I'm dealing with something like $\sum_{j=2}^{n-1}j^2$. I know I can do this $\sum_{j=1}^{n-1}j^2 - \sum_{j=1}^{1}j^2$.  Would that be equal to $\frac{j(j+1)(2j+1)}{6} - j^2$ or I'm missing some properties with $n-1$?  If so, which ones?
I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:  $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$  I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
Render Glass like materials with Alpha Channel to composite in external app Hope you can help me with some issues I've been dealing these days.  As I'm rendering products, most of the time I need to render some transparent materials in order to do a post-pro in Photoshop, change the backgrounds, superpose images, etc.  I've been looking for an answer but yet I've couldn't get a solution for this problem.  I need to render materials with transparency and no background in Cycles.  I've found that there are two ways to do this:   Render with transparency in the materials.  Render and get two renders, the original and another one only with the alpha channel to edit in Photoshop.   For the first option, I've achieved to render a glass-like material, as PET plastic using this node tree:    For the second option, I haven't found a way to get the alpha channel in cycles that includes the object with transparency and, if possible, the shadows in the background| but without the background. I need to remove every line that has a value of 2 or less in the 8th field (column).   My data looks like this:  12-31   Airport 189 379 41  49.70946503 -124.91377258   2   2880    30.8 01-01   AlberniElementary   165 331 16  49.26100922 -124.80662537   4   5760    26.1 01-09   BamfieldMarine  161 323 23  48.83490372 -125.13572693   2   2875    27.4 01-10   BamfieldMarine  161 323 23  48.83490372 -125.13572693   3   3068    38.6   I understand that using awk I can strip off the values desired and print them to another file, and I understand that sed would edit the current file. In either case, I need to retain the original file.   Note: Please provide thorough explanations with your solutions. It is not suffice to just write the command, please annotate suggestions.   Further note: The data has a header line, so most likely solution will need to      awk 'FNR >1'   I suppose? Can I install Microsoft Office? If so, how? I want to know how to install Microsoft Office (specifically, Microsoft Office 2010) in Ubuntu?   I am a die hard fan of MS Office and I want to know if there is a way I can use it in Ubuntu. In the midst of attempting to prove that for a commutative ring $A$ with identity, and an ideal $I$ of $A$, $I = \operatorname{rad}(I)$, where $\operatorname{rad}(I) = \{x: x^m \in I, m &gt;0\}$, implies that $I$ is an intersection of prime ideals, I've run into a little bit of confusion.   An answer to another question on this site, , indicates a strategy that I am attempting to use. I know that if $x$ is an element of $A\setminus I$, then so too is $x^m$ for any positive integer $m$. I use this fact to consider a subset $X = \{x^m : m &gt; 0\}$, which is closed under multiplication, and use Zorn's lemma to find an ideal of $A$ containing $I$ maximal with respect to the constraint that said ideal, $J$, is disjoint from X.  I want to prove $J$ is prime by contradiction. I suppose that $J$ were not prime; this also means that $J$ is not maximal in $A$. Then there would exist two elements, $f$ and $g$, of $A$ so that $fg$ is in $J$, but neither $f$ nor $g$ is in $J$. Following the prompting of the aforementioned answer, I know that $J + \langle a\rangle$ and $J + \langle b\rangle$ would be ideals strictly containing $J$, and since J is the maximal ideal disjoint from $X$, both $J +  \langle a\rangle$ and $J + \langle b\rangle$ contain at least some elements of $X$ (I am tempted to say all of them, but I'm not entirely sure that's true, in case for example $\langle a\rangle$ started containing powers of $x$ at $x^2$, in which case it isn't necessarily true that $\langle a\rangle$ contains $x$ itself), and so both $\langle a\rangle$ and $\langle b\rangle$ must themselves contain elements of $X$.  And this is where I am stuck. I feel like there is something I should know about multiplicative subsets of rings that would help me arrive at the desired contradiction, but I can't remember what it might be. I am a beginner using QGIS (1.8.0 Lisboa) on a Mac (OS X 10.6.8 Snow Leopard). I work on a file with two layers in the following sequence (top to bottom):    a vector polygon layer with a few outlines of various buildings in France and Italy (CRS: EPSG:4326 - WGS 84 - "on the fly" re projection is activated ) a Google satellite layer uploaded via open layer plugin.   Sometimes when changing the scale (scrolling in and out) the Google layer updates a little wrong in size or position compared to the polygon layer. But as soon as I "pan" the map everything reconstructs right. Now the problem I have concerns the print composer.  The result of any kind of export - pdf, or image or svg - of my map (vector layer + Google layer in the background) differs from what is shown in the preview of the print composer: the vector layer and the Google layer positions are mismatching and in a different scale. I can send some pictures showing screenshots of the print composer preview and the final results if required.  I tried almost everything:    new files with different CRS settings "cache" update preview "render" update preview different paper sizes,  different scales,   The problem appears only using an open layer map (Google, OpenStreetMap...). If I try the same using two vector layer (shapefile) - one with the buildings, another one with the city in background - everything works out great.  ...exasperating! Since it is a real important research project I would be very grateful for some help and tips. In any finite graph with at least two vertices, there must be two vertices with the same degree Show that, in any finite graph with at least two vertices, there must be two vertices with the same degree.  