Polynom to linear factors above the complex numbers
What is the square root of complex number i?
How to show that $\sum\limits_{k=0}^n (-1)^k\tfrac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}} = \frac{x}{x+n}$ I'm trying to show that      $$\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}} = \frac{x}{x+n}$$   I've tried expanding this using definition of binomial coefficient, then binomial expansion, now I'm using induction to simplify $\sum_{k=0}^m (-1)^k\frac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}}$, but it looks very inelegant, is there a better way? The moderator agreement is visible at the following URL:       I acknowledge and agree that as a moderator for Stack Exchange Inc.      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Stack Exchange Inc. reserves the   right to terminate my privileges as a moderator at any time without   warning.   It hasn't been updated with the new company name and still refers to Stack Exchange Inc. instead of Stack Overflow (Inc.?). Microphone doesn't work in Hangouts I have a DELL inspiron 5537 i5 laptop with Ubuntu 14.04. When I use Google Hangouts on Google Chrome, my mic just doesn't work. When I open the secure info setting it shows that the mic is allowed.  Works in Windows.  I have tried to add the Hangouts plugin and it also doesn't work.  Is there any to solve the problem? On finding polynomials that approximate a function and its derivative (extensions of Stone-Weierstrass?) The Stone-Weierstrass Theorem tells us that we can approximate any continuous $f:\mathbb{R}^n\to\mathbb{R}$ arbitrary well on a compact subset of $\mathbb{R}^n$ by some polynomial. Suppose that $f$ is continuously differentiable, under what conditions can we guarantee we can find a polynomial $p$ such that $p$ approximates arbitrarily well $f$ and the partial derivatives approximate of $p$ arbitrarily well those of $f$ (on a compact subset of $\mathbb{R}^n$)? References are welcome. See below for my (probably naive) approach to the question.    It is straightforward to find conditions if $n=1$: Let $K:=[a,b]\subseteq\mathbb{R}$ and let $K_+$ denote some open cover of $K$. Suppose that $f:K_+\to\mathbb{R}$ is continuously differentiable and let $f'$ denote its derivative. Then, by Weierstrass's approximation Theorem we have that there exists a polynomial $q$ such that   $$||f'(x)-q(x)||\leq \varepsilon$$  for all $x\in K$. Choose some $y\in K$. For any $x\in K$ define  $$p(x):=f(y)+\int_y^xq(t)dt.\quad\quad(*)$$  By the Fundamental Theorem of calculus we have that $p'=q$. So for any $x\in K$  $$||f(x)-p(x)||=\left|\left|\int_y^x (f'(t)-q(t))dt\right|\right|\leq \int_y^x ||f'(t)-q(t)||dt\leq \varepsilon||x-y||\leq \varepsilon (a-b).$$  However, the above argument doesn't work in higher dimensions because the analogue of $(*)$ does not imply that $\nabla p = q$ (since not every vector of polynomials spans a gradient field). Can anyone suggest a way around this? That is, a way to to show that for continuously differentiable $f:\mathbb{R}^n\to\mathbb{R}$ and for any $\varepsilon&gt;0$, we can always find a polynomial $p$ such that   $$||f(x)-p(x)||+||\nabla f(x)-\nabla p(x)||\leq \varepsilon$$  for all $x$ in some compact set $K$? Feel free to add extra conditions on $K$, if it makes life easier. Is voting to balance in the spirit of the site? The other day I saw an answer which was ok. Not great but not bad. So no reason to upvote or downvote. However the answer already had a single downvote. I felt this was unfair so upvoted the answer to keep it at what I felt was a fair score. It wasn't until afterward that it occurred to me that I should have perhaps solely voted on the content of the answer.  Is this a reasonable reason for an upvote or should I have left the answer alone? How can a path animation be used as an action in the game engine? I animated a rollercoaster using a Bezier Curve and a Follow Path Constraint affecting the vehicle.  Then I tried to use the automatically created action in the game engine but the Action Actuator has no effect .    How can this action be used in the BGE? Can the Path Animation alternatively be converted into keyframes? How to make a package manager wait if another instance of APT is running? I have seen many softwares such as Update Manager and Synaptic Package Manager, they wait if some other program is using the /var/lib/dpkg/lock and is locked. How can we do this through the Terminal? I saw apt-get's manual but didn't find anything useful. Custom fields not useable in Approve/Reject Approval Request I'm trying to get through my first SalesForce mod and it's definitely been a rocky path. I have one task I'm trying to do, but this section is holding me back.  I have two custom fields.  One a pick list and the other a text box. I have added them to the Approve/Reject Approval Request section in the Opportunity page. I added the fields through the Select Fields to Display on Approval Page Layout and they DO show up on the Approve/Reject Approval Request section. I added them in like ().  