Not able to install dropbox Ubuntu 12.04
How do I resolve unmet dependencies after adding a PPA?
$$\frac{1}{z-\sin z}$$ $$\frac{1}{z(1-\frac{\sin z}{z})}$$  the laurent's transform is also used to simplify the function.  the residue is found a $\frac{2*36*2}{5}$ but the actual answer is given as $\frac{3}{10}$ the residue is given as The solution is incorrect and also the pole is of the order 3. I have just migrated from Windows environment. I have installed Python 3.2 in a separate directory. How can I get the python installation path in Ubuntu shell?  Is there any way I can let the shell know/choose at runtime which python version is to be used for further code execution?  Are there any environment variables and search path kind of things in Ubuntu Linux as well? A proof by strong induction that $a_n\le3^n$ where $a_n=a_{n-1}+a_{n-2}+a_{n-3}$ I am not sure whether this is right. Can anyone verify, whether this proof is valid? Thanks!        Define a sequence $\{a_n\}_{n\ge0}$ as follows:   $$a_0=1,\qquad,a_1=3,\qquad,a_2=9,\qquad,a_n=a_{n-1}+a_{n-2}+a_{n-3}\text{ for }n\ge3.$$   Prove that for any positive integer $n$, $a_n\le3^n$.        Proof. Let $P(n)$: $a_n\le 3^n$, $n$ is a non-negative integer.      (i) Base case:         Consider when $n=0$. LHS$=a_0=1$, RHS=$3^0=1$ $\therefore$ LHS$\le$RHS, then $P(0)$ holds.   Consider when $n=1$. LHS$=a_1=3$, RHS=$3^1=3$ $\therefore$ LHS$\le$RHS, then $P(1)$ holds.   Consider when $n=1$. LHS$=a_2=9$, RHS=$3^2=9$ $\therefore$ LHS$\le$RHS, then $P(2)$ holds.         (ii) Inductive case:   Assume $P(i)$ is true for $0 \le i \le k$, $k\ge2$.      (iii) Inductive conclusion: Consider $n=k+1$.   $$ \begin{align*} \mathrm{LHS}=a_{k+1}&amp;=a_k+a_{k-1}+a_{k-2} \qquad\text{(by definition, since $k\ge 2$)}\\                     &amp;\le 3^k+3^{k-1}+3^{k-2} \qquad\text{(by Induction Hypothesis)}\\                     &amp;=3^k(3^{-1}+3^{-2}+3^{-3})\\                     &amp;=3^k\left(\frac13+\frac19+\frac1{27}\right)\\                     &amp;=\frac{13}{27} 3^k\\                     &amp;\le 3^{k+1} = \mathrm{RHS} \end{align*} $$   Therefore, by the Principle of Strong Induction, $P(n)$ is true for all non-negative positive integers $n$. We currently have some indexes where ALLOW_PAGELOCKS is set to off. This presumably was done in order to reduce deadlocks. However I doubt that it would really had an effect back then.   Now I am trying to understand WHEN SQL Server actually chooses to start locking pages rather than keys in a clustered index. I asked Jonathan Keyhaisas recently and he told me that this could happen if I am touching rows on several subsequent pages. However I didn't manage to get any exclusive page locks by updating rows in a clustered index with a sample query.   Could you help me understand page locks better with a sample query and table? I am running SQL Server 2008 SP4.  Thanks in advance  Martin How to create text file and insert data to that file on Android Is it possible to write a label expression in qgis that displays the label at a given coordinate? I am using the QGIS expression builder for the label properties of my data.  Is it possible to put different parts of the label in different coordinates on the map?  For instance I have a linestring and I would like to have "A" displayed in the beginning of the linestring and "B" in the end of it. How to install a dictionary (danish) in Texmaker Column and row padding in tables Using global variables in a function Order of products of elements in a finite Abelian group solution for apex cpu time limit exceeded exception Another layman blackhole question, pulling one end of a string out from behind the event horizon No long explanation is needed,  What would happen if I were to allow one end of a rope to fall past the event horizon of a black hole while I held the other end?  Would I be able to pull it out? Would the rope feel extremely (infinitely?) heavy? I'm training a neural network but the training loss doesn't decrease. How can I fix this?  I'm not asking about overfitting or regularization. I'm asking about how to solve the problem where my network's performance doesn't improve on the training set.    This question is intentionally general so that other questions about how to train a neural network can be closed as a duplicate of this one, with the attitude that "if you give a man a fish you feed him for a day, but if you teach a man to fish, you can feed him for the rest of his life." See this Meta thread for a discussion:   If your neural network does not generalize well, see: Irreducibility of a polynomial if it has no root (Capelli) Let $F$ be a field of arbitrary characteristic, $a\in F$, and $p$ a prime number. Show that $$f(X)=X^p-a$$ is irreducible in $F[X]$ if it has no root in $F$.   This  to a related question mentions the result is due to Capelli.  I can prove the result if $F$ has characteristic $p$ as follows.  Suppose $f$ is reducible: $f(X)=g(X)h(X)$ with $g(X)$ an irreducible factor of degree $m$, $1\le m&lt;p$. Then if $\alpha$ is a root of $g$ in some extension field $K$ of $F$, we have $$f(X)=X^p-\alpha^p=(X-\alpha)^p$$ so its divisor $g(X)$ must be of the form $(X-\alpha)^m$. Since the coefficient of $X^{m-1}$ in $g$ is in $F$, we have $m\alpha\in F$.  So $\alpha\in F$ because $m$ is invertible modulo $p$.  