doesn't have to be with multibib, but I'd like to avoid a separate .bib file and a bibtex call)  EDIT: I'd also like separate cite keys (i.e. rendered as [a1] and [b1]; not yet sure if multibib can do that; edit: no it cant,  says "You can also edit the .bbl files created by BibTEX and add 'a', 'b',. . . to the labels, but your changes will be overwritten by subsequent BibTEX runs." and recommends \cite[a]{key}) - how possible would that be?  &nbsp;  The code is this:   \documentclass{letter}  % Environment thebibliography undefined when using letter: http://www.latex-community.org/forum/viewtopic.php?f=4&amp;t=3359 \makeatletter \setlength \labelsep {.5em} \newcommand\newblock{\hskip   .11em\@plus.33em\@minus.07em} \let\@openbib@code\@empty \newcommand\refname{References} \newcommand\section{\@startsection   {section}{1}{\z@}%   {-3.5ex \@plus -1ex \@minus -.2ex}%   {2.3ex \@plus.2ex}%   {\normalfont\Large\bfseries}} \newenvironment{thebibliography}[1] {\section*{\refname}%   \@mkboth{\MakeUppercase\refname}{\MakeUppercase\refname}%   \list{\@biblabel{\@arabic\c@enumiv}}%   {\settowidth\labelwidth{\@biblabel{#1}}%     \leftmargin\labelwidth \advance\leftmargin\labelsep \@openbib@code     \usecounter{enumiv}%     \let\p@enumiv\@empty \renewcommand\theenumiv{\@arabic\c@enumiv}}%   \sloppy \clubpenalty4000 \@clubpenalty \clubpenalty   \widowpenalty4000%   \sfcode`\.\@m} {\def\@noitemerr {\@latex@warning{Empty       `thebibliography' environment}}%       \endlist} \makeatother  \usepackage{multibib} \newcites{pri}{Primary sources} \newcites{sec}{Secondary sources}  \begin{document}   \begin{letter}{Addressee} Blah, blah - content...  Primary cite: \citepri{lamport94} ...  Secondary cite: \citesec{lamport94} ...   \begin{thebibliography}{9}  \bibitem{lamport94}   Leslie Lamport,   \emph{\LaTeX: A Document Preparation System}.   Addison Wesley, Massachusetts,   2nd Edition,   1994.  \bibitem{lamport95}   Leslie Lamport,   \emph{\LaTeX: A Document Preparation System}.   Addison Wesley, Massachusetts,   2nd Edition,   1995.  \end{thebibliography}  \end{letter}   \end{document} Characterizing all ring homomorphisms $C[0,1]\to\mathbb{R}$. This is something I've been trying to work out this evening.      Let $R$ be the ring of continuous real-valued functions on $[0,1]$ with pointwise addition and multiplication. For $t\in [0,1]$, the map $\phi_t\colon f\to f(t)$ is a ring homomorphism of $R$ to $\mathbb{R}$. I'm trying to show that every ring homomorphism of $R\to\mathbb{R}$ has this form.   Suppose otherwise, that there is some $\phi\neq\phi_t$, and thus there is some $f_t\in R$ such that $\phi(f_t)\neq \phi_t(f_t)=f_t(t)$. Define $g_t=f_t-\phi(f_t)1\in R$. Here $\phi(f_t)1$ is the constant function sending $[0,1]$ to $\phi(f_t)$. Then $g_t(t)\neq 0$. My first small question is why does $\phi(g_t)=0$? It seems only that $\phi(g_t)=\phi(f_t)-\phi(\phi(f_t)1)$.  I would like to conclude that there are only finitely many $t_i$ such that $g(x)=\sum g_{t_i}^2(x)\neq 0$ for all $x$. Then $g^{-1}=1/g(x)\in R$, but $\phi(g)=0$, contradicting the fact that homomorphisms map units to units. How can we be sure there are only finitely many $g_{t_i}$ such that the sum of their squares is never $0$? Thanks. Question  The following is from Bartle:  Give an example of a function $f$ on $X$ to $\mathbb{R}$  which is not $X-$measurable but is such that $|f|,$ and $f^2$ are measurable.  Relavant Information  A family $\textbf{X}$ of subsets of $X$ is a $\sigma$ algebra in case:   $\phi, \mathbb{R} \in \textbf{X}$ $X \setminus A \in \textbf{X}$ if $A \in \textbf{X}$ If $(A_n) \in \textbf{X}$, then the union $\bigcup_{n=1}^{\infty} A_n \in \textbf{X}$  Any set in $\textbf{X}$ is called $X-$ measurable.   