Prove by induction: $b^n-a^n=(b-a)\cdot \sum_{k=0}^{n-1}{a^k\cdot b^{n-1-k}}$ $\forall n\in \mathbb{N}$ and $a,b\in \mathbb{R}$  $\exists n\in \mathbb{N}:b^n-a^n=(b-a)\cdot \sum_{k=0}^{n-1}{a^k\cdot b^{n-1-k}}$  $\Rightarrow b^{n+1}-a^{n+1}=(b-a)\cdot \sum_{k=0}^{n}{a^k\cdot b^{n-k}}$  Base case: $n_0=1: b^1-a^2=(b-a)\cdot \underbrace{\sum_{k=0}^{1-1}{a^k\cdot b^{1-1-k}}}_{=1} \quad \checkmark$  Inductive step:  $b^{n+1}-a^{n+1}=(b-a)\cdot \sum_{k=0}^{n}{a^k\cdot b^{n-k}}$  The sigma sign is what stresses me out, how can I deal with it?
This question is from [Number Theory George E. Andrews 1-1 #3].     Prove that $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}).$$   This problem is driving me crazy. $$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dots +xy^{n-2}+y^{n-1)}$$  $(x^n-y^n)/(x-y) =$ the sum for the first $n$ numbers and then I added $(xy^{(n+1)-2}+y^{(n+1)-1})$ which should equal $(x^{n+1}-y^{n+1})/(x-y)$ but I can't figure it out    This is a similar problem in the book and I tried this method but it wasn't working out      $\quad$Thereom $\bf1$-$\bf2$: $\,\,\,\,$ If $\,x$ is any real number other than $1$, then $$\sum_{j=0}^{n-1}x^j=1+x+x^2+\ldots+x^{n-1}=\dfrac{x^n-1}{x-1}.$$    $\quad$Remark: $\displaystyle\sum_{j=0}^{n-1}A_j$ is shorthand for $A_0+A_1+A_2+\ldots+A_{n-1}.$     $\quad$Proof: Again we proceed by mathematical induction. If $n=1$ then $\displaystyle\sum_{j=0}^{1-1}x^j=x^0=1$ and $(x-1)/(x-1)=1$. Thus the theorem is true for $n=1$.  $\quad$ Assuming that $\displaystyle\sum_{j=0}^{k-1}x^j=(x^k-1)/(x-1)$, we find that $$ \eqalign{ \sum^{(k+1)-1}_ {j=0}x^j &amp; = \sum^{k-1}_ {j=0}x^j+x^k=\dfrac{x^k-1}{x-1}+x^k \\ &amp;= \dfrac{x^k-1+x^{k+1}-x^k}{x-1}\\ &amp;= \dfrac{x^{k+1}-1}{x-1}. }$$   Hence condition $(\rm ii)$ is fulfilled, and we have established the theorem.    $\quad$Corollary $\bf1$-$\bf1$: $\,\,$ If $\,m$ and $n$ are positive integers and if $m&gt;1$, then $n&lt;m^n.$
I have a directory which contains numbered image files, something like this:  01.png 02.png 03.png 03.svg 04.png 05.png 06.jpg 07.png 08.png 09.png 09.svg 10.png   Sometimes there may be multiple versions of a file in different formats (eg. a png and svg version of the 03 and 09 files above) but the numbers are otherwise consecutive.  Typically there are 40-80 such files in each directory.  The numbers correspond to the order these images appear in a manuscript (a Word document, but that's not important).  There is no other way to determine the order of the images.  If I add a new image to the manuscript I need to place a copy of the image in this directory with the correct numbering.  So if the new image is the fifth in the manuscript I need to rename the files in the directory to this in order to make room for it:  01.png 02.png 03.png 03.svg 04.png 06.png 07.jpg 08.png 09.png 10.png 10.svg 11.png   What is the most straightforward way from the command line, or from a script or macro to renumber all the files starting at a certain number?  I have a standard Fedora Linux install using bash. What properties do natural isomorphisms between functors preserve? If $F,G: C \rightarrow D$ are functors between regular categories and $F \Rightarrow G$ is a natural isomorphism, is it true that if $F$ is faithful {resp. full) then $G$ is full?  I believe this is true but would like some confirmation.  What about if $F$ is essentially surjective?  Is it necessarily true that $G$ is essentially surjective?  Thanks. The value of $\lim_{n \to \infty}\int_{0}^{1}nx^ne^{x^2}dx$ Square version of \cdot (small black square)? I'm looking for the symbol that's similar to \cdot in math mode, but instead of a small circle, it's a square. Like this: What is the difference between DX format and FX format lenses, and which to choose for what purpose? What is the difference between DX format and FX format lenses, and which to choose for what purpose?  What are the pros and cons of both? The escape velocity of earth is roughly $11 kms^{-1}$. However, what if a long ladder was built extending out of Earth's atmosphere and considerably more. Then if something was to climb up at much less than the escape velocity, what would happen when it reached the end?  And what if the object that climbed the "ladder" then fired some kind of thruster/rocket and was going fast enough so that it orbited the Earth. It would mean less energy required to get into orbit? To be honest I'm not sure how to run LAStools from within QGIS 2.8. I'm trying to learn by experimentation.  QGIS shows all the LAStools in the toolbox  If I double click on a LAStool in the toolbox I get the following message:     Missing dependency. This algorithm cannot run :-(   LAStools folder is not configured. Please configure it before running LAStools algorithms.   I can't find any LAStools *.exe files anywhere within my QGIS Wien file structure.  That makes me wonder how I could have LAStools show in the toolbox.  Could it be a *.dll somewhere?  What am I doing wrong? DOS of Van Hove singularity in 2D square lattice tight binding model For the simplest example, 2D square lattice tight binding model gives the energy band as $$\varepsilon_k=-2t(\cos k_x+\cos k_y) \, .$$ We know that $\mathbf{k}=(0,\pi)$ and related momentum points are saddle points which give $|\nabla_k \varepsilon_k|=0$ and thus some kind of singularity in density of state (DOS) since $$\rho(\varepsilon)\propto \int_{\varepsilon=\text{const}} \frac{d S}{|\nabla_k \varepsilon_k|} \, .