How would you show the result in other characteristics? I just sent a text to a friend, who didn't understand an acronym I used for a game:      "Ah, I said it in a previous text so thought it was ok to abbreviate   it."   Though, since it was an acronym I had created, could I have used the word 'acronymise/ze' to be more specific?   I'm pretty sure it isn't a (formally used) word, but I wondered, as you often refer to acronyms specifically as acronyms, rather than abbreviations. Can you be more specific with a verb, too?  It sounds too odd to me, so I'm guessing generally no. But I wondered—so perhaps I'm wrong. Or half wrong. I was recently invited to join InitiativeQ via a friend on Facebook. The underlying idea makes rough sense to me (as a non finance expert). However their 'hook' seems to be that early adopters will get free money later down the line when the currency launches (the earlier you adopt, the more currency you will receive). Looking at the website, there is a big button with the text Reserve $128,957 (I assume that's their idea of what the dollar value of my reservation will be if it launches).  The website's production quality suggests it is not a scam, however the language suggests strongly that it is. Is this a scam, a get rich quick scheme? Or am I being overly paranoid? I am trying to get an intuitive idea of how the  affects Quantum computation. My understanding is that given a qubit $Q$ in superposition $Q_0 \left| 0 \right&gt; + Q_1 \left| 1 \right&gt;$, NCT states another Qubit $S$ cannot be designed such that $S$ is equivalent to the state of $Q$.   Now the catch is, what does Equivalent mean? It could mean either that:   $S = S_0 \left| 0 \right&gt; + S_1 \left| 1 \right&gt;$ such that $S_0 = Q_0, S_1 = Q_1$. Or it could mean that $S = Q$, meaning that if $S$ is observed to be some value ( for example 0) then $Q$ MUST be that same value, and vice versa.    So it seems that point 2, occurs anyways in entangled systems (particularly cat-states), so I can eliminate that option and conclude that that No Cloning states, given a qubit $Q$, it's impossible to make another qubit $S$ such that:  $S = S_0 \left| 0 \right&gt; + S_1 \left| 1 \right&gt;$ such that $S_0 = Q_0, S_1 = Q_1$.  Is this correct? Constrains relationship I have the following design. Each user has it's own contacts, it's own hosts and assign those to a domain which belongs only to him.   I want to insure that the host you assign to the domain belongs to the user that domain is assigned to.  Is it something that I have to insure in my application logic?  If I set domains.user_id foreign key to reference host.user_id and user_id.id does it eliminate my issue?    By the way can anyone suggest me a better design for my schema? The bounty I offered on  is now in its grace period and the information box tells me:     you must award this bounty as soon as possible   But there are no answers - what do I do? Convergence of sequence of real numbers and (sub)subsequence Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers. A well known result is:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ converges to $a\in \mathbb{R}$.   Now, I'm working on an exercise which essentialy asks to prove the following:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if for every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ there is a (sub)subsequence $(x_{n_{k_j}})_{j\in \mathbb{N}}$ which converges to $a$.   Now, one direction is trivial. If in fact $(x_n)_{n\in \mathbb{N}}$ converges to $a$, then by the result I've mentioned, every subsequence of $(x_{n})_{n\in \mathbb{N}}$ converges to $a$. Let then $(x_{n_k})_{k\in \mathbb{N}}$ be an arbitrary subsequence. In that case, we know that it is a subsequence of itself, and hence we've found a (sub)subsequence which converges to $a$.  The other way around seems quite complicated for me. The reason is as follows: I must assume that every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ has a (sub)subsequence which converges to $a$ and the proof that the original one converges to $a$.  My problem is exactly that "a". If it had all (sub)subsequences converging to $a$, then by the result i've mentionted, the subsequence $(x_{n_k})_{k\in \mathbb{N}}$ would converge to $a$. In that case, we would have every subsequence converging to $a$ and the result would follow.  My idea to prove the claim was then:  First remember that a subsequence is the restriction of the original one $x: \mathbb{N}\to \mathbb{R}$ to an infinite subset $\mathbb{N}'\subset \mathbb{N}$. In that case a (sub)subsequence is the restriction of $\tilde{x} = x|\mathbb{N}'$ to an infinite subset $\mathbb{N}''\subset \mathbb{N}$. But since $\tilde{x}$ is the restriction of $x$ to $\mathbb{N}'$ and since $\mathbb{N}''\subset \mathbb{N}'$, the restriction of $\tilde{x}$ to $\mathbb{N}''$ is itself the restriction of $x$ to $\mathbb{N}''$.  In other words, every (sub)subsequence can be viewed as a subsequence of the original one. Considering this, by hypothesis every subsequence has a (sub)subsequence which converges to $a$. But by this argument, this means that we can replace every subsequence with a convergent one. And then we can use the result I stated.  Is my idea correct? Is this really the way to prove the claim? My only doubt is that I need every subsequence converging to $a$ to use the result. In this case I can't see how I can argue that this implies every subsequence converges to $a$.  How can I finish this proof?