Suppose I'd like to find the coefficient of $x^{l}$ in the expansion of  $(1+x+x^{2}+...+x^{n})^{m}$, where $n$ and $m$ are given positive integers, for some given integer $l$ such that $n &lt; l &lt; mn$. Is there a straightforward formula (in terms of some multinomial coefficient, or similar)?
The literature defines multinomial coefficients (or extended bnomial coefficients) as   $$ \binom{n}{r_1,r_2,\cdots,r_l} = \frac{n!}{r_1!r_2!\cdots r_l!}$$ where $$ r_1+r_2+\cdots+r_l = n$$  Which is the coefficient of $x_1^{r_1}x_2^{r_2}\cdots x_l^{r_l}$ in expansion of $(x_1+x_2+\cdots+x_l)^n$.  Question: How do we call the coefficients of $x^r$ in expansion of $(x+x^2+\cdots+x^l)^n$. Is it bad to name them multinomial coefficients?  In case of binomial coefficients the two are same in expansion of $(x_1+x_2)^n$ and the expansion of $(x+x^2)^n$. Obviously the two are different for multinomial expansion.   Example: As $(x_1+x_2+x_3)^2= x_1^2+2 x_2 x_1+2 x_3 x_1+x_2^2+x_3^2+2 x_2 x_3$. The coefficients are $\{ 1,2,2,1,1,2 \}$. Are these MULTI-NOMIAL?  $ As (x+x^2+x^3)^2=x^2 + 2 x^3 + 3 x^4 + 2 x^5 + x^6$. The coefficients are $\{1,2,3,2,1\}$. Are these too the MULTI-NOMIAL?  I prefer different terminologies to avoid confusion. Therefore want to know how these are distinguished in existing literature.
Why does ice form on bridges even if the temperature is above freezing? So with this "arctic blast" continuing, I've noticed that for my area, the temperature drops below freezing just long enough to cause freezing rain, but then the sun comes out and the temperature rises immediately. However, on bridges, ice continues to form.   How can ice form even if the temperature is above freezing? What does this symbol mean? (looks like a 1 with double vertical line) I'm studying a course on probability and statistics and at some point this symbol comes up without introduction. It looks like the number one, but slightly bigger and with a double vertical line.  First time it comes up is when discussing stochastic/random variables that are neither continuous nor discreet in an example:   And somewhat later in a proof:    Anyone got an idea what this symbol represents?  EDIT: Thanks for the fast answers, I was still editing the question for better (larger) images :) How do I create a master workbook in excel? In powerpoint we have something called a master slide.  I want the same idea for excel.  I want to create a template that I can work on with formulas and I want it to update all children files.  Example: I have a budget template.  This template has a budget listing all required equipments and their cost for a project.  We then have a formula that calculates the total cost of all equipment. I want to update only the template file with equipment and formulas, but I want the actual files (children) to contain the costs (because they might be different costs for different locations, but the rest of the budget and its formulas are shared).  Its pretty much exactly what a master slide in powerpoint is, if you've ever used that you should understand what I'm trying to do pretty easily.  One master file to store formulas, and a bunch of children files that store the data to be accessed by those formulas.  This way if a formula needs to be changed, the templates is adjusted and all other files update accordingly.  Any ideas?  I'd prefer not to have to use VB, and I'd also like a solution that works for both mac and windows.  Found  question but no answers.  I've already created a template, but that doesn't seem to update the other files when I modify it. I no longer use the service that I originally used to register, and would like to change the service I use to log in, or at least add a second account.  Is this possible? How do I change my login provider or associate a backup method to log in?     For more information, see "" in the . I can't figure out how to turn on the Bluetooth on my Nikon D5600. I have updated the firmware to the latest 1.01 and still the same problem.   Bluetooth is dimmed out on the camera and will not turn on. I get an error message saying, "This option is not available at current settings or in the camera's current state".  I have tried everything I can think of and no go. I can't find help for this problem anywhere on the internet and still no reply from Nikon.  