I'm an aspiring LaTeX user and have so far managed the first steps and created my first few documents. Coming from a programming background I've been wondering how you could use \newcommand to create a command that formats numbers or strings in a hard-coded way.  For example, let's say I have the string AB1234567890. Now what I'd like to do is create a command, that used like this  \applyformat{AB1234567890}   should create the output  AB12 3456 7890   I realize something like that can be done with \,  AB12\,3456\,7890   but I would like to define a fixed "layout" in the command to make sure that every string gets formated the same way as I use that command with each string later.  In a programming language I could split strings and such but I'm unaware how to solve that with LaTeX.
I have a block of text that I would like to automatically break in blocks of equal length. Something like what \num does with numbers.  What I need is a definition for what I called \makeBlocks. The MWE below should output: aaaaa bbbbb ccccc ddddd followed by abc def ghi  \documentclass{article}  \newcommand{\makeBlocks}[1]{???}  \begin{document} \makeBlocks{5}{aaaaabbbbbcccccddddd}  followed by \makeBlocks{3}{abcdefghi} \end{document}
Diagonalisation and characteristic polynomial Let $A\in M_3(\Bbb R)$ which is not a diagonal matrix. Let $P$ be a polynomial (in one variable), with real coefficients and of degree 3 such that $P(A) = 0$. Pick out the true statements:  a. $P = cP_A$ where $c \in\Bbb R$ and $P_A$ is the characteristic polynomial of $A$;  b. if $P$ has a complex root (i.e. a root with non-zero imaginary part), then $P = cP_A$, with $c$ and $P_A$ as above;  c. if $P$ has a complex root, then $A$ is diagonalizable over $\Bbb C$.  I guess b and c are true although I am unable to disprove a. Because A could satisfy $(X-A)$ but this is not a polynomial. Now if $P$ has a complex root, then its conjugate too is its root as the polynomial has  real coefficients. As we all know An odd degree polynomial has always a real root, what we have is three distinct roots. Since $A$ satisfies the polynomial the only difference between the characteristic polynomial and $P$ has to be a constant. Integral $\int\frac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$ Integrate  $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$   I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$. How to let `dpkg -i` install dependencies for me? Does photon have size measurement because of its particle nature Does photon have size measurement because of its particle nature like electron's 3.86*10^-13m etc.. Distorted Cabin & Oily Sheen: What is going on in this paragraph? Trying to find a Sci-Fi movie, probably from the 50's or 60's I’m trying to identify a movie I saw when I was very young.  It was a “Saturday Matinee” playing on one of the local stations, and I likely watched it in the late 60’s.  So I expect that the movie is from the 50’s or 60’s.  I watched it in black and white, but it could have been a color movie on a black and white set. I clearly recall a scene where scientists enter a lab and some see something move past them in a blur.  Later in the movie, they come upon themselves in the same lab where they appear to be frozen and not moving.  Then they notice that their “other selves” are moving, just almost inperceptibly slowly.  They are encountering themselves but at a completely different time scale. Unfortunately, I don’t recall much else about the movie, just the idea that they encounter themselves once with the others flashing past, and again seemingly not moving at all.  If this rings any bells with anyone, I’d love some pointers to try to track this movie down. Essentially as the title suggests - in some commutative ring $K$ (with 0,1), if we have 2 distinct proper prime ideals $\mathfrak{p}_1 \neq \mathfrak{p}_2$, is it necessarily the case (or if not, when is it the case?) that $\mathfrak{p}_1 + \mathfrak{p}_2 = K$ ?  I ask because I'm working through some lecture notes which frequently take some $K$ complete with respect to a discrete valuation, a prime ideal in a valuation ring  $\mathfrak{p} \subset \mathcal{O}_K$, a finite extension $L/K$, and then use the fact $\mathfrak{p}\mathcal{O}_L$ factors uniquely as some $\mathcal{P}_1^{a_1} \ldots \mathcal{P}_r^{a_r}$ (where $\mathcal{O}_L$ is the integral closure of $\mathcal{O}_K$ in $L$). Then we say $\mathcal{P}_i + \mathcal{P}_j = \mathcal{O}_L$ $(\forall$ $ i \neq j)$.   However it isn't clear to me, because we've made so many different assumptions along the way, what the reason is that $\mathcal{P}_i + \mathcal{P}_j = \mathcal{O}_L$. Is it because these primes occur in the factorisation of $\mathfrak{p}$, is it because of some property of Dedekind domains or fields, or is it not really possible to say out of context? Many thanks, and if anything needs clarification please ask. Complex Exponent of Complex Numbers How does one find the algebraic solution of a Complex number raised to the power of another Complex number?  Here is the work I have done so far, if there are any mistakes please inform me.  