However, just the label for the fields show, the field itself is nowhere to be found and can't be used.    You can see and even use the fields when you open the Opportunity page itself, but when you go to use the Approve/Reject, you can't do anything with them or see any values.  I was never a SalesForce user and got this tossed in my lap and really don't know my way around yet. I appreciate any help on this!!  Thanks!!  Kurt Google Chrome will take up my memory to the point where it causes my computer to freeze to a near halt. What can I do to prevent this? It gets so horrible that I can not even switch to TTY1 to force a reboot. I have to hard reset, and I lose work.  Ideally, I would like to be able to use the chrome browser on Ubuntu without having to keep reminding myself to close tabs.   Are there any extensions for Chrome, or tools for Ubuntu that can help prevent me going past the point where Ubuntu eats up all the RAM and becomes incapacitated? Proof that $\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} $ over a closed curve is zero. Prove that $$\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} $$ over a closed  smooth simple curve that contains all of the roots of the polynomial $P(z)$ is either zero if $n \geq 2$ or equals $\frac{1} {a_0} $ if it's degree is $1$.   Proof: $n\geq 2$ $$\frac {1} {P(z)=(z-z_0)(z-z_1)\dots(z-z_n)}$$ So after partial decomposition of the fraction I'll have something like $$\frac {A} {z-z_0} +\frac{B} {z-z_1}+\cdots$$ where $A,B \in \mathbb{C}$ (Is this fact true that $A,B$ wont be polynomials and just coefficients I can see it why in an intuitive way but cannot write a full proof of it for the decomposition trying to get a linear system or something) then use Cauchy's theorem and integrate all of these on  different circles with their center at each root  and argue that  the integral is the same as the sum of those circle integrals. Now since $\frac {1}{z-z_n}$ will be analytic inside the circle when I'm integrating on circles that do not contain $z_n$  some  of those integrals  will be zero. and now since my curves are of the form $\gamma(t)=z_n+e^{it}$ , $t \in [0,2\pi]$ the rest of the integrals  will look like $$\int_0^{2\pi}  \frac{Aie^{ti}} {e^{it}}\,{\rm d}t $$  and all of them will be $A_i$.   Is my proof correct? Only part I cannot show is that I won't have any problem with decomposing the fraction. And I'm also using the assumption that after the decomposition I'll have as numerators simple complex numbers not not functions of any kind. How will Cryptography be changed by Quantum Computing? I realise this isn't a 'yes or no' question, and I apologise for asking something that could be seen as a discussion thread, but I had to ask.  I'm currently doing an EPQ in CS (specifically how QC will change Cryptography). I'm trying to gather up topics to cover, and so far I've scribbled down -  Effects -   National security, classical cryptographic methods (RSA, DSA, AES-256). Including Shor's algorithm. Communications Online services, e.g. BitCoin.   How would we counter this (not sure how better to phrase this) -   Lattice-based Multivariate Hash-based signatures Code-based cryptography   Is there anything else you'd recommend me including (as well as the above), or anything in the above you don't think I should include?   Honestly, the list above is a rough, first draft the main bulk of it... so don't it as gospel.  Best regards,  Cameron. I want to open a file for reading, the C++ way. I need to be able to do it for:   text files, which would involve some sort of read line function. binary files, which would provide a way to read raw data into a char* buffer. How to set up Dolphin as default file manager? RegEx for Javascript to allow only alphanumeric I need to find a reg ex that only allows alphanumeric. So far, everyone I try only works if the string is alphanumeric, meaning contains both a letter and a number. I just want one what would allow either and not require both. Terminal not displaying colors for directory listings Are battery packs allowed in hand luggage? Gummy residue from baking spray/oil I'm sure everyone's seen this - that brownish stuff left behind when you grease an area of a baking dish that gets left exposed to the heat. Soap doesn't bother it too much, and it likes to gum up my scrubber. What's going on there, and are there any easy ways to get it off? How can I prove $n - 2\sqrt{n} = \Theta(n)$ Prove that if $f$ is a real continuous function such that $|f|\le 1$ then $|\int_{|z|=1} f(z)dz| \le 4$ I've encountered the following problem in complex analysis:  Prove that if $f$ is a real continuous function such that $|f|\le 1$ then $|\int_{|z|=1} f(z)dz| \le 4$.  I've decomposed $f$ to it's real and imaginary parts and got that it's enough to prove that:  $(\int_0^{2\pi}cos(\theta)g(\theta)d\theta)^2 +(\int_0^{2\pi}sin(\theta)g(\theta)d\theta)^2 \le 16$   Where $g(\theta)=f(e^{i\theta})$.   But now I'm stuck. Any ideas?  Thanks plotting 3d points: z-axis is squashed