How do I turn on my camera's Bluetooth? I hope this qualifies as a programming question, as in any programming tutorial, you eventually come across 'foo' in the code examples. (yeah, right?)  what does 'foo' really mean?  If it is meant to mean nothing, when did it begin to be used so? Biblatex: submitting to a journal Blender 2.78 with Windows 10 lags when interacting with the UI Update  This has been confirmed resolved in a later version 376.33 and possibly the update prior.     Okay, I recently did a computer update. I used to run Blender 2.77 on a Dell desktop (5 GB RAM, 2.7 GHZ, Windows 8, build in graphics card). Since that computer wasn't fast enough for my preference, I saved up and got a new gaming laptop - the Asus ROG GL552 (Windows 10, 16 GB DDR4 RAM, Nvidia GeForce GTX 960M with 4 GB DDR5 RAM, 2.6-3.5 Ghz, and quad-core intel processor). That laptop has been known to have outdated drivers, so when I got it, I updated it to up to date Windows 10 drivers.  As soon as I started Blender, I immediately noticed its speed. Even doing simple things, like clicking buttons, or rotating the view with the middle mouse button, took too long to register in the program. I tried changing some User Preference System settings, but that didn't really help much. And the hardware seems to be fine - the render is lightning fast (renders the startup scene in under .1 second; it's just laggy when loading the uv screen), so I don't know why it's like this, or what to do. Also, I ran Blender 2.77 on this computer, and it was also slow (although it was fine on my previous computer).  It could be an issue with drivers perhaps (or maybe OpenGL?), but I ran Unreal Engine 4, and had none of the same issues - the speed was pretty decent on that; so I'm not sure why Blender would be so slow. Any help is appreciated - thanks guys! I want to do:  int a = 255;  cout &lt;&lt; a;   and have it show FF in the output, how would I do this? Spaces and em-dash In using the em-dash, where do you put the spaces? Which of the following is correct?     School is based on the three R’s —reading, writing, and ’rithemtic.   or     School is based on the three R’s— reading, writing, and ’rithemtic.   or     School is based on the three R’s — reading, writing, and ’rithemtic. WMS and WFS queries needs SRS and BBOX parameters.  http://example.com/geoserver/wfs?   service=WFS&amp;   version=2.0.0&amp;   request=GetFeature&amp;   typeNames=namespace:featuretype&amp;   srsName=CRS&amp;   bbox=a1,b1,a2,b2&amp;   For example I have GeoServer to publish WFS. My data SRS is EPSG:4326 and my map SRS is EPSG:3857. So I send a WFS query.  http://example.com/geoserver/wfs?   service=WFS&amp;   version=2.0.0&amp;   request=GetFeature&amp;   typeNames=namespace:featuretype&amp;   srsName=EPSG:3857&amp;   bbox=978393.9621,2680799.4560,-430493.3433,3326539.4710&amp;   Does GeoServer or ArcGIS server give an error? Because my data has EPSG:4326 . Or does GeoServer convert these coordinates internally and find data and send me as EPSG:3857 Do we want emojis in posts? I am looking for a canonical question and answer on this issue.   (see ) on  has prompted me to ask this question.        In all honesty I think we just don't need a comic strip to explain what a user is saying. Currently I edit out emoticons I see, but this post presents a whole new type of .... ? what I'm not sure.  Is there are community consensus on this either way?  These questions are related — but different:        Please try and answer sensibly (rolls eyes at the prospect of a flood of emojis in answers) — so we can actually have something that we can link to new users (or old users) when editing this stuff out — or (even worse), defending its existence. Is there a test to determine whether solder is leaded or lead-free? Perhaps conductivity/resistance? CPU usage is too high with Yosemite and an external monitor I use Yosemite 10.10.4 on my MacBook Pro (2012).  I used to work with an external monitor: Apple Thunderbolt 27" monitor.  The issue is that, according to Activity Monitor, the kernel_task process consumes more than 600% of CPU even when no tasks are running! The computer becomes totally unusable.   