How can I calculate the determinant of this type of matrix? Given the matrix  $$ A = \begin{bmatrix} 1&amp;1&amp;1&amp;1\\ a&amp;b&amp;c&amp;d\\ a^{2}&amp;b^{2}&amp;c^{2}&amp;d^{2}\\ a^{3}&amp;b^{3}&amp;c^{3}&amp;d^{3} \end{bmatrix} $$  How can I prove that $\det(A)=(d-c)(d-b)(d-a)(c-b)(c-a)(b-a)$ ?  And as a second question, what is the general formula for this type of matrix and how can I prove i?    I'm quite certain (though I could be wrong) that the general formula is as follows:  $$ \det( \begin{bmatrix} 1&amp;\dots&amp;1\\ a_{1}&amp;\dots&amp;a_{n}\\ \vdots&amp;&amp;\vdots\\ a_{1}^{n-1}&amp;\dots&amp;a_{n}^{n-1} \end{bmatrix} ) = \prod_{i&gt;j}(a_{i}-a_{j}) $$ It definitely out for $n\leq4$.    I have tried a proof by induction over $n$. It's easy to prove that it holds for $n=1$:  $$ \det(\begin{bmatrix}1\end{bmatrix})=1\ \checkmark $$  For the inductive step, I've tried using row reduction on the last row and got to this:  $$ \det( \begin{bmatrix} 1&amp;\dots&amp;1\\ a_{1}&amp;\dots&amp;a_{n+1}\\ \vdots&amp;&amp;\vdots\\ a_{1}^{n}&amp;\dots&amp;a_{n+1}^{n} \end{bmatrix} ) = \sum_{k=1}^{n+1}(-1)^{n+k+1}\prod_{\begin{align}i&amp;&gt;j\\i,j&amp;\neq k\end{align}}(a_{i}-a_{j}) $$  but I can't figure out how to prove that equals $\prod_{i&gt;j}(a_{i}-a_{j})$. Is there a way to prove this. And if not, how should I go about solving this exercise?  If some of the mathematical notation I used is incorrect or very unusual,  I would also appreciate it if that was pointed out to me.
Vandermonde Determinant This is an exercise from Ian Stewart's Galois Theory, $3^{rd}$ edition:     If $z_1,z_2,\ldots,z_n$ are distinct complex numbers, show that the determinant $$D=\left|\begin{array}[cccc]  11&amp;1&amp;\cdots&amp;1\\ z_1&amp;z_2&amp;\cdots&amp;z_n\\ z_1^2&amp;z_2^2&amp;\cdots&amp;z_n^2\\ \vdots&amp;\vdots&amp;\ddots&amp;\vdots\\ z_1^{n-1}&amp;z_2^{n-1}&amp;\cdots&amp;z_n^{n-1} \end{array}\right|$$   is nonzero.       Hint: Consider the $z_j$ as independent indeterminates over $\Bbb C$. Then $D$ is a polynomial in the $z_j$, of total degree $0+1+\cdots+(n-1)=\frac{1}{2}n(n-1)$. Moreover, $D$ vanishes whenever $z_j=z_k$ for $k\neq j$, as it then has two identical rows. Therefore $D$ is divisible by $z_j-z_k$ for all $j\neq k$, hence it is divisible by $\prod_{j&lt;k}(z_j-z_k)$. Now compare degrees.   My question is how to do this. I follow the hint and I'm thinking of $D$ as a cofactor expansion; is that the idea? Also, the fact that $D$ vanishes whenever $z_j=z_k$ for $k\neq j$ follows since it would have two identical columns, right? I don't see why $z_j=z_k$ would imply it has two identical rows, as the hint suggests. The fact that it is divisible by $(z_j-z_k)$ is easy since that would make $z_k$ a root for $k=j$, but I'm not quite sure I follow why the degrees would be different. I think the degree of the product would be $1+2+\cdots+(k-1)$, but then add these up for all $k&lt;n$?   I appreciate any clarification that you can provide!