$$ How can I get the $\ln$ divergence for DOS near $\varepsilon=0$? Should I only care about those singularity points and omit the integral from normal parts and do Taylor expansion near those saddle points?   Moreover, why are points like $(\pi/2,-\pi/2)$ or $(2\pi/3,-\pi/3)$ not called Van Hove points when those points are also lie in the $\varepsilon=0$ line and give $|\nabla_k \varepsilon_k|=0$? How to concatenate text from multiple rows into a single text string in SQL server? Simplest or nicest proof that $1+x \le e^x$ REST API composite query: How do I use the result of a query subrequest in the referenceId of another subrequest? Is there a way to determine how many digits a power of 2 will contain? Is there a direct way to determine how many digits a power of 2 will contain without actually performing the multiplication?  An estimation would help as well if there is no absolute solution.  EDIT: In both decimal and binary bases. Are Fidelius charms stackable? Can Yoda pilot a starfighter? In a trailer for Kinect Star Wars, I saw Yoda driving a Jungle Speeder and that got me wondering. Can and has Yoda ever piloted a starfighter?  During the Clone Wars, it seems plausible that they could commission a small-sized green Jedi starfighter. Is their any indication that he has piloting experience? Possible Duplicate:             I faced this issue today when I was replying to a comment posted which used @Transaction annotation.     @Trevor: @Transaction and ... @Transaction   When typing this comment, I got a warning. Intuition behind logarithm change of base I try to understand the actual intuition behind the logarithm properties and came across  that explains the multiplication and thereby also the division properties very nicely:     Suppose you have a table of powers of 2, which looks like this: (after revision)      $$\begin{array}{rrrrrrrrrr} 0&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9&amp;10\\ 1&amp;2&amp;4&amp;8&amp;16&amp;32&amp;64&amp;128&amp;256&amp;512&amp;1024 \end{array}$$      Each column says how many twos you have to multiply to get the number in that column.  For example, if you multiply 5 twos, you get $2\cdot2\cdot2\cdot2\cdot2=32$, which is the number in column 5.      Now suppose you want to multiply two numbers from the bottom row, say $16\cdot 64$.  Well, the $16$ is the product of 4 twos, and the $64$ is the product of 6 twos, so when you multiply them together you get a product of 10 twos, which is $1024$.   I found that very helpful to understand the actual proofs for this property.  I still struggle to get the idea behind the change of base rule. I'm familiar with the proof that goes like:  $$\log_a x = y \implies a^y = x$$ $$\log_b a^y = \log_b x$$ $$y \cdot \log_b a = \log_b x$$ $$y = \frac{\log_b x}{\log_b a}$$  But can somehow provide a explanation in the style of the quoted answer why this actually works? Why is the inclusion of the tensor product of the duals into the dual of the tensor product not an isomorphism? Let $V$ and $W$ be vector spaces (say over the reals). There is a linear injection $V^* \otimes W^* \to (V \otimes W)^*$ which sends $\sum_i f_i \otimes g_i \in V^* \otimes W^*$ to the unique functional in $(V \otimes W)^*$ sending $v \otimes w \mapsto \sum_i f_i(v) \cdot g_i(w)$ for all $(v,w) \in V \times W$. In the finite dimensional case it is easy to see this is an isomorphism by comparing dimensions on both sides. I'm aware that this is not an isomorphism in the infinite-dimensional case (see Relation with the dual space in the  on tensor products) but I must admit that I'm not sure why. Cardinal arithmetic does not seem help here and, even if it did, I would be much happier to see an explicit example of a functional in $(V \otimes W)^*$ outside the range of this map. I'm having trouble cooking one up myself.  Added: Let me elaborate on my comment about cardinal arithmetic. Suppose that $X$ and $Y$ are infinite dimensional vector spaces over a field $k$. I'm confident that $\dim (X \otimes Y) = \dim X \cdot \dim Y$ holds. It is clear that $|X^*| = |k|^{\dim X}$ since we may identify a functional on $X$ with a function from a basis for $X$ to $k$. Also I think I've convinced myself that if $\dim X \geq |k|$ then in fact $|X| = \dim X$ ie. the cardinality and dimension of a vector space agree when the dimension is larger than the cardinality of the ground field. Consequently, assuming say that $|k| \leq \dim X \leq \dim Y$ it seems we have $$ \dim(X \otimes Y)^* = |k|^{ \dim X \cdot \dim Y} = |k|^{ \dim Y} = \dim Y^* = \dim X^* \cdot \dim Y^* = \dim (X^* \otimes Y^*)$$  which implies that in fact $(X \otimes Y)^*$ and $X^* \otimes Y^*$ are isomorphic but, rather frustratingly, the obvious map does not do the job. Did I make a mistake here? If not, is it generally true (ie without making assumptions about $|k|$) that there exists some isomorphism $(X \otimes Y)^* \to X^* \otimes Y^*$? Travel alone on accompanied visa Google Chrome: Is there a keyboard shortcut to open a link in a new tab? Is is possible to open a link in a new background tab in Google Chrome with a keyboard shortcut?  (I know it's possible with Ctrl+click and middle click, and I know that Shift+Enter opens the link in a new browser window.) For all $a, m, n \in \mathbb{Z}^+$,   $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$