The workaround is to disconnect the external monitor, but I really need this secondary screen to work.  Why didn't Apple fix this issue in the latest Yosemite update?  Is there a fix for this? Find all continuous functions satisfying $\int_0^xf=(f(x))^2+C$ for some constant $C \neq 0$. 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. Why do Oreo crumbs float to a single glob at the very center in a glass of milk? I had Oreos and milk a while ago and left my half-full cup of milk out on the counter. Afterwards I noticed that the crumbs had surfaced in a circular coin-sized glob, and just now I looked again to see all of the crumbs seem to have compressed into a more dense glob at the very center of the cup.   Why does this happen? how can i automatically record every terminal session? Evaluate the integral   $$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx$$   My Attempt:  Let $x = \frac{1}{t}$. Then $dx = -\frac{1}{t^2}\,dt$. Then the integral converts to   $$ -\int \frac{t^3}{(1+t^4)^{3/4}}\,dt $$  Now Let $(1+t^4) = u$. Then $t^3\,dt = \frac{1}{4}du$. This changes the integral to  $$ \begin{align} -\frac{1}{4}\int t^{-3/4}\,dt &amp;= -u^{1/4}+\mathcal{C}\\ &amp;= -\left(1+t^4\right)^{1/4}+\mathcal{C} \end{align} $$  So we arrive at the solution  $$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx = - \left(\frac{1+x^4}{x^4}\right)^{1/4}+\mathcal{C.}$$  Question: Is there any other method for solving this problem? How to interpret regression coefficients when response was transformed by the 4th root? I'm using fourth root (1/4) power transformation on my response variable, as a result of heteroscedasticity. But now I'm not sure how to interpret my regression coefficients.   I assume that I would need to take the coefficients to the fourth power when I back-transform (see below regression output). All of the variables are in units of dollar in millions, but I would like to know the change in dollar in billions.    While holding the other independent variable constant, a billion dollar change in fees, on average, leads to a change of 32 (or 32,000 dollars) in collections. I take 0.000075223 * 1000 (to get to billions) ^ 4 = 0.000032. Now do I multiply this number by 1 million or 1 billion (the original unit of the dependent variable is in millions)?  lm(formula = (Collections^(1/4)) ~ Fees + DIR)                   Estimate      Std. Error  t value            Pr(&gt;|t|) (Intercept)   2.094573355     0.112292375   18.653  0.0000000000000151 Fees        **0.000075223   **0.000008411    8.943  0.0000000131878713 DIR           0.000022279     0.000004107    5.425  0.0000221138881913 If GCD $(a_1,\ldots, a_n)=1$ then there's a matrix in $SL_n(\mathbb{Z})$ with first row $(a_1,\ldots, a_n)$ Let $n \geq 2$. Let $a_1, a_2, \ldots, a_n$ be $n$ integers such that $\gcd\left(a_1, a_2, \ldots, a_n\right) = 1$. Prove that there exists a matrix in $\operatorname{SL}_n\left(\mathbb{Z}\right)$ whose first row is $\left(a_1, a_2, \ldots, a_n\right)$.   Since the gcd of the integers $a_1,\ldots, a_n$ is $1$, there exists weights $x_i \in \mathbb{Z}$ such that $a_1x_1+\cdots+ a_nx_n=1$. My two ideas are (a) to brute force construct an $n\times n$ matrix with first row $a_1,\ldots ,a_n$ and to construct the remaining rows such that the determinant is $\sum a_ix_i=1$ or (b) to use induction.  (a) (Constructive) This is tedious since once I find a way to construct the remaining $n-1$ rows to ensure that $a_1x_1$ appears in the determinant, I am not sure how to modify these $n-1$ rows to ensure  that only the terms $a_ix_i$ appear in the cofactor expansion. If such a matrix exists, I'd like to see it.  (ii) (Non-constructive) If I proceed by induction then the base case $n=2$ is settled since I can choose the 2nd row to be $-x_2, x_1$ so that the determinant is $a_1x_1-a_2(-x_2)=1$. However, I'm not sure how to use the inductive hypothesis to show that if I can construct such an $n\times n$-matrix then I can construct an $\left(n+1\right) \times \left(n+1\right)$-matrix with the desired property. In particular, if the gcd $(a_1,\ldots ,a_{n+1})$ is $1$, it is not necesarry that the gcd of any $n$ of these terms is $1$, so induction may not even apply here.  How can I construct such a matrix or prove that one exists (without necessarily constructing it)?