Why is $c+q$ still a primitive root modulo $q$? Question: Let $p$ and $q$ be distinct odd prime numbers.  By considering primitive roots, we need to show $\exists c\in\mathbb{Z}$ with the property that:  $\bullet$ $c^n\equiv 1\pmod{p}$ whenever $n$ is a multiple of $p-1$, and   $\bullet$ whenever $n$ is not divisible by $q-1$, then $c^n\not\equiv 1\pmod{q}$  I am only interested with the beginning of this answer.  Answer: We let $c$ be a primitive root modulo $q$.  We are going to use Fermat's little theorem so we need $(c,p)=1$  If $(c,p)\neq 1$, then we have $p\mid c$, but then we have $(c+q, p)=1$  and since $c+q$ is a primitive root modulo $q$, we can replace $c$ with $c+q$ for $(c,p)=1$ for using Fermat's little theorem.  I understand the general method and argument, but...  Contention: Why is $c+q$ still a primitive root modulo $q$?  When is this generally true? Is there an identity for when things remain a primitive root?  Why is $(c+q)^{\phi(q)}\equiv 1\pmod{q}$? 2 different passports with different surnames Methods to work around the problem of missing data in machine learning Virtually any database we want to make predictions using machine learning algorithms will find missing values ​​for some of the characteristics.  There are several approaches to address this problem, to exclude lines that have missing values ​​until they fill with the mean values ​​of the characteristics.  I would like to use for a somewhat more robust approach, which would basically run a regression (or another method) where the dependent variable (Y) would be each of the columns that have missing values ​​but only with the rows of the table that contain all the data , and predict the missing values ​​with this method, complete the table by the table and move to the next 'column' with missing values ​​and repeat the method until everything is filled.  But that gives me some doubts.  Why any column start? I believe that the one with the smallest missing values ​​until the one with the most  Is there any threshold of missing values ​​that is not worth trying to complete it? (for example, if this characteristic only has 10% of the values ​​filled would not it be more interesting to exclude it)  Is there any kind of implementation in traditional packages or other methods that are robust to missings? What are the increased possibilities (if any) of being able to crack a private key given the following:   The associated bitcoin (ECDSA Secp256k1-based) public key is known. The private key has been used to generate a public key from a different ECDSA curve and that public key is also known. It is known that the two public keys are derived from the same private key. It is known which ECDSA curves were used to generate the public keys. It is known which ECDSA curve was used to generate which public key.   Does it make any more difference if there is data that is known to have been signed by the private key and the signature(s) are known and the raw data is known?  EDIT: I should have pointed out that the conditions are ANDed, I.e. All conditions are known. Sorry, I should have been clearer the first time. Add an image directly from the clipboard I've tried to find other questions covering this, but (surprisingly) I did not find any.  When I want to add a screenshot to a post, as far I know the only way to do it is to save it as a file and then upload the file, using Stack Overflow's image hosting. OK, much better than finding a hosting site yourself, but sometimes I wish I could just paste the image from the clipboard.  It's not as straightforward as it may seem, I know. , and it needs a Java applet to get it done. How to create wifi hotspot in Ubuntu 17.10 Finding $\lim_{n\to\infty}\int_0^{\frac{1}{2^n}}(1+\cos(x))^n\,dx$ During our studies on Riemann Integrals we were given the following question in my Calculus II course: $$ \forall n\in\mathbb{N}: \text{let } f_n(x)=(1+\cos(x))^n, f_n:[0,1]\to \mathbb{R}$$ $$\text{ Find}: \lim_{n\to\infty}\int_0^{\frac{1}{2^n}}f_n(x)\, dx $$ I have tried finding the limit using the squeeze theorem. I noticed that the function is defined on $[0,1]$, where $\cos(x)$ is positive and decreasing. I used the following boundaries: $$ 1\leq f_n\leq 2^n \implies \frac{1}{2^n} = \int_0^{\frac{1}{2^n}}1 \leq \int_0^{\frac{1}{2^n}}f_n(x) \leq \int_0^{\frac{1}{2^n}}2^n = 1  $$ but $ \displaystyle{\lim_{n\to\infty}}\frac{1}{2^n} = 0 $ which means I'm unable to apply the squeeze theroem. I've tried using tighter lower boundaries such as $1+\cos(1)$ But it didn't seem to work. I did some trial and error by calculaing the integrals manually for different values of $n$ and it does seem like $\displaystyle{\lim_{n\to\infty}}\int_0^{\frac{1}{2^n}}f_n(x) dx = 1 $ This means that somehow I should find a lower bundary which tends to 1 when $n\to\infty$. Any ideas on how to go forward from here are welcome. Is the cube root of a prime number rational? How many arrangement of the letters in MISSISSIPPI have at least 2 adjacent S's?  