A real number with a Complex Exponent: $$A \in \Bbb R, \space z \in \Bbb C,\space z = x + iy$$ $$A^z = A^{x+iy} = A^xA^{iy}$$ $$\implies e^{iy\ln A} = A^{iy}$$ $$\exp(iy\ln A) = \cos (y\ln A)+i\sin(y\ln A)$$ $$A^z = A^x[\cos (y\ln A)+i\sin(y\ln A)]$$  A Complex number with a Complex Exponent: [Using previous variables] $$C \in \Bbb C,\space C = a +bi\space|\space re^{i\theta}, \theta =\arg C$$ $$C^z = (a+ib)^z$$ [After previous mistake the following notes are inaccurate] $$C^z = a^x[\cos (y\ln a)+i\sin(y\ln a)] + ib^xib^{iy}$$  And this is as far as I could go as the expansion of $ib^xib^{iy}$ is being problematic. Could I have some help with this expansion? How to convert URL parameters to a JavaScript object? Should a "thank you" comment, for instance Thanks for your answer, be flagged as non-constructive?  If a "thank you" comment is flagged as non-constructive what would the moderator reaction to this flag be? Would it be marked as helpful, or not? Prove that $\sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$ Prove that $\displaystyle \sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$   $\bf{My\; Try::}$ We can write $$\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{n!}{r!\times (n-r)!}\times \frac{r!\times 3!}{(r+3)!} = \frac{3!\times n!}{(n-r)!\times (r+3)!}$$  So our sum is $$ 3!\sum^{n}_{r=0}(-1)^r\frac{n!}{(n-r)!\times (r+3)!} = \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\frac{(n+3)!}{(r+3)!\times (n-r)!}$$  So we get Sum $$ \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\binom{n+3}{r+3} = -\frac{3!}{(n+1)(n+2)}\sum^{n}_{r=0}(-1)^{r+3}\binom{n+3}{r+3}$$  $$  = -\frac{3!}{(n+1)(n+2)(n+3)}\left[-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]$$  $$=-\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]+\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}\right]$$  So we get $$ = \frac{3!}{(n+1)(n+2)(n+3)}\left[1-(n+3)+\frac{(n+3)(n+2)}{2}\right] = \frac{3!}{(n+1)(n+2)}\times \frac{n^2+3n+2}{2} = \frac{3!}{(n+1)(n+2)(n+3)}\times \frac{(n+1)(n+2)}{2} = \frac{3!}{2(n+3)}$$  Can we prove it Using Combinatorial way,bcz above method is very lengthy  If yes then plz explain here  Thanks Inductive proof of a formula for Fibonacci numbers May someone help me? I am trying to use induction to prove that the formula for finding the $n$-th term of the Fibonacci sequence is:   $$F_n=\frac{1}{\sqrt{5}}⋅\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}⋅\left(\frac{1-\sqrt{5}}{2}\right)^n.$$  I tried to put $n=1$ into the equation and prove that if $n=1$ works then $n=2$ works and it should work for any number, but it didn't work. I need to prove that this formula gives the $n$th Fibonacci number. If you give a user sudo privileges how can I remove the sudo privileges and make the user become just a regular user?  I used:  sudo adduser username sudo   Now I've changed my mind. Canonical Isomorphism Between $\mathbf{V}$ and $(\mathbf{V}^*)^*$ For the finite-dimensional case, we have a canonical isomorphism between $\mathbf{V}$, a vector space with the usual addition and scalar multiplication, and $(\mathbf{V}^*)^*$, the "dual of the dual of $\mathbf{V}$." This canonical isomorphism means that the isomorphism is always the same, independent of additional choices.   We can define a map $I : \mathbf{V} \to (\mathbf{V}^*)^*$ by $$x \mapsto I(x) \in (\mathbf{V}^*)^* \ \text{ where } \ I(x)(f) = f(x) \ \text{for any } \  f \in \mathbf{V}^*$$  My Question: what can go wrong in the infinite-dimensional case? The notes I am studying remark that if $\mathbf{V}$ is finite-dimensional, then $I$ is an isomorphism, but in the infinite-dimensional case we can go wrong? How? What exactly is flash sync speed and should it be a factor in a buying decision? I've heard that special relativity makes the concept of magnetic fields irrelevant, replacing them with relativistic effects between charges moving in different velocity frames. Is this true? If so, how does this work? When can a hyphen ("-") be used, and what does it mean? I've seen the hyphen (-) used in phrases a lot, for example:     audience-knowing      moves-sometimes      do-gooder   When and why is the hyphen used, and what does it mean when it's used? Unable to modify files/folders in NTFS partition Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = \gcd(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$. Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = hcf(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$.   We know that if $a|c$ and $b|c$ then $a\cdot b\cdot s=c$ (for some positive integer $s$). $(ab|c)$  Then doesn't $ab|dc$ since $ab|c$?  I feel like I'm misunderstanding my givens.    Can we say $\operatorname{lcm}(a,b)=c$,  $\operatorname{hcf}(a,b)=d$, and $\operatorname{lcd}(a,b) \operatorname{hcf}(a*b)=a*b$? Thus, $ab|dc$  as $dc = ab$. Computing matrix-vector calculus derivatives $x, a$ in $\mathbb R^n$, $A$ in $\mathbb  R^{n\times n}$. Compute $d(x^T a)/dx$ and $d(x^T A x)/dx$.  I'm not sure about how to think about these and how to do these. Can someone explain how to derive the expressions for the two?  Finally, what happens when we have $A$ and $X$, BOTH in $\mathbb R^{n\times n}$, and we want to find $dTrace(XA)/dX$?