Accidentally deleted Ubuntu installation, how can I reinstall it cleanly on a dual boot system?
Dual boot. How do I reinstall the Ubuntu partition?
Using induction prove $n^3-n$ is divisible by 3 whenever n is a positive number. I am not sure if I am doing this right, but I have this:  There exists an integer $k$. $2k =$ positive number  $(2k)^3 - 2k$ [*And this is where I get lost. How does one prove this?] Recommend a statistics fundamentals book To give you some background, I have a grasp on the basics of statistics and probability theory and even remember touching Bayes theorem at the university data mining course. But being a few years away from the university made my math got extremely rusty (so much for last-minute pre-exam cramming). While I remember various random basic concepts, there are a lot of gaps in my understanding of them.  What would be a good material (a book, a site, or otherwise equally accessible medium) to revise the fundamentals and go beyond basics? I'd like a book that can be actually read as a book (most statistics books are really dry and are close to being reference material, rather than a book). If $(A-2I)^3(A+2I)^2=0$, then what are the possible Jordan canonical forms of $A$? Here is the exercise:     Let $A$ be a $5\times5$ complex matrix such that $(A-2)^3(A+2)^2=0$, where we define $A-\mu:=A-\mu I$ for scalar $\mu$. Assume that $\lambda=2$ is an eigenvalue of $A$ and its geometric multiplicity is at least $2$. What are the possibilities for the Jordan canonical form (JCF)?   What I know so far from the assumption is that    the minimal polynomial of $A$ is of the form $f(x)=(x-2)^i(x+2)^j$ where $1\leq i\leq 3$ and $0\leq j\leq 2$.  The number of blocks in the Jordan segment $J(2)$ is at least $2$.    One can write the possible minimal polynomial one by one, which gives the information of the size of the largest block in each Jordan segment, and use the possible geometric multiplicity of $\lambda=2$ to find JCF.   Here are my questions:   Is there an alternative approach?  Can we use the characteristic polynomial of $A$ here? Merging dictionaries in C# What's the best way to merge 2 or more dictionaries (Dictionary&lt;T1,T2&gt;) in C#? (3.0 features like LINQ are fine).  I'm thinking of a method signature along the lines of:  public static Dictionary&lt;TKey,TValue&gt;                  Merge&lt;TKey,TValue&gt;(Dictionary&lt;TKey,TValue&gt;[] dictionaries);   or  public static Dictionary&lt;TKey,TValue&gt;                  Merge&lt;TKey,TValue&gt;(IEnumerable&lt;Dictionary&lt;TKey,TValue&gt;&gt; dictionaries);   EDIT: Got a cool solution from JaredPar and Jon Skeet, but I was thinking of something that handles duplicate keys. In case of collision, it doesn't matter which value is saved to the dict as long as it's consistent. I have a particular situation here, In March of 1997 I travelled to the US as a tourist and lived there illegally for almost 9 years. I was never arrested, or had any trouble with the law, I was never deported. I paid my taxed (had a TFN) had a job, bank account bought a house, etc. About 9 years ago I decided to leave as I was tired of living like that, not able to be legal, in fear of deportation and loosing everything. I also got married (to a person in the exact situation as me) we were able to legally migrate to Australia and now we are Australian citizens. We never returned our I-94s and in 2005 left through Mexico (walked through the border) so there are no records of us ever leaving. We have lived here in Australia for over 9 years.  OK so my question is... We have been thinking we would like to visit our friends and family back in the US now that we have a little girl. Is there any chance they won't let us in because they know we overstayed? We would be travelling with our Australian passports. In my bedroom the bottom half of all receptacles are controlled by one switch. Is it possible to have just one receptacle with top and bottom hot all the time?  The top half is hot all the time the bottom half is controlled by one switch.  Keep in mind there are six other receptacles in this room that are the same way controlled by the same switch. When I took the receptacle out the right side tab had been removed.  The top right screw had a black wire, the bottom right had a red wire, the top and bottom left screws had separate white wires.  I changed the receptacle out as suggested, I capped off the red wire and I capped off the white wire that was attached to the bottom left screw. I attached the other white wire to the top left screw and attached the black wire to the top right screw. I now have none of the receptacles working at all. What did I do wrong? How does one prove that the Klein bottle cannot be embedded in $R^3$? I'm talking about smooth embeddings. In my textbook it is given that      'The wave function describes the position and state of the electron and its square gives the probability density of electrons.'    Can someone give me a very simple example of a wave function with explanation?  (Note: This question is not a duplicate. I have searched for other questions of this type but the answers were overwhelmingly difficult to understand.) Word for killing off or attempting to kill off an entire bloodline? Update #2: Seems that it was my secondary screen the one that didn't show the color! Tried with another screen and I could see the progress clearer... perhaps a stronger color will make it easier to see across different devices?  Update: Seems to be an issue in Firefox, as I tried on Chrome and the progress is visible.    I just noticed that the badge progress on your user profile had it's style changed (don't know when but it seems that recently).  However, seems that the progress of gold badges is not working properly. If we see the Bronze and Silver badges, they are "filled with color" as you progress on the badge, however golden ones are not colored (just the border).   Here a picture illustrating what I say:    Is there a reason for this behavior? Bootstrapped confidence intervals: different values at every computation? I am running an ANOVA and computing bootstrapped CIs for my effect sizes. I am using for this the Measures of Effect Size Toolbox for Matlab.  To my surprise, if I re-run the same line of code, i.e. compute the same thing for the same data, the lower&amp;upper margins of the CI are slightly different each time.  I know that bootstrapped CIs are built from permutation analyses based on random generations of distributions. Still, I do not remeber reading anywhere this odd property of CIs of being never twice the same! An explanation for the Landauer's principle Has anyone understood the ? What is the current status?  In specific, is there a theoretical derivation of the Landauer's Principle?(not the heuristic one based on Salizard's engine) How does one account for thermodynamic concepts applicable to situations involving measurement? What does it mean to reset the state of the apparatus why is free energy expended in the process? Is your MAC address revealed when you hit up a website? When you hit up a site, such as cnn.com, does the TCP/IP packet contain the MAC address of your network card?  Could cnn.com theoretically record my MAC address? In what order should the Star Wars movies be watched? Please don't lynch me, but I've never sat through an entire Star Wars film. I'm aware that the 3 (new) films are technically prequels, but I feel strange watching 3 new films then 3 old films?  Do I need to watch the new ones then the old ones? Are there any differences between Skyrim & Skyrim: Special Edition? Gtk skip_taskbar_hint (seen in Glade-ui-Designer) equivalent in Qt With Glade Designer that lets you set up Gtk-Ui's you can check the option that windows are not shown in the launcher. If I take a look at the ui-file generated by Glade, it says &lt;property name="skip_taskbar_hint"&gt;True&lt;/property&gt;. In gtk code this hint is  window.set_property('skip-taskbar-hint', True). However I believe that this option is not exclusively Gtk-specific, and want to apply this one on my PyQt4-app. The Ubuntu window manager must be able to receive that hint, otherwise I would see the window of a gtk-application i use (stickynotes-indicator). So there MUST be a way of doing this in Qt or any other language, right?  Does anybody know such an option? Python or Qt syntax, doesn't matter! Find all continuous functions $f$ satisfying   $$\int_0^xf=(f(x))^2+C$$ for some constant $C \neq 0$, assuming that $f$ has at most one $0$.  I have a question about the solution to this problem. It says that   Clearly $f^2$ is differentiable everywhere, its derivative at $x$ is $f(x)$ (I'm assuming that this is because the left hand side is differentiable, the right hand side must be differentiable as well). So $f$ is differentiable at $x$ whenever $f(x) \neq 0$, and  $$f(x)=2f(x)f'(x),$$ so $f'(x) = \frac{1}{2}$ at such points. Thus, $f= \frac{1}{2}x+b$ for some $b$ on any interval where it is non-zero; if $f$ has a zero, this gives two possible solutions, with two possible values of $b$, but since $f$ is assumed continuous, they must be the same. So we need $$\int_0^x(\frac{1}{2}t+b)dt=(\frac{1}{2}x+b)^2+C$$  so we must have $b=\sqrt{-C}$ or $-\sqrt{-C}$, leading to two solutions for $C \lt 0$.  I don't understand the bolded statements. First, how does the differentiability of $f^2$ guarantee that $f$ is differentiable, and why only at $x$ where $f(x) \neq 0$?  Next, why are there two possible solutions if $f$ has a zero, but since $f$ is continuous they must be the same?  Finally, does the last statement mean that if $C \gt 0$, then there is no solution, since we're restricted to real numbers here?  I would appreciate it if anyone answers my questions. How to get Ubuntu download for a Mac? Given the sentence:     Most people who live and work near Washington, DC, would have trouble      imagining dinosaurs walking around the area.   Grammatically, what would ‘imaging’ be called in this case? Present participle or gerund or something?  I mean, if it is a present participle, ‘imagining~’ might be a description of the subject. And if it is a gerund, ‘imagining~’ might be an object of the verb phrase of ‘have trouble’. How do they explain this ‘imagining’ after ‘have trouble’? Why can we always drop any constant term in a Lagrangian density in quantum field theory?  This issue is somehow related to the constant term being some kind of cosmological constant.  Can you please explain this issue?