HINTS ONLY! I used the inbuilt dataset stackloss here. I used following R code for creating the multiple regression model.  &gt; head(stackloss)   Air.Flow Water.Temp Acid.Conc. stack.loss   1        80         27         89         42   2        80         27         88         37   3        75         25         90         37   4        62         24         87         28   5        62         22         87         18   6        62         23         87         18  &gt; fit &lt;- lm(stack.loss ~ .,data=stackloss) &gt; plot(fit)   And when I plot the model, the four following graphs are displayed.      How to interpret these graphs? What are the significance of these graphs? How to use these graphs? What is the difference between <%# and <%= in ASP.NET MVC? 70s UK science fiction TV show with time travel and telepathic cat I'm trying to identify a science fiction series on UK TV during the 70s, but I can only recall the following:   A party of adults and children. There was a woman with a telepathic bond to her cat: she could look into its eyes to see what it had seen. Time travel was involved. I think it may have been inadvertent or uncontrolled time travel. On one occasion the party encountered an Elizabethan seaman who demanded to know if Elizabeth was still queen. They party replied "err... yes" because they didn't want to have to explain time travel. How many $4$ digit even numbers have all $4$ digits distinct ?   This question has been asked if we replace even with positive.   I was solving this Problem like: For first position $1-9$ (No $0$) for next $0-9$ (leaving out the digit which has been choose) for next $0-9$ (leaving out $2$ digits), but on last digit I get stuck. From what I should choose. Will it be $0, 2, 4, 6, 8$, because number needs to be even? How should I solve it? What is the easiest way to resolve apt-get BADSIG GPG errors? How do I programmatically log in a user with a POST request? What is the best way to dump entire objects to a log in C#? So for viewing a current object's state at runtime, I really like what the Visual Studio Immediate window gives me.  Just doing a simple  ? objectname   Will give me a nicely formatted 'dump' of the object.    Is there an easy way to do this in code, so I can do something similar when logging? Keyboard Layout indicator doesn't display current layout in Ubuntu 18.04 (empty drop down box) I see an empty drop down box where the language code of the  should be.  To reproduce the bug: Lock screen or log out and log in again.    I also get the same behavior on the lock screen at the time when I try to log in with my password:    After I tried the workaround (below) the language code on the login screen disappeared as soon as I typed the first character of my password.  Workaround (not permanent): Switching with keyboard (Super+space) or mouse.  System info: Ubuntu 18.04.1 LTS, GNOME 3.28.2.    Please  if also affects you. How can I automatically enable large text when logging in to the ubuntu session? If $D$ is critical dimension of Bosonic strings, a particular derivation goes like the following, where we arrive finally at  $$ \frac{D-2}{2}\sum_{n=1}^\infty n + 1 = 0. $$ Now mathematically this is clearly a divergent series, but using zeta function regularization here we are taking  $$ \sum_{n=1}^\infty n = \zeta(-1) = -\frac{1}{12}. $$ And obtain $ D = 26 $ where $\zeta $ is the analytic continuation of the zeta function we know. But it makes no sense in putting $ s = -1 $ in the formulae $$ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}. $$ As the above is only valid for $ Re(s) &gt; 1 $. So what is going on in here? Can anyone give me a reasonable explanation about obtaining $ -1/12 $? Given Newton's third law, why are things capable of moving? Exactly half of the elements of $\mathcal{P}(A)$ are odd-sized Evaluate $\lim_{n \to \infty} \frac{1}{n}[(n +1)(n+2) \cdots (n+n)]^{\frac{1}{n}}$ Evaluate $$\lim_{n \to \infty} \frac{1}{n}[(n +1)(n+2) \cdots (n+n)]^{\frac{1}{n}}$$  We can write as $\displaystyle\lim_{n \to \infty} \frac{1}{n}[(n +1)(n+2) \cdots (n+n)]^{\frac{1}{n}}$ $$= \lim_{n \to \infty}\left[\frac{(n +1)(n+2) \cdots (n+n)}{n^n}\right]^{\frac{1}{n}} = \lim_{n \to \infty}\left[(1+\frac{1}{n})(1+\frac{2}{n})\cdots (1+\frac{n}{n})\right]^{\frac{1}{n}}$$  I have used this result if $\{a_n\}$ is a sequence of positive term which is convergent to $l$ and $b_n = (a_1. a_2 . \cdots a_n)^{1/n}$ ,thhen $b_n$ converges to $l$  here $a_n = 2$, but i think it is not right,how to use this result in this question. Give me hint how to proceed . Thank you How does Cantor's diagonal argument work? I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the  article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):  \begin{align} s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.  My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:  \begin{align} s_0 &amp;= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\ s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ s_4 &amp;= (0,1,1,\mathbf{1},\dots)\\ s_4 &amp;= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\ s_5 &amp;= (1,0,0,0,\dots)\\ s_6 &amp;= (1,0,0,1,\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:     For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,      $$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$      ...snip...      Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.      Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.   I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:  $$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 &amp; \mathrm{if\ } s_{m,m} = 1\\1 &amp; \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$  This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because...   Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.  Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".  What am I missing?