How to determine roots of a complex polynomial
Show that the equation, $x^3+10x^2-100x+1729=0$ has at least one complex root $z$ such that $|z|>12.$
Why are PATH variables different when running via sudo and su? On my fedora VM, when running with my user account I have /usr/local/bin in my path:  [justin@justin-fedora12 ~]$ env | grep PATH  PATH=/usr/kerberos/sbin:/usr/kerberos/bin:/usr/local/bin:/usr/bin:/bin:/usr/local/sbin:/usr/sbin:/sbin:/home/justin/bin   And likewise when running su:  [justin@justin-fedora12 ~]$ su - Password:  [root@justin-fedora12 justin]# env | grep PATH PATH=/usr/kerberos/sbin:/usr/kerberos/bin:/usr/local/bin:/usr/bin:/bin:/usr/local/sbin:/usr/sbin:/sbin:/home/justin/bin   However, when running via sudo, this directory is not in the path:  [root@justin-fedora12 justin]# exit [justin@justin-fedora12 ~]$ sudo bash [root@justin-fedora12 ~]# env | grep PATH PATH=/usr/kerberos/sbin:/usr/kerberos/bin:/sbin:/bin:/usr/sbin:/usr/bin   Why would the path be different when running via sudo? How to solve "sudo: /etc/sudoers.d is world writable" Documents required by Indian citizens wishing to travel to Nepal How it this possible? Even if the gif is fake, the Mythbusters did it and with a large sail it really moves forward. What is the explanation? Can a massless rope accelerate? How to change boot priority? I have a laptop with Ubuntu installed as the only OS. I also have a USB stick with a Windows 7 installation on it. I want to install Windows from the USB stick on my laptop but when I try to restart the laptop to get access to BIOS, I can't seem to find the command for it, its like it skips everything and just says Ubuntu and then takes me to the login screen.  How do I change the boot priority to USB in Ubuntu? Suppose $X,Y$ and $Z$ are multivariate normal with means and full covariance matrix. The conditional expectation $E(X | Y)$ is well know. What is the conditional expectation of $E(X | Y,Z)$ if $Y$ and $Z$ (and $X$) are correlated? Standard textbooks only seem to cover the case when $Y$ and $Z$ are uncorrelated. I need to cast single figures (1 to 9) to (01 to 09). I can think of a way but its big and ugly and cumbersome. I'm sure there must be some concise way. Any Suggestions Distribution of orbital velocities in a disk galaxy for N-body simulation? solving third-order nonlinear ordinary differential equation Contraction mapping in the context of $f(x_n)=x_{n+1}$. I'm interested in convergence of $f(x_n)=x_{n+1}$ and often hear this term referenced.  What does it mean to be a contraction mapping in the context of the sequence of real numbers given by $f(x_n)=x_{n+1}$? And what does it tell us about such sequence?  A search online gives the answer given by Wikipedia, that it is a function $f$ defined on a metric space $(M,d)$ from $M$ to itself with the property that for some real number $k \in [0,1)$,  $$d(f(x),f(y)) \leq k d(x,y)$$  This definition is quite unhelpful because I barely know what a metric space is. I feel like this definition is to generalized, and I'm interested in in the specific case of recursively defined real sequences $f(x_n)=x_{n+1}$ that there should be a more specific answer to my question. My phone was stolen last night. I made sure it was in Lost Mode in iCloud and it was immediately in Offline status when I open iCloud. I am positive the thief switched it off. Now I am 100% certain the Find my iPhone option was turned on on my iPhone, but will the "Lost Mode" be of any help if it was not turned on on the iPhone? Would it be showing "Offline" if I did not turn on the option? I've seen this Kuratowski definition for ordered pairs, but can't fathom why it implies an order to $x$ and $y$  $(x,y):=\{\{x\}, \{x,y\}\}$  As I understand sets, $\{\{x\}, \{x,y\}\}$ is also $\{\{x,y\}, \{x\}\}$. Only when I think about the Axiom of Union does $\{\{x\}, \{x,y\}\}$ "collapse down" to $S = \{x, y\}$, but that doesn't give me much either. All I can see is some as yet hidden message in the set saying "I am the set $\{x,y\}$ and my order of $x$ first is indicated by having $\{x\}$ along for the ride." Prove by induction $\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$ for $n\ge1$ Prove the following statement $S(n)$ for $n\ge1$:  $$\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$$  To prove the basis, I substitute $1$ for $n$ in $S(n)$:  $$\sum_{i=1}^11^3=1=\frac{1^2(2)^2}{4}$$  Great. For the inductive step, I assume $S(n)$ to be true and prove $S(n+1)$:  $$\sum_{i=1}^{n+1}i^3=\frac{(n+1)^2(n+2)^2}{4}$$  Considering the sum on the left side:  $$\sum_{i=1}^{n+1}i^3=\sum_{i=1}^ni^3+(n+1)^3$$  I make use of $S(n)$ by substituting its right side for $\sum_{i=1}^ni^3$:  $$\sum_{i=1}^{n+1}i^3=\frac{n^2(n+1)^2}{4}+(n+1)^3$$  This is where I get a little lost. I think I expand the equation to be  $$=\frac{(n^4+2n^3+n^2)}{4}+(n+1)^3$$  but I'm not totally confident about that. Can anyone provide some guidance? MWE for externalize + automatic recompilation after tikz figure modification Pre & post increment operator behavior in C, C++, Java, & C# ChaCha with 64bit integers as a PRNG with streams Can I use ChaCha with four rounds as a good non-cryptographic PRNG with different streams if I use 64 bit integers instead of the standard 32 bit integers? I need a bigger state and this seems the easiest way to do it. Ubuntu 20.04 run ok on Lenovo ThinkPad E15 laptop? What can you say about $f$ and $g$ in the case that $fg$ is 1) Injective, 2) Surjective - Cohn - Classic Algebra Page 15 Question: Are my proofs below valid?   In both cases we are using: $f:A\to B, g: B\to C$  Notation of your type converted: $(g\circ f)(x)=g(f(x))=xfg$    If $fg$ is injective what can be said about $f$ and $g$?  If $fg$ is injective then $f$ is injective:  Proof:  $(x_1fg = x_2fg \implies x_1=x_2)$  $x_1fg=x_2fg$ is clearly true when $x_1f=x_2f$  Which requires $x_1=x_2$, hence $f$ is injective.  If $f$ were not injective, then $x_1fg\ne x_2fg$  $\blacksquare$    If $fg$ is surjective what can be said about $f$ and $g$?  If $fg$ is surjective then $g$ is surjective:  Proof:  $(\forall c\in C)(\exists a\in A) |( a(fg))=c$  This means for $g$ every element must have had been mapped to. How can this be proved Mathematically?  well $(af)g=c$ is surjective, hence $(af)=(b\in B)$ and $bg=c$  $(\forall c\in g)(\exists b\in B)|(bg=c)$  Hence $g$ is surjective.  $\blacksquare$ Rewrite code to make it faster