I was thinking that I can glue two of the S's together, so there will be 9 letters plus the special letter SS, and the number of arrangements will be $10!/(2!2!4!)$, this is not the correct answer, but what is wrong with my reasoning.  (the correct answer is $11!/(4!4!2!)-7!/(4!2!)\binom{8}{4}$, I understand the solution) Which queue for an unmarried EU/non-EU couple travelling through UK Immigration Control together? Cascading Drop down in SharePoint Online I'm trying to implement cascading drop down for list in SharePoint online  I have 3 different list with name Countries , States, cities.  and the master list Address in which I have Fields Country , State, City.  I'm trying to display Sates for selected country. and City for selected States in SharePoint Online list Unable to make recursive call in Javascript What is the center of $D_{2n}/Z(D_{2n})$ What is the center of $D_{2n}/Z(D_{2n})$. I see that when $n=2^k$ then I have a $p$-group so the center is not trivial. But when $n$ is not power of $2$ how can i know what is the center of this group?. Surely when $n$ is odd then the center is trivial. How to make GRASS and SAGA tools capable to use in QGIS？ I installed QGIS, GRASS and GDAL by OSGeo4w package, and as I finish the installation, I want to use GRASS tool, I face this problem：    But I already installed GRASS and SAGA, how to make those tools in QGIS to be used？ Ubuntu 12.04 (Wubi) not starting - root.disk corrupted I have Ubuntu 12.04 installed through Wubi on my laptop. For some reason when i click on Ubuntu on start up (when you have to either select windows or Ubuntu) I get the following screen, the grub cursor keeps flashing.  Gnu grub version 1.99-21 Ubuntu 3.4   Minimal bash-like line editing is supported. For the first word, tab lists possible command completions. Anywhere else tab lists possible device or file completions.   grub&gt;   Could you tell me what I can do to fix this problem and to start Ubuntu normally? External hard drive and Recycle Bin behavior When my Seagate external hard drive is plugged in, its $RECYCLE.BIN folder automatically contains all the files contained in the Recycle Bin of the PC. Is it true that actually we don't have any Recycle Bin of external hard drive, rather it is the Recycle Bin of the PC? Why is it so? Manganese has more unpaired electrons than Iron so why is Iron ferromagnetic Manganese paramagnetic? Manganese has five unpaired electrons, but Iron has four, then why is Iron ferromagnetic and Manganese paramagnetic? What's that property I'm missing? How to create Mist in Blender Java map with values limited by key's type parameter Is there a way in Java to have a map where the type parameter of a value is tied to the type parameter of a key? What I want to write is something like the following:  public class Foo {     // This declaration won't compile - what should it be?     private static Map&lt;Class&lt;T&gt;, T&gt; defaultValues;      // These two methods are just fine     public static &lt;T&gt; void setDefaultValue(Class&lt;T&gt; clazz, T value) {         defaultValues.put(clazz, value);     }      public static &lt;T&gt; T getDefaultValue(Class&lt;T&gt; clazz) {         return defaultValues.get(clazz);     } }   That is, I can store any default value against a Class object, provided the value's type matches that of the Class object. I don't see why this shouldn't be allowed since I can ensure when setting/getting values that the types are correct.  EDIT: Thanks to cletus for his answer. I don't actually need the type parameters on the map itself since I can ensure consistency in the methods which get/set values, even if it means using some slightly ugly casts. Mavericks input source is set to US keyboard in login screen I'm from Belgium, and on my own user account, the input source is set to a Belgian (actually French) keyboard. The input sources menu is enabled in the menu bar, so I can switch it to the Belgian keyboard before entering my password, but how can I permanently change the input source used for login? I'm the system admin, and I know how to get root access if necessary.  It worked perfectly in Mountain Lion.