A function is $X-$ measurable if for every real $\alpha$, the set $\{x \in X : f(x)&gt;\alpha\}$ belongs to $\textbf{X}$  Attempt  Consider the following function: $$f=\begin{cases} 1 &amp;\mbox{ if} \ x \not\in\mathbb{Q} \\                   -1 &amp;\mbox{ if} \ x \in \mathbb{Q}  \end{cases}$$ defined from $\mathbb{R}$ to $\mathbb{R}$ with $\textbf{X}$ being the Borel algebra (the $\sigma$ algebra generated by all open intervals $(a,b)$) Then if $\alpha \in [-1,1)$, the set $\{x: f(x)&gt;\alpha\}$ consists of the irrational numbers. If I can show that the irrationals are not in the Borel Algebra, then I would be done since $|f|=f^2$ is a constant and easily shown to be measurable. Unfortunately I do not know how to show that the irrationals are not in the Borel Algebra.  Edit it has been brought to my attention that the irrationals are in the Borel algebra. I think this is because the rationals, being countable, can be shown to be in the Borel algebra by constructing appropriate small neighborhoods around each rational and then taking the intersections. I now seek to construct a set that is not in the Borel algebra  Alternatively I have tried to define a set other than the irrationals that is not in the Borel Algebra and define a similar function but to no avail. I am currently self studying this book so it would be of great help to me if the answer could contain heuristics on how to tell when a set if borel measurable (although I do not require this in the answer). gain admin access to sql express 2008 How can I get suffix matching in a search in SharePoint 2013? I'm new to SharePoint 2013 and I'm quite puzzled, that it seems to be impossible to search for the middle part of a string.  Simple example:  I've added a lot of documents (PDFs, ...) from our Windows Server to the SharePoint Library and I want to add metadata. I try to filter the view of the document library, to only get those documents which I currently want to edit.  The names of the documents are very systematic, composed of naming conventions with abbreviations like:  Superthing2000-b_Suppl_REF_13580_note_1108_04en which means e. g.: Document about Product "Superthing2000", variant "b", Supplement to the manual, ID "13580", from 08/2011 in english language.  Now, I'd like for example to filter out all documents containing ID 13580 or "en" for english, but no chance, as suffix matching is not possible, even if I explicitly search for e. g. *en*.  Is there a way to define a search with kind of a "contains" operator which also finds files when the search string is only a part in the middle of the file name? Even the Windows Explorer search feature is able to do that!! Several years ago I heard of something called dummy subjects in high school. It was then stated that, for example, it is a dummy subject when it starts many instances of sentences, e.g.        It is rainy today.   It is hard to imagine how hard is to live alone.   It is not my university degree that matters here.     My question is if we can use it with plural subjects? For example in       If some think that in Mahmoud Dowlatabadi’s works it is only social themes that matter and the writer is ignoring fictional arrays and artistic techniques, at least, endings of his stories show wrongness of this idea.     My question deals with the italicized part. If this usage is OK, can we also write they are social themes that matter? I mean can they also play the role of a dummy subject? Vandermonde identity corollary $\sum_{v=0}^{n}\frac{(2n)!}{(v!)^2(n-v)!^2}={2n \choose n}^2$ Using matplotlib in pygame I tried to use sudo cd name_of_dir but am getting the error message:   sudo: cd: command not found   Is there any other way to enter a directory owned by another user which has 700 permission? I'm new to QGIS and have no programming skills.  Is there an easy/uncomplicated way to create multiple buffers around a line feature? What are metaclasses in Python? In Python, what are metaclasses and what do we use them for? Unfortunately MyApp has stopped. How can I solve this? A probability theory question about independent coin tosses by two players Say Bob tosses his $n+1$ fair coins and Alice tosses her $n$ fair coins. Lets assume independent coin tosses. Now after all the $2n+1$ coin tosses one wants to know the probability that Bob has gotten more heads than Alice.    The way I thought of it is this : if Bob gets $0$ heads then there is no way he can get more heads than Alice. Otherwise the number of heads Bob can get which allows him to win is anything in the set $\{1,2,\dots,n+1\}$. And if Bob gets $x$ heads then the number of heads that Alice can get is anything in the set $\{0,1,2,..,x-1\}$. So\begin{align}P(\text{Bob gets more heads than Alice})&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} P( \text{Bob gets x heads }\cap \text{Alice gets y heads }) \\[0.2cm]&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \left(C^{n+1}_x \frac{1}{2}^{x} \frac{1}{2}^{n+1-x}\right)\left( C^n_y \frac{1}{2}^y \frac {1}{2}^{n-y}\right)\\[0.2cm]&amp; = \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \frac{C^{n+1}_x C^n_y}{2^{2n+1}}\end{align}   How does one simplify this?      Apparently the answer is $\frac{1}{2}$ by an argument which looks like this, Since Bob tosses one more coin that Alice, it is impossible that they toss both the same number of heads and the same number of tails. So Bob tosses either more heads than Alice or more tails than Alice (but not both). Since the coins are fair, these events are equally likely by symmetry, so both events have probability 1/2. How to make traditional output for derivatives How do you add a timed delay to a C++ program? I am trying to add a timed delay in a C++ program, and was wondering if anyone has any suggestions on what I can try or information I can look at?  I wish I had more details on how I am implementing this timed delay, but until I have more information on how to add a timed delay I am not sure on how I should even attempt to implement this. How to build a custom kernel? Coefficient growth in the power series $\sum u_n z^n = e^{1/(1-z)}$? I have a DisplayPort based monitor, and a laptop with a DP out.  I recently got a second laptop with HDMI out, and went looking for a HDMI -> DP converter.  I was very surprised to find that such a thing effectively doesn't seem to exist.   and the  seem to imply that it's just not possible to go HDMI -> DP, without a very expensive converter box (not just an adaptor).  However, I can't find anything that explains why this is the case.  Is DP designed to be backwards compatible but can't interpret HDMI?  Is it a case of HDMI-quality signal not being high enough quality to be convertible?  Some other reason?  Note that this question is not looking for an adaptor - I'm looking for the underlying cause of the lack of them. Show that $\mu (E \setminus A)>0$ if $A$ is a nonmeasurable subset of $X$ and $A \subseteq E$, $E$ is measurable. This question was asked , but the question wasn't asking for an answer, just a clarification on the proof by contradiction they already had, so I'm reposting it to ask for help arriving at the answer.  Because the asker above said they proved it by contradiction, that is what I'm attempting. But I'm having issues wrapping my brain around the concept of set measurability. So let's give it a good college try with what I can understand.  For clarity, $\mu$ is an outer measure.  A is nonmeasurable, so $\exists B \subseteq X$ such that $\mu (B) \ne \mu (B \bigcap A) + (B \bigcap A^c)$.  E is measurable, so $\mu (B) = \mu (B \bigcap E) + (B \bigcap E^c)$.  This essentially means $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$ (Since monotonicity is alredy implied by the definition of outer measure).  So we can attempt a proof (or the start of one) by contradiction.  Assume $\mu (E \setminus A)=0$. So $E \setminus A$ is a null set.  So $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$. I assume I'm supposed to show a contradiction using that A is nonmeasurable. I'm unsure how.  Another way I think could work would be to use: $\mu (E \setminus A)= \mu (E \bigcap A^c) = 0$ and arrive at a contradiction given that A is nonmeasurable, but again, I'm unsure how.  I just need some direction from here. Thank